1 Affine Varieties

1 Aﬃne Varieties

We will begin following Kempf’s Algebraic Varieties, and eventually will do things more like in Hartshorne. We will also use various sources for commutative algebra. What is algebraic geometry? Classically, it is the study of the zero sets of polynomials. We will now fix some notation. k will be some fixed algebraically closed field, any ring is commutative with identity, ring homomorphisms preserve identity, and a k-algebra is a ring R which contains k (i.e., we have a ring homomorphism ι : k → R). P ⊆ R an ideal is prime iff R/P is an integral domain. Algebraic Sets n n We define affine n-space, A = k = {(a1, . . . , an): ai ∈ k}. n Any f = f(x1, . . . , xn) ∈ k[x1, . . . , xn] defines a function f : A → k : (a1, . . . , an) 7→ f(a1, . . . , an). Exercise If f, g ∈ k[x1, . . . , xn] define the same function then f = g as polynomials.

Deﬁnition 1.1 (Algebraic Sets). Let S ⊆ k[x1, . . . , xn] be any subset. Then V (S) = {a ∈ An : f(a) = 0 for all f ∈ S}. A subset of An is called algebraic if it is of this form.

e.g., a point {(a1, . . . , an)} = V (x1 − a1, . . . , xn − an). Exercises 1. I = (S) is the ideal generated by S. Then V (S) = V (I). 2. I ⊆ J ⇒ V (J) ⊆ V (I). P 3. V (∪αIα) = V ( Iα) = ∩V (Iα). 4. V (I ∩ J) = V (I · J) = V (I) ∪ V (J).

Definition 1.2 (Zariski Topology). We can define a topology on An by defining the closed subsets to be the algebraic subsets. U ⊆ An is open iff An \ U = V (S) for some S ⊆ k[x1, . . . , xn]. Exercises 3 and 4 imply that this is a topology.

The closed subsets of A1 are the ﬁnite subsets and A1 itself. Deﬁnition 1.3 (Ideal of a Subset). If W ⊂ An is any subset, then I(W ) = {f ∈ k[x1, . . . , xn]: f(a) = 0 for all a ∈ W } Facts/Exercises

1. V ⊆ W ⇒ I(W ) ⊆ I(V )

2. I(∅) = (1) = k[x1, . . . , xn]

3. I(An) = (0).

1 Deﬁnition 1.4 (Aﬃne Coordinate Ring). W ⊂ An is algebraic. Then A(W ) = k[W ] = k[x1, . . . , xn]/I(W ) We can think of this as the ring of all polynomial functions f : W → k.

Deﬁnition 1.5 (Radical Ideal)√ . Let R be a ring and I ⊆ R be an ideal, then i the radical of I is the ideal I = {f√∈ R : f ∈ I for some i ∈ N} We call I a radical ideal if I = I.

Exercise √ If I is an ideal, then I is a radical ideal.

Proposition 1.1. W ⊆ An any subset, then I(W ) is a radical ideal. Proof. We have that I(W ) ⊆ pI(W ). Suppose f ∈ pI(W ). Then f i ∈ I for some i. That is, for all a ∈ W , f i(a) = 0. Thus, f(a)m = 0 = f(a). And so, f(a) ∈ I. Exercises

1. S ⊆ k[x1, . . . , xn], then S ⊆ I(V (S)).

2. W ⊆ An then W ⊆ V (I(W )).

3. W ⊆ An is an algebraic subset, then W = V (I(W )). √ √ 4. I ⊆ k[x1, . . . , xn] is any ideal, then V (I) = V ( I) and I ⊆ I(V (I))

Theorem 1.2 (Nullstellensatz)√. Let k be an algebraically closed field, and I ⊆ k[x1, . . . , xn] is an ideal, then I = I(V (I)). √ Corollary 1.3. k[V (I)] = k[x1, . . . , xn]/ I. To prove the Nullstellensatz, we will need the following: Theorem 1.4 (Nöther’sNormalization Theorem). If R is any finitely generated k-algebra (k can be any field), then there exist y1, . . . , ym ∈ R such that y1, . . . , ym are algebraically independent over k and R is an integral extension of the subring k[y1, . . . , ym]. Proof is in Eisenbud and other Commutative Algebra texts. Theorem 1.5 (Weak Nullstellensatz). Let k be an algebraically closed field, and I ( k[x1, . . . , xn] any proper ideal, then V (I) 6= ∅. Proof. We may assume without loss of generality that I is actually a maximal ideal. Then R = k[x1, . . . , xn]/I is a field. R is also a finitely generated k- algebra, and so by Normalization, ∃y1, . . . , ym ∈ R such that y1, . . . , ym are algebraically independent over k and that R is integral over k[y1, . . . , ym]. −1 Claim: m = 0. Otherwise, y1 ∈ R is integral over k[y1, . . . , ym], and so then −p 1−p −1 y1 +y1 f1 +...+y1 fp−1 +fp = 0 for fi ∈ k[y1, . . . , ym]. Multiplying through

2 p p−1 p by y1 gives 1 = −(y1f1 + ... + y1 fp−1 + y1 fp) ∈ (y1), which contradicts the algebraic independence. Thus, the ﬁeld R is algebraic over k. As k is algebraically closed, R = k. k ⊆ k[x1, . . . , xn] → R = k Let ai = the image in k of xi. Then xi −ai ∈ I. Thus, the ideal generated by (x1 −a1, . . . , xn −an) ⊆ I ( k[x1, . . . , xn], and so they I = (x1 −a1, . . . , xn −an), as it is a maximal ideal. V (I) = V (x1 − a1, . . . , xn − an) = {(a1, . . . , an)}= 6 ∅

Note: Any maximal ideal of k[x1, . . . , xn] is of the form (x1 −a1, . . . , xn −an) with ai ∈ k. This is NOT true over R, look at the ideal (x2 + 1) ⊆ R[x]. It is, in fact, maximal. Now, we can prove the Nullstellensatz. √ Proof. Let I ⊆ k[x1, . . . , xn√] be any ideal. We will prove that I(V (I)) = I. It was an exercise that I ⊆ I(V (I)). √ Let f ∈ I(V (I)). We must show that f ∈ I. n+1 Looking at A , we have the variables, x1, . . . , xn, y. Set J = (I, 1 − yf) ⊆ k[x1, . . . , xn, y]. n+1 Claim: V (J) = ∅ ⊂ A . This is as, if p = (a1, . . . , an, p) ∈ V (J), then (a1, . . . , an) ∈ V (I), then (1 − yf)(p) = 1 − bf(a1, . . . , an). But f(a1, . . . , an) = 0, so (1 − yf)(p) = 1, and so p∈ / V (J). By the Weak Nullstellensatz, J = k[x1, . . . , xn, y]. Thus 1 = h1g1 + ... + hmgm + q(1 − yf) where g1, . . . , gm ∈ I and h1, . . . , hm, q ∈ k[x1, . . . , xn, y]. Set y = f −1, and multiply by some big power of f to get a polynomial equation once more. N ˜ ˜ ˜ N Then f = h1g1 + ... + hmgm where the√hi = f hi(x1, . . . , xn). And so, we have f N ∈ I, and thus, f ∈ I, by deﬁnition. exercise: V (I(W )) = W in the Zariski Topology. Irreducible Algebraic Sets Recall: V (y2 − xy − x2y + x3) = V (y − x) ∪ V (y − x2), and V (xz, yz) = V (x, y) ∪ V (z).

Deﬁnition 1.6 (Reducible Subsets). A Zariski Closed subset W ⊆ An is called reducible if W = W1 ∪ W2 where Wi ( W and Wi closed. Otherwise, we say that W is irreducible.

Proposition 1.6. Let W ⊆ An be closed. Then W is irreducible iﬀ I(W ) is a prime ideal.

Proof. ⇒: Suppose I(W ) is not prime. Then ∃f1, f2 ∈/ I(W ) such that f1f2 ∈ W . Set W1 = W ∩ V (f1) and W2 = W ∩ V (f2). As fi ∈/ I(W ), W 6⊆ V (fi), and so Wi ( W . Now we must show that W = W1 ∪ W2. Let a ∈ W . Assume a∈ / W1. Then f1(a) 6= 0, but f1(a)f2(a) = 0, so f2(a) = 0, thus a ∈ W2. ⇐: Exercise

3 This gives us the beginning of an algebra-geometry dictionary. Algebra Geometry n k[x1, . . . , xn] A radical ideals closed subsets prime ideals irreducible closed subsets maximal ideals points In fact, this is an order reversing correspondence. So I ⊆ J ⇐⇒ V (I) ⊇ V (J), but this requires I,J to be radical. Definition 1.7 (NötherianRing). A ring R is called Nötherianif every ideal I ⊆ R is finitely generated.

Exercise A ring R is Nötherianiff every ascending chain of ideals I1 ⊆ I2 ⊆ ... stabilizes, that is, ∃N such that IN = IN+1 = .... Theorem 1.7 (Hilbert’s Basis Theorem). If R is Nötherian, then R[x] is Nötherian.

Corollary 1.8. k[x1, . . . , xn] is Nötherian. Definition 1.8 (NötherianTopological Space). A topological space X is Nötherian if every descending chain of closed subsets stabilizes.

Corollary 1.9. An is Nötherian. n W1 ⊇ W2 ⊇ ... closed in A , then I(W1) ⊆ I(W2) ⊆ ... ideals in k[x1, . . . , xn], and so must stabilize. Theorem 1.10. Any closed subset of a NötherianTopological Space X is a union of finitely many irreducible closed subsets. Proof. Assume the result is false. ∃W a closed subset of X which is not the union of finitely many irreducible closed sets. As X is Nötherian,we may assume that W is a minimal counterexample. W is not irreducible, and so W = W1 ∪W2, where Wi ( W and Wi closed. The Wi can’t be counterexamples, as W is a minimal one, but then W = W1 ∪ W2 and each Wi is the union of finitely many irreducible closed sets. Thus, W cannot be a counterexample.

Corollary 1.11. Every closed W ⊆ An is union of finitely many irreducible closed subsets. Example: V (xy) = V (x) ∪ V (y) ∪ V (x − 1, y). Recall: X is a topological space, then if Y ⊆ X is any subset, it has the subspace topology, that is, U ⊆ Y is open iff ∃U 0 ⊆ X open such that U = U 0 ∩ Y . Note: 1. W ⊆ Y is closed iff W = W ∩ Y , where the closure is in X. 2. X is Nötherianimplies that Y is Nötherianin the subspace topology.

4 Definition 1.9 (Zariski Topology on X ⊆ An). If X ⊆ An, then the Zariski Topology on X is the subspace topology. Definition 1.10 (Components of X). If X is any NötherianTopological Space, then the maximal irreducible closed subsets of X are called the (irreducible) components of X. Exercises

1. X has ﬁnitely many components. 2. X = the union of its irreducible components. 3. X 6= union of any proper subset of its components.

4. A topological space is Nötherianif and only if every subset is quasicompact. 5. A NötherianHausdorff space is finite.

n Recall: X ⊆ A closed. Then A(X) = k[x1, . . . , xn]/I(X). Deﬁnition 1.11. If f ∈ A(X), set D(f) = {a ∈ X : f(a) 6= 0}. Proposition 1.12. The sets D(f) form a basis for the Zariski Topology on X.

Proof. Let p ∈ U ⊆ X, U open. Show that p ∈ D(f) ⊆ U for some f ∈ A(X). Z = X \U a closed subset of X, and Z ( Z ∪{p} implies that I(Z) ) I(Z ∪{p}). Take any f ∈ I(Z) \ I(Z ∪ {p}). Then f vanishes on Z but not at p, so p ∈ D(f). Regular Functions Let X ⊆ An be an algebraic subset, and U ⊆ X is a relatively open subset of X. Deﬁnition 1.12 (Regular Function). A function f : U → k is called regular if f is locally rational. That is, ∃ open cover U = ∪αUα and functions pα, qα ∈ A(X) such that ∀a ∈ Uα, qα(a) 6= 0 and f(a) = pα(a)/qα(a). We deﬁne k[U] to be the set of regular functions from U to k. Note: 1. k[U] is a k-algebra. 2. A(X) ⊆ k[X].

Example Let X = V (xy − zw) ⊆ A4. f : U → k can be deﬁned by f = x/w on D(w) and f = z/y on D(y). Thus, f ∈ k[U]. Exercise; 6 ∃p, q ∈ A(X) such that q(a) 6= 0 and f(a) = p(a)/q(a) for all a ∈ U.

5 Lemma 1.13. Let q1, . . . , qn ∈ A(X). Then D(q1) ∪ ... ∪ D(qn) = X iﬀ (q1, . . . , qm) = (1) = A(X). P Proof. ⇐: 1 = hiqi, hi ∈ A(X), then the qi cannot all vanish at any point, and so we are done. ⇒: Take Qi ∈ k[x1, . . . , xn] such that qi = Qi ∈ A(X). D(q1)∪...∪D(qm) = X, so X ∩ V (Q1) ∩ ... ∩ V (Qm) = ∅ = V (I(X),Q1,...,Qm) = ∅, and so, by the weak nullstellensatz, (I(X),Q1,...,Qm) = (1) ⊆ k[x1, . . . , xn], and so (q1, . . . , qm) = (1) = A(X).

Theorem 1.14. Let X ⊆ An be an algebraic set. Then k[X] = A(X).

Proof. Let f ∈ k[X]. Then X = U1 ∪ ... ∪ Um and there are pi, qi ∈ A(X) such that qi 6= 0 and f = pi/qi on Ui. We can reﬁne the open cover such that each Ui = D(gi) for some gi. Note: pigi f = pi/qi = on Ui = D(gi) = D(giqi). We can replace pi with pigi and qi qigi by qigi. 2 2 Then we can assume that Ui is D(qi) = D(qi ). Thus, X = D(q1) ∪ ... ∪ 2 Pm 2 2 D(qm). By the lemma, we know that 1 = i=1 hiqi , hi ∈ A(X). Note qi f = qipi on Ui, and qi = 0 outside of Ui. Pm 2 Pm f = 1f = i=1 hiqi f = i=1 hiqipi, so f ∈ A(X). Deﬁnition 1.13 (Spaces With Functions). A space with functions (SWF) is a topological space X together with an assignment to each open U ⊆ X of a k−algebra k[U] consisting of functions U → k. These are called regular functions. It must also satisfy the following:

1. If U = ∪Uα is an open cover and f : U → k any function, then f is

regular on U iﬀ f|Uα is regular on Uα for all α. 2. If U ⊆ X is open, f ∈ k[U], then D(f) = {a ∈ U : f(a) 6= 0} is open and 1 f ∈ k[D(f)].

Note: OX (U) = k[U] is another common notation. Examples:

1. Algebraic sets. These are called Aﬃne Algebraic Varieties

2. M is a diﬀerentiable manifold, k = R, k[U] = {C∞ functions U → R}.

3. X is a SWF, U ⊆ X open subset, then U is a SWF. OU (V ) = OX (V ).

Deﬁnition 1.14 (Morphism of SWFs). Let X,Y be SWFs, then a morphism ϕ : X → Y is a continuous map which pulls back regular functions to regular ∗ −1 functions. i.e., if V ⊆ Y is open, f ∈ OY (V ), then ϕ (f) ∈ OX (ϕ (V )), ϕ∗(f) = f ◦ ϕ. Deﬁnition 1.15 (Isomorphism). ϕ : X → Y is an isomorphism if ϕ is a morphism and ∃ a morphism ψ : Y → X such that ϕ◦ψ = idY and ψ ◦ϕ = idX .

6 Exercises

1. The id function of a SWF is a morphism. 2. Compositions of morphisms are morphisms.

3. Let X be any SWF and Y ⊆ An closed, that is, an aﬃne variety. Then f = (f1, . . . , fn): X → Y , is a morphism iﬀ fi ∈ k[X] for all i.

Example: A1 \{0} is (isomorphic to) an affine variety defined by V (1 − xy). Localization Let R be a ring and S ⊆ R multiplicatively closed subset. That is, s, t ∈ S ⇒ st ∈ S and 1 ∈ S. We define S−1R = {f/s : f ∈ R, s ∈ S}. We consider f/s to be the same element as g/t iff there exists u ∈ S such that u(ft − sg) = 0. This is a ring by f g fg f g ft+gs s t = st and s + t = st . Exercise: Check these assertions. −1 n Special Case: If f ∈ R, then Rf = S R where S = {f : n ∈ N}. In fact, ∼ Rf = R[y]/(1 − fy). Definition 1.16 (Reduced Ring). R is a reduced ring iff f n = 0 implies f = 0 for all f ∈ R. Equivalently, (0) = p(0). Facts:

1. R reduced implies S−1R is reduced √ 2. R/I reduced iﬀ I = I

Proposition 1.15. Let X ⊆ An be a closed affine variety and f ∈ A(X). Then D(f) if an affine variety, with affine coordinate ring A(X)f .

Proof. Let I = I(X) ⊆ k[x1, . . . , xn] define J = (I, yf − 1) ⊆ k[x1, . . . , xn, y]. n+1 −1 Let φ : D(f) → V (J) ⊆ A by (a1, . . . , an) 7→ (a1, . . . , an, f(a1, . . . , an) ). Note: φ is an isomorphism. What remains is to compute the coordinate ring. A(X) is reduced, and so A(√X)f is reduced. A(X)f = k[x1, . . . , xn, y]/J, so J is a radical ideal, so J = J = I(V (J)). Therefore, k[D(f)] = k[V (J)] = k[x1, . . . , xn, y]/J = A(X)f . Definition 1.17 (Prevariety). A prevariety is a space with functions X such that X has a finite open cover X = U1 ∪ ... ∪ Um where Ui is an affine variety. Example: Any affine variety is a prevariety. Exercise: Any prevariety is a NötherianTopological Space Example: An open subset of a prevariety is a prevariety. This follows from the previous proposition and the fact that principle open sets are a basis for the topology of any affine variety.

7 Proposition 1.16. X is a space with functions, Y ⊆ An an aﬃne variety, we have a 1-1 correspondence: {morphisms X → Y } ⇐⇒ {k-algebra homomorphism A(Y ) → k[X]} by ϕ ⇐⇒ ϕ∗

∗ n Proof. Note φ 7→ φ is a well defined map. Write A(A ) = k[y1, . . . , yn], then I(Y ) ⊆ k[y1, . . . , yn]. Theny ¯i is the image of yi in A(Y ). Assume that α : A(Y ) → k[X] is a k-algebra homomorphism. We define φ : X → An by φ(x) = (φ1(x), . . . , φn(x)). If f ∈ I(Y ) then f(y ¯1,..., y¯n) = 0, so f(φ1, . . . , φn) = α(f(y ¯1,..., y¯n)) = 0, ∗ ∗ and so φ(X) ⊆ Y . Note, φ (y ¯i) = yi ◦ φ = φi = α(y ¯i). Thus, φ = α. ∗ If φ : X → Y is a morphism, then φi = yi ◦ φ = φ (y ¯i) Thus, φ is the morphism that we construct from φ∗. Corollary 1.17. Two affine varieties are isomorphic iff their affine coordinate rings are isomorphic as k-algebras

Exercise: An \{(0,..., 0)} is not affine for n ≥ 2. Proposition 1.18. We have a one-to-one correspondence between affine varieties and reduced finitely generated k-algebras, up to isomorphism, by X 7→ k[X]. Proof. Last time, we proved that two affine varieties are isomorphic iff their coordinate rings are isomorphic. Thus, X 7→ k[X] is injective. Let R be a finitely generated reduced k-algebra generated by r1, . . . , rn ∈ R. There is a k-alg homomorphism φk[x1, . . . , xn] → R by xi 7→ ri which is surjective. Set I = ker φ and let X = V (I) ⊆ An. I is radical, as R is reduced, so k[X] = k[x1, . . . , xn]/I ' R.

Note: Assume m ⊆ R is a maximal ideal, then φ : k[x1, . . . , xn] → R as −1 in proof, then M = φ (m) is maximal, and M = (x1 − a1, . . . , xn − an). R/m = k[x1, . . . , xn]/M = k. Canonical Construction Let R be a finitely generated reduced k-algebra. Then define Spec −m(R) = {m ⊆ R max ideals}. The topology will be that the closed sets V (I) = {m ⊇ I|I ⊆ R and ideal }. Let f ∈ R. We define f : Spec −m(R) → k by f(m) =image of f in R/m = k. So f is a function from Spec −m to k. I.E., f(m) ∈ k ⊆ R is the unique element such that f − f(m) ∈ m. Finally, if U ⊆ Spec −m is open, f : U → k is some function, then f is regular if f is locally of the form f(m) = p(m)/q(m) where p, q ∈ R. Exercise: Spec −m(R) =∼ X, where X is the affine variety with coordinate ring R, as spaces with functions. Subspaces of SWFs Let X be any space with functions, and Y ⊆ X any subset. Then give Y an ”inherited” SWF structure as follows: We give Y the subspace topology, and if U ⊆ Y is open and f : U → k is a function, then f is regular iff f can be locally extended to a regular function on

8 X. That is, for every point y ∈ U, there is an open subset U 0 ⊆ X containing 0 0 y and F ∈ OX (U ) such that f(x) = F (x) for all x ∈ U ∩ U . Exercises 1. Y is a SWF

2. i : Y → X the inclusion map is a morphism. 3. Let Z be a SWF, φ : Z → Y function. Then φ is a morphism iﬀ i ◦ φ is a morphism. 4. The SWF structure on Y is uniquely determined by (2) and (3) together.

5. Let Z ⊆ Y ⊆ X. Then Z inherits the same structure from Y and X.

Example: X ⊂ An an algebraic set inherits structure from An. If Y ⊆ X is closed, then Y inherits structure from X (or An). Proposition 1.19. A closed subset of a prevariety is a prevariety.

Proof. Let X be a prevariety, and Y ⊆ X a closed subset. X = U1 ∪ ... ∪ Um where Ui are open aﬃne subsets of X. Ui ∩ Y is a closed subset of Ui, which implies that Ui ∩ Y is aﬃne, and so Y has the open cover (U1 ∩ Y ) ∪ ... ∪ (Un ∩ Y ), and so is a prevariety. Projective Space Theorem 1.20. Two distinct lines in the place intersect in exactly one point. (except when parallel) Theorem 1.21. A line meets a parabola in exactly two points. (except when false)

We need projective space to remove the bad cases.

Definition 1.18 (Projective Space). Define an equivalence relation on An+1 \ {0} by ∗ (a0, . . . , an) ∼ (λa0, . . . , λan) where λ ∈ k = k \{0}. So Pn = (An+1 \{0})/ ∼ and π : An+1 \{0} → Pn the projection. There is a topology on Pn given by U ⊆ Pn is open iff π−1(U) ⊆ An+1 is open. The regular functions on Pn are f : U → k such that π∗(f) = π ◦ f : π−1(U) → k is regular. Thus, Pn is a SWF called Projective Space. Note: Pn = {lines through the origin in An+1}, and this method of thinking is often very helpful. We will use the notation (a0 : ... : an) for the image π(a0, . . . , an) ∈ n P . If f ∈ k[x0, . . . , xn] is a homogeneous polynomial of total degree d, then d f(λa0, . . . , λan) = λ f(a0, . . . , an). Thus it is well-defined to ask if f(a0 : ... : an) = 0 or not.

9 n Deﬁnition 1.19. D+(f) = {(a0 : ... : an) ∈ P : f(a0 : ... : an) 6= 0}. Theorem 1.22. Pn is a prevariety. n Proof. Let Ui = D+(xi) ⊆ P for 0 ≤ i ≤ n. n Claim: Ui ' A .

n φ : A → Ui :(a0, . . . , ai−1, ai+1, . . . , an) 7→ (a0 : ... : ai−1 : 1 : ai+1 : ... : an)

n a0 aˆi an ψ : Ui → A :(a0 : ... : an) 7→ ,..., ,..., ai ai ai

n ` n ` n−1 n Note:P = D+(x0) V+(x0) = A P , that is, A = {(1 : a1 : . . . an)} n−1 the usual n-space and P = {0 : a1 : ... : an} points at ∞. The points at ∞ correspond to lines through the origin in An, that is, we can think of them as being directions. Example: P2 has ”homogeneous coordinate ring” k[x, y, z]. We can think of 2 2 2 that as A = D+(z) ⊂ P and we know that k[A ] = k[x/z, y/z]. We want to intersect a parabola with a line. 2 The vertical line is V (x/z − 1) = V+(x−z) and the parabola is V (y/z − (x/z) ) = 2 V+(yz − x ). 2 And so, V+(x − z) ∩ V+(yz − x ) = {(1 : 1 : 1), (0 : 1 : 0)}, where (0 : 1 : 0) is the point at infinity in the direction ”up”. Exercise: k[Pn] = k. Exercise: X a SWF, φ : Pn → X a function, then φ is a morphism iff φ ◦ π : An+1 \{0} → X is a morphism. n Definition 1.20 (Projective Coordinate Ring of P ). We define k[x0, x1, . . . , xn] n to be the coordinate ring of P . An ideal I ⊆ k[x0, . . . , xn] is homogeneous if it is generated by homogeneous polynomials. Equivalently, f ∈ I iff each homogeneous component is in I.

n −1 Deﬁnition 1.21. If W ⊆ P is a subset, then I(W ) = I(π (W )) ⊆ k[x0, . . . , xn]. Notice that I(W ) is homogeneous. Let f = f0 + ... + fd ∈ I(W ), fi a form of degree i, then (a0 : ... : an) ∈ W , so 0 = f(λa0, . . . , λan) = f0(a0, . . . , an) + d ... + λ fd(a0, . . . , an). As this is true for all λ, fi(a0, . . . , an) = 0 for all i, and so fi ∈ I(W ).

Deﬁnition 1.22. If I ⊆ k[x0, . . . , xn] is a homogeneous ideal, then deﬁne n V+(I) = {(a0 : . . . , an) ∈ P : f(a0, . . . , an) = 0 for all f ∈ I}

Theorem 1.23 (Projective Nullstellensatz). If I ⊆ k[x0, . . . , xn] is a homogeneous ideal, then √ N 1. V+(I) = ∅ ⇒ (x0, . . . , xn) ⊆ I for some N > 0. That is, I = (1) or (x0, . . . , xn).

10 √ 2. V+(I) 6= ∅ then I(V+(I)) = I.

Proof. 1. V√+(I) = ∅ ⇐⇒ V (I) = ∅ or V (I) = {0}. By the regular nullstellensatz, I = I(V (I)) = (1) or (x0, . . . , xn).

−1 −1 n+1 2. V+(I) 6= ∅. Then π √(V+(I)) = π (V+(I)) ∪ {0} = V (I) ⊆ A . So I(V+(I)) = I(V (I)) = I.

This gives us a 1-1 correspondence between closed subsets of Pn and radical homogeneous ideals in k[x0, . . . , xn] except for (x0, . . . , xn). This ideal is often called the irrelevant ideal. Definition 1.23 (Locally Closed subset). X is a topological space, W ⊆ X a subset is locally closed if it is the intersection of an open set in X and a closed set in X. Note: A locally closed subset of a prevariety is a prevariety. Terminology: a projective variety is any closed subset of Pn considered as a space with functions. A Quasi-projective variety is a locally closed subset of Pn. An affine variety is a closed subset of An. A quasi-affine variety is a locally closed subset of An. We notice that anything affine is also quasi-affine and anything quasi-affine is quasi-projective. Something that is projective will also be quasi-projective. Exercise: Pn is not quasi-affine for n ≥ 1. Later: If X is both projective and quasi-affine, then X is finite.

Deﬁnition 1.24 (Projective Coordinate Ring). X ⊆ Pn is a closed projective variety, then the projective coordinate ring of X = k[x0, . . . , xn]/I(X).

Warning: This deﬁnition depends on the embedding of X in Pn. Example: φ : P1 → P2 by φ(a : b) = (a2 : ab : b2). This is a morphism. In 1 2 1 fact, it is an isomorphism of P and V+(xz − y ), but the coordinate ring of P 2 2 is k[s, t] and the coordinate ring of V+(xz − y ) is k[x, y, z]/(xz − y ). These two rings are NOT isomorphic as k-algebras.

Definition 1.25 (Projective Closure of an affine variety). X ⊆ An is affine. n n n n Then we know that A = D+(x0) ⊆ P , and X ⊂ A ⊆ P makes X a quasi- projective variety, so we take X¯ = the closure of X in Pn.

I = I(X) ⊆ k[x1, . . . , xn] if f = f0 + ... + fd ∈ k[x1, . . . , xn] where fi is a ∗ d d−1 form of degree i. Then we deﬁne f = x0f0 + x0 f1 + ... + fd ∈ k[x0, . . . , xn]. ∗ ∗ And I is the ideal generated by {f : f ∈ I} in k[x0, . . . , xn]. Exercise: I(X¯) = I(X)∗. 2 2 3 Example: I = (y−x , z −x ) ⊆ k[x, y, z]. Then X = V (I) ⊆ A = D+(w) ⊂ Pn. I(X) = I∗ = (yw − x2, y − z) ⊇ (wy − x2, wx − x2) 2 2 So V+(wy − x , wz − x ) = X¯ ∪ V+(x, w). We now recall that a graded ring is a ring R with decomposition R = ⊕d≥0Rd as an abelian group such that Rd · Re ⊆ Rd+e.

11 e.g., R = k[x0, . . . , xn]. f ∈ Rd ⇒ Rf is a Z-graded ring g ∈ Rp implies that m g/f ∈ Rf is homogeneous of degree p − md.

Deﬁnition 1.26. R(f) = {homogeneous elements of degree zero} = (Rf )0 = m {g/f : g ∈ Rdm}.

Exercise: f ∈ k[x0, . . . , xn] homogeneous implies that k[x0, . . . , xn](f) is a ﬁnitely generated reduced k-algebra.

n Theorem 1.24. f∈ / k ⇒ D+(f) ⊆ P is aﬃne and in fact k[D+(f)] = k[x0, . . . , xn](f). ∗ Proof. k[D+(f)] = {h ∈ k[D(f)] : h(λx) = h(x), ∀λ ∈ k , x ∈ D(f)}. m If h ∈ k[D(f)] = k[x0, . . . , xn]f , h = g/f , g ∈ k[x0, . . . , xn].

g g (λa , . . . , λa ) = (a , . . . , a ) ⇐⇒ g homogeneous of degree md f m 0 n f m 0 n

Therefore, k[D+(f)] = k[x0, . . . , xn](f). The identity map k[D+(f)] → k[D+(f)] gives a morphism φ : D+(f) → Spec −m(k[D+(f)]) by φ(x) = Mx where Mx = I({x}) ⊆ k[D+(f)] Observe that if x, y ∈ D+(f), x 6= y then ∃ homogeneous g ∈ k[x0, . . . , xn] g such that deg(g) = d and g(x) = 0 with g(y) 6= 0. So Mx 6= My. f ∈ Mx,/∈ My. d xi Thus, φ is injective. Set hi = f ∈ k[D+(f)] for 0 ≤ i ≤ n. Ui = D(hi) ⊆ D+(f), Vi = D(hi) ⊆ Spec −m(k[D+(f)]) = {m 63 hi}. n n Now we must check that D+(f) = ∪i=0Ui and Spec −m(k[D+(f)]) = ∪i=0Vi. It is enough to prove that φ : Ui → Vi is an isomorphism for all i. n D+(xi) ⊆ P is affine. k[D+(xi)] = k[x0/xi, . . . , xn/xi] = k[x0, . . . , xn](xi). d Thus, Ui = D(f/x ) ⊆ D+(xi) is affine, so k[Ui] = (k[x0, . . . , xn](x )) d , i i f/xi which is k[x0, . . . , xn](x,f) = k[D+(f)]hi = k[Vi]. Thus, Ui ' Vi. 2 2 Example: f = xz − y ∈ k[x, y, z]. X = D+(f) ⊆ P , R = k[x, y, z](f). R is generated by A = x2/f, B = y2/f, C = z2/f, D = xy/f, E = yz/f, and F = xz/f. So X ' V (AB − D2, AC − F 2,BC − E2,F − B − 1) ⊆ A6 by (x : y : z) 7→ (A, B, C, D, E, F ). n Exercise:X ⊆ P a projective variety f ∈ R = k[x0, . . . , xn]/I(X) is homogeneous, then D+(f) ⊆ X is affine with affine coordinate ring k[D+(f)] = R(f).

2 Algebraic Varieties

Products Let X,Y be two sets. Then X × Y , the cartesian product, is the set {(x, y): x ∈ X, y ∈ Y }. What is X ×Y , really? Well, it is a set with projection πX : X ×Y → X and πY : X × Y → Y . This set with the projections satisﬁes a universal property in the category of sets.

12 For any set Z with arbitrary functions f : Z → X and g : Z → y, there exists a unique function φ : Z → X × Y such that f = πX ◦ φ and g = πY ◦ φ.

...... X ...... f ...... π ...... X ...... ∃! ...... Z ...... X × Y ...... g ..... π ...... Y ...... Y Definition 2.1 (Product of SWFs). Let X,Y be spaces with functions. A product of X and Y is a SWF called X × Y with morphism πX : X × Y → X and πY : X × Y → Y which satisfies the above universal property except with ”morphisms” rather than ”functions”. 0 0 0 Exercise: Assume that (P, πX , πY ) and (P , πX , πY ) are two products of X and Y . Then they are isomorphic by unique isomorphism. (See homework problem) Example: A1 × A1 = A2. NOTE: A2 does not have the product topology! General Construction: X,Y spaces with functions. Then X × Y = {(x, y): x ∈ X, y ∈ Y } is the point set. If U ⊂ X and V ⊂ Y are open, then U × V ⊂ −1 −1 X × Y is open, as it is πX (U) ∩ πY (V ). Pn Let g1, . . . , gu ∈ OX (U) and h1, . . . , hn ∈ OY (V ) set f(u, v) = i=1 gi(u)hi(v). Then f : U × V → k must be regular. Thus, DU×V (f) = {(u, v) ∈ U × V : f(u, v) 6= 0} must be open in X × Y . And so, we define our topology by S ⊆ X × Y is open iff it is a union of sets DU×V (f). The regular functions F : S → k are the functions that can locally be written as f 0(u, v)/f(u, v) on some DU×V (f). That is, ∀(x, y) ∈ S ∃U ⊆ X open and V ⊆ Y open and f(u, v) = Pn 0 Pm 0 0 0 0 i=1 gi(u)hi(v) and f (u, v) = j=1 gj(u)hj(u) with gi, gj ∈ OX (U) and hi, hj ∈ 0 OY (V ) such that (x, y) ∈ DU×V (f) ⊆ S and F (u, v) = f (u, v)/f(u, v) for all (u, v) ∈ DU×V (f). Exercises (for X × Y above) 1. X × Y is an SWF

2. πX : X × Y → X and πY : X × Y → Y are morphisms. 3. X × Y is the product of X and Y . Remark: X,Y SWFs, and U ⊆ X, V ⊆ Y are arbitrary subsets, U, V have inherited structure as SWFs. Then U × V has the product space with functions structure and subspace SWF structure U × V ⊆ X × Y . These are in fact the same, due to the universal properties. For now, we call U × V the product. We obtain the following diagram. ... U ...... X ...... πU ...... πX ...... φ ...... i ... Z ...... U × V ...... X × Y ...... πV .....πY ...... V ...... Y

13 φ is a morphism iﬀ i ◦ φ is one, and so we see that the two structures are the same. Deﬁnition 2.2 (Separated SWF). A SWF X is separated if ∀ SWFs Y and morphisms f, g : Y → X the set {y ∈ Y : f(y) = g(y)} ⊆ Y is closed.

1 1 Example: Let X = (A \{0}) ∪ {O1,O2}. We can define φi : A → X by a a 6= 0 taking a 7→ Oi a = 0 −1 1 We define a topology by U ⊆ X is open iff φi (U) ⊆ A is open for all i.A ∗ −1 function f : U → k is regular iff φi (f) = f ◦ φi : φi (U) → k is regular for all i. 1 1 1 1 X is a prevariety as X = φ1(A ) ∪ φ2(A ) and φi(A ) ' A . However, it is 1 1 1 not separated, as {a ∈ A : φ1(a) = φ2(a)} = A \{0} is not closed in A . Definition 2.3 (Algebraic Variety). An algebraic variety is a separated prevariety. Exercise: 1. Any subspace of a separated SWF is separated. 2. A product of separated SWFs is separated.

Remark: If X is any SWF, then ∆ : X → X ×X : x 7→ (x, x) is a morphism. Now we set ∆X = ∆(X) ⊆ X × X. Then ∆ : X → ∆X is an isomorphism.

Lemma 2.1. X is separated iﬀ ∆X ⊆ X × X is closed.

Proof. ⇒: πi : X × X → X be the projections. ∆X = {z ∈ X × X : π1(z) = π2(z)} is closed. ⇐: Let Y be a SWF, f, g : Y → X maps. Deﬁne φ : Y → X × X by −1 φ(y) = (f(y), g(y)) is a morphism. Now {y ∈ Y : f(y) = g(y)} ⇒ φ (∆X ) is closed.

Exercise: A topological space X is Hausdorff iff ∆X ⊆ X × X is closed. NB: Product topology! Exercise: An × Am = An+m. Proposition 2.2. All affine varieties are varieties.

Proof. Enough to show that An itself is separated. n n 2n 2n n ∆A ⊆ A × A = A is closed, as k[A ] = [x1, . . . , xn, y1, . . . , yn], so n ∆A = V ({xi − yi}). Products of Aﬃne Varieties

Lemma 2.3. Let A, B be ﬁnitely generated reduced k-algebras. Then A ⊗k B is a ﬁnitely generated reducted k-algebra and k algebraically closed.

Recall: ring structure on A ⊗ B by (a1 ⊗ b1)(a2 ⊗ b2) = a1a2 ⊗ b1b2.

14 Proof. If A is generated by a1, . . . , an and B is generated by b1, . . . , bm then A ⊗ B is generated by a1 ⊗ 1, . . . , an ⊗ 1, 1 ⊗ b1,..., 1 ⊗ bm. For example, k[x1, . . . , xn] ⊗ k[y1, . . . , ym] = k[x1, . . . , xn, y1, . . . , ym]. Let X,Y be affine varieties such that k[X] = A, k[Y ] = B. We define φ : A ⊗k B →the set of all functions X × Y → k by f ⊗ g 7→ (x, y) 7→ f(x)g(x) is a k-algebra homomorphism. Pn φ is injective (which implies that A⊗B is reduced): Suppose φ ( i=1 fi ⊗ gi) = 0. WLOG we can assume g1, . . . , gn are linearly independent. Let x ∈ X. Then Pn i=1 fi(x)gi = 0 ∈ B, but {gi} is linearly independent so fi(x) = 0 for all i and x, thus fi = 0 ∈ A. Theorem 2.4. 1. If X,Y are affine then X × Y is affine and k[X × Y ] = k[X] ⊗k k[Y ]. 2. A product of prevarieties is a prevariety.

Proof. 1 ⇒ 2: X,Y prevarieties, X = ∪Ui, Y = ∪Vj with Ui,Vj affine. Then X × Y = ∪i,jUi × Vj affine. Now, we must only prove 1. Set P = Spec-m(k[X]⊗k[Y ]), We have k-algebra homomorphisms k[X] → k[X] ⊗ k[Y ]: f 7→ f ⊗ 1 and k[Y ] → k[X] ⊗ k[Y ]: g 7→ 1 ⊗ g. These give morphism πX : P → X and πY : P → Y . Claim: P = X × Y . Let Z be a SWF, p : Z → X and q : Z → Y . Define k-alg homomorphism k[X] 7→ k[Y ] → k[Z] by f ⊗ g 7→ p∗(f)q∗(g). This gives us a morphism φ : Z → P demonstrating that P is the product. Remark: If Y is affine and X ⊆ Y is closed, then k[X] ' k[Y ]/I(X). On the other hand, assume X,Y are affine, ϕ : X → Y is a morphism and ϕ∗ : k[Y ] → k[X] is surjective. Then set I = ker(ϕ∗) ⊆ k[Y ], we get ϕ∗ ...... k[Y ] ...... k[X] YX...... ϕ ...... ' ...... inj ...... ' k[Y ]/I V (I) Therefore ϕ is an embedding of X as a closed subset of Y . Recall: ∆ : X → X × X : x 7→ (x, x) gives X ' ∆X = ∆(X) = {(x, x): x ∈ X}. Proposition 2.5. A prevariety X is separated iff ∀ open affine U, V ⊆ X, U ∩V is affine and k[U × V ] = k[U] ⊗ k[V ] → k[U ∩ V ] = k[∆U∩V ] is surjective.

Proof. ⇒: U ∩ V ' ∆U∩V = ∆X ∩ (U × V ) ⊆ U × V is closed, thus U ∩ V is aﬃne and k[U × V ] → k[U ∩ V ] is surjective. ⇐: If U, V, U ∩ V are aﬃne and k[U × V ] → k[U ∩ V ] is surjective, then ∆ : U ∩V → U ×V is an inclusion of closed subsets. Thus, ∆X ∩(U ×V ) ⊆ U ×V closed. So if X × X = ∪U × V is an open cover then ∆X ⊆ X × X is closed. Exercises

1. X is a prevariety such that ∀x, y ∈ X there is an open aﬃne U ⊆ X such that x, y ∈ U. Then X is separated.

15 2. Pn has this property. Corollary 2.6. Quasi-projective varieties are separated. We want to show that the products of projective varieties are again projective. Let X ⊆ Pn and Y ⊆ Pm are closed. Then X × Y ⊆ Pn × Pm is closed. It is enough to show that Pn × Pm is projective, that is, Pn × Pm ⊂ PN is closed. Segre Map: If N = (n + 1)(m + 1) − 1 = nm + n + m, then we can deﬁne n m N s : P × P → P :(x0 : ... : xn) × (y0 : ... : ym) 7→ (x0y0 : x0y1 : ... : x0ym : x1y0 : ... : xnym). N We call the projective coordinates on P as zij for 0 ≤ i ≤ n, 0 ≤ j ≤ m. n m N Exercise: s : P × P → V+({zijzpq − ziqzpj}) ⊆ P 2 Note: Pn × Pn ⊆ Pn +2n is closed. N n Exercise: ∆P = V+({zij − zji}) ⊆ P . Complete Varieties: analogues of compact manifolds Deﬁnition 2.4 (Complete). A variety X is complete if for any variety Y , the projection πY : X × Y → Y is closed. (i.e.: Z ⊆ X × Y is closed implies that πY (Z) ⊆ Y is closed) Note: 1) closed subsets of complete varieties are complete. 2) Products of complete varieties are complete. Examples:Points are complete. A1 is not complete, as Z = V (xy − 1) ⊆ A1 × A1 = A2 = X × Y is such that 1 πY (Z) is not closed, as it is A \{0} Proposition 2.7. Let ϕ : X → Y be a morphism of varieties. If X is complete, then ϕ(X) is closed in Y and is complete. Proof. Γ(ϕ) = {(x, ϕ(x)) ∈ X × Y : x ∈ X} ⊆ X × Y = (ϕ × 1)−1(∆Y ) As Y is separated, Γ(ϕ) ⊆ X × Y is closed. X is complete implies that ϕ(X) = πY (Γ(ϕ)) ⊆ Y is closed. Now, let Z ⊆ ϕ(X) × Y 0 be closed. Then ϕ × 1 0 .... 0 X × Y ...... ϕ(X) × Y ...... 0 ...... π˜Y ...... 0 ...... πY ...... Y 0 −1 0 W = (ϕ×1) (Z) ⊆ X ×Y is closed, πY 0 (Z) = πY 0 ((ϕ×1)(W )) =π ˜Y 0 (W ) is closed. Exercise: ϕ : X → Y is a continuous map of topological spaces then X is irreducible implies that ϕ(X) is irreducible. Proposition 2.8. If X is an irreducible complete variety then k[X] = k.

16 Proof. Let f ∈ k[X]. f : X → A1 a morphism. As X is complete, f(X) must be is irreducible, closed and complete. Thus, it must be a point. Thus, f is constant. Proposition 2.9. A complete quasi-aﬃne variety is ﬁnite.

Proof. X is such a variety, without loss of generality X is irreducible. X ⊆ An is n locally closed, then xi : X ⊆ A → k must be constant, and so x is a point. Theorem 2.10. Pn is complete.

Note: I ⊆ S = k[x0, . . . , xn] a homogeneous ideal, then V+(I) is not empty iff Id ( Sd for all d ∈ N. n Proof. Let Y be a variety and Z ⊆ P × Y be closed. Show that πY (Z) ⊆ Y is closed. Y = ∪Yi and open affine cover. It is enough to show that πY (Z) ∩ Yi ⊆ Yi n n is closed, that is, πYi (Z ∩ (P × Yi)) is closed. So we take πYi : P × Yi → Yi. Thus, WLOG, we assume Y is affine. Then let C(Z) = (π × id)−1(Z) ⊆ n+1 n+1 A × Y . k[A × Y ] = S ⊗ k[Y ] = k[Y ][x0, . . . , xn] = ⊕d≥0Sd ⊗k k[Y ], so it is a graded ring. Note that (y, (a0, . . . , an)) ∈ C(Z) implies that (y, (λa0, . . . , λan)) ∈ C(Z) for λ ∈ k. Thus, I(C(Z)) ⊆ k[An+1 × Y ] is a homogeneous ideal. We write I(C(Z)) =

(f1, . . . , fm) with fi ∈ Sdi ⊗ k[Y ]. n For y ∈ Y , fi(y) = fi(−, y) ∈ Sdi . We observe that y ∈ πY (Z) iff ∃x ∈ P n such that (x, y) ∈ Z. This happens iff V+(f1(y), . . . , fm(y)) 6= ∅ ⊆ P . This is true iff (f1(y), . . . , fm(y))d 6= Sd for all d ≥ 0. m Fix d ≥ 0, then y ∈ Y defines a linear map ΦY : ⊕i=1Sd−di → Sd : Pm (g1, . . . , gm) 7→ i=1 fi(y)gi. Note: Every entry of the matrix ΦY is a regular function of Y . n + d Now, (f (y), . . . , f (y)) 6= S iff rank(Φ ) < dim(S ) = which 1 m d d Y d n n + d holds iff all minors in Φ of size vanish. Therefore, W = {y ∈ Y : Y n d (d1(y), . . . , fm(y))d 6= Sd} ⊆ Y is closed. Finally, πY (Z) = ∩d≥0Wd, and so is closed. Challenge: Find a complete variety that is not projective. Exercise:

1. Let X be a topological space and W ⊆ X is a subset. W = X iﬀ W ∩U 6= ∅ for all nonempty open sets U ⊆ X. 2. f : X → Y continuous and W = X and f(X) = Y then f(W ) = Y 3. X is irreducible and ∅= 6 U ⊆ X is open. Then U = X and U is irreducible.

17 Rational Map: X and Y are irreducible varieties. The ideal is that a morphism f : X → y is uniquely determined by restriction to any non-empty open subset of X. Consider pairs (U, f) where U ⊆ X nonempty and open and f : U → Y is a morphism. Relation: (U, f) ∼ (V, g) iﬀ f = g on U ∩ V because Y is separated and X is irreducible, this is an equivalence relation. Checking this is an exercise.

Definition 2.5 (Rational Map). A rational map f : X 99K Y is an equivalence class for ∼. X irreducible implies that U ∩ W ⊇ U ∩ V ∩ W dense, and Y separated implies that f = h on a closed subset of U ∩ W . Remark: If f : X 99K Y then there is a maximal open U ⊆ X where f is defined as a morphism. U = ∪(V,g)∼f V ⊂ X. 2 2 2 Example: f : A 99K A :(x, y) 7→ (x/y, y/x ) defined as a morphism of D(xy). 2 2 2 2 2 Exercise: f : A 99K A ⊆ P . Find the max open where f : A 99K P is defined. Definition 2.6 (Rational Function). A rational function on X is a rational 1 map f : X 99K A = k. f is given by a regular function f : U → k where ∅= 6 U ⊆ X open. k(X) = {f : X 99K k} is the field of rational functions on X. Note: If (U, f), (V, g) ∈ k(X) then f + g, f − g, fg : U ∩ V → k define rational functions on X. If f 6= 0 in k[U] then ∅ 6= D(f) ⊆ U is open, and 1/f : D(f) → k is regular. Thus, 1/f = (D(f), 1/f) ∈ k(X). n n Examples: k(A ) = k(x1, . . . , xn). k(P ) = k(x1/x0, . . . , xn/x0). Proposition 2.11. Let X be an irreducible variety

1. If ∅= 6 U ⊆ X open, then k(X) = k(U).

2. If X is aﬃne, then k(X) = k[X]0 =ﬁeld of fractions of k[X].

Proof. 1. k(X) → k(U):(V, g) → (V ∩ U, g|V ∩U ) is isomorphism.

2. Deﬁne k[X]0 → k(X): f/g 7→ (D(g), f/g). Injective: As this is a homomorphism of ﬁelds, it is enough to say that it is not identically zero, and it maps 1 to (X, 1), which is not the zero function. Surjective: If f : U → k is regular, ∅ 6= U ⊆ X open. Find 0 6= g ∈ k[X]

such that ∅= 6 D(g) ⊆ U. Then f ∈ k[D(g)] = k[X]g ⊆ k[X]x0 .

Deﬁnition 2.7 (Dominant). (U, f): X 99K Y is dominant if f(U) = Y .

18 Indep. of rep.: If ∅= 6 V ⊆ U open, then f(V ) = Y by the homework. Suppose (U, f): X 99K Y is dominant and (V, g): Y 99K Z is any rational map, then we can compose g ◦ f : X 99K Z, as f(U) = Y so f(U) ∩ V 6= ∅, so f −1(V ) 6= ∅ ⊆ U. g ◦ f = (f −1(V ), g ◦ f). Exercise: If f, g both dominant, then g ◦ f dominant. Proposition 2.12. X,Y irreducible varieties, then there is a one to one correspondence between {φ : X → Y with φ dominant} and field extensions k(Y ) ⊆ k(X) by φ 7→ φ∗ = [h 7→ h ◦ φ] Proof. WLOG X,Y are affine. The map φ 7→ φ∗ is: ∗ ∗ Injective: So we let ψ : X 99K Y dominant and ψ = φ . Take D(h) ⊆ X such that φ and ψ are both defined on D(h). φ∗ = ψ∗ ... k(Y ) ...... k(X) ...... ⊆ . ⊆ ...... k[Y ] ...... k[X]h This diagram commutes, and so φ = ψ on D(h). Surjective: Let α : k(Y ) → k(X) be a k-algebra homomorphism. k[Y ] is generated by f1, . . . , fn. α(fi) = gi/hi, gi, hi ∈ k[X]. Let h = h1 . . . hn ∈ k[X]. So α : k[Y ] → k[X]h is a k-alg hom. This gives us a morphism φ : D(h) → Y and φ∗ = α.

Deﬁnition 2.8 (Birational). Let f : X 99K Y be a rational map. It is birational if f is dominant and there is a dominant g : Y 99K X such that f ◦ g = idY and g ◦ f = idX as rational maps. Deﬁnition 2.9 (Birationally Equivalent). X and Y are birationally equivalent (often X and Y are birational) written X ≈ Y if there exists f : X 99K Y a birational map.

Examples: A2 ≈ P2 ≈ P1 × P1. If U, V ⊆ X open and X irred., then U ≈ Y . Theorem 2.13. The following are equivalent

1. X ≈ Y 2. k(X) ' k(Y ) as k-algebras 3. ∃∅= 6 U ⊂ X,V ⊂ Y open such that U ' V as varieties.

Proof. 3 ⇒ 2 is clear from the ﬁrst prop on this topic. 2 ⇒ 1 is clear from the second prop on this topic. 1 ⇒ 3: Let (U, f): X 99K Y and (V, g): Y 99K X be inverses. Set −1 U0 = f (V ) ⊆ U. Then g ◦ f : U0 → V → X must be the inclusion of U0 into

19 −1 −1 X. Thus, g(f(U0)) ⊆ U0, so f(U0) ⊆ g (U0). Set V0 = g (U0) ⊆ V . Then U0 ' V0 as varieties by f, g. Deﬁnition 2.10 (Rational Variety). An irreducible variety is rational if it is birational to An for some n.

Examples: Any curve C ⊆ P2 of degree 2 is rational. 2 3 2 2 1 Let C = V (y −x −x ) ⊆ A is rational. Let φ : C 99K A by φ(x, y) = y/x. The inverse should be ψ : A1 → C by ψ(t) = (1 − t2, t − t3). So φ ◦ ψ(t) = 3 2 1 (t − t )/(1 − t ) = t = idA , and the opposite is also an identity (Exercise, show this). Challenge: Show that E = V (y2 − x3 + x) ⊆ A2 is not rational. Big Challenge: If C is any irreducible variety and there exists a dominant 1 rational map A 99K C then C is rational. Transcendence Degree Let k ⊆ L a ﬁeld extension. Then L is algebraic over k if for all f ∈ L, there n n−1 is a polynomial equation f + a1f + ... + an = 0 for ai ∈ k. S ⊆ L subseteq, then S is algebraically independent over k if for all s1, . . . , sn ∈ S with si 6= sj for i 6= j, then k[x1, . . . , xn] → L by xi 7→ si is injective. Deﬁnition 2.11 (Transcendence Basis). A transcendence basis for L over k is a set B ⊆ L such that B is algebraically independent over k and L is algebraic over k(B). Theorem 2.14. 1. All transcendence bases have the same cardinality.

2. If S ⊆ Γ ⊆ L subsets such that S is alg indep over k and L is alg over k(Γ), then there exists a transcendence basis B for L over k such that S ⊆ B ⊆ Γ. Proof. The idea is as ”any vector space has a basis.” Exercise: Prove where L is a ﬁnitely generated extension of k. Lang’s Algebra contains the proof.

Definition 2.12 (Transcendence Degree). The transcendence degree tr degk(L) = tr deg(L) = the number of elements in any transcendence basis for L over k. Definition 2.13 (Dimension). Let X be an irreducible variety. Then define dim(X) = tr deg(k(X)). Examples:

n 1. dim(A ) = tr degk k(x1, . . . , xn) = n 2. If X is irreducible and dim(X) = 0, then k ⊆ k(X) is algebraic extension, then k(X) = k. Thus, X is a point.

20 Some terminology: a curve is a variety of dimension 1, a surface is a variety of dimension 2, and an n-fold is a variety of dimension n. Notation: If R is a finitely generated domain over k, then we write tr deg(R) = tr deg(R0). We will state the following without proof. Theorem 2.15 (Principle Ideal Theorem). If R is a finitely generated domain over k and 0 6= f ∈ R and P ⊆ R is a minimal prime, then P is a minimal prime containing f. Then tr deg(R/P ) = tr deg(R) − 1. Geometric Statement: If X is any irreducible variety and 0 6= f ∈ k[X] and if Z ⊆ V (f) is an irreducible component then dim Z = dim X − 1. Proof. Take U ⊆ X open affine such that U ∩ Z 6= ∅. Then Z ∩ U = V (P ) ⊆ U, P ⊆ k[U] prime ideal. Z is a component of V (f) ⇒ P is minimum over (f) ⊆ k[U]. Thus, dim(Z) = tr deg(k[U]/P ) = tr deg k[U] − 1 = dim X − 1.

Theorem 2.16. Let X be an irreducible variety, and let ∅= 6 X0 ( X1 ( ... ( Xn = X be a maximal chain of irreducible closed subsets. Then dim(X) = n.

Proof. WLOG, X is aﬃne. Take 0 6= f ∈ I(Xn−1). Then Xn−1 ⊆ V (f) is a component. PIT says that dim Xn−1 = dim X − 1. Induction implies that dim Xn−1 = n − 1, so dim X = n. Deﬁnition 2.14. If X is any variety, set dim(X) =the supremum of all n such that ∃ a chain ∅= 6 X1 ( X1 ( ... ( Xn ⊆ X where Xi ⊆ irreducible and closed for all i. Exercises

1. X = X1 ∪ ... ∪ Xm and Xi ⊆ X closed, then dim X = max dim(Xi) 2. dim(X × Y ) = dim X + dim Y Recall: If R is a ring, then dim(R) is the supremum of all n such that ∃ Pn ( Pn−1 ( ... ( P0 ⊆ R where Pi is a prime ideal. Note: If X is aﬃne then dim X = dim k[X]. Theorem 2.17 (PIT For Several Equations). If X is an irreducible variety and f1, . . . , fr ∈ k[X] and Z ⊆ V (f1, . . . , fr) are components, then dim Z ≥ dim X − r. Proof. Enough to show that if W ⊆ X is a closed subset and each component of W has dim ≥ d, then each component of W ∩ V (f) has dim ≥ d − 1 for all f ∈ k[X]. Let Z ⊆ W be a component. If f|Z = 0 then V (f) ∩ Z = Z. If f|Z 6= 0 then every component of Z ∩ V (f) has dim= dim(Z) − 1 ≥ d − 1. Therefore W ∩ V (f) = union of ﬁnitely many irreducible closed subsets of dim ≥ d − 1.

21 Lemma 2.18 (Prime Avoidance). X is an aﬃne variety, Z ⊆ X an irreducible closed subset and X1,...,Xm ⊆ X are also irreducible closed subsets, then if Xi 6⊆ Z then ∃f ∈ I(Z) such that f∈ / I(Xi). Proof. Induction on m. If m = 1, then X1 6⊆ Z ⇒ I(Z) 6⊆ I(X1) ⇒ ∃f ∈ I(Z) \ I(X1). For m ≥ 2, take fi ∈ I(Z) such that fi ∈/ I(Xj) for j 6= i. If any fi ∈/ I(Xj), then done. Take f = fi. If fi ∈ I(Xi) for all i, then f = f1 + f2f3 . . . fm ∈ I(Z). Deﬁnition 2.15 (Codimension). If X is any variety, Z ⊆ X closed and irreducible, let X1,...,Xm be the components of X containing Z. Set codim(Z; X) = dim(X1 ∪ ... ∪ Xm) − dim Z. E.g. X is the union of a line and a plane, Z is a single point of X. Then codim(Z; X) is 2 if it is a point in the plane, 1 otherwise.

Theorem 2.19 (Reverse PIT). X affine, Z ⊆ X irreducible closed and c = codim(Z; X). Then ∃f1, . . . , fc ∈ k[X] such that Z ⊆ V (f1, . . . , fc) irreducible component. Proof. If Z is a component of Z, then c = 0. Otherwise, no components of X are contained in Z, so the lemma implies that there exists f1 ∈ k[X] such that f1 ∈ I(Z) and f1 does not vanish on any component of X. PIT implies that codim(Z; V (f)) < c. Induction on c gives us that there are f2, . . . , fc ∈ I(Z) such that Z is a component of V (f1, . . . , fc). Resultants n Let K be an arbitrary field, and f(T ) = anT + ... + a1T + a0 and g(T ) = m bmT + ... + b1T + b0 ∈ K[T ]. Q: Do f(T ) and g(T ) have a common factor?   an a2 a1 a0 0  an a2 a1 a0    Set A =  bm b1 b0 .    bm b1 b0  bm b1 b0 Definition 2.16 (Resultant). We define Res(f, g) = det A ∈ K.

n+m Let ~v = (cm−1, c0, dn−1, d1, d0) ∈ K , then ~v · A = (rn+m−1, . . . , r1, r0) ∈ n+m m−1 n−1 K . Then (cm−1 + ... + c1T + c0)f(T ) + (dn−1T + ... + d1T + d0)g(T ) = m+n−1 rm+n−1T + ... + r1T + r0.

Proposition 2.20. Suppose an 6= 0, then Res(f, g) 6= 0 ⇐⇒ (f, g) = 1 ∈ K[T ]. Proof. Res(f, g) = 0 iff ∃~v ∈ Km+n such that ~v · A = 0 iff ∃p(T ), q(T ) of deg ≤ m − 1, n − 1 such that p(T )f(T ) = q(T )g(T ), iff (f, g) 6= 1.

22 Pn i 0 If f(T ) = i=0 aiT then we allow formal diﬀerentiation, that is, f (T ) = Pn i−1 i=1 iaiT ∈ K[T ]. Note that (fg)0 = f 0g + fg0, and similar rules still hold.

n Corollary 2.21. If f(T ) = anT + ... + a1T + a0 then f(T ) has n diﬀerent roots in K iﬀ Res(f, f 0) 6= 0.

Qr di 0 Proof. f(T ) = an i=1(T − αi) , where αi 6= αj for i 6= j. Then f (T ) = Pr di−1 Q dj an i=1 di(T − αi) j6=i(T − αj) . 0 0 0 Res(f, f ) = 0 iff (f, f ) 6= 1 iff f (α`) = 0 for some ` iff d` ≥ 2 for some `. Definition 2.17 (Discriminant). The discriminant of f(T ) is Res(f, f 0). Exercise: a 6= 0 and f(T ) = aT 2 + bT + c then discriminant= −a(b2 − 4ac). Remark: If char(K) = 0 and if f(T ) ∈ k[T ] is an irreducible polynomial, then (f(T ), f 0(T )) = 1 so Res(f, f 0) 6= 0. In char(K) = p, then f 0(T ) may be zero, for example (T p + 1)0 = 0. Remark: ϕ : X → Y is a morphism, then dim(ϕ(X)) ≤ dim(X). This is as k(ϕ(X)) ⊆ k(X). Theorem 2.22. φ : X → Y is a dominant morphism of irreducible varieties, such that k(Y ) ⊆ k(X) is a finite extension of degree d. Suppose that char(k) = 0 or k(X)/k(Y ) is separable. Then ∃ dense open V ⊆ Y such that |φ−1(y)| = d for all y ∈ V .

d Proof. Assume X,Y affine and k[X] = k[Y ][f]. Let P (T ) = adT + ... + a1T + a0 ∈ k(Y )[T ] be the minimum polynomial for f ∈ k(X) over k(Y ). ie P (f) = 0 ∈ k(X). WLOG, ai ∈ k[Y ] for all i and we can replace Y with D(ad) and X with −1 φ (D(ad)) ⊆ X. We may assume that ad = 1. Now k[X] = k[Y ][T ]/(P (T )). πY This implies that X ' V (P ) ⊆ Y × A1 → Y , and φ : X → Y goes through this path. 1 Pd i If (y, t) ∈ Y × A then we set Py(T ) = i=0 ai(y)T ∈ k[T ]. (y, t) ∈ X iff 0 Py(t) = 0. Let ∆ = Res(P,P ) ∈ k[Y ]. P (T ) irreducible and char(k) = 0 imply 0 that ∆ 6= 0. Note that Res(Py,Py) = ∆(y). Thus, if y ∈ D(∆), then Py(t) = 0 has exactly d solutions. Now the general case: X and Y are irreducible varieties and φ : X → Y dominant. Let V ⊆ Y and U ⊆ φ−1(V ) ⊆ X be open affines. Then dim(φ(X \ U)) ≤ dim(X \ U) < dim(X) = dim(Y ). Thus φ(X \ U) ( Y , and so ∃h ∈ k[V ] such that D(h) ∩ φ(X \ U) = ∅. We can replace X with D(φ∗h) and Y with D(h). And so, WLOG, X,Y affine. φ dominant implies that k[Y ] ⊆ k[X] and k[X] generated by f1, . . . , fn. Then k[Y ] ⊆ k[Y ][f1] ⊆ ... ⊆ k[Y ][f1, . . . , fn] = k[X] gives X = Xn → Xn−1 → ψ ... → X1 → Y a sequence of dominant maps. Induction on n: ∃ a dense open U ⊆ X1 such that all points of U are hit by d1 = [k(X): k(X1)] pts of X.

23 As above: ψ(X1 \ U) ( Y implies ∃h ∈ k[Y ] such that D(h)∩ψ(X1 \U) = ∅, and so ψ−1(D(h)) = D(φ∗h) ⊆ U. ψ : D(φ∗h) → D(h) gives k[D(φ∗h)] = k[X]φ∗h = k[X]h. Thus, k[Y ][f1]h = k[Y ]h[f1] = k[D(h)][f1]. Thus, the ﬁrst case implies that ∃∅= 6 V ⊆ D(h) ⊆ Y open such that −1 −1 −1 |ψ (y)| = [k(X1): k(Y )]. Since ψ (D(h)) ⊆ U we have |φ (y)| = [k(X1): k(Y )] · d1 = d.

Exercise: πY : X × Y → Y is an open map. That is, if U ⊆ X × Y is open then πY (U) is open in Y . Corollary 2.23. φ : X → Y is a dominant morphism of irreducible varieties. Then φ(X) contains a dense open subset of Y .

Proof. We can assume that X,Y are aﬃne. Choose B = {f1, . . . , fn} ⊆ k[X] such that B is a transcendence basis for k(X)/k(Y ). ψ n πY Then k[Y ] ⊆ k[Y ][f1, . . . , fn] ⊆ k[X] gives X → Y × A → Y and φ is the composition. The theorem says that there is a open subset U ⊆ Y × An such that U ⊆ ψ(X). As πY is an open mapping, πY (U) is open. Deﬁnition 2.18. Let X be a variety.

1. W ⊆ X is locally closed if W = open ∩ closed. 2. W ⊆ X is constructible if W =the union of ﬁnitely many locally closed subsets.

Example: W = D(xy) ∪ {0} ⊆ A2. Notice: φ : A2 → A2 :(x, y) 7→ (x2y, xy), then φ(A2) = W . Exercise: φ : X → Y is an arbitrary morphism of varieties, then φ(X) is constructible.

3 Nonsingular Varieties

Local Rings Definition 3.1 (Local Ring at a point). If X is an irreducible variety and x ∈ X then OX,x = {f ∈ k(X): f(x) defined}. This is a local ring with maximal ideal mx = {f ∈ OX,x : f(x) = 0}. T Note: If U ⊆ X is any open subset, then k[U] = x∈U OX,x ⊆ k(X). Let U ⊆ X open affine, x ∈ U then M = I({x}) ⊆ k[U]. f ∈ OX,x then f is defined on D(h) ⊆ U for some h ∈ k[U] \ M, so f = g/hn where g ∈ k[U]. And so, OX,x = {g/h : g, h ∈ k[U], h(x) 6= 0} = k[U]M . Remark: X not irreducible implies that OX,x = lim OX (U). If U ⊆ X −→U3x open affine, x ∈ U we still have that OX,x = k[U]I({x}). Note: If X is irreducible then dim(X) = dim OX,x for any x ∈ X. If we let X be any variety, then dim(X) = maxx∈X dim OX,x.

24 Definition 3.2 (Regular Local Ring). If (R, m) is a local ring, then F = R/m is a field called the residue field of R and m/m2 is an F -vector space. 2 R is a regular local ring if dimF (m/m ) = dim R.

2 Exercise: If R is a N¨otherianLocal Ring then dimF (m/m ) =min number of generators for m ≥ dim(R). The equality is from Nakayama, the inequality from PIT. Deﬁnition 3.3. Let X be a variety and x ∈ X.

1. X is nonsingular at x if OX,x is a regular local ring. 2. Otherwise, X is singular at x. 3. X is nonsingular if all points x ∈ X are nonsingular.

Exercise: S is a commutative ring, and M ⊆ S is a maximal ideal, then set R = SM a local ring. Then unique maximal ideal m = MSM . Show that S/M ' R/m and M/M 2 = m/m2. Example: Let C = V (f) ⊆ A2 a curve. Let P = (0, 0) ∈ C. So f = ax+by+ HOT . And k[C] = k[x, y]/(f). Let M = I({P }) = (¯x, y¯) = (x, y)/(f) ⊆ k[C]. 2 2 2 2 2 2 mP /mP = M/M = (x, y)/(x , xy, y , f) = (x, y)/(x , xy, y , ax + by). So 2 P ∈ C is nonsing iff dimk(M/M ) = dim(C) = 1. This is iff ax + by 6= 0 ∈ k[x, y], note that ax + by 6= 0 implies that V (f) looks like V (ax + by) close to the point. Exercise: Find all singular points of V (y2 − x3 − x2),V (y2 − x3). n X ⊂ A a closed affine variety. Then I = I(X) = (f1, . . . , ft) ⊆ S = k[x1, . . . , xn]. Idea: X is nonsing at P ∈ X iff X has a tangent space at P . h i ∂fi Definition 3.4 (Jacobi Matrix). Let JP = (P ) be a t × n matrix, we call ∂xj this the Jacobi matrix.

n t Note: If ~v ∈ k then Jp · ~v ∈ k is the partial derivative of (f1, . . . , ft) at p in the direction ~v. n ker(JP ) = {~v ∈ k : Jp · ~v = ~0} is a candidate for a tangent space.

n 2 Lemma 3.1. Let P ∈ X ⊆ A , then rank(Jp) + dimk(mP /mP ) = n. Proof. Set M = I({P }) ⊆ S. Deﬁne d : M → kn by d(f) = ( ∂f (P ),..., ∂f (P )). ∂x1 ∂xn d is surjective as d(xi − pi) = ei. Note: f, g ∈ S that d(fg) = f(P )d(g) + d(f)g(P ). So d(M 2) = 0. Thus, d : M/M 2 → kn is an isomorphism if it is injective. It is injective as the two vector spaces are of the same dimension and it is surjective. th P P d(fi) = i row of JP and d( gifi) = gi(P )d(fi) so d(I) =row span of n 2 2 JP in k . Thus d : I + M /M → row span of JP is an isomorphism. Thus 2 2 rank(Jp) + dim(M/I + M ) = dim(M/M ) = n. Finally OX,P = (S/I)M and 2 2 2 mP /mP = (M/I)/(M/I) = M/I + M .

25 n Theorem 3.2. P ∈ X ⊆ A , then rank(JP ) ≤ n − dim(OX,P ) and rank JP = n − dim OX,P ⇐⇒ P nonsingular.

2 Proof. rank(Jp) = n − dim(mP /mP ) ≤ n − dim OX,P . 2 We have equality iﬀ dimk(mP /mP ) = dim OX,P .

Example: X = V (z2 − x2y2) ⊆ A3 where char(k) 6= 2, 3. Then J = 2 2 2 [−2xy , −2x y, 3z ]. p ∈ X is a nonsingular point iff rank(Jp) = 3 − x = 1, 3 2 2 2 2 2 so Xsing = V (z − x y , xy , x y, z ) = V (xy, z). Exercise: X is a variety and p ∈ X. Then OX,P is a domain iff p is in only one component. Theorem 3.3. Any Nötherianregular local ring is a domain. (in fact, a UFD, and even Macaulay)

Conclude: The points on an intersection of two components are singular.

Proposition 3.4. Xsing ⊆ X is a closed subset of X.

Proof. Let X = X1 ∪ ... ∪ Xm be the components of X. Then Xsing = Sm S i=1(Xi)sing ∪ i6=j Xi ∩ Xj. The latter are closed, so without loss of generality, X irreducible and aﬃne. n X ⊆ A closed, I(X) = (f1, . . . , ft) ⊆ k[x1, . . . , xn], P ∈ Xsing iﬀ p ∈ X and rank(Jp) < n − dim OX,p = n − dim(X). ∂fi Let m1, . . . , mN be all of the minors of size n−dim(X) in J = [ ]. Xsing = ∂xj X ∩ V (m1, . . . , mN ).

Fact: Xsing 6= X.

n Lemma 3.5. If p ∈ X = V (g1, . . . , gr) ⊆ A and if rank Jp(g1, . . . , fr) = r then OX,p is regular local of dimension n − r.

Proof. PIT implies that dim OX,p ≥ n − r. I(X) = (f1, . . . , ft) ⊇ (g1, . . . , gr). Thus, row span Jp(f1, . . . , ft) ⊇ row span Jp(g1, . . . , gr), so r = rank Jp(g1, . . . , gr) ≤ rank Jp(f1, . . . , ft) ≤ n−dim OX,p ≤ r.

Theorem 3.6 (Implicit Function Theorem). If f1, . . . , fc are holomorphic func- n ∂fi tions in a classical nbhd of p ∈ C . Suppose det ∂x (p) 6= 0. Then ∃ j 1≤i,j≤c n−c holomorphic functions w1, . . . , wc on classical open subset of C and classical n open subset V ⊆ C such that p ∈ V and so that for all z ∈ V , f1(z) = ... = fc(z) = 0 iﬀ z − i = wi(zc+1, . . . , zn) for all 1 ≤ i ≤ c.

Theorem 3.7. X ⊆ Cn a complex aﬃne variety. p ∈ X a nonsingular point. Then a classical neighborhood of p in X is holomorphic to a classical open subset d of C where d = dim OX,p.

26 Proof. WLOG, X is irreducible. I(X) = (f1, . . . , ft) and rank Jp(f1, . . . , ft) = ∂fi n n − d = c. WLOG, det( (p)) 6= 0. Set Y = V (f1, . . . , fc) ⊆ , then ∂xj C rank Jp(f1, . . . , fc) = c implies that p is a nonsingular point of Y . So OY,p is regular local of dimension d. Now, only one component of Y contains p, p ∈ X and X ⊆ Y . dim X is the same as the dimension of the component of Y containing p, and as X is irreducible, X is the component of Y containing p. Then there exists open n U ⊆ A such that X ∩ U = Y ∩ U = V (f1, . . . , fc) ∩ U. We apply the IFT, let V ⊂ U, p ∈ V , and w1, . . . , wc be as in the IFT. n d Define π : C → C by π(z) = (zc+1, . . . , zn). Then π : X ∩V → π(X ∩V ) ⊆ Cd is an holomorphism, as we can get an inverse map π−1 : π(X ∩ V ) → X ∩ V by (zc+1, . . . , zn) 7→ (w1(z), . . . , wc(z), zc+1, . . . , zn). Corollary 3.8. Every nonsingular complex variety variety is a complex manifold. If X is an affine variety, recall that pts in X are in 1-1 correspondence with max ideals in k[X]. So X an irreducible variety, p ∈ X corresponds to local rings OX,P = {f ∈ k(X): f(P ) defined }.

Lemma 3.9. X an irred var, x, y ∈ X, if OX,x ⊆ OX,y then x = y. Proof. Take open affine x ∈ U ⊆ X, y ∈ V ⊆ X. X is separated, so U ∩ V is affine and k[X] ⊗ k[Y ] → k[U ∩ V ] is surjective, and k[U] ⊆ OX,x ⊆ OX,y with k[V ] ⊆ OX,y, thus, k[U ∩ V ] ⊆ OX,y. k[V ] ⊆ k[U ∩ V ] ⊆ OX,y proves that y ∈ U ∩ V . Then k[U ∩ V ] ∩ my is a max ideal in k[U ∩ V ] so it corresponds to a point in U ∩ V which maps to y under U ∩ V → V the inclusion. Thus, x, y ∈ U ⊆ X. U is affine. If x 6= y, then there is an f ∈ k[U] such 1 1 that f(x) 6= 0, f(y) = 0. Then f ∈ OX,x and f ∈/ OX,y. Nonsingular Curves Definition 3.5 (Curves). A curve is an irreducible variety of dimension one.

Example: C = A1. k[C] = k[t] and k(C) = k(t). Let 0 6= f ∈ k(t). What is the order of vanishing of f at 0? n m f = p/q, p, q ∈ k[t], we can write p = t p0 and q = t q0 with p0(0) 6= 0 and n−m q0(0) 6= 0. f = (p0/q0)t , so v(f) = n − m is the order of vanishing. Note that OC,0 = k[t](t) with m0 = (t) ⊆ OC,0 is a maximal ideal. Then n−m f = ut where u = p0/q0 is a unit in OC,0. Deﬁnition 3.6 (Discrete Valuation Ring). A discrete valuation ring, or DVR, is a N¨otherian regular local ring of dimension 1.

Examples are OC,P where C is a curve and P a nonsingular point. Let (R, m) be a DVR, m = (t), t is a uniformizing parameter, and K = R0 is the ﬁeld of fractions of R.

27 Claim: Any f ∈ K∗ = K \{0} can be written f = utn where u ∈ R \ m a unit in R and n ∈ Z. WLOG, f ∈ R \{0}. Assume that the claim is false, then choose a counterexample such that (f) is maximal. f is not a unit implies that f ∈ m, so f = gt for some g ∈ R. (f) ( (g) as (f) = (g) ⇒ g = hf ⇒ f = thf ⇒ th = 1, contradiction. So g = utn, u ∈ R a unit implies that f = utn+1. Check that if f = utn, u a unit, then u, n are unique. n = min{p ∈ Z : f ∈ (tp) ⊆ K}.

Definition 3.7 (Valuation Map). v : K∗ → Z is a valuation map v(f) = n if f = utn with u ∈ R \ m. Note that R = {f ∈ K∗ : v(f) ≥ 0}∪{0} and m = {f ∈ K∗ : v(f) > 0}∪{0}. Rules: v(fg) = v(f) + v(g). v(f + g) ≥ min(v(f), v(g)) if f, g, f + g ∈ K∗. If f = utn, g = vtm and n ≤ m, then f + g = (u + vtm−n)tn = u0trtn. Example: C a curve, p ∈ C nonsing, then R = OC,p is a DVR with K = ∗ ∗ R0 = k(C). vp : K(C) → Z is a valuation, f ∈ k(C) vp(f) =the order of vanishing of f at p. vp(f) > 0 iff f ∈ mp, vp(f) = 0 iff f(p) 6= 0 and vp(f) < 0 iff f is not defined at p.

Lemma 3.10. Let R be a DVR, R0 = K and S any ring such that R ⊆ S ⊆ K. Then S = R or S = K.

Proof. If S 6= R, then take f ∈ S, f∈ / R. m = (t), f = utn, n < 0, S ⊇ R[f] = K.

Recall: A domain R is called integrally closed iff for all f ∈ R0, f is integral over R implies that f ∈ R. Theorem 3.11. If R is any NötherianLocal Domain of dimension one, then R is regular iff R is integrally closed. Exercise: Prove ⇒ (easy, as DVR ⇒ PID ⇒ UFD ⇒ int closed, so prove the last one)

Proposition 3.12. Let A be a domain. A is integrally closed iﬀ AP is integrally closed for all maximal ideals P ⊆ A.

Proof. ⇒: Easy ⇐: A = ∩P ⊆AAP over maximal ideals P . Note: A a f.g. domain over k, X = Spec −m(A), then A = k[X] = ∩p∈X OX,p = ∩Ap. Definition 3.8 (Dedekind Domain). A Dedekind Domain is an integrally closed Nötheriandomain of dimension 1. Note: If X is an irred affine variety, X nonsingular curve iff k[X] is a Dedekind domain.

28 OX,p a DVR for all p ∈ X iff k[X]p integrally closed for all max ideals p by the theorem, each of these is integrally closed iff k[X] is, and that is the def of a Dedekind domain. Finiteness of Integral Closure Let R be a finitely generated domain over k, K = R0, and K ⊆ L a finite field extension. Then R is defined to be the integral closure of R in L. That is, {f ∈ L : f integral over R}. Fact R is an integrally closed domain with field of fractions L. n n−1 Let f ∈ L. Then f + a1f + ... + an = 0 with ai ∈ K. Take b ∈ R such n n−1 n that bai ∈ R for all i. Then (bf) + ba1(bf) + ... + b an = 0. Thus bf ∈ R. b ∈ R, so f ∈ (R)0. Theorem 3.13 (Finiteness of Integral Closure). R is a finitely generated R- module. In particular: R is a finitely generated K algebra. Definition 3.9 (DVR of K/k). k ⊆ K is a field extension, a DVR of K/k is a subring R ⊆ K such that: 1. R is a DVR

2. R0 = K 3. k ⊆ R. Let K be a function field of dimension 1 over k. i.e., k ⊆ K is finitely generated as a field extension and it has transcendence degree 1. Q: ∃ a nonsingular curve C such that k(C) = K? Key construction: Let f ∈ K \ k. Then k(f) ⊆ K is a finite field extension. It is finitely generated by assumption, and {f} must be a transcendence basis for K/k, thus it is algebraic, and so it is a finite field extension. Set B = k[f] ⊆ K. Finiteness of integral closure says that B is a finitely generated k-algebra, and so B is a Dedekind domain with B0 = K. Proposition 3.14. X = Spec −m(B) is a nonsingular curve. 1. Points in X are in 1-1 correspondence with DVRs R of K/k such that f ∈ R.

2. Points in V (f) correspond to the DVRs R of K/k such that f ∈ mR. Proof. X → {DVRs R 3 f} is well-deﬁned and injective. Let R be a DVR of K/k with f ∈ R. k[f] ⊆ R ⇒ B ⊆ R. Set M = B ∩ mR ⊆ B a prime ideal. Then BM ⊆ R 6= K ⇒ M 6= 0 and so BM is a DVR of K/k. And so, the lemma implies that BM = R. Now we prove (b). M ∈ V (f) ⇐⇒ f ∈ M ⇐⇒ f ∈ MBM = m.

29 Corollary 3.15. Every DVR R of K/k is the local ring of a nonsingular curve at some point. In particular, R/mR = K. Proof. Let f ∈ R \ k, B = k[f] ⊆ K. Then R = BM for some maximal ideal M ∈ Spec −m(B). So R = OX,p. Corollary 3.16. Given f ∈ K∗, there are only ﬁnitely many DVRs R of K/k such that f ∈ mR. Also have ﬁnitely many R such that f∈ / R.

Proof. WLOG, f∈ / k. Set B = k[f] ⊆ K. {R : f ∈ mR} is in correspondence with V (f) ⊆ Spec −m(B). PIT implies that dim V (f) = 0. Thus, V (f) is a ﬁnite set. Note: f∈ / R iﬀ 1/f ∈ mR.

Deﬁnition 3.10. CK = {DVRs of K/k} Elements of CK will be called ”points” P . DVR given by P is RP with maximal ideal mP .

We deﬁne a topology on CK to have as closed sets the ﬁnite sets and all of CK . We also let f ∈ K and P ∈ CK , assume f ∈ RP .

Deﬁnition 3.11. f(P ) is deﬁned to be the image of f by RP → RP /mP = k. ∼ i.e. f(P ) = f mod mP . The regular functions on a nonempty open subset U ⊆ CK are then the set k[U] = ∩p∈U RP ⊆ K.

This makes CK a SWF. Note: If U ⊆ CK open, f ∈ k[U] then D(f) = {P ∈ U : f(P ) 6= 0} = {P ∈ U : f∈ / mP } is open by the second corollary. Example Let K = k(t). The DVRs of K/k are k[t](t−a) for a ∈ k and k[1/t](1/t). 1 Then CK and P are in 1-1 correspondence, with k[t](t−a) corresponding to a ∈ A1 and the other point corresponding to the point at inﬁnity. ∼ Note: If f ∈ k[t](t−a) then f(t) = f(a) mod (t − a), f(k[t](t−a)) = f(a).

Theorem 3.17. CK is a non-singular curve and k(CK ) = K.

Proof. Take any f ∈ K \ k. B = k[f] ⊆ K. U = {p ∈ CK : f ∈ RP } ⊆ CK is open. We define φ : Spec −m(B) → U by M 7→ BM . The prop implies that this is bijective. φ is a homeomorphism, as the closed sets are the finite sets in both. It remains to show that this is a morphism of spaces with functions. For V ⊆ Spec −m(B) open, then k[V ] = ∩M∈V BM = ∩P ∈φ(V )RP = k[φ(V )]. Thus, φ : Spec −m(B) → U is an isomorphism of spaces with functions. Note: If P ∈ CK then f ∈ RP or 1/f ∈ RP . Thus, Ck = Spec −m(k[f]) ∪ −1 Spec −m(k[f ]). This is an open affine cover, and so Ck is a prevariety.

30 Let P,Q ∈ CK . Enough to find an open affine U ⊆ CK such that P,Q ∈ U. Take f ∈ RP \ mP and f∈ / k.(f = 1 + t,(t) = mP ). If f ∈ RQ then P,Q are both in Spec −m(k[f]). Otherwise, 1/f ∈ RQ, so both are in Spec −m(k[1/f]). Thus CK is a variety. It is nonsingular and of dimension one by construction. We must show that it is irreducible. CK is, in fact, irreducible because its proper closed sets are finite and CK is infinite. Proposition 3.18. Let C be an irreducible curve and P ∈ C a nonsingular point. Let Y be any projective variety and φ : C \{P } → Y is any morphism of varieties. Then ∃! extension φ : C → Y . Note: No points in Y are ”missing”.

Proof. Y ⊆ Pn closed subset. It is enough to make φ : C → Pn. n WLOG φ(C \{p}) 6⊆ V+(xi) for all i. Set U = D(x0, x1, . . . , xn) ⊆ P . φ(C \{p}) ∩ U 6= ∅. −1 Set fij = xi/xj ◦ φ ∈ k(C). Deﬁned on φ (U) 6= ∅. ∗ vP : k(C) → Z is the valuation given by OC,P . Set ri = vP (fi0) for 0 ≤ i ≤ n. Choose j such that rj minimal. xi xi/x0 As = , fij = fi0/fj0, so vP (fij) = ri − rj ≥ 0. Thus, fij ∈ C,P for xj xj /x0 O −1 all i. Note that if Q ∈ φ (U), then φ(Q) = (f0j(Q): f1j(Q): ... : fjj(Q) = 1 : ... : fnj(Q)). We can then extend φ to P by this expression. Then φ is a morphism on φ−1(U) ∪ {p} and so φ : C → Pn is a morphism. Lemma 3.19. R ⊆ K is a local ring, k ⊆ R, R is not a ﬁeld. Then R is contained in some discrete valuation ring of K/k. Proof. Set B = R ⊆ K. Lying over implies that there exists some maximal ideal M ⊆ B such that M ∩ R = mR. Claim: BM is a DVR of K/k. Let 0 6= f ∈ mR. S = k[f] is a Dedekind domain, S ⊆ B. M˜ = M ∩ S is a maximal ideal of S.

Thus SM˜ is a DVR of K/k. SM˜ ⊆ BM ( K and a lemma from before says that we have equality of SM˜ and BM .

Theorem 3.20. CK is a projective curve.

−1 Proof. Let f ∈ K \ k. U = Spec −m(k[f]), V = Spec −m(k[f ]), and CK = U ∪ V an open aﬃne cover. U ⊆ AN closed. U ⊆ PN projective closure. The proposition implies that the inclusion U → U extends to a morphism ϕ1 : CK → U. Similarly, we take V to be the projective closure of V . Then V → V extends to ϕ2 : CK → V .

31 We now deﬁne ϕ : Ck → U × V by ϕ(P ) = (ϕ1(P ), ϕ2(P )). Set Y = ϕ(CK ) ⊆ U × V . Y is a projective variety. Claim: ϕ : CK → Y is an isomorphism. Note: ϕ(U) ⊆ U × V is a closed subset. Let ψ = ϕ2 × id : U × V → V × V . −1 ϕ(U) = {(u, v) ∈ U × V : ϕ2(u) = v} = ψ (∆V ) closed. Thus, ϕ(U) = Y ∩ (U × V ), which implies that ϕ : U → Y ∩ (U × V ) is bijective, isomorphism. So πU : Y ∩ (U × V ) → U is the inverse. Similarly, ϕ : V → Y ∩ (U × V ) is an isomorphism. Note: k(Y ) = k(CK ) = K, and forall P ∈ CK , OY,ϕ(P ) = RP ⊆ K. Thus, ϕ is injective. For surjective, let y ∈ Y . Then k ⊆ OY,y ⊆ K is a local ring. By the lemma, OY,y ⊆ RP for some P ∈ CK . OY,y ⊆ RP = OY,ϕ(P ) so y = ϕ(P ). Corollary 3.21. Any curve is birational to some nonsingular projective curve.

Corollary 3.22. X is any nonsingular curve, then X =∼ some open subset of CK , K = k(X).

Proof. ϕ : X → CK by ϕ(x) = P where P ∈ CK such that RP = OX,x ⊆ K. Injectivity is clear. Claim: ϕ(X) ⊆ CK is open. Take U ⊆ X open affine. k[U] is generated by f1, . . . , fn. P ∈ ϕ(U) iff k[U] ⊆ RP , iff fi ∈ RP for all i. n Thus, ϕ(U) = ∩i=1 Spec −m(k[fi]) open. Thus ϕ(X) ⊆ CK is open. ϕ : X → ϕ(X) is a homeomorphism. To check that ϕ is an isomorphism, then U ⊆ X open gives k[U] ∩x∈U OX,x = ∩P ∈ϕ(U)Rϕ(x) = k[ϕ(U)]. Exercise: Two nonsingular projective curves are isomorphic iff they have the same function field. Degree of Projective Varieties in Pn Bezout: f1, . . . , fn ∈ k[x0, . . . , xn] homogeneous of degrees d1, . . . , dn. Then V+(f1, . . . , fn) has cardinality at most d1 . . . dn or is infinite. If it is finite and counted with multiplicity, then it is equal to d1 . . . dn. Classical Definition: X ⊆ Pn closed, then deg(X) = #(X∩V ) where V ⊆ Pn is a linear subspace with dim V + dim X = n. e.g. f ∈ S = k[x0, . . . , xn] a square-free homogeneous polynomial. Then #(V+(f)∩ general line) = deg f. 2 2 Warning: V+(xz − y ), V+(x) ⊆ P . These are isomorphic but have different degrees. So degree is not a property of a projective variety, but rather one of the embedding into projective space. Example: f ∈ S is square-free, deg f = d. Then X = V+(f), I(X) = (f), R = S/(f) is the projective coordinate ring. m + n Note: dim (S ) = , where S is all forms of degree m. This is k m n m 1 n! (m + n)(m + n − 1) ... (m + 1), which is actually a polynomial in m of degree 1 n with lead coefficient n! .

32 f Consider 0 → S → S → R → 0 implies that 0 → Sm−d → Sm → Rm → 0 is m + n m + n − d exact. Then dim R = dim S −dim S , which is − , m m m−d n n 1 1 which is n! (m + n) ... (m + 1) − n! (m + n − d) ... (m + 1 − d). Which is a polynomial of degree n − 1. 1 P P nd d This has lead coeﬃcient n! ( i − (i − d)) = n! = (n−1)! . Recall: A graded S-module is a module M with a decomposition M =

⊕d∈ZMd as abelian group such that SmMd ⊂ Mm+d. Definition 3.12. Ann(M) = {f ∈ S : fM = 0} is a homogeneous ideal, n Supp(M) = V+(Ann(M)) ⊆ P . Reason for Supp is x ∈ Pn, P = I({x}) ⊆ S is a homogeneous prime ideal. MP 6= 0 iff P ⊇ Ann(M), iff x ∈ Supp(M). Example: X ⊆ Pn closed, then Ann(S/I(X)) = I(X), so Supp(S/I(X)) = V+(I(X)) = X. Note: Md is a k-vector space for all d ∈ Z because S0Md ⊆ Md. If M is a finitely generated graded S-module, then dimk Md < ∞ forall d.

Deﬁnition 3.13 (Hilbert Function). HM (d) = dimk Md is the Hilbert function of M.

Theorem 3.23. If M is a ﬁnitely generated graded S-module then ∃!PM (z) ∈ Q[z] such that PM (d) = dimk(Md) for all d suﬃciently large. We call PM (z) the Hilbert Polynomial.

Deﬁnition 3.14 (Numerical Polynomial). P (x) ∈ Q[z] is a numerical polynomial if P (d) ∈ Z for all d suﬃciently large in Z. z Example: = 1 z(z − 1) ... (z − m + 1). m m! z Note: : m ∈ is a basis over for [z]. m N Q Q z z Lemma 3.24. P (z) = c + c + ... + c ∈ [z], c ∈ . Then TFAE 0 1 1 r r Q i Q

1. P (z) is a numerical polynomial.

2. P (d) ∈ Z for all d ∈ Z.

3. ci ∈ Z. Proof. 3 ⇒ 2 ⇒ 1 are easy, so we need 1 ⇒ 3. z + 1 z z We know that − = . Thus P (z + 1) − P (z) = m m m − 1 z z c + c + ... + c . 1 2 1 r r − 1 We perform induction on r: P (z) is numeric, then P (z + 1) − P (z) is also numeric. Thus, c1, . . . , cr ∈ Z, and so c0 must also be an integer.

33 Theorem 3.25 (Aﬃne Dimension Theorem). X,Y ⊆ An closed and irreducible, then Z ⊆ X ∩ Y component has dim Z ≥ dim X + dim Y − n.

n n n Proof. X ∩ Y = (X × Y ) ∩ ∆A = (X × Y ) ∩ V (x1 − y1, . . . , xn − yn) ⊆ A × A , use PIT. WARNING: Does not prevent X ∩ Y = ∅.

Theorem 3.26 (Projective Dimension Theorem). X,Y ⊆ Pn closed irreducible, Z ⊆ X ∩Y a component, then dim Z ≥ dim X +dim Y −n. If dim X +dim Y −n is nonnegative, then X ∩ Y is nonempty. Proof. First statement follows from ADT. Set s = dim X, t = dim Y . C(X) = π−1(X) ⊆ An+1, then dim C(X) = s+1, dim C(Y ) = t + 1. Every component of C(X) ∩ C(Y ) has dimension ≥ s + 1 + t + 1 − n − 1 = s + t − n + 1 ≥ 1, and 0 ∈ C(X) ∩ C(Y ), so such a component exists.

Definition 3.15 (Twisted Module). Let ` ∈ Z, M(`) is the twisted module given by M(`)d = M`+d. i.e., we are shifting the grading, but doing nothing else. Definition 3.16 (Homogeneous Homomorphism). A homomorphism ϕ : M → N of graded S-modules is homogeneous if ϕ(Md) ⊆ Md for all d. Definition 3.17 (Homogeneous Submodule). A submodule N ⊆ M is homogeneous if N = ⊕d∈Z(N ∩ Md).

This implies that N and M/N = ⊕Md/(N ∩ Md) are graded and 0 → N → M → M/N → 0 is a short exact sequence of homogeneous homomorphisms. Note: Let m ∈ M be homogeneous of degree `, that is, m ∈ M`, then I = Ann(m) ⊆ S is a homogeneous ideal. Set N = S · m ⊆ M, then 0 → I → S → N → 0 is a short exact sequence, but it is not quite of homogeneous maps, unless we change S, I to S(−`),I(−`). Thus, (S/I)(−`) is isomorphic to N = S · m ⊆ M. Exercise: M f.g. module over a Nötherianring, then M is a Nötherian module. Lemma 3.27. M a f.g. graded module over a Nötheriangraded ring S, then ∃ filtration 0 = M 0 ⊆ ... ⊆ M r = M such that M i ⊆ M is a homogeneous i i−1 submodule and M /M ' S/Pi(ì) where Pi is a homogeneous prime ideal and ì ∈ Z. Proof. Let N ⊆ M be maximal homogeneous submodule such that the lemma is true for N. Claim: N = M. Else M 00 = M/N 6= 0, Take 0 6= m ∈ M 00 homogeneous such that Ann(m) ⊆ S is as large as possible. Claim: P = Ann(m) prime ideal. P 6= S. Let f, g ∈ S \ P . Enough to show that fg∈ / P . Note: gm 6= 0 and Ann(gm) ) Ann(m), thus Ann(gm) = Ann(m). So we must have fg∈ / Ann(m), else f ∈ Ann(gm).

34 Thus, Sm ' (S/P )(−`), ` = deg m. So M → M/N = M 00 ⊇ Sm, N˜ ⊆ M is the inverse image of Sm, and N ( N˜, and the lemma is true for N˜. Deﬁnition 3.18 (Eventually Polynomial). f : Z → Z is eventually polynomial if ∃P (z) ∈ Q[z] such that f(n) = P (n) for all n >> 0. Set ∆f(n) = f(n + 1) − f(n) Lemma 3.28. f is eventually polynomial of degree r iﬀ ∆f is eventually polynomial of degree r − 1. Proof. ⇒: Obvious. z ⇐: Assume ∆f(n) = Q(n) for all n >> 0 where Q(z) = c + c + ... + 1 2 1 z z z c . Set P (z) = c + ... + c . r r − 1 1 1 r r Then ∆P = Q, so ∆(f − P )(n) = 0 for all n >> 0. Thus f(n) − P (n) = c0 a constant for n >> 0.

Set S = k[x0, . . . , xn], M a f.g. graded S-module. n Hilbert Function: HM (d) = dimk(Md), and Supp(M) = V+(Ann(M)) ⊆ P . Note: If 0 → M 0 → M → M 00 → 0 is a short exact sequence of graded S-modules, then Supp(M) = Supp(M 0) ∪ Supp(M 00). ⊇: Ann(M) ⊆ Ann(M 0) ∩ Ann(M 00). ⊆: Let x ∈ Pn. If x∈ / Supp(M 0) ∪ Supp(M 00), then ∃f ∈ Ann(M 0) such that f(x) 6= 0 and ∃g ∈ Ann(M 00) such that g(x) 6= 0, then fg ∈ Ann(M), but fg(x) 6= 0. Ann(M 0) Ann(M 00) ⊆ Ann(M) ⊆ Ann(M 0) ∩ Ann(M 00).

Theorem 3.29. M f.g. graded module over S = k[x0, . . . , xn] then HM (d) is eventually equal to a polynomial PM (d) of degree dim(Supp(M)). 0 r i i−1 Proof. ∃ filtration 0 = M ⊂ ... ⊂ M = M. such that M /M ' (S/Pi)(ì) where Pi ⊆ S is a homogeneous prime. r r P i i−1 P Note: HM (d) = i=1 HM /M (d) = i=1 HS/Pi (d + ì). r i i−1 r Supp(M) = ∪i=1 Supp(M /M ) = ∪i=1V+(Pi). WLOG, M = S/P , P a homogeneous prime. Induction of V+(P ): If P = (x0, . . . , xn), then the theorem is true if we take dim ∅ = deg 0 = −1. Otherwise, some xi ∈/ P . Set I = P + (xi). Then V+(I) ( V+(P ), so dim V+(I) = dim V+(P ) − 1 by the projective dimension theorem. By induction, HS/I (d) is eventually polynomial and deg(PS/I ) = dim V+(P )− 1 0 → (S/P )(−1) → S/P → S/I → 0, so ∆HS/P (d − 1) = HS/P (d) − HS/P (d − 1) = HS/I (d), so HS/P (d) is eventually polynomial of degree equal to dim V+(P ). z z Note: P (z) = c + c + ... + c ∈ [z] is a numeric polynomial, M 0 1 1 r r Q so ci ∈ Z. r = dim Supp(M), so r!(lead coef of PM (z)) ∈ Z ≥ 0

35 Deﬁnition 3.19 (Hilbert Polynomial of a Variety). Let X ⊆ Pn be a closed subvariety of dimension r, then PX (z) = PS/I(X)(z) is the Hilbert Polynomial for X. We now deﬁne deg(X) = r!(lead coef of PX (z)). Examples

1. deg V+(f) = deg f

n 2. V ⊆ P linear subspace. WLOG, V = V+S(xr+1, . . . , xn). S/I(V ) ' r + d k[x , . . . , x ], so P (d) = , so lead coef is 1/r!, so we get 0 r S/I(X) r degree 1.

n Proposition 3.30. X1,X2 ⊆ P closed, dim X1 = dim X2 = r, no components in common, then deg(X1 ∪ X2) = deg X1 + deg X2

Proof. I1 = I(X1), X2 = I(X2). I(X1 ∪ X2) = I1 ∩ I2, so 0 → S/(I1 ∩ I2) → S/I1 ⊕ S/I2 → S/(I1 + I2) → 0. The ﬁrst takes f 7→ (f, f) and the second takes (f, g) 7→ f − g. They are injective and surjective, and so we see this this is a short exact sequence.

Thus PX1 (d) + PX2 (d) = PX1∪X2 (d) + PS/(I1+I2)(d), so Supp(S/(I1 + I2)) =

V+(I1 + I2) = X1 ∩ X2. dim < r, so LC(PX1 ) + LC(PX2 ) = LC(PX1∪X2 ).

Corollary 3.31. If X ⊆ Pn has dim zero, then deg(X) =the number of points in X. Definition 3.20 (Simple Module). If R is a ring and M an R-module, then M is simple if M has no nontrivial submodules and M 6= 0. This is equivalent to M ' R/P where P ⊆ R is a maximal ideal. Definition 3.21 (Decomposition Series). A decomposition series for M is a filtration 0 = M0 ( M1 ( ... ( Mr = M such that Mi/Mi−1 is simple for all i. Definition 3.22 (Artinian Module). If there exists a decomposition series, then M is an Artinian module and we define length(M) = r. Assume that R is Nötherianand that M is a finitely generated R-module. Then there exists a filtration 0 = M0 ( M1 ( ... ( Mr = M. such that Mi/Mi−1 is isomorphic to R/Pi where Pi ⊆ R is maximal. Note: Ann(M) ⊆ Pi for all i.

Lemma 3.32. If P ⊆ R is a minimal prime over Ann(M) then MP is an

Artinian RP -module which has lengthRP (MP ) = |{i : Pi = P }| in the ﬁltration of M.

Proof. 0 = (M0)P ⊆ (M1)P ⊆ ... ⊆ (Mr)P = MP , and (Mi)P /(Mi−1)P = (Mi/Mi−1)P = (R/Pi)P . If P = Pi, then we get RP /P RP , else we get 0.

36 n X ⊆ P is a closed subvariety, S = k[x0, . . . , xn], then dimk(S/I(X))d = PX (d) for all d >> 0. dim(X) = deg(PX ) and deg(X) = (dim X)! LC(PX ) Assume X ⊆ Pn has pure dimension r. That is, all components of X have dimension r. n Y = V+(f) ⊆ P a hypersurface such that no component of X is contained in Y . Also assume that f ∈ S is square-free. Let Z ⊆ X ∩ Y be a component. Then dim Z = r − 1. Set M = S/(I(X) + (f)). Supp(M) = X ∩Y , do if P = I(Z), then P is minimal prime over Ann(M).

Deﬁnition 3.23 (Intersection Multiplicity). I(X · Y ; Z) = lengthSP (MP ) P Theorem 3.33. deg(X) deg(Y ) = Z⊆X∩Y I(X · Y ; Z) deg(Z). Proof. Set d = deg(X), e = deg(Y ). f Then we have a short exact sequence 0 → S/I(X) → S/I(X) → M → 0. This gives us 0 → (S/I(X))`−e → (S/(I(X)))` → M` → 0 will be a short exact sequence for each `, where this is breaking it into homogeneous parts. Thus PM (`) = PX (`)−PX (`−e). Thus LC(PM ) = r ·e·LC(PX ) = red/r! = de (r−1)! . Take ﬁltration 0 = M0 ( M1 ( ... ( Mt = M, Mi ⊆ M is homogeneous and Mi/Mi−1 is isomorphic to S/Pi where Pi ⊆ P is a homogeneous prime ideal. So P (`) = Pt P (`) = P P (`)|{i : Q = P }|, M i=1 Mi/Mi−1 Q⊆S hom prime S/Q i P and from this we get Z⊆X∩Y ;component |{i : Pi = I(Z)}|PZ (`) + LOT . So we 1 P have LC(PM ) = (r−1)! Z⊆X∩Y I(X · Y ; Z) deg(Z).

Corollary 3.34 (Bezout’s Theorem). Let X,Y ⊆ P2 be curves of degree d and P e such that X ∩ Y is a finite set, then P ∈X∩Y I(X · Y ; P ) = de. Exercise: P ∈ X ∩ Y ⊆ P2 then I(X · Y ; P ) = 1 ⇐⇒ P is a nonsingular point of both X and Y and X and Y have different tangent directions at P . Bezout’s Theorem for Pn Idea: If X ⊆ Pn is closed and irreducible and Y ⊆ Pn is a hypersurface, then P [X] · [Y ] = Z⊆X∩Y I(X · Y ; Z) · [Z]. n Suppose that Y1,Y2,...,Yn ⊆ P are hypersurfaces such that their inter- n PN section is finite. Then (... ([P ] · [Y1]) · [Y2] ...) · [Yn] = i=1 ci[Pi] where P Qn Y1 ∩ ... ∩ Yn = {P1,...,PN }. Then ci = j=1 deg(Yj). n PNm (m) In fact: (... ([P ] · [Y1]) ...) · [Ym] = i=1 ci [Zi] where Y1 ∩ ... ∩ Ym = Qm PNm (m) Z1 ∪ ... ∪ ZNm components, then j=1 deg(Yj) = i=1 ci deg(Zi). (m) Fact: ci, ci do not depend on the order of multiplication. Q Corollary 3.35. Y1 ∩ ... ∩ Yn finite implies that |Y1 ∩ ... ∩ Yn| ≤ deg(Yj).

n Useful Fact: X1,...,Xm ⊆ P irreducible closed, d ∈ N then there exists n irreducible hypersurface Y ⊆ P of degree d such that Xi 6⊂ Y for all i.

37 n + d v : n → N , N = − 1 the veronese map, then Y = n ∩ H, d P P n P H ⊆ PN hyperplane. n Intrinsic: {hypersurfaces of degree d in P } ↔ P(Sd)−nonsquarefree by V+(f) ↔ [f] {irreducible hypersurfaces} ↔ P(Sd)\∪p+q=dϕp,q(P(Sp)×P(Sq)) where ϕp,q : P(Sp) × P(Sq) → P(Sp+q):[f] × [g] 7→ [fg]. n + p n + q n + p + q + ≤ so U ⊆ (S ) is a dense open subset. n n n P d n Now, if X ⊆ P is closed, then X ⊆ V+(f) iﬀ f ∈ I(X)d. Thus {hypersurfaces6⊇ X} ↔ WX = P(Sd) \ P(I(X)d). Now, we can conclude n that {irreducible hypersurfaces Y in P of degree d such that Xi 6⊆ Y for all i correspond to U ∩ WX1 ∩ ... ∩ WXm ⊆ P(Sd) is still a dense open subset.

4 Sheaves

Deﬁnition 4.1 (Presheaf). Let X be a topological space. A presheaf F of abelian groups on X is an assignment U 7→ F (U) of an abelian group F (U) to each open subseteq U of X plus group homomorphisms ρUV : F (U) → F (V ) whenever V ⊆ U ⊆ X open with

1 ρUU = id

2 ρVW ◦ ρUV = ρUW when W ⊆ V ⊆ U

Notation: elements of F (U) are called sections of F over U, ρUV are called restriction maps and s ∈ F (U), V ⊆ U, then s|V = ρUV (s). Also, say Γ(U, F ) = F (U). Deﬁnition 4.2 (Sheaf). A sheaf is a presheaf F such that

3 If s ∈ F (U) and U = ∪Vi an open cover, then s|Vi = 0 iﬀ s = 0.

4 If U = ∪Vi an open cover and have sections si ∈ F (Vi) and si|Vi∩Vj =

sj|Vi∩Vj for all i, j, then there exists s ∈ F (U) such that s|Vi = si.

Note: The section s of axiom 4 is unique by axiom 3. Remark: We can easily deﬁne sheaves of rings, sets, modules, etc. Examples:

1. X is a SWF, then we can deﬁne OX of k-algebras by OX (U) = k[U]. This sheaf is called the structure sheaf of X.

∞ 2. Let M be a manifold, OM (U) = {C f : U → R}. Then there is π : −1 TM → M the tangent bundle, π (x) = TxM. Then TM ↔ a sheaf T of OM -modules. T (U) = {s : U → TM such that π(s(x)) = x for all x ∈ U}. T (U) is a OM (U)-module.

38 Deﬁnition 4.3 (Morphism of Sheaves). A morphism ϕ : F → G of (pre)sheaves consists of homomorphisms ϕU : F → G for all U ⊆ X open such that s ∈ F , V ⊆ U open, then ϕU (s)|V = ϕV (s|V ) ∈ G (V ) and ... G (V ) ...... G (U) ...... ϕV . ϕU ...... F (V ) ...... F (U) If ϕ : F → G and ψ : G → H are morphisms, we deﬁne ψ ◦ ϕ : F → H by (ψ ◦ ϕ)U = ψU ◦ ϕU . An isomorphism is a morphism with an inverse morphism.

Deﬁnition 4.4 (The stalk of F at x). If F is a presheaf on X, x ∈ X then the stalk of F at x is Fx = lim F (U). This means −→U3x

1. Fx is an abelian group (or whatever)

2. If x ∈ U we have homomorphism ρUx : F (U) → Fx, if x ∈ V ⊆ U then ρUx = ρV x ◦ ρUV .

3. If G is any abelian group with homomorphisms ΘU : F (U) → G such that ΘU = ΘV ◦ ρUV for all x ∈ V ⊆ U, then there exists a unique group homomorphism Θx : Fx → G such that ΘU = Θx ◦ ρUx

Example: X a variety, x ∈ X. Then OX,x = lim OU =local ring of X at −→U3x x. L Construction Fx = U3x F (U) /h(0,..., 0, s, 0,..., 0, −s|V , 0,..., 0) : ∀s ∈ F (U), x ∈ V ⊆ Ui. Notation: If s ∈ FU and x ∈ U write sx = ρUx(s) ∈ Fx. Exercise Let F be a presheaf.

1. All elements of Fx can be written as sx for some s ∈ F (U), x ∈ U.

2. s ∈ F (U), x ∈ U, sx = 0 ∈ Fx iﬀ s|V = 0 ∈ F (V ) for x ∈ V ⊂ U.

Exercise: F is a sheaf. s ∈ F (U). s = 0 ⇐⇒ sx = 0 for all x ∈ U. Note: A morphism ϕ : F → G gives a homomorphism ϕx : Fx → Gx, ϕx(sx) = ϕU (s)x, s ∈ F (U), x ∈ U. x ∈ U: .... Gx ...... G (U) ...... Fx ...... F (U)

Proposition 4.1. ϕ : F → G is a morphism of sheaves. ϕ is an isomorphism iﬀ ϕx : Fx → Gx are isomorphisms for all x ∈ X.

39 Proof. ⇒: Clear ⇐: We must show that ϕ : F (U) → G (U) are isomorphisms. ϕU is injective as s ∈ F (U), if ϕU (s) = 0 ∈ G (U) then ϕx(sx) = ϕU (s)x = 0, but ϕx is injective, and so sx = 0 for all x ∈ U, so s = 0. To see that ϕU is surjective, take t ∈ G (U). ϕx surjective implies that tx = ϕx(s(x)x) ∈ Gx for some s(x) ∈ F (Vx) where Vx ⊆ U are open subsets containing x.

Now tx = ϕVx (s(x))x. We can make Vx smaller such that t|Vx = ϕVx (s(x)) ∈

Patch: There exists a unique s ∈ F (U) such that s|Vx = s(x) ∈ F (Vx) for all x ∈ U. ϕU (s)|Vx = ϕVx (s|Vx ) = ϕVx (s(x)) = t|Vx . Thus ϕU (s) = t ∈ G (U).

Remark: F is a sheaf on X, U ⊆ X is open. Define F |U to be the sheaf on U by Γ(V, F |U ) = Γ(V, F ) Example: Let S1 = {(x, y) ∈ R2 : x2 + y2 = 1}. Define sheaves F , G by ∂g F (U) = {f : U → R : f is locally constant} and G (U) = {g : U → R : ∂θ = 1}. 1 If U ( S open then F |U ' GU by f(x, y) 7→ f(x, y) + Arg(x, y), so Fx ' Fx for all x ∈ S1 but F and G are not isomorphic as F (S1) = R and G (S1) = ∅. Sheafification Let F be a presheaf on X, define a sheaf F + as follows: U ⊆ X open set + ` F (U) to be the set of all functions s : U → x∈U Fx such that

1. s(x) ∈ Fx for all x 2. ∀x ∈ U there exists an open set V with x ∈ V ⊆ U and t ∈ F (V ) such that s(y) = ty ∈ Fy forall y ∈ V .

+ Deﬁnition 4.5. We deﬁne a morphism Θ: F → F by t ∈ F (U), ΘU (t) = + [x 7→ tx] ∈ F (U). + Exercise: Θx : Fx → Fx . Proposition 4.2. Let F be a presheaf and G be a sheaf. ϕ : F → G is any morphism. Then there exists a unique ϕ+ : F + → G such that ϕ = ϕ+ ◦ Θ. + ` Proof. Let s ∈ F (U). i.e. s : U → x∈U Fx. For x ∈ U choose t(x) ∈ F (Vx) s.t. x ∈ Vx ⊂ U open and s(y) = t(x)y for all y ∈ Vx.

Set τ(x) = ϕVx = ϕVx (t(x)) ∈ G (Vx). If y ∈ Vx, then τ(x)y = ϕ(t(x))y = ϕy(t(x)y) = ϕy(s(y)) ∈ Gy. Thus, τ(x )| = τ(x )| , which gives that {τ(x)} glue to τ ∈ (U). Set 1 Vx1 ∩Vx2 2 Vx1 ∩Vx2 G ϕU (s) = t. Exercise: Check the details! Deﬁnition 4.6. ϕ : F 0 → F is a morphism of sheaves. We say ϕ is injective if ϕU is injective for all U ⊆ X. If ϕ is injective, then we say F 0 ⊆ F is a subsheaf as F 0(U) ⊆ F (U). 0 Exercise: ϕ injective iﬀ ϕx : Fx → Fx is injective for all x ∈ X. Consequence: If ϕ : F → G is a morphism of presheaves such that ϕU : F (U) → G (U) injective for all U ⊆ X open, then ϕ+ : F + → G + is injective.

40 Proof. Check: ϕp : Fp → Gp injective for all p ∈ X. If ϕp(sp) = 0 ∈ Gp for s ∈ F (U), p ∈ U then ϕU (s)p = 0 so ϕU (s)|V = 0 for p ∈ V ⊆ U, so ϕV (s|V ) = 0. Thus s|V = 0, so sP = 0. In particular: If a presheaf F is a subpresheaf of a sheaf G then F + is a subsheaf of G . Definition 4.7 (Kernel and Image). Let ϕ : F → G be a morphism of sheaves of abelian groups. Then ker ϕ = the sheaf U 7→ ker(ϕU ) ⊆ F (U). Im ϕ = the sheafification of the presheaf U 7→ Im(ϕU ) ⊂ G (U). Note, ker ϕ ⊂ F , Im ϕ ⊂ G are subsheaves, and ϕ injective iff ker ϕ = 0. Definition 4.8 (Surjective). ϕ is surjective if Im(ϕ) = G .

WARNING: ϕ surjective DOES NOT IMPLY that ϕU is surjective. Exercise: ϕ surjective iff ϕp : Fp → Gp surjective for all p ∈ X. Definition 4.9 (Quotient Sheaf). F 0 ⊂ F a subsheaf, then F /F 0 = the sheafification of [U 7→ F (U)/F 0(U)] We have a surjective morphism F → F /F 0 which has kernel F 0. φ φi+1 Notation: A sequence of sheaves → F i →i F i+1 → F i+2 → ... is a complex if φi+1 ◦ φi = 0 for all i and is exact if Im φi = ker φi+1 for all i. Equivalently, complexes and exact sequences of the stalks in the category of abelian groups. Example, 0 → F 0 → F → F 00 → 0 is exact iff F 0 ⊂ F and F 00 ' F /F 0.

Definition 4.10. f : X → Y a continuous map, F a sheaf on X, then f∗F is −1 a sheaf on Y defined by f∗F (V ) = F (f (V )). Example: X a variety, Y ⊆ X closed, i : Y → X the inclusion, U ⊆ X open, then IY (U) = {f ∈ OX (U)|f(y) = 0, ∀y ∈ U ∩ Y }, IY ⊆ OX is a subsheaf of ideals. Then OX (U)/IY (U) are the regular functions U ∩ Y → k which can be extended to all of U. OX /IY ' i∗OY because we can extend locally. We have exact sequence 0 → IY → OX → i∗OY → 0, which will often be written 0 → IY → OX → OY → 0. ∗ Example, f : X → y a morphism of SWFs, we get morphism f : OY → −1 ∗ f∗OX by OY (V ) → f∗(OX (V )) = OX (f (V )), h 7→ h ◦ f = f h. ∗ Exercise: Find a morphism f : X → Y of varieties such that f : OY → f∗OX is an isomorphism, but f is NOT an isomorphism. Definition 4.11 (Inverse Image Sheaf). Let f : X → Y continuous, G a sheaf on Y . U ⊆ X open, define pre − f −1G (U) = lim G (V ). −→V ⊇f(U) We have maps G (V ) → pre − f −1G (U), s 7→ f −1s, f(U) ⊆ V . pre − −1 −1 −1 f G (U) = {f s|s ∈ G (V ),V ⊇ f(U)}. f s = 0 ⇐⇒ s|W = 0 where f(U) ⊆ W ⊆ V . This is a presheaf on X. We define f −1G to be the sheafification of this presheaf.

41 −1 Special Case: X ⊆ Y is a subset, i : X → Y the inclusion, G |X = i G . Example: X ⊆ Y open, pre − i−1G (U) = lim G (V ) = U . It is already −→V ⊇U a sheaf, and so the special case just mentioned above holds. −1 Exercise: (f G )p = Gf(p). Adjoint Property f : X → Y continuous, F a sheaf on X, G a sheaf on Y , let ϕ : G → f∗F ϕ be a morphism of sheaves on Y . U ⊆ X open, V ⊇ f(U), then G (V ) →V F (f −1(V )) → F (U) and maps are compatible with restrictions of G , so this −1 −1 induces ψU : pre−f G (U) → F (U), which gives a morphism ψ : pre−f G → F , we sheaﬁfy to get ψ+ : f −1G → F . −1 + Exercise: hom(G , f∗F ) → hom(f G , F ): ϕ 7→ ψ is an isomorphism of abelian groups. −1 Category Theory Interpretation: f is a left adjoint functor to f∗ and f∗ is a right adjoint functor to f −1. Let X be a variety (ringed-space)

Deﬁnition 4.12 (OX -module). An OX -module is a sheaf F on X such that F (U) is an OX (U)-module for all open U ⊆ X, such that if V ⊆ U open then F (U) → F (V ) is OX (U)-homomorphism (ie, f ∈ OX (U), m ∈ F (U), (f · m)|V = f|V · m|V .) An OX -homomorphism ϕ : F → G is a morphism of sheaves such that ϕU : F (U) → G (U) is an OX (U)-homomorphism for all U ⊆ X open.

Note ker ϕ ⊆ F and Im ϕ ⊆ G are sub OX -modules, F is an OX -module implies that FP is an OX,P -module.

Deﬁnition 4.13 (Tensor Product). F , G , OX -modules, F ⊗OX G = [U 7→ + F (U) ⊗OX (U) G (U)]

Exercise: (F ⊗OX G )p = Fp ⊗OX,P Gp S Deﬁnition 4.14 (Locally Free). An OX -module F is locally free if ∃ α Uα = X an open cover such that | ' ⊗r. F Uα OUα Example: π : An+1 \{0} → Pn, let m ∈ Z.

n n Deﬁnition 4.15. A sheaf O(m) = OP (m) of OP -modules is Γ(U, O(m)) = {h ∈ k[π−1(U)] : h(λx) = λmh(x)∀x ∈ π−1(U), λ ∈ k∗}. −1 Note: f ∈ S = k[x0, . . . , xn] homogeneous of degree > 0. k[π (D+(f))] = k[D(f)] = Sf . So Γ(D+(f), O(m)) = (Sf )m.

n O(m) is an invertible OP -module (invertible means locally free of rank 1). m n n On Ui = D+(xi): OUi → O(m)|Ui , h 7→ xi h. Note that OP (0) = OP . n Γ(P , O(m)) = Sm when m ≥ 0 and 0 else.

n Lemma 4.3. O(m) ⊗P O(p) ' O(m + p). Proof. U ⊆ Pn open. Γ(U, O(m)) ⊗ Γ(U, O(p)) → Γ(U, O(m + p)), f ⊗ g 7→ fg. Sheaﬁﬁcation gives a map O(m) ⊗ O(p) → O(m + p), restricting to U we have

OUi ⊗ OUi → OUi .

42 Consequence: O(m) ' O(p) iﬀ m = p. If m ≤ p and O(m) ' O(p) implies that O(m − p) ' O(p − p), on the right we have nonzero global sections, on the left we only do if m − p nonnegative, so m = p. Later: Any invertible sheaf on Pn is isomorphic to O(m), m ∈ Z. Coherent Sheaves Let X be an aﬃne variety and A = k[X], let M be an A-module.

˜ + Deﬁnition 4.16 (Quasi-Coherent Sheaf). M = [U 7→ M ⊗A OX (U)] is called a quasi-coherent OX -module.

Note: M ⊗A OX (D(f)) = M ⊗A Af = Mf . ˜ ˜ Examples A = OX , Y ⊆ X closed, I = I(Y ) ⊂ A. I = IY ⊂ OX . ˜ Exercise: A/I = i∗OY . ˜ Claim: Mp = MI(p) for p ∈ X, I(p) = I({p}) ⊂ A. ˜ ˜ MI(p) → Mp by m/f 7→ (m ⊗ 1/f)p, this tensor product is in M(D(F )). Surjectivity is clear. ˜ Injective: if m/f 7→ 0 ∈ Mp then (m ⊗ 1/f)|D(h) = 0 for some h ∈ A \ I(p), n so m⊗1/f = 0 ∈ M ⊗A Ah = Mh, thus h m = 0 ∈ M, and so m/f = 0 ∈ MI(p). ˜ ` Consequence M(U) =set of functions s : U → p∈U MI(p) such that ∀p ∈ U, there exists p ∈ V ⊂ U and m ∈ M, f ∈ A such tthat s(p) = m/f ∈ MI(q) for all q ∈ V .

Proposition 4.4. M˜ (D(f)) ' Mf .

Corollary 4.5. 1. OX (D(f)) = Af

2. Γ(X, M˜ ) = M.

We now prove the proposition.

Proof. ψ : Mf → M˜ (D(f)) the obvious m n n It is injective, as if ψ(m/f ) = 0 then m/f = 0 ∈ MI(p) for all p ∈ D(f). Thus, for all p ∈ D(F ), there exists hp ∈ A \ I(p) with hpm = 0 ∈ M. D(f) ⊆ S N P D(hp) ⇒ V (f) ⊇ V ({hp}), so f ∈ I(V ({hp})) means f = p aphp for N P n ap ∈ A. f m = aphpm = 0, thus m/f = 0 ∈ Mf . ˜ Sr Surjectivity: Let s ∈ M(D(f)), there exists a cover D(f) = i=1 Vi, mi ∈ M, hi ∈ A such that s = mi/hi on Vi. WLOG Vi = D(gi) and Vi = D(hi) (replace mi 7→ gimi by hi 7→ gihi). On D(hihj) = D(hi)∩D(hj), s = mi/hi = mj/hj ∈ M˜ (D(hihj)), injectivity N for D(hihj): mi/hi = mj/hj ∈ Mhihj . So (hihj) (hjmi − himj) = 0 ∈ M. N N+1 Replace mi 7→ hi mi, hi 7→ hi , hjmi = himj. S n P P D(f) ⊆ D(hi) so f = aihi where ai ∈ A. Set m = aimi ∈ M. Claim: s = ψ(m/f n). P P n For all j, hjm = i hjaimi = ( i aihi) mj = f mj. n So m/f = mj/hj = s on D(hj).

43 Definition 4.17 (Quasi-Coherent and Coherent). Let X be any variety. An S OX -mdule F is quasi-coherent if there exists an open affine cover X = Ui ˜ and k[Ui]-modules Mi such that F |Ui ' Mi as OUi -modules. F is coherent if Mi finitely generated k[Ui]-modules for all i. Examples:

˜⊗r ⊗r 1. All locally free OX -modules are coherent, U = Spec −m(A), A = OU .

2. Y ⊆ X closed, IY and i∗OY are coherent OX -modules. O 1 (U) 0 ∈/ U Example: X = 1, (U) = A . This is called the exten- A F 0 0 ∈ U

1 sion of the OA \{0} by zeros. F is an OX -module, but is not quasi-coherent, as if U ⊆ X is open affine and 0 ∈ U, then Γ(U, F ) = 0 but F |U is not the zero sheaf. Exercises: ˜ 1. F is quasi-coherent iff F |U ' F (U) for all open affine U ⊆ X. 2. F is coherent implies F (U) is finitely generated k[U]-module for all affine open U.

3. f : X → Y a morphism of varieties.

(a) f aﬃne implies that f∗OX quasi-coherent.

(b) f ﬁnite f∗OX coherent.

Example: M a finitely generated A-module, X = Spec −m(A). Then M˜ is locally free OX -module of rank r iff M is a projective A-module of const. rank r. S ⊗ ⊗r (M˜ loc free iff X = D(fi), M˜ ' r iff Mf = A iff M projec- D(fi) OD(fi) i fi tive.) Recall: X is complete implies Γ(X, OX ) = k. More general fact: F a coherent sheaf on a complete variety X then dimk Γ(X, F ) < ∞. We will use this without proof. (Projective case in Hartshorne, General in EGA.) 1 1 Note that Γ(A , OA ) = k[t] which has infinite dimension. Pushforward, Pullback Let f : X → Y be a morphism of varieties. F an OX -module, then f∗F is ∗ a f∗OX -module, we have ring homomorphism f : OY → f∗OX , and so f∗F is an OY -module. −1 −1 −1 −1 −1 Let G be a OY -module f G is an f OY -module. (f h·f s = f (hs)) −1 OY → f∗OX is the same as f OY → OX by the adjoint property.

∗ −1 −1 Deﬁnition 4.18 (Pullback). Deﬁne f G = f G ⊗f OY OX , we call it the pullback. Examples

44 ∗ 1. f OY = OX .

∗ −1 −1 2. (f G )p = (f (G))p ⊗(f OY )p OX,p = Gf(p) ⊗OY,f(p) OX,p. ∗ 3. U ⊆ Y open, i : U → Y inclusion, i G = G |U ⊗OY |U OU = G |U .

Adjoint Property −1 f : X → Y a morphism of SWFs. G an OY -module. We have f OY - −1 ∗ homomorphism f G → f G by s 7→ s⊗1. This gives an α : OY -homomorphism ∗ F → f∗f G .

∗ Lemma 4.6. G is an OY -module, F is an OX -module. Then homOX (f G , F ) ' homOY (G , f∗F ).

∗ α ∗ f∗ψ Proof. Given ψ : f G → F we obtain φ : G → f∗f G → f∗F . ˜ −1 −1 Given φ : G → f∗F , we obtain φ : f G → F which is an f OY -hom. ∗ −1 −1 ˜ Take ψ : f G = f F ⊗f OY OX → F by s ⊗ h 7→ h · φ(s).

f g Functoriality X → Y → Z morphism of SWFs. If F is a sheaf on X, g∗(f∗F ) = (gf)∗F .

Proposition 4.7. 1. G on Z implies that (gf)−1G = f −1(g−1G )

∗ ∗ ∗ 2. G an OZ -module implies that (gf) G = f (g G ). Proof. We will prove case 2. ∗ ∗ ∗ ∗ ∗ id : (gf) G → (gf) G gives G → (gf)∗(gf) G = g∗f∗(gf) G gives g G → ∗ ∗ ∗ ∗ f∗(gf) G which gives f (g G ) → (gf) G . ∗ ∗ We have a global homomorphism, so enough to check stalks. f (g G )p = ∗ (g G )f(p) ⊗OY,f(p) OX,p. This is (Gg(f(p)) ⊗OZ,gf(p) OY,f(p)) ⊗OY,f(p) OX,p = ∗ Ggf(p) ⊗OZ,gf(p) OX,p = ((gf) G )p

−1 Let f : X → Y be a morphism of SWFs. G is an OY -module, and f G is −1 an f OY -module.

∗ −1 −1 Deﬁnition 4.19 (Pullback). f G = f G ⊗f OY OX is an OX -module. ∗ −1 f : OY → f∗OX induces a map f OY → OX . Some sections: σ ∈ G (V ), we set f ∗σ = f −1σ ⊗ 1 ∈ Γ(f −1(V ), f ∗G ). The ∗ stalks (f G )p = Gf(p) ⊗OY,f(p) OX,p. g f If Z → X → Y , then (fg)∗G = g∗(f ∗G ).

∗ Corollary 4.8. If G is a locally free OY -module, then f G is a locally free OX -module of the same rank. Proof. Let Y = S V be an open cover such that | ' ⊕r. Set U = i G Vi OVi i −1 f (Vi) ⊂ X.

45 f ... XY...... p . q ...... f 0 . .... Ui ...... Vi f ∗ | = p∗f ∗ = f 0∗q∗ ' f 0∗q∗( ⊕r) = ⊕r G Ui G G OVi OUi Lemma 4.9. f : X → Y a morphism, 0 → G 0 → G → G 00 → 0 is a short exact ∗ 0 ∗ ∗ 00 sequence of OY -modules. Then f G → f G → f G → 0 is an exact sequence 00 of OX -modules. If G is locally free, then the ﬁrst map is injective. 0 00 Proof. On stalks we start with 0 → Gf(p) → Gf(p) → Gf(p) → 0 exact. Tensor produce is right exact and gets us to f ∗, ad so we have the ﬁrst part of the theorem immediately. 00 00 If G is locally free, then Gf(p) is a free OY,f(p)-module, and so the original sequence is split-exact.

Definition 4.20 (Generated by Finitely Many Global Sections). The OX - module F is generated by finitely many global sections iff ∃ a surjective map ⊕m OX → F . Equivalently, ∃s1, . . . , sm ∈ Γ(X, F ) such that Fp generated by (s1)p,..., (sm)p as an OX,p-module.

Example: Any quasi-coherent OX -module, if X is aﬃne (this is just generated by global sections, requires coherent to be generated by ﬁnitely many) n n Example: OP (1) is generated by x0, . . . , xn ∈ Γ(P , O(1)). n ⊕n+1 Suppose that f : X → P is a morphism, then O n → O(1) → 0 exact ⊕n+1 ∗ ∗ P implies that OX → f O(1) → 0 is exact, so f O(1) is generated by global ∗ ∗ sections f (x0), . . . , f (xn).

Proposition 4.10. X a variety, L invertible OX -module generated by global n ∗ sections s0, . . . , sn ∈ Γ(X, L ). Then ∃!f : X → P such that f O(1) ' L and ∗ f (xi) ↔ si.

Proof. Set Ui = {p ∈ X|(si)p ∈/ mpLp}. Ui open: If V ⊆ X open with L |V ' OV , then L |V is generated by t ∈ Γ(V, L ) so we write si = hit on V with hi ∈ k[V ]. Then Ui ∩ V = {p ∈ V |hi(p) 6= 0} is open in V . Sn Note: L is generated by s0, . . . , sn implies that X = i=1 Ui, and OUi '

L |Ui implies that 1 7→ si|Ui . On Ui, we can write si = hijsj for some hij ∈ k[Ui]. We define a map n g : Ui → P by f(p) = (hi0(p): ... : hin(p)). The maps are compatible: On Ui ∩ Uj, h`js` = sj = hijsi = hijhìs`, so h`j = hìhij. The map on U` : p 7→ (h`0(p): ... : h`n(p)) = hì(p)hi0(p): ... : n hì(p)hin(p). Thus, we have a morphism f : X → P . ∗ ∗ Claim: ∃ isomorphism L → f O(1) by si 7→ f (xi). On Ui, we define ∗ ∗ ∗ L |Ui → f O(1)|Ui by hsi 7→ hf (xi). This means that s` = hi`si 7→ hi`f (xi),

46 ∗ ∗ we must check that f (x`) = hi`f (xi). The definition of f implies that (x`/xi)◦ hi` ∗ ∗ x` ∗ x` ∗ ∗ f = = hi`, so f (x`) = f ( xi) = f ( )f (xi) = hi`f (xi). hii xi xi And now we prove uniqueness: If f : X → Pn is any morphism such ∗ ∗ ∗ ∗ x` that ' f (1) and si ↔ f (xi) on Ui, hi`si = f (x`) = f ( xi) = L O xi ∗ x` ∗ ∗ x` ∗ f ( )f (xi) = f ( )si, so f (x`/xi) = hi` on Ui. xi xi Definition 4.21 (Very Ample Sheaf). Let L be an invertible sheaf on X. L is very ample iff L is generated by (finitely many) global sections and the n map f : X → P given by generators s0, . . . , sn ∈ Γ(X, L ) is an isomorphism f : X → W ⊆ Pn locally closed.

n Exercise: OP (m) is very ample iff m ≥ 1. Definition 4.22 (PGL). P GL(n) = GL(n + 1)/k∗. Exercise: P GL(n) is an affine algebraic group. FACT: Every invertible sheaf on Pn is isomorphic to O(m) for some m.

Corollary 4.11. Aut(Pn) ' P GL(n). Proof. P GL(n) ≤ Aut(Pn) is trivial. Let f : Pn → Pn be an automorphism. The fact implies that f ∗O(1) ' n + m (m) for some m. In fact, = dim Γ( n, (m)) = dim Γ(f ∗ (1)) = O m k P O O dim Γ(O(1)) = n + 1, so m = 1. ∗ ∗ n ∗ Pn f x0, . . . , f xn ∈ Γ(P , O(1)) form a basis. We write f (xi) = i=0 aijxj for aij ∈ k. Then A = (aij) ∈ GL(n + 1). n n P ∗ x` Deﬁne ϕ : P → P by ϕ(x0 : ... : xn) = ( a0jxj : ... : anjxj). ϕ ( x ) = P i a`j xj ∗ x` P = f ( ). aij xj xi Thus, f = ϕ ∈ P GL(n).

Corollary 4.12. Pn is not an algbriac group. Normal Varieties Definition 4.23 (Normal Variety). X is irreducible. Then X is normal iff OX,P is normal (integrally closed) for all p. Example: Nonsingular varieties. Note: X affine, then X is normal iff k[X]m normal for all maximal m iff k[X] is normal. Exercise: If A is a domain, S ⊆ A multiplicative, then S−1A = S−1A.

Definition 4.24 (Normalization). If f is an affine variety, k[X] ⊂ k(X), then k[X] ⊆ k(X) is the integral closure. The normalization of X is X = Spec −m(k[X]). As we have the inclusion k[X] → k[X], we get a projection map X → X which is finite.

47 Lemma 4.13. ϕ : U → X morphism of affines, ϕ an open embedding iff ∗ ∗ ∗ ∃f1, . . . f −n ∈ k[X] such that (ϕ f1, . . . , ϕ fn) = (1) ⊂ k[U] and ϕ : k[X]fi → ∗ k[U]ϕ fi is an isomorphism for all i. Sr Proof. ⇒: Take open cover U = i=1 D(fi), fi ∈ k[X]. Sr ∗ ∗ ⇐: Set V = i=1 D(fi) ⊆ X.(ϕ f1, . . . , ϕ fr) = (1) ⊆ k[U] implies that −1 ϕ(U) ⊆ V . Thus, we have that ϕ : ϕ (D(fi)) → D(fi) is an isomorphism, so ϕ : U → V isomorphism. Lemma 4.14. Assume X affine, U ⊆ X is an open affine, then U¯ ⊆ X¯ open affine. Proof. k[X] ⊆ k[U] ⊆ k(X). Thus, k[X] ⊆ k[U], so we have a morphism ϕ : U¯ → X¯. Take f1, . . . , fn ∈ k[X] as in lemma wrt U ⊂ X. ¯ ¯ (f1, . . . , fn) = (1) ⊆ k[U], and k[U]fi = k[U]fi = k[X]fi = k[X]fi . And so, the lemma implies that ϕ is an open embedding.

Exercise: Given pre-varieties X1,...,Xn, open subsets Uij ⊆ Xi and isomorphisms ϕij : Uij → Uji such that Uii = Xi, ϕii = id, for all i, j, k, ϕij(Uij ∩ Uik) = Uji ∩ Ujk and ϕik = ϕjk ◦ ϕij on Uij ∩ Uik, then ∃! prevariety X with morphisms ψi : Xi → X such that ψi : Xi → open⊂ X is an Sn isomorphism, X = i=1 ψi(Xi), ψi(Uij) = ψi(Xi) ∩ ψj(Xj) and ψi = ψj ◦ ϕij on Uij.

Deﬁnition 4.25 (Normal). A variety X is normal iﬀ X is irreducible and OX,P are normal for all p.

Construction: X an irreducible variety, X = X1 ∪... ∪Xn open affine cover, set Uij = Xi ∩ Xj. Then Uij is affine, have ψij : Uij → Uji nbe the identity. Now U¯ij ⊆ X¯i and still have φij : U¯ij → U¯ji is the identity, so the satisfy the hypotheses of the exercise. Thus, there exists a prevariety X¯ = X¯1 ∪ ... ∪ X¯n. We call this the normalization of X. Note: k[Xi] is a finitely generated k[Xi]-module, so we have finite π : X¯i → Xi, which we can glue to a morphism π : X¯ → X. −1 Exercise: π : X¯ → X is finite. (Check that π (Xi) = X¯i) Exercise: ϕ : X → Y affine morphism of pre-varieities. Then Y is separated implies that X is serpated, and so X¯ is an irreducible normal separated variety. Example: X irred curve implies π : X¯ → X resoluton of singularities. Definition 4.26 (Local Ring along Subvariety). Let X be a variety, V ⊆ X irreducible and closed. Then OX,V = lim OX (U). −→U⊆X,U∩V 6=

If X is irreducible, then OX,V = {f ∈ k(X): f is deﬁned at one point of V }. General Case: U ⊆ X open aﬃne, U ∩ V 6= ∅, P = I(U ∩ V ) ⊆ k[U], OX,V = k[U]P .

Deﬁnition 4.27 (Regular along V ). X is regular along V if OX,V is a regular local ring. i.e., the maximal ideal in OXV is generated by dim(OX,V ) = codim(V ; X) elements. This happens iﬀ V 6⊆ Xsing.

48 5 Divisors

Let X be a normal variety. Deﬁnition 5.1 (Prime Divisor). A prime divisor on X is a closed irreducible subvariety of codimension 1.

Note: Y a prime divisor implies that OX,Y is a normal N¨otherianlocal ring of dimension 1. That is, OX,Y is a DVR. Consequences

1. codim(Xsing; X) ≥ 2.

∗ 2. Have valuation map vY : k(X) → Z for each prime divisor Y ⊆ X.

∗ Lemma 5.1. Let f ∈ k(X) . Then vY (f) = 0 for all but ﬁnitely many Y .

Proof. Show vY (f) < 0 for finitely many Y . Set U ⊆ X open set where f defined. Z = X \ U. vY (f) < 0 iff f∈ / OX,Y iff f is not defined at any point of Y iff Y ⊆ Z component.

Definition 5.2 (Divisor Group). Define Div(X) =free abelian group generated by all prime divisors. P An element D = ni[Yi] is a finite sum and is called a Weil Divisor.

∗ P Definition 5.3. For f ∈ k(X) , set (f) = Y vY (f) · [Y ] ∈ Div(X). Note (f −1) = −(f), (fg) = (f) + (g) Thus k(X)∗ → Div(X) by f 7→ (f) is a group homomorphism. Definition 5.4 (Class Group). Define C`(X) = Div(X)/{(f): f ∈ k(X)∗}.

Example:C`(An) = 0, as every hypersurface corresponds to a prime divisor. Remark: X complex, dim(X) = n, then C`(X) ↔ H2n−2(X; Z). Remark: X irreducible but not normal, we can still deﬁne C`(X), use vY (f/g) = lengthOX,Y (OX,Y /(f)) − lengthOX,Y (OX,Y /(g)). Divisors on Pn Note: All prime divisors are hypersurfaces Y = V+(h) where h ∈ S = k[x0, . . . , xn] is an irreducible form.

n P Deﬁnition 5.5. Degree of a Divisor deg : Div(P ) → Z by deg( mi[Yi]) = P mi deg Yi.

n ∗ Qr mi Let f ∈ k(P ) , mi ∈ Z, g = i=1 hi , hi ∈ S irreducible form. Then P n mi deg(hi) = 0, so Yi = V+(hi) ⊆ P is a prime divisor, so vYi (hi) = 1, vYi (f) = mi. Pr P (f) = i=1 mi[Yi] implies that deg(f) = mi deg(hi) = 0, thus, deg : C`(Pn) → Z is well-deﬁned. Claim: Isomorphism.

49 Surjective: H ⊆ Pn hyperplane, deg(m[H]) = m. Injective: Let D = P n mi[Yi] ∈ Div(P ), suppose deg(D) = 0, then Yi = V+(hi), hi ∈ S irred form, P Q mi n ∗ so mi deg(hi) = deg(D) = 0, so f = hi ∈ k(P ) and D = (f). Later: X nonsingular implies C`(X) ' Pic(X), thus Pic(Pn) = Z. X normal, Y ⊆ X prime divisor implies that OX,Y is a DVR.

Theorem 5.2. R normal N¨otheriandomain implies R = ∩ht p=1Rp where ht p = dim Rp, the max m such that ∃0 ( p1 ( ... ( pm = p. ∗ Corollary 5.3. If X is normal, f ∈ k(X) , then f ∈ k[X] iﬀ vY (f) ≥ 0 for all Y ⊆ X prime divisors.

Lemma 5.4. R Nötherian,then R is a UFD iff all prime ideals of height one are principal. Proof. ⇒: Assume R a UFD. P ⊆ R prime of height 1, let x ∈ P be an irreducible element, 0 ( (x) ⊆ P , so P = (x). ⇐: Exercise: R any Nötheriandomain then every element of R is a product of irreducible elements. Unique Factorization: Show x ∈ R irred and x|fg implies that x|f or x|g, ie, (x) ⊆ R is prime, let P ⊇ (x) min prime, PIT implies ht(P ) = 1 implies P = (y), x = ay, a ∈ R a unit. Proposition 5.5. X irreducible affine variety, k[X] a UFD iff X normal and C`(X) = 0. Proof. ⇒: UFD implies normal. Let Y ⊆ X a prime divisor, P = I(Y ) ⊆ k[X] prime of height 1. P = (h) ⊆ k[X], h ∈ k[X]. So (h) = [Y ] implies [Y ] = 0 ∈ C`(X). ⇐: Let P ⊆ k[X] prime of height 1, Y = V (P ) ⊆ X a prime divisor, ∗ [Y ] = 0 ∈ C`(X) so [Y ] = (h) ∈ Div(X), h ∈ k(X) . vZ (h) ≥ 0 for all Z ⊆ X prime divisors implies h ∈ k[X]. Claim: P = (h) ⊆ k[X]. ⊇ is clear. Let g ∈ P , then vY (g) ≥ 1, so vZ (g/f) ≥ 0 for all Z, so g/f ∈ k[X], and g = af ∈ (f). Proposition 5.6. X normal, Z ⊆ X is a proper closed subset, U = X \ Z. Then

1. C`(X) → C`(U) by [Y ] 7→ [Y ∩ U] if Y ∩ U 6= ∅ and 0 else is surjective 2. If codim(Z; X) ≥ 2, then C`(X) = C`(U).

3. If Z prime divisor, then Z → C`(X) → C`(U) → 0 is exact.

∗ Proof. 1. Well deﬁned Div(X) → Div(U). f ∈ k(X) ,(f) 7→ (f|U ) (because if Y ⊆ X is a prime divisor, Y ∩ U 6= ∅ then OX,Y = OU,U∩Y ). Thus, C`(X) → C`(U) is well deﬁned. Surjective: If V ⊆ U a prime divisor, V¯ ⊆ X is a prime divisor [V¯ ] 7→ [V ].

2. Div(X) = Div(U), so (f) = (f|U ).

50 P ∗ 3. If D = nY [Y ] 7→ 0 ∈ C`(U), then ∃f ∈ k(X) : vY (f) = nY for all Y 6= Z. D − (f) = m[Z] ⇒ D = m[Z] ∈ C`(X).

Example: X = V (xy − z2) ⊂ A3. Exercise: X (above) is normal. L = V (y) ∩ X = V (y, z) is a prime divisor on X. z2 Max ideal of OX,L is generated by z, y = x ∈ OX,L. 2 Set U = X \ L aﬃne, k[U] = (k[x, y, z]/(xy − z ))y = k[y, z]y is a UFD, so C`(U) = 0. Z → C`(X) → C`(U) = 0. So C`(X) = {m[L]: m ∈ Z}, y ∈ k(X)∗. 2 (y) = vL(y)[L] = vL(z )[L] = 2[L]. Thus, C`(X) = Z/2Z or C`(X) = 0. As k[x, y, z]/(xy − z2) is not a UFD, C`(X) = Z/2Z. Picard Group Invertible OX -module = line bundle.

Let X be any variety, L1 and L2 are line bundles, then L1 ⊗OX L2 is a line bundle. L is invertible implies that we can define L −1 = [U 7→ homOU (L |U , OU )]. −1 −1 Exercise: L is an invertible OX -module and L ⊗ L ' OX . Definition 5.6 (Picard Group). Pic(X) = {isomorphism classes of invertible sheaves on X}. This is a group under tensor product. Notation: X irreducible variety, L an invertible sheaf on X and s ∈ L (U), t ∈ L (V ) are nonzero sections. Take W ⊂ U ∩ V open such that L |W ' OW generated by u ∈ L (W ). Then s|W = fu and t|W = gu, f, g ∈ k[W ]. Define s/t = f/g ∈ k(X)∗. n n Example: s0, . . . , sn ∈ Γ(X, L ). Define f : X 99K A ⊂ P . f(x) = (s1/s0(x), . . . , sn/s0(x)) = (s1/s0(x): ... : sn/s0(x) : 1). If s0, . . . , sn generate L , then f extends to a morphism f : X → Pn. X is a normal variety, s ∈ L (U), s 6= 0, Y ⊆ X is a prime divisor, take V ⊆ X open such that L |V ' OV generated by t ∈ L (V ) and V ∩ Y 6= ∅.

Deﬁnition 5.7. VY (s) = VY (s/t).

0 0 0 0 Well defined, if V ⊆ X, V ∩ Y 6= ∅, L |V 0 generated by y ∈ L (V ) implies 0 0 0 that t/t nowhere vanishing function on V ∩ V , so t/t is a unit in OX,Y . 0 0 Thus, VY (s/t ) = VY (s/t · t/t ) = VY (s/t) + 0. P Definition 5.8. (s) = Y VY (s)[Y ] ∈ Div(X). Note: If s0 ∈ L (U 0) then (s0) = (s) + (s0/s). Then (s0) = (s) ∈ C`(X), so all nonzero sections of a line bundle are equivalent in the class group. Thus, we have a well defined map Pic(X) → C`(X) by L 7→ (s). Check: s1 ∈ L1(U), s2 ∈ L2(U), then s1 ⊗s2 ∈ L1 ⊗L2(U) and (s1 ⊗s2) = (s1) + (s2), so this map is a group homomorphism. Cartier Divisors X normal.

51 P Deﬁnition 5.9 (Cartier Divisor). A Cartier is a Weil Divisor D = ni[Yi] Sn which is locally principal. I.E. there exists an open covering i=1 Ui = X such that D|Ui = 0 ∈ C`(Ui) for all i.

∗ Note: D|Uj = (fj) where fj ∈ k(Uj) . We can think about D as the collection {fj} of these generators. Deﬁnition 5.10 (Cartier Class Group). CaC`(X) = {Cartier Divisors on X}/{(f): f ∈ k(X)∗}.

Recall: L is an invertible OX -module, s ∈ L (U) nonzero section, then P (s) = Y prime vY (s) · [Y ] ∈ Div(X) where vY (s) = vY (s/t) for t ∈ L (V ) generator of L |V ' OV , Y ∩ V 6= ∅. Note: (s) is Cartier. Thus, we have a group homomorphism Pic(X) → CaC`(X) ⊂ C`(X). Line Bundles from Divisors P Let D = nY [Y ] ∈ Div(X).

∗ Deﬁnition 5.11. OX -module L (D) or OX (D) Γ(U, L (D)) = {f ∈ k(X) |vY (f) ≥ −nY for all prime divisors Y such that Y ∩ U 6= ∅} ∪ {0}.

Example: L (0) = OX (0) = OX . 1 Example: X = P , Q = (a : b) ∈ X, D = n[Q], Q = V+(h), h = bx0 − ax1 ∈ n 1 1 k[x0, x1]. So OP (n[Q]) ' OP (n) by f 7→ h f. ∗ Note: If h ∈ k(X) then OX (D + (h)) → OX (D) by f 7→ hf is an isomorphism. vY (f) ≥ −nY − vY (h) ⇐⇒ vY (fh) ≥ −nY . Consequence: D is a Cartier Divisor implies that OX (D) is an invertible OX -module. If D|U = (h) ∈ Div(U), then OX (D)|U = OU ((h)) ' OU . Note, this says we have a map CaC`(X) → Pic(X) by D 7→ OX (D). WARNING: If f ∈ Γ(U, OX (D)) then f a rational function. The notation (f) means two things! Proposition 5.7. Pic(X) ' CaC`(X) as abstract groups. Proof. Will check that Pic(X) →← CaC`(X) are inverse maps. P Let D = nY [Y ] a Cartier Divisor. Set V = X \ (∪nY <0Y ) ⊆ X open. ∗ Then 1 ∈ k(X) is a section of OX (D) over V .(vY (1) ≥ −nY ⇐⇒ nY ≥ 0). Claim: (1) = D ∈ Div(X). If D|U = (g) ∈ Div(U), then OX (D)|U ' OU −1 −1 gneerated by h ∈ Γ(U, OX (D)), Y ∩U 6= ∅ implies that vY (1) = vY (1/h ) = vY (h). Thus, (1)|U = (h)|U = D|U . Thus, CaC`(X) → Pic(X) → CaC`(X) is the identity. Let L be a line bundle on X. t ∈ Γ(U, L ) a non-zero section. Note: If 0 6= s ∈ L (V ), then Y ∩ V 6= ∅ implies vY (s/t) = vY (s) − vY (t) ≥ −vY (t), and so s/t ∈ Γ(V, OX ((t))). Claim: L ' OX ((t)) by s 7→ s/t. If L |V ' OV gneerated by u ∈ L (V ) then (t)|V = (t/u) ∈ Div(V ) implies that OX ((t))|V is generated by u/t as u 7→ u/t we get L |V ' OX ((t))|V .

52 Examples:

1. Pic(An) = CaC`(An) ⊂ C`(An) = 0, so all line bundles on An are trivial.

n Fact: Any locally free OA -module of ﬁnite rank is trivial. n Sn 2. P = i=0 D+(xi), C`(D+(xi)) = 0, so all Weil divisors are Cartier, thus Pic(Pn) = CaC`(Pn) = C`(Pn) = Z.

n By the maps we have, any line bundle is isomorphic to OP (m[H]), where n m n n H ⊂ P is a hyperplane. I(H) = (h), OP (m[H]) ' OP (m) by f 7→ h f. Thus, Pic(Pn) = {O(m)}. 3. X = V (xy − z2) ⊂ A3. L = V (y) ∩ X, I(L) = (y, z) ⊂ k[A3]. Claim: [L] is not Cartier. Otherwise there exists open aﬃne U ⊂ X such that P = (0, 0, 0) ∈ U with [L ∩ U] = (f|U ) ∈ Div(U). Thus f ∈ k[U] and I(L ∩ U) = (f) ⊂ k[U], so I(L) · OX,P = (y, z) ⊂ OX,P is principal. 2 But P ∈ X is a singular point, so dimk(mP /(m)P ) = 3, so {x, y, z} is a 2 2 basis, and so dim((y, z) + mP /mP ) = 2, so (y, z) ⊂ OX,P is not principal, which is a contradiction. Thus CaC`(X) = Pic(X) = 0 6= C`(X).

Note: OX,P is not a UFD, as (y, z) ⊂ OX,P is height 1 prime but not principal.

Deﬁnition 5.12 (Locally Factorial). An irreducible variety X is locally factorial if OX,P is a UFD for all p ∈ X. Example: Nonsingular implies locally factorial implies normal. Proposition 5.8. X locally factorial implies Pic(X) = C`(X).

Proof. Show that any prime divisor [Y ] is Cartier. First: U = X \Y ,[Y ]|U = 0. Let P ∈ Y , then I(Y ) · OX,P ⊂ OX,P is a height 1 prime, so I(Y ) · OX,P = (f) ⊂ OX,P , f ∈ OX,P ⊂ k(X). Note: vY (f) = 1, if Z 6= Y prime divisor, p ∈ Z, then f ∈ OX,Z (deﬁned at P P ) and f∈ / I(Z) · OX,P . Thus, vZ (f) = 0, and we have (f) = [Y ] + ni[Zi] where p∈ / Zi for all i. Set U = X \ (∪Zi) open in X, p ∈ U. Then [Y ]|U = (f)|U ∈ Div(U) principal, so [Y ] is Cartier.

2 3 Example: X = V (xy − z ) ⊂ A , X0 = X \{0, 0, 0}, X0 is nonsingular, so Pic(X0) = C`(X0) = C`(X) = Z/2Z, so there exists a unique nontrivial line bundle on X0 which is NOT equal to the restriction of a line bundle on X. Definition 5.13 (Affine, Finite). Let f : X → Y be a morphism of varieties. f is affine if f −1(V ) ⊂ X is affine for all V ⊂ Y is open affine. f is finite if it is affine and k[f −1(V )] is a finitely generated k[V ]-module.

53 Exercise: Enough that this is true for an open affine cover of Y . Examples: X,Y affine, f : X → Y morphism is affine. X ⊂ Y closed, then the inclusion is finite. Divisors on Non-Singular Curves Recall that X nonsing complete curve implies that X is projective. X ⊂ CK open, K = k(X), X = CK . Lemma 5.9. Let X be a complete, nonsingular curve, then any nonconstant morphism f : X → Y is finite. Proof. WLOG: Y is a curve. Thus, f ∗ : k(Y ) ⊂ k(X) is a finite field extension. Take V ⊆ Y open affine, k[V ] ⊂ k(Y ). Set A = k[V ] ⊂ k(X), then A is a finitely generated k[V ]-module. U = Spec −m(A) a nonsingular curve, k(U) = k(X) ⇒ we have diagram f ... X ...... Y ...... ⊂ . ⊂ ...... UV...... −1 −1 Claim: f (V ) = U. x ∈ f (V ) ⇒ k[V ] ⊆ OX,x, so A ⊂ OX,x, thus OX,x = AP for some P ⊂ A prime. Thus, x = P ∈ U = Spec −m(A). Definition 5.14 (Degree of f). Let f : X → Y be a finite, dominant morphism, then deg(f) = [k(X): k(Y )]. Pullback of Divisors on Curves f : X → Y a finite morphism of nonsingular curves. Q ∈ Y , mQ = (t) ⊆ ∗ ∗ OY,Q. If f(P ) = Q then f : OY,Q → OX,P , f t ∈ mP . ∗ P Definition 5.15. f : Div(Y ) → Div(X):[Q] → P ∈f −1(Q) vP (t)[P ]. Alternatively, if D ∈ Div(Y ), set V = Y −Supp(D), then s = 1 ∈ Γ(V, L (D)). Note: (s) = D. Then f ∗s ∈ Γ(f −1(V ), f ∗L (D)) is the pullback. Exercise: f ∗D = (f ∗s) ∈ Div(X). Definition 5.16 (Torsion Free). Let R be a domain, M an R-module. M is torsion free if ∀a ∈ R, x ∈ M, then ax = 0 implies a = 0 or x = 0. Fact: Any f.g. torsion-free module over a PID is free. P Definition 5.17 (Degree of a Divisor). X a nonsingular curve, D = ni[Pi] = P P niPi ∈ Div(X). Set deg(D) = ni Warning: If X is not complete, then deg is not defined on C`(X). Proposition 5.10. f : X → Y is a finite morphism of nonsingular curves, D ∈ Div(Y ). Then deg(f ∗D) = deg(f) deg(D).

54 Proof. ETS if Q ∈ Y a point, then deg(f ∗Q) = deg(f). V ⊆ Y open aﬃne with Q ∈ V . Then f −1(V ) = Spec −m(A) ⊂ X, A = k[V ] ⊂ k(X). −1 Q ⊂ k[V ] a max ideal, set B = AQ = (k[V ] \ Q) A. A ﬁnitely generated k[V ]-module implies B f.g. k[V ]Q = OY,Q-module. OY,Q a DVR, B torsion free, so B is free OY,Q-module.

rankOY,Q (B) = dimk(Y ) k(X) = deg(f). mQ = (t) ⊂ OY,Q. OY,Q/tOY,Q = k. Thus, dimk(B/tB) = deg(f). Note: points in f −1(Q) correspond to max ideals in P ⊂ A such that P ∩ k[V ] = Q, which correspond to max ideals in AQ = B. −1 s Write f (Q) = {P1,...,Ps}, Pi ⊆ A max ideals, B = ∩i=1BPi ⇒ tB = s s ∩i=1tBPi = ∩i=1(tBPi ∩ B). s By the Chinese Remainder Theorem, B/tB ' ⊕i=1B/(tBPi ∩ B). Injective: clear ni Surjective: t ∈ Pi for all i, BPi DVR, so tBPi = (PiBPi ) , so tBPi ∩B ⊆ PiB and tBPi ∩ B 6⊆ PjB for j 6= i. th Thus, this is an isomorphism after ⊗BPi (only the i summand survives).

Now: B/(tBPi ∩ B) = (B/(tBPi ∩ B))Pi = (B/tB)Pi = BPi /tBPi =

OX,Pi /(t). Thus dimk B/(tBPi ∩ B) = vPi (t). ∗ P Thus, deg(f Q) = vPi (t) = dimk(B/tB) = deg(f). Lemma 5.11. h ∈ k(Y )∗ implies f ∗((h)) = (f ∗h). f ∗h = h ◦ f ∈ k(X).

Proof. Let P ∈ X, Q = f(P ) ∈ Y . mP = (s) ⊆ OX,P . mQ = (t) ⊂ OY,Q. m ∗ n h = ut , u ∈ OY,Q a unit. f t = vs , v ∈ OX,P a unit. ∗ Coef of [P ] in f ((h)) = vQ(h)vP (t) = mn. f ∗h = f ∗(utm) = (f ∗u)(f ∗t)m = (f ∗u)vmsnm, so coef of [P ] in (f ∗h) is nm. So, we have a group homomorphism f ∗ :C`(Y ) → C`(X). Corollary 5.12. X complete nonsing curve, f ∈ k(X)∗ implies deg((f)) = 0. Proof. f is defined on open U ⊂ X. Then f : U → A1 ⊂ P1 is a regular function. As P1 is complete, f extends to f : X → P1. As X is complete, f is finite. k[A1] = k[t], f ∗(t) = f ∈ k(X). (f) = (f ∗t) = f ∗((t)), so deg((f)) = deg(f ∗((t))) = deg(f) deg(t). (t) = [0] − [∞] ∈ Div(P1) so deg((t)) = 0. So if X is a complete nonsingular curve, there exists deg : C`(X) → Z. Notation: D,D0 ∈ Div(X), D ∼ D0 iff D = D0 ∈ C`(X). Proposition 5.13. If X complete nonsingular curve, then X rational iff ∃P 6= Q ∈ X such that P ∼ Q. Proof. ⇒: X = P1, then P ∼ Q for all P,Q ∈ P1. ⇐: ∃f ∈ k(X)∗ such that (f) = [P ] − [Q] ∈ Div(X). f : X → P1 a morphism. Then (f) = (f ∗t) = f ∗((t)) = f ∗([0] − [∞]). This tells us that f ∗([0]) = [P ] and f ∗([∞]) = [Q]. So deg(f ∗[0]) = 1 = deg(f) · 1, so f is degree 1, so it is birational. Thus isomorphism.

55 Deﬁnition 5.18 (Elliptic Curve). An elliptic curve is a nonsingular closed plane curve E ⊂ P2 such that deg(E) = 3. 2 3 2 2 Example: V+(zy − x + z x) ⊂ P . Claim: No elliptic curve is rational.

2 Exercise: Set OE(1) = OP (1)|E. Then Γ(E, OE(1)) = (S/I(E))1, S = k[x, y, z]. Therefore, dimk Γ(E, OE(1)) = 3. 2 P 0 0 Let L ⊂ P be a line. L.E = P 0∈L∩E I(L·E; P )[P ] = P +Q+R ∈ Div(E). 2 Exercise: OE(1) = OP ([L])|E = OE([L.E]), so ∃D(= L.E), D ∈ Div(E) such that deg(D) = 3 and dimk Γ(E, OE(D)) = 3. 1 1 1 Let D ∈ Div(P ) such that deg(D) = 3. Then OP (D) = OP (3), this gives 1 1 1 1 us that Γ(P , OP (D)) = Γ(P , OP (3)) = k[x0, x1]3. The dimension of this is 4. Conclude: E is not rational. Let X ⊂ P2 be any nonsingular curve. deg : C`(X) → Z, [P ] 7→ 1. Deﬁnition 5.19. C`0(C) = ker(deg). So we have a short exact sequence 0 → 0 0 C` (X) → C`(X) → Z → 0 that splits, so C`(X) = C` (X) ⊕ Z. Fact: C`0 corresponds to a nonginular complete abelian algebraic group, the Jacobi Variety of X. 2 Let L = V+(f), M = V+(g) be lines in P . P X.L = P I(X · L; P )P = P1 + ... + Pn where n = deg(X). X.M = Q1 + ... + Qn. Exercise: f/g ∈ k(X)∗ and (f/g) = X.L − X.M ∈ Div(X). X.L − X.M = P1 + P2 + P3 − Q1 − Q2 − Q3 = 0 ∈ C`(X) for X = E.

0 Theorem 5.14. Let P0 ∈ E be any point, then E → C` (E) by P 7→ P − P0 is bijective.

0 Proof. Injective: If P − P0 = Q − P0 ∈ C` (E) then P ∼ Q so P = Q as E is not rational. 2 Surjective: Let M ⊂ P be tangent line to E at P0. M.E = 2P0 + R, R ∈ E. 0 P Let D ∈ C` (E). Write D = ni(Qi − P0) for Qi ∈ E, ni ∈ Z. 0 Assume that ni < 0. Then L =line through Qi and R, L.E = Qi + R + Qi, 0 0 0 0 = L.E − M.E = Qi + R + Qi − 2P0 − R so Qi − P0 = −(Qi − P0) ∈ C` (E). 0 Replace Qi by Qi, ni 7→ −ni, WLOG, ni ≥ 0. 0 P Claim: D = P − P0 ∈ C` (E), P ∈ E. Induction on ni. P If ni ≥ 2, then Q1 − P0, Q2 − P0 have positive coeﬃcients. L =line 0 through Q1,Q2, L.E = Q1 + Q2 + Q ∈ Div(E). 0 0 0 0 00 Let L be the line through Q and P0. Then L .E = Q + P0 + Q . L.E − 0 00 00 0 L .E = Q1+Q2−P0−Q = 0, so (Q1−P0)+(Q2−P0) = (Q −P0) ∈ C` (E).

2 2 Example: char k 6= 2, λ ∈ k, λ 6= 0, 1. Eλ = V+(zy −x(x−z)(x−λz)) ⊂ P . 0 Take P0 = (0 : 1 : 0) ∈ EEλ corresponds to C` (Eλ) by P 7→ P − P0. Let ⊕ 00 be a group op on W . Q1 ⊕ Q2 = Q (Picture omitted) Fact: Any elliptic curve is isomorphic to Eλ by P0 ↔ (0 : 1 : 0).

56 Theorem 5.15. E is an algebraic group.

Proof. (char k 6= 2): WLOG, E = Eλ, P0 = (0 : 1 : 0). Deﬁne ϕ : E × E → E by ϕ(P,Q) = R the unique point such that ∃ a line L with L.E = P + Q + R. It is enough to show that ϕ : E × E → E is a morphism. P ⊕ Q = ϕ(P0, ϕ(P,Q)), −P = ϕ(P1,P0). Set U1 = D+(z) and U2 = D+(y) subsets of E. E = U1 ∪ U2. −1 Show that (Ui ×Uj)∩ϕ (U`) → U` by ϕ is a morphism for all i, j, ` ∈ {0, 1}. 2 2 −1 U1 = V (y −x(x−1)(x−λ)) ⊂ A .(U1×U1)∩ϕ (U1) → U1 → k is the regu- 2  (y2−y1) 2 − (x1 + x2) + 1 + λ x1 6= x2  (x2−x1) lar function (x1, y1)×(x2, y2) 7→ 2 2 2 x1+x1x2+x2+λ−(1+λ)(x1+x2) + 1 + λ − (x1 + x2) y1 + y2 6= 0  y1+y2

Diﬀerentials R is a ring, S is a commutative R-algebra, M an S-module.

Deﬁnition 5.20 (R-derivation). A function D : S → M is an R-derivation if D(fg) = fD(g)+gD(f) for all f, g ∈ S, D(f +g) = D(f)+D(g), and D(f) = 0 for all f ∈ R. Remark: the third conditioon holds iﬀ D is an homomorphism of R-modules. ⇒: f ∈ R ⇒ D(fg) = fD(g) and ⇐ is an exercise (use D(1) = 0).

Definition 5.21 (Module of Kählerdifferentials). F =free S-module with basis 0 {d(f)|f ∈ S} = ⊕f∈SS · d(f). F =submodule generated by d(f) for f ∈ R, d(fg) − fd(g) − gd(f), d(f + g) − d(f) − d(g). 0 We define ΩS/R = F/F is the module of Kählerdifferentials of S over R

0 We deﬁne d = dS = dS/R : S → ΩS/R by f 7→ d(f) + F . This is the universal R-derivation of S. It has the universal property that given any R-derivation D : S → M, there ˜ ˜ exists a unique map S-homomorphism D :ΩS/R → M such that D = D ◦ dS. Exercise: Let P (x1, . . . , xn) ∈ R[x1, . . . , xn] and f1, . . . , fn ∈ S, and D : S → Pn ∂P M is an R-derivation. Then D(P (f1, . . . , fn)) = (f1, . . . , fn)D(fi). i=1 ∂xi Consequence: If S generated by f1, . . . , fn as an R-algebra, then ΩS/R is gererated by dS(f1), . . . , dS(fn) as an S-module.

Proposition 5.16. S = R[x1, . . . , xn]. Then ΩS/R is the free S-module on dx1, . . . , dxn.

n Proof. Have a surjectve S-hom from S → ΩS/R which sends ei 7→ dxi. This n ∂P ∂P is surjective. We deﬁne D : S → S by P (x1, . . . , xn) 7→ ,..., . By ∂x1 ∂xn ˜ n the universal property, there is a unique S-homomorphism D :ΩS/R → S , by deﬁnition, d(xi) 7→ D(xi) = ei, so this is an inverse. Proposition 5.17. Given ring homomorphisms R → S → Y then we have an exact sequence of T -modules ΩS/R ⊗S T → ΩT/R → ΩT/S → 0.

57 dT Proof. S → T → ΩT/R is an R-deriv of S. So we get S-hom ϕ :ΩS/R → ΩT/R ϕ˜→ via ϕ(dS(f)) = dT (f). Thus, we have a T -hom ΩS/R ⊗ T Ω T/R by ω ⊗ h 7→ hϕ(ω). Note: Image(ϕ ˜) =submodule of ΩT/R generated by dT (f) for f ∈ S. Thus ΩT/R/ Im(ϕ ˜) = ΩT/S. Note: I ⊂ S an ideal, T = S/I, then I/I2 is a T -module, T × I/I2 → I/I2 by (f + I) · (h + I2) = fh + I2 ∈ I/I2. Proposition 5.18. T = S/I. We have an exact sequence of T -modules I/I2 → 2 ΩS/R ⊗S T → ΩT/R → 0, where the ﬁrst map is given by h + I 7→ ds(h) ⊗ 1.

2 Proof. Set M equal to the image of I/I in ΩS/R ⊗S T . Then M is generated by {ds(h) ⊗ 1|h ∈ I}. We deﬁne D : T → (ΩS/R ⊗ T )/M by D(f + I) = (dS(f) ⊗ 1) + M. This is an R-derivation. ˜ Thus, there is a unique T -hom D :ΩT/R → (ΩS/R ⊗S T )/M by dT (f + I) 7→ (ds(f) ⊗ 1) + M

Example: S = R[x1, . . . , xn], I = (f1, . . . , fp) ⊂ S. T = S/I.ΩS/R ⊗S T = n ⊕n ⊕ T dxi = T . i=1 2 ⊕n ∂f¯i ∂f¯i The image of fi under I/I → T : ds(fi) ⊗ 1 = ,..., . Set ∂x1 ∂xn ∂f¯ J = i ∈ Mat(p × n; T ) is the Jacobi matrix. ∂xj 2 p J So the image of fi is eiJ, so ΩT/R = coker(I/I → ΩS/R ⊗T ) = coker(T → T n). 2 3 2 e.g. T = k[x, y]/(y − x + x), so J = [1 − 3x , 2y], so ΩT/k = T ⊕ T/h(1 − 2 3x )e1 + (2y)e2i. Proposition 5.19. S an R-algebra, U ⊆ S multiplicatively closed subset, then −1 ΩU −1S/R = U ΩS/R

−1 Proof. S → U S → ΩU −1S/R is an R-derivation. Thus, it induces an S- homomorphism ΩS/R → ΩU −1S/R, dS(f) 7→ d(f), where d is the universal derivation of U −1S. −1 −1 −1 This induces U S-hom U ΩS/R → ΩU −1S/R by ds(f)/u 7→ u d(f). −1 −1 ud(s)−sd(u) We deﬁne D : U S → U ΩS/R by D(s/u) 7→ u2 . Exercise: D is well deﬁned R-derivation. ˜ −1 This induces D :ΩU −1S/R → U ΩS/R is the inverse map. Let X be a topological space. R, S sheaves of rings on X, R → S a ring hom.

Definition 5.22. pre − ΩS /R(U) = ΩS (U)/T (U) for U ⊆ X open. For V ⊂ U d open, S (U) → S (V ) → pre − Ω(V ) is an R(U)-derivation. So, we get S (U)- hom pre − Ω(U) → pre − Ω(V ). + We define ΩS /R = (pre − ΩS /R) , the sheafification.

58 ∗ −1 Let ϕ : X → Y morphism of varieties, then we have ring hom ϕ : ϕ OY → OX .

−1 Deﬁnition 5.23 (Relative cotangent sheaf). ΩX/Y = ΩOX /ϕ OY is called the relative cotangent sheaf

Special case: X → {pt},ΩX = ΩX/k = ΩX/{pt}. This is called the cotangent sheaf.

Proposition 5.20. ϕ : X → Y a morphism of affine varieties, then ΩX/Y = ˜ Ωk[X]/k[Y ] Proof next time. As a consequence, ΩX/Y is always coherent. Lemma 5.21. If (A, m) is a local Nötheriandomain, N a finitely generated

A-module, then we set r = dimA/m(N/mN). If r ≤ dimA0 (N0), then N is free of rank r. Proof. Nakayama’s Lemma implies that N can be generated by r elements. Thus, there exists an exact sequence 0 → K → Ar → N → 0, localization r is exact, so 0 → K0 → A0 → N0 → 0 is exact, so the last morphism is an r isomorphism of vector spaces, so A0 ' N0, so K0 = 0. Thus, K = 0 as it is torsion free and localizes to zero, so Ar ' N.

Recall: Let X ⊂ An be a closed irreducible variety. Let I = I(X) = n 2 n 2 (f1, . . . , fs). Let P ∈ X. Set M = I({p}) ⊂ k[A ], M/M ' k via h + M 7→ ∂h (P ),..., ∂h (P ) ∂x1 ∂xn 2 2 2 2 So we have an exact sequence (I + M )/M → M/M → mP /mP → 0. s n 2 mP ⊂ OX,P a max ideal, therefore k → k → mP /mP → 0 is also exact where the first map is J(P ), and we call the second φ. If h ∈ M, then φ ∂h (P ),..., ∂h (P ) = h + m2 . ∂x1 ∂xn P J Note: rank(k(X)s → k(X)n) ≤ c = codim(X; An). (If h is any (c + 1) × (c + 1)-minor of J, then h ∈ k[X], and h(P ) = (c + 1) × (c + 1)-minor. J(P ) = 0 for all P ∈ X. So h = 0 ∈ k[X].) Theorem 5.22. Assume X is an irreducible variety of dimension r, let P ∈ ⊕r X. Then P is a nonsingular point iff ΩX,P ' OX,P . If mP is generated by h1, . . . , hr ∈ mP , then dh1, . . . , dhr ∈ ΩX,P is a basis for ΩX,P . n ∂fi Proof. WLOG: X ⊆ affine. I(X) = (f1, . . . , fs), J = . A ∂xj s J n s J n k[X] → k[X] → Ωk[X]/k → 0 yields OX,P → OX,P → ΩX,P → 0, which we will call (∗). s J(P ) n We mod out by mP , and get k → k → ΩX,P /mP ΩX,P → 0. 2 2 Pn ∂h Thus, mP /m ' ΩX,P /mP ΩX,P by h + m 7→ (P )dxj. P P j=1 ∂xj 2 Assume that ΩX,P is free of rank r, then dimk(mP /mP ) = r, thus P is a nonsingular point.

59 2 s J n Assume that dimk(mP /mP ) = r.(∗) ⇒ k(X) → k(X) → (ΩX,P )0 → 0 is exact. Note: r = dimk(ΩX,P /mP ΩX,P ) ≤ dimk(X)((ΩX,P )0). ⊕r The lemma implies that ΩX,P ' OX,P . Lemma 5.23. ϕ : X → Y is a morphism of aﬃne varieties. Then Γ(X, pre − ΩX/Y ) = Ωk[X]/k[Y ]

−1 Proof. S = Γ(X, ϕ OY ), ring homomorphisms k[Y ] → S → k[X]. Thus, ΩS/k[Y ]⊗S k[X] → Ωk[X]/k[Y ] → Ωk[X]/S → 0 where the last is Γ(X, pre−ΩX/Y ), so enough to show that the ﬁrst map is zero. Let f ∈ Im(S → k[X]). We must show that df = 0 ∈ Ωk[X]/k[Y ]. There n −1 exists open cover X = ∪i=1Ui such that f|Ui ∈ image of Γ(Ui, pre − ϕ OY ) = lim OY (V ) −→V ⊃ϕ(Ui)

WLOG, Ui = Xgi where gi ∈ k[X]. Enough to show that df = 0 in N N N Ωk[X]/k[Y ] for each i, since g df = 0 ∈ Ωk[X]/k[Y ] and (g , . . . , g ) = gi i i n (1) = k[X]. But Ωk[X]/k[Y ] = Ωk[U ]/k[Y ]. So we replace X with Ui, we may assume gi i −1 that f ∈ image of Γ(X, pre − ϕ OY ) = lim OY (V ). I.E. there exists −→V ⊃ϕ(X) 0 ∗ 0 V ⊂ Y open, f ∈ OY (V ) such that ϕ(X) ⊂ V and f = ϕ (f ) ∈ k[X]. Now m 0 N 0 N+1 V = ∪i=1Yhi , hi ∈ k[Y ], f ∈ k[Y ]hi , so hi f ∈ k[Y ] for all i, so hi df = N+1 d(hi f) = 0 ∈ Ωk[X]/k[Y ]. N N Now, X = ∪Xhi ⇒ (h1 , . . . , hm) = (1) ⊂ k[X], so df = 0. ˜ Proposition 5.24. ϕ : X → Y morphism of aﬃnes. Then ΩX/Y = Ωk[X]/k[Y ]

Proof. Set Ω = Ωk[X]/k[Y ]. We have Ω ' Γ(X, pre−ΩX/Y ) → Γ(X, ΩX/Y ), this ˜ gives an OX -homomorphism Ω → ΩX/Y . ˜ Let f ∈ k[X]. Γ(Xf , Ω) = Ωf = Ωk[Xf ]/k[Y ] = Γ(Xf , pre − ΩX/Y ). {Xf } is a basis for the top, and so they have the same stalks.

Corollary 5.25. ϕ : X → Y any morphism of varieties. Then ΩX/Y is coherent.

−1 Proof. Let Y = ∪Vi open aﬃne cover. ϕ (Vi) = ∪Uij ⊆ X is an open aﬃne ˜ cover of X.ΩX/Y |Uij = Ωk[Uij ]/k[Vi].

Corollary 5.26. If X irreducible, then X is nonsingular iﬀ ΩX is a locally free OX -module.

1 1 Example: X = P ,ΩP is a line bundle. The projective coordinate ring is x1 1 ∗ k[x0, x1]. Set t = ∈ k( ) . t ∈ 1 (D+(x0)), dt ∈ Γ(D+(x0), Ω 1 ). Find x0 P OP P (dt) ∈ Div(P1). 1 1 1 Ui = D+(xi), U0 = A ⊂ P . If p ∈ U0, then t − p generated mp, so ΩP ,p is generated by d(t − p) = dt, so vp(dt) = 0 for all p ∈ U0. −1 −1 −2 −2 k[U1] = k[s], s = t , dt = d(s ) = −s ds, v∞(dt) = v∞(s ) = −2. 1 1 1 1 Thus (dt) = −2[∞] ∈ Div(P ), and so ΩP ' OP (−2[pt]) = OP (−2).

60 2 Example: E = Eλ ⊂ P an elliptic curve. Then ΩE ' OE. Linear Systems P Let X be a nonsingular projective variety. D = nY [Y ] ∈ Div(X). We say that D is effective if nY ≥ 0 for all Y . If D is effective, we write D ≥ 0. Definition 5.24 (Complete Linear System of D). Given any D ∈ Div(X), define |D| = {D0 ∈ Div(X): D0 ∼ D and D0 ≥ 0}.

Theorem 5.27. If X is projective and F is a coherent OX -module, then dimk Γ(X, F ) < ∞.

Deﬁnition 5.25. Let `(D) = dimk Γ(X, OX (D)).

Theorem 5.28. P(Γ(X, OX (D))) → |D| by s 7→ (s) is bijective.

Proof. Let s ∈ Γ(X, OX (D)), then (s) ≥ 0 and (s) ∼ D. So the map is well defined. Injective: Suppose s1, s2 ∈ Γ(X, OX (D)), assume that (s1) = (s2) ∈ Div(X). Then (s1/s2) = (s1) − (s2) = 0, so s1/s2 ∈ k[X] = k. Surjective: Let D0 ∈ |D|. Then D0 ∼ D, so D0 = D + (f) where f ∈ k(X)∗. We define s to be the section given by f ∈ Γ(X, OX (D)). This is a global section, P because vY (f) ≥ −nY for all Y ,(D = nY [Y ]). Set s0 = 1 ∈ Γ(U, OX (D)). 0 (s) = (f · s0) = (f) + (s0) = (f) + D = D . Lemma 5.29. X is a complete nonsingular curve, D ∈ Div(X), if `(D) 6= 0 then deg(D) ≥ 0 and if `(D) 6= 0 and deg(D) = 0 then D ∼ 0. Proof. `(D) 6= 0, then |D|= 6 ∅, so D ∼ D0 ≥ 0. So deg(D) = deg(D0) ≥ 0. If deg(D) = 0, then deg(D0) = 0, but as D0 is effective, D0 = 0. Riemann-Roch Theorem Let X be a complete nonsingular curve. Definition 5.26 (Canonical Divisor). K ∈ Div(X) is a canonical divisor on X if ΩX ' O(K).

Deﬁnition 5.27 (Genus). The genus g = `(K) = dimk Γ(X, ΩX )

1 1 Example: X = P ,ΩP = O(−2), so g = 0. 2 Example: E = Eλ ⊂ P elliptic curve, ΩE ' OE. So g = 1. Theorem 5.30 (Riemann-Roch). For any D ∈ Div(X) where X is a complete nonsingular curve, we have `(D) + `(K − D) = deg(D) + 1 − g.

Example: X = P1, K = −2P for some P ∈ P1. The RRT theorem says that `(nP ) + `(−2P − nP ) = n + 1 − 0, so if n ≥ 0, we have that `(nP ) = n + 1. If n = −1, then 0 + 0 = −1 + 1 = 0. If n ≥ −2, we also see that it works. Example: Set D = K, then `(K) − `(K − K) = deg(K) + 1 − g, so g − 1 = deg(K) + 1 − g, so deg(K) = 2g − 2. Corollary 5.31. A nonsingular curve is either aﬃne or projective.

61 Proof. C nonsingular curve implies that C = CK \{P1,...,Pn} where K = k(C), X = CK . If m >> 0, then `(mPi) = m+1−g ≥ 2. 1, fi ∈ Γ(X, OX (mPi)), fi ∈/ k. Pn ∗ (fi) = −ri[Pi] effective divisor in Div(X). Set f = i=1 fi ∈ k(X) . f is defined exactly on C ⊆ X, so C ' Spec −m(k[f]) is affine. Corollary 5.32. X is rational iff g = 0.

Proof. ⇒: genus of P1 is 0. ⇐: Let P 6= Q ∈ X. Set D = P − Q ∈ Div(X). By Riemann-Roch, `(D) ≥ deg(D) + 1 − g, so `(D) ≥ 1. Thus |D|= 6 ∅, so there is D0 ≥ 0 such that D ∼ D0, but deg(D0) = 0, so D0 = 0.

Corollary 5.33. X complete nonsingular curve of g = 1, P0 ∈ X, char(k) 6= 2. 2 2 Then ∃ isomorphism X ' Eλ = V+(zy − x(x − z)(x − λz)) ⊂ P for some λ not 0 or 1, that sends P0 7→ (0 : 1 : 0).

Proof. deg(K) = 2g − 2 = 0, so Riemann-Roch implies `(nP0) = n + 1 − 1 = n for all n ≥ 1, k = Γ(OX ) = Γ(OX (P0)) ( Γ(OX (2P0)) ( ... ( k(X).

Take x ∈ Γ(OX (2P0)) \ k. vP0 (x) = −2 (x) = A + B − 2P0 for A, B ∈ X. 1 ∗ ∗ x : X → P is a morphism, x ([0] − [∞]) = A + B − 2P0, so x ([0]) = A + B, thus [k(X): k(x)] = deg(x) = 2.

Take y ∈ Γ(OX (3P0)) \ Γ(OX (2P0)). vP0 (y) = −3, but as this is odd, y∈ / k(x). Thus k(X) = k(x, y). 2 2 3 2 {1, x, y, x , xy} is a basis for Γ(OX (5P0)), 1, x, y, x , xy, x , y ∈ Γ(OX (6P0)), dim = 6. 2 3 2 So there exists a linear relations. Rescale x, y: y +b1xy +b0y = x +a2x + 1 2 a1x + a0. Replace y with y + 2 (b1x + b0), y = (x − a)(x − b)(x − c) where a, b, c ∈ k. Claim: a 6= b 6= c 6= a. 2 2 ϕ : X \{P0} → C = V (y − (x − a)(x − b)(x − c)) ⊂ A , P 7→ (x(P ), y(P )). This is birational, as k(X) = k(x, y). Assume a = b = c. Then C is a curve with a cusp, and is rational, so X would be rational, contradiction. Assume a = b 6= c, then C is the nodal curve, which is also rational. x−a 2 c−a Replace x with b−a , rescale y, y − x(x − 1)(x − λ) where λ = b−a 6= 0, 1. 2 X \{P0} → C ⊂ A extends to an isomorphism X → Eλ.

Lemma 5.34. X complete nonsingular curve of genus g. Let P0,Q0,...,Qg ∈ Pg Pg X, then there exists P1,...,Pg ∈ X such that i=0 Pi ∼ i=0 Qi.

Proof. WLOG P0 6= Qi for all i. P Set D = Qi. `(D) ≥ deg(D) + 1 − g = 2. Thus, ∃h ∈ Γ(X, OX (D)) such ∗ that h∈ / k. Set f = h − h(P0) ∈ k(X) .(f) = −D + P0 + P1 + ... + Pg = 0 ∈ C`(X). Corollary 5.35. The Map Xg = X × ... × X with g factors to C`0(X) by Pg (P1,...,Pg) 7→ i=1(Pi − P0) is surjective.

62 Proof. Note: Let Q ∈ X, the lemma implies that there are P1,...,Pg ∈ X such Pg 0 that (g + 1)P0 ∼ Q + P1 + ... + Pg, so −(Q − P0) = i=1(Pi − P0) ∈ C` (X). P 0 Let D = nQ(Q − P0) ∈ C` (X). The note implies that we can assume P nQ ≥ 0, and the lemma implies that we may assume nQ ≤ g. Blow-Up of Varieties Y aﬃne variety, X ⊆ Y closed, I = I(X) = (f0, . . . , fn) ⊂ k[Y ]. Let n ϕ : Y \ X → P by ϕ(y) = (f0(y): ... : fn(y)).

Definition 5.28 (Blowup of Y along X.). B`X (Y ) = (y, ϕ(y)) : y ∈ Y \ X ⊂ Y × Pn. n Note: If Y \ X ⊆ Y is dense, then π : B`X (Y ) → Y is surjective because P is complete. And π : π−1(Y \ X) → Y \ X is an isomorphism. Point: If Y is singular along X, then usually B`X (Y ) is less singular. Example: Y = V (y2 −x2(x+1)) ⊂ A2. I(X) = (x, y) ⊂ k[Y ]. ϕ : Y \{0} → P1, P 7→ line through O and P . B`X (Y ) = {(P, ϕ(P )), (0, (1 : −1)), (0, (1 : 1))}. −1 Note: π (X) ⊂ B`X (Y ) is an effective Cartier divisor. i.e. codim = 1 and the ideal of π−1(X) is locally generated by a single element. ∗ ∗ n In fact: L = π n (OP (1)). Define si = π n (zi) ∈ Γ(B`X (Y ), L ), zi ∈ n P nP n P Γ(P , OP (1)) for 0 ≤ i ≤ n. We define s = i=0 fisi ∈ Γ(B`X (Y ), L ). Then −1 π (X) = Z(s) ⊂ B`X (Y ). Note: B`X (Y ) is independent of the generators fi of I(X). B`X (Y ) ⊂ n Y × P , set J = I(B`X (Y )) ⊂ k[Y ][z0, . . . , zn], a graded ideal. It is a fact that one can recover B`X (Y ) from k[Y ][z0, . . . , zn]/J. d d d Claim: k[Y ][z0, . . . , zn]/J ' ⊕d≥0I and this is by definition ⊕I t ⊂ k[Y ][t] and it is the subring of k[Y ][t] generated by k[Y ] as well as tf0, . . . , tfn. 1 n+1 Morphism ψ : Y × A → Y × A by ψ(y, t) 7→ (y, (tf0(y), . . . , tfn(y))). 1 1 ∗ ψ(Y × A ) =cone over B`X (Y ), so J = I(ψ(Y × A )) ⊂ k[Y ][z0, . . . , zn]. ψ : k[Y ][z0, . . . , zn] → k[Y ][t] with zi 7→ tfi. ∗ ∗ d J = ker(ψ ), so k[Y ][z0, . . . , zn]/J ' Im(ψ ) = k[Y ][tf0, . . . , tfn] = ⊕d≥0I . Now let Y be any variety, X ⊂ Y any closed subset. Take and open affine cover Y = ∪Yi. B`Yi∩X (Yi) can be glued together to get B`X (Y ) =

∪iB`Yi∩X (Yi). d IX ⊂ OY a sheaf of ideals. ⊕d≥0IX is a sheaf of graded OY -algebras, which can be turned into a variety, B`X (Y ) (see next semester).

6 Schemes

We will try to state definitions and theorems from commutative algebra, but will not prove many of them, as our focus is geometry. Let A be a ring Definition 6.1 (Spec A). We define Spec A = {prime ideals P ⊂ A}. If I ⊆ A, we set V (I) = {P ∈ Spec A : I ⊆ P }.

63 Lemma 6.1. 1. V (IJ) = V (I ∩ J) = V (I) ∪ V (J) P 2. V ( Iα) = ∩V (Iα) 3. V (0) = Spec A, V (A) = ∅ √ √ 4. V (I) ⊆ V (J) ⇐⇒ I ⊇ J √ Proof.√ We begin with 4: This follows from the fact that V (I) = V ( I) and I = ∩P ∈V (I)P . 1: IJ ⊂ I ∩ J ⊆ I, so V (IJ) ⊇ V (I ∩ J) ⊇ V (I) ∪ V (J). Let P ∈ V (IJ). Assume it is not in V (I). Then there exists a ∈ I with a∈ / P . Then for any b ∈ J, ab ∈ IJ ⊆ P , so b ∈ P , thus J ⊂ P , so P ∈ V (J). 2 and 3 are straightforward exercises. Topology: Let U ⊂ Spec(A) be open iff Spec(A) \ U = V (I) for some I ⊂ A. We note that P ∈ Spec(A), then {P } = V (P ), so P is a closed point iff P is maximal. Example: Take k = k¯ an alg closed field, A = k[x, y]. Then Spec(A) = {(x − a, y − b): a, b ∈ k} ∪ {(f): f(x, y) irreducible} ∪ {0}, and these correspond to {(a, b) ∈ k × k} ∪ {irred curves ⊂ k × k} ∪ {k × k}. The points (a, b) are the closed points, the others are called generic points for curves or the plane, because their closure is either everything (generic point of the plane) or are all of the points that lie on the curve. Structure Sheaf: For P ∈ Spec(A), set AP = {a/f : a, f ∈ A, f∈ / P }. For ` U = Spec(A), open define, O(U) = {s : U → P ∈U AP : s(P ) ∈ AP and s is locally a quotient}, that is, ∀P ∈ U, there exists an open neighborhood V , P ∈ V ⊂ U and a, f ∈ A such that s(Q) = a/f ∈ AQ for all Q ∈ V . Note: (1) O is a shead of rings on Spec(A) (2) The O(U) are not ”really” functions, so care must be taken. Definition 6.2 (Spectrum of a Ring). The spectrum of A is (Spec(A), O). Definition 6.3. For f ∈ A, set D(f) = {P ∈ Spec(A): f∈ / P } = Spec(A) \ V (f).

Note: {D(f)} is a basis for the topology on Spec(A). Let U ⊆ Spec(A) be open, P ∈ U. Write Spec(A) \ U = V (I). Then P/∈ V (I), so I 6⊂ P , thus ∃f ∈ I, f∈ / P , so P ∈ D(f) ⊂ U

Proposition 6.2. 1. OP = AP for all P ∈ Spec(A)

2. O(D(f)) = Af , for all f ∈ A. 3. O(Spec(A)) = A.

Proof. 1. For P ∈ U open, we have a ring homomorphism O → AP by s 7→ s(P ). This induces ϕ : OP → AP by ϕ(sP ) = s(P ). This is surjective, as we can let a/f ∈ AP . Then a/f ∈ O(D(f)) and ϕ(a/f) = a/f. It is injective, as we can assume that ϕ(sP ) = 0 ∈ AP , where s ∈ O(U), U

64 open, P ∈ U. WLOG, s = a/f for all Q ∈ U, a/f = 0 ∈ AP , so f∈ / P ag and there exists g ∈ A \ P such that ga = 0 ∈ A. s|U∩D(g) = fg = 0 ∈ O(U ∩ D(g)) ⇒ sP = 0 ∈ OP . n 2. Deﬁne ψ : Af → O(D(f)) by ψ(a/f ) = [P 7→ a/fn ∈ AP ]. For injectivity, assume that ψ(a/f n) = 0 ∈ O(D(f)). Set I = Ann(a) ⊂ A, for n P ∈ D(f), we have a/f = 0 ∈ AP , so there exists H ∈ A \ P such that ha = 0. Thus h ∈ I, h∈ /√P , so P/∈ V (I). Therefore, D(f) ∩ V (I) = ∅, so m n V (I) ⊂ V (f), thus f ∈ I, so f ∈ Ann(a), so a/f = 0 ∈ Af .

For surjectivity, let s ∈ O(D(f)). There exists an open cover D(f) = ∪Vi such that s = ai/gi on Vi for all i. WLOG, Vi = D(hi) for hi ∈ A. D(hi) ⊂ p ni D(gi) implies that V (hi) ⊇ V (gi), so hi ∈ (gi). Thus hi = cigi with ni ci ∈ A. Note that ai/gi = ciai/cigi = ciai/hi . So we replace ai with ciai ni and hi with hi , and WLOG, s = ai/hi pm D(hi) for all i. We claim that D(f) =union of finitely many D(hi). D(f) ⊆ ∪D(hi) iff V (f) ⊇ ∩V (hi) = p m m P V (hhii) iff f ∈ hhii iff f ∈ hhii iff f = bihi for a finite sum with bi ∈ A. Note that s = ai/hi = aj/hj on D(hi) ∩ D(hj) = D(hihj). ψ n injective implies ai/hi = aj/hj ∈ Ahihj , so (hihj) (hjai − hiaj) = 0 That n+1 n n+1 n n is, hj (hi ai) = hi (hj aj) for all i, j. Replace ai with hi ai and hi with n+1 hi . WLOG, hjai = hiaj ∈ A. m P P P P m f = bihi, set a = biai, hja = i biaihj = biajhi = f aj, so s = m m aj/hj = a/f on D(hi), the D(hi) cover D(f), so s = a/f ∈ O(D(f)). 3. Follows from part 2.

Deﬁnition 6.4 (Ringed Space). A Ringed Space is a pair (X, OX ) where X is a topological space and OX is a sheaf of rings on X. (Usually just denoted by X). A morphism of ringed spaces f : X → Y is a pair f = (f, f ]) where f : X → ] Y is continuous and f : OY → f∗OX a ring homomorphism. Let f : X → Y be a morphism of ringed spaces. P ∈ X. For V ⊂ Y open, ] f −1 f(P ) ∈ V , then we have OY (V ) → f∗OX (V ) = OX (f (V )) → OX,P induces ] fP : OY,f(P ) → OX,P .

Definition 6.5 (Locally Ringed Space). X = (X, OX ) is a locally ringed space if it is a ringed space such that OX,P is a local ring for all P ∈ X. We call the maximal ideal mP ⊆ OX,P . f : X → Y is a morphism of locally ringed spaces iff it is a morphism of ] ringed spaces such that fP : OY,f(P ) → OX,P is a local homomorphism for all ] ] −1 P ∈ X, that is, fP (mf(P )) ⊆ mP . (iff (fP ) (mP ) = mf(P )). Example: (Spec A, O) is a locally ringed space. Why do we want locally ringed spaces? Look at all the morphisms Spec(A) → Spec(B), there are a lot of morphisms of ringed spaces, however, the morphisms of locally ringed spaces are in correspondence with ring homomorphisms B → A.

65 Set X = Spec(A), Y = Spec(B). Let f : A → B be a ring homomorphism. Deﬁne ϕ : Y → X by ϕ(Q) = f −1(Q). Continuous: I ⊂ A an ideal, ϕ−1(V (I)) = {Q ∈ Y : f −1(Q) ⊃ I} = V (f(I)B) ⊂ Y For Q ∈ Y , let fQ : Af −1(Q) → BQ be a map on local rings. Let U ⊂ X open, ` ] OX (U) = {s : U → p∈U Ap|s is locally a quotient}. Deﬁne ϕ : OX (U) → −1 ϕ∗OY (U) = OY (ϕ (U)) by s 7→ [Q 7→ fQ(s(ϕ(Q))) ∈ BQ] ] Check: ϕQ = fQ : OX,ϕ(Q) → OY,Q. Proposition 6.3. There is a bijective correspondence between ring homomorphisms f : A → B and morphisms of LRS ϕ : Y → X by f 7→ ϕ.

] Proof. Note: f = ϕ : Γ(X, OX ) = A → Γ(Y, OY ) = B. Must Show: Any morphism of LRS ϕ : Y → X is determined by ϕ] : A → B. Let Q ∈ Y , set P = ϕ(Q) ∈ X. The following commutes: ϕ] ... A ...... B ...... ] ..... ϕQ ... AP = OX,P ...... OY,Q = BQ ] ] −1 ] −1 ϕQ is a local ring homomorphism so mP = (ϕQ) (mQ), so P = (ϕ ) (Q)

Remark: If X is an LRS, U ⊆ X open, then (U, OX |U ) is an LRS and U → X the inclusion is a morphism of LRS.

Definition 6.6 (Affine Scheme). An affine scheme is an LRS (X, OX ) such that (X, OX ) ' Spec(A) for some ring A.

Definition 6.7 (Scheme). A scheme is an LRS (X, OX ) such that there is an open cover X = ∪Uα such that (Uα, OX |Uα ) is an affine scheme for all α. Example: If k is a field, then n = Spec k[x , . . . , x ], n = Spec [x , . . . , x ], Ak 1 n AZ Z 1 n and the dimension of n is n + 1. AZ Example: Assume that char(k) = p > 0, and A any k-algebra. Then there is a ring homomorphism A → A : a 7→ ap. This gives the Absolute Frobenius Morphism F : Spec(A) → Spec(A), F (P ) = {a ∈ A : ap ∈ P } = P , so it is the identity on points, but is NOT the identity of schemes. Example: Suppose k = k¯, X a (pre)variety over k. Define X˜ = {closed irreducible subvarieties in X}. Then i : X → X˜ is the inclusion. Open subsets ˜ ˜ of X are U for U ⊆ X open. Define OX˜ = i∗OX . ˜ Exercise: (X, OX˜ ) is a scheme. Note: There exists a unique morphism of varieties X → {pt} = Spec(k), and this gives a structure morphism from X˜ → Spec(k). (X˜ is a scheme over Spec(k))

66 ϕ ... ˜ ...... ˜ X .. Y ...... If ϕ : X → Y is a morphism of varieties, then Spec(k) commutes. Exercise: Morphisms of varieties X → Y correspond to morphisms of schemes as above. Example: The absolute Frobenius map F : X˜ → X˜ is NOT a morphism of varieties.

Definition 6.8 (Reduced Scheme). A scheme X is reduced if OX (U) is a reduced ring for all U ⊂ X open. Definition 6.9 (Finite Type). A morphism ϕ : X → Y of schemes is of finite type if for all open affine V ⊂ Y there exists a finite open affine cover −1 ϕ (V ) = ∪Uα such that OX (Uα) is a finitely generated OY (V )-algebra for all α. Exercise: The category of prevarieties over an algebraically closed field k is equivalent to the reduced schemes of finite type over Spec(k). Projective Schemes S = ⊕d≥0Sd graded ring, Sd · Se ⊆ Sd+e. Exercise: 1 ∈ S0. I ⊆ S a homogeneous ideal, if I = ⊕(I∩Sd) iff I is generated by homogeneous elements. EG S+ = ⊕d>0Sd.

Definition 6.10. Proj(S) = {P ⊆ S homogeneous prime such that S+ 6⊂ P } For I a homogeneous ideal, set V (I) = {P ∈ Proj(S)|P ⊇ I}. P Check V (I) ∪ V (J) = V (IJ) = V (I ∩ J). ∩V (Iα) = V ( Iα) Topology: U ⊂ Proj(S) open iff Proj(S) \ U = V (I). Let P ∈ Proj(S), set T = {homogeneous elements of S \ P }, T −1S = −1 {a/f : a ∈ S, f ∈ T } = ⊕d∈Z(T S)d, deg(a/f) = deg a − deg f. Define −1 S(P ) = (T S)0 ` For U ⊂ Proj(S) open, define O(U) = {s : U → p∈U S(p)|s(p) ∈ S(p) and is locally a quotient }. ie, for all P ∈ U, there is an open nbhd P ∈ V ⊂ U and homogeneous elements a, f ∈ S of the same degree such that s(Q) = a/f ∈ S(Q) for all Q ∈ V . −1 Note: S(P ) is a local ring, S(P ) ⊂ T S → SP , max ideal S(P ) ∩ PSP .

Proposition 6.4. S graded ring, (Proj(S), O) as above, OP = S(P ) for all P ∈ Proj(S).

Proof. OP → S(P ), tP 7→ t(P ), and this is injective and surjective, following the proof for Spec A.

67 For f ∈ S+ homogeneous, deﬁne D+(f) = {P ∈ Proj(S): f∈ / P } = Proj(S) \ V (f) Exercise: {D+(f)} is a basis for the topology on Proj(S). n Deﬁne: S(f) = (Sf )0 = {a/f : deg(a) = n deg(f)}.

Proposition 6.5. (D+(f), O|D+(f)) ' Spec(S(f)).

Proof. S → Sf ⊇ S(f). Deﬁne ϕ : D+(f) → Spec(S(f)) by P 7→ PSf ∩ S(f). p Exercise: For Q ∈ Spec(S(f)) we have hQi ⊂ Sf prime ideal. Inverse map ϕ−1(Q) = phQi ∩ S. ϕ deg(f) deg(h) Homeomorphism: D+(h) ⊆ D+(f) → Spec(S(f)) sends D+(h) to D(h /f ). Note: (S(f))ϕ(P ) ' S(P ). −1 −1 For V ⊂ Spec(S(f)) open, then ϕ∗OProj(V ) = OProj(ϕ (V )) = {s : ϕ (V ) → ` ` P ∈ϕ−1(V ) S(P )} = {s : V → Q∈V (S(f))Q} = OSpec S(f) (V ).

n Example: A a ring, PA = Proj A[x0, . . . , xn], covered by

n D+(xi) = Spec A[x0, . . . , xn](xi) = Spec A[x0/xi, . . . , xn/xi] = AA Deﬁnition 6.11 (Properties of Schemes). A scheme X is

1. connected if it is connected as a topological space 2. irreducible if it is irreducible as a topological space

3. reduced if OX (U) is a reduced ring for all open U ⊂ X iﬀ OX,P reduced for all P ∈ X.

4. integral if OX (U) a domain for all U ⊂ X open

5. noetherian if X = ∪ Spec(Ai) where Ai Noetherian and the cover is ﬁnite

6. locally noetherian if X = Spec(Ai) for Ai noetherian, not necessarily a finite cover √ √ Example: X = Spec(A) is irreducible iff 0 ⊂ A is prime, it is reduced iff 0 = 0 and it is integral iff A is a domain.

Proposition 6.6. A scheme X is integral iff X is irreducible and reduced. Note: An open subscheme of (locally) Noetherian scheme is (locally) Noethe- rian Proposition 6.7. Spec(A) is locally noetherian iff notherian iff A is noetherian

Proof. Assume that Spec(A) is locally noetherian. Let U = Spec(B) ⊂ Spec(A) be open, B is noetherian. There exists f ∈ A such that D(f) ⊆ U, f|U ∈

O(U) = B. Spec(Af ) = Spec(Bf|U ) = D(f). Af = Bf|U Noetherian. We can write Spec(A) = ∪i Spec(Af ), Af Noetherian, so ∅ = V ({fi}) ⇒ P i i (fi) = (1) = A ⇒ 1 = aifi with ai ∈ A a ﬁnite sum.

68 Thus, there are f1, . . . , fr ∈ A such that Afi noetherian and (f1, . . . , fr) = A.

Let I ⊆ A ideal, choose g1, . . . , gm ∈ I such that Ifi = IAfi = (g1, . . . , gm) ⊆ N Afi for all i. Claim: I = (g1, . . . , gm). b ∈ I, fi b ∈ (g1, . . . , gm) for all i. So N N (f1 , . . . , fr ) = (1) ⊆ A, so A is Noetherian. Deﬁnition 6.12 (Properties of Morphisms). A morphism f : X → Y is

1. locally of finite type if for all open affine V ⊆ Y there exists open affine −1 cover f (V ) = ∪Ui s.t. OX (Ui) is finitely generated OY (V )-algebra. 2. of finite type if ∃ a finite cover of f −1(V ) as above. 3. affine if for all open affine V ⊆ Y , f −1(V ) ⊆ X is open affine.

−1 4. is finite if affine and OX (f (V )) is a finitely generated OY (V )-module whenever V is affine.

Exercise: In all cases, it is enough to know the property on a single open affine cover. Definition 6.13 (Open Immersion). f : X → Y is an open immersion if it can be factored f : X →' U ⊆ Y open. Definition 6.14 (Closed Immersion/Closed Subscheme). f : X → Y is a closed immersion if

1. f is a homeomorphism of X with closed subset of Y .

] 2. f : OY → f∗O∗ is surjective.

Example: Y = Spec(A), I ⊆ A ideal, A → A/I gives a closed immersion Spec(A/I) → Spec(A) with image V (I). Exercise: All closed immersions X → Spec(A) have this form. Exercise: Y scheme, V ⊆ Y a closed subset, then there exists a unique closed immersion X → Y with image V such that X is reduced.

Deﬁnition 6.15 (Dimension). dim(X) =supremum of n such that ∃ ∅ 6= Z0 ( Z1 ( ... ( Zn ⊆ X with Zi closed irreducible subset. Z ⊆ X closed irreducible. Then codim(Z; X) is the supremum of n such that ∃Z = Z0 ( Z1 ( ... ( Zn ⊆ X with Zi closed irreducible. Y ⊆ X is any closed subset, then codim(Y ; X) = inf{codim(Z; X): Z ⊆ Y closed irreducible}. WARNING: Z ⊆ X closed and irreducible, dim(Z)+codim(Z; X) = dim(X) does not always hold! Products

69 Deﬁnition 6.16 (Product). Let X,Y,S be schemes with morphisms α : X → S and β : Y → S. A product of X and Y over S is a scheme X ×S Y with morphisms X ×S Y ...... q ...... p ...... XY ...... α ...... β S along with the universal property that for any scheme Z with morphisms f, g to X and Y s.t. αf = βg, there exists a unique morphism ϕ : Z → X ×S Y such that f = pϕ and g = qϕ.

Exercise: X ×S Y is unique up to unique isomorphism. Exercise: X any scheme, A is a commutative ring, then there is a correspondence between {morphisms X → Spec(A)} and {ring hom A → OX (X)}. Consequence: X = Spec(A), Y = Spec(B) and S = Spec(R), then α, β make A and B into R-algebras, and X ×S Y = Spec(A ⊗R B). Observe that if S ⊆ T is an open subscheme, then X ×T Y = X ×S Y , as if j : S → T is the inclusion, then αf = βg ⇐⇒ jαf = jβg. Also observe that if −1 U ⊆ X open, then U ×S Y = p (U) ⊆ X ×S Y , and so it is an open subscheme of X ×S Y , because if h : Z → X is a morphism, so that h(Z) ⊆ U, then we can factor h : Z → U ⊆ X. Consequence: If X0 ⊆ X and Y 0 ⊆ Y , S0 ⊆ S are all open such that 0 0 0 0 0 0 0 −1 0 −1 0 α(X ), β(Y ) ⊆ S , then X ×S0 Y = X ×S Y and this is p (X ) ∩ q (Y ) ⊆ X ×S Y . Construction Assume that S is affine. Take open affine covers X = ∪Xi, Y = ∪Yj. We S glue X ×S Y := i,j Xi ×S Yj by (Xi ×S Yj)∩(Xk ×S Y`) = (Xi ∩Xk)×S (Yj ∩Y`). −1 If S is any scheme, we take an open affine cover S = ∪Si. Xi = α (Si), −1 Yi = β (Si) in X and Y are open sets. And here we glue X ×S Y := ∪iXi ×Si Yi by (Xi ×Si Yi) ∩ (Xj ×Sj Yj) = (Xi ∩ Xj) ×Si∩Sj (Yi ∩ Yj) Examples: X is a scheme

1. U, V ⊆ X open. U ×X V = U ∩ V .

2. Y,Z ⊆ X closed subschemes. We deﬁne Y ∩ Z := Y ×X Z, the scheme theoretic intersection. This is still a closed subscheme of X.

2 2 2 X = Ak = Spec k[x, y], Y = V (y − x ) = Spec k[x, y]/(y − x ), Z = V (y) = Spec k[x, y]/(y). We have a diagram of commutative rings k[x, y] ...... k[x, y]/(y − x2) k[x, y]/(y) ...... k[x, y]/(y − x2, y)

70 2 2 2 So Y ×X Z = Y ∩Z = Spec k[x, y]/(x , y) = Spec k[x]/(x ). dimk k[x]/(x ) = 2. Let X be a scheme and p ∈ X a point.

Deﬁnition 6.17 (Residue Field at p). k(P ) = OX,P /mP is called the residue ﬁeld at P .

If P ∈ U = Spec(A) ⊆ X open, then k(P ) = AP /P AP , the field of fractions of A/P . Note: A → k(P ) gives a morphism Spec(k(P )) → Spec(A) → X with image {P }. Examples: X an irreducible algebraic variety over k. If P ∈ X is a closed point, we get k(P ) = k. If P0 ∈ X is a generic point, we get k(P0) = k(X). For p ∈ U ⊆ X, f ∈ OX (U), set f(P ) =the image of f under OX (U) → OX,P → k(P ). Note: V (f) = {P ∈ U : f(P ) = 0 ∈ k(P )} is relatively closed in U. ] Let ϕ : Y → X a morphism, ϕ(Q) = P , P ∈ U. Then ϕQ : OX,P → OY,Q ] is a local ring homomrophism so it inducesϕ ¯Q : k(P ) → k(Q) a field extension. ] −1 ] ] ] ϕ (f) ∈ OY (ϕ (U)), ϕ (f)(Q) =ϕ ¯Q(f(P )) =ϕ ¯Q(f(ϕ(Q))) ∈ k(Q). E.G. X,Y varieties over k = k¯, Q ∈ Y a closed point implies that ϕ](f)(Q) = f(ϕ(Q)) ∈ k. Exercise: A rational map of irreducible varieties f : X 99K Y is the same as a morphism of schemes over k f : Spec k(X) → Y P ∈ X a closed point, then {rational maps X 99K Y defined at P } correspond to {morphismsSpec OX,P → Y over k}. Note: Spec(k(X)) → Spec(OX,P ) → X. Examples of Products X,Y varieties over k, the product X × Y from last semester corresponds to X ×k Y = X ×Spec(k) Y . −1 Fibers: ϕ : X → Y morphism of schemes, Q ∈ Y , then XQ = ϕ (Q) := X ×Y Spec k(Q) with the following diagram: ϕ ... XY...... XQ ...... Spec(k(Q)) Example: ϕ : A1 → A1, ϕ(t) = t2. For a ∈ A1 a closed point, ϕ−1(a) = 1 2 2 2 × 1 {a} = Spec(k[t] ⊗ 2 k[t ]/(t − a)) = Spec(k[t]/(t − a)), if a 6= 0, we A A √ √ k[t ] get ϕ−1(a) = { a, − a}, if a = 0, ϕ−1(0) = Spec k[t]/(t2). 2 1 Example: Pk ×k Ak = Proj k[x, y, z] × Spec k[t] = Proj A[x, y, z] where A = k[t]. E = V (zy2 − x(x − z)(x − tz)) ⊆ P2 × A1, ϕ : E → A1 projection, then for 1 2 2 λ ∈ A a closed point, Eλ = V (zy − x(x − z)(x − λz)) ⊆ Pk. Separated Morphisms

71 Let f : X → Y a morphism, ∆ : X → X ×Y X the diagonal morphism (the unique map into this product which, when composed with either projection, is the identity)

Deﬁnition 6.18 (Separated). f is separated if ∆ : X → X ×Y X is a closed immersion. X is separated if X → Spec(Z) is separated. Note: A a ring, then there exists a unique Z → A. 1 Example: X = (A \{0}) ∪ {01, 02} the doubled line to Spec(k) is not separated (See last semester). Proposition 6.8. Any morphism f : X → Y of aﬃne schemes is separated. ] Proof. X = Spec(A), Y = Spec(B), f : B → A. Then X×Y X = Spec(A⊗BA). ] ∆ : X → X ×Y X corresponds to ∆ : A ⊗B A → A, a1 ⊗ a2 7→ a1a2. ] ∆ is surjective, so A = A ⊗B A/I, so X = V (I) ⊂ X ×Y X.

Corollary 6.9. f : X → Y separated iﬀ ∆(X) ⊆ X ×Y X closed. Proof. Assume that ∆(X) is closed. It is a homeomorphism as ∆ : X → ∆(X) has continuous inverse ∆(X) → X × X → X by p1. ] We must now check that ∆ : OX×Y X → ∆∗OX is surjective.

If Q ∈ X ×Y X \ ∆(X), then OX×Y X,Q → (∆∗OX )Q = 0 is surjective. Let P ∈ X. Choose V ⊆ Y open aﬃne such that f(P ) ∈ V . Choose U ⊆ f −1(U) ] open aﬃne such that P ∈ U. Then ∆(P ) ∈ U ×V U ⊆ X ×Y X so ∆ is surjective in a nbhd of ∆(P ), because ∆ : U → U ×V U is separated.

Let (G, ≤) be a totally ordered abelian group. IE, g1 ≤ g2 implies g1 + g3 ≤ × g2 + g3. K a field, and K = K \{0}. Definition 6.19 (Valuation). A valuation of K with values in G is a map v : K× → G s.t. v(xy) = v(x) + v(y), v(x + y) ≥ min{v(x), v(y)}. e.g. v : k(t) → Z by v(tmf(t)) = m if f is defined at 0 and f(0) 6= 0 Note: {x ∈ K× : v(x) ≥ 0} ∪ {0} ⊆ K is a subring. Definition 6.20 (Valuation Ring). R is a valuation ring if R is a domain and there exist a valuation v : K(R)× → G for some G such that R = {x ∈ K(R)× : v(x) ≥ 0} ∪ {0}. R ⊆ K(R) gives us a morphism i : Spec K(R) → Spec R. Theorem 6.10. f : X → Y is a morphism, then f is separated iff the following condition holds: For any valuation ring R and morphisms α : Spec R → Y and β : Spec K(R) → X such that αi = fβ f ... XY...... β . α ...... i... Spec K(R)...... Spec(R)

72 Then there is at most one γ : Spec R → X making this all commute, that is, fγ = α and γi = β. Intuition: X,Y are prevarieties, C a curve and P ∈ C a nonsingular point. R = OC,P is a DVR. K = k(C). Then β : C 99K X is a rational map, and α : C 99K Y is also a rational map defined at P ∈ C. Then α = fβ is the commutativity condition on the square. If f is separated, then there is at most one possible value of β(P ). 1 e.g. If X is the doubled affine line, Y is Spec k then β : A 99K X defined on A1 \{0}. There are two ways to define β(0), so this is not separated. Corollary 6.11. 1. open and closed immersions are separated 2. compositions of separated morphisms are separated. 3. separated morphisms are stable under base extensions f ... XY...... f 0 . 0 0 ...... 0 X = X ×Y Y .... Y Definition 6.21 (Base Extension). Then f 0 is the base extension of f by Y 0. Proof. We will prove (b). f g Let X → Y → Z are separated morphisms. Let R be a valuation ring, take β : Spec K(R) → X and α : Spec(R) → Z. Assume that γ1, γ2 : Spec R → X are morphisms satisfying the diagram. Proof by diagram chasing. Exercise: If g is separated, then gf is separated iff f is separated. Let X be a scheme, x1 ∈ X.

Deﬁnition 6.22 (Specialization). x0 ∈ X is a specialization of x1 if x0 ∈ {x1}.

This gives a ring homomorphism OX,x0 → OX,x1 which is NOT local.

Note: Spec k(x0) → X has image {x0}, Spec k(x0) → Spec OX,x0 → X has image {x1 ∈ X : x0 is a specialization of x1}. Assume R is a local domain, K = K(R) i : Spec K → Spec R, t0 = mR ∈ Spec R and t1 = 0 ∈ Spec R. Let β : Spec K → X a morphism, x1 = β(0), ] β : k(x1) → K

Lemma 6.12. There is a correspondence {γ : Spec R → X|γi = β} to {x0 ∈ β] {x1}|image OX,x0 → OX,x1 → k(x1) → K is a subset of R} by γ 7→ γ(t0).

73 Proof. Assume γi = β. t0 ∈ {t1} implies that γ(t0) ∈ {x1}. mx0 7→ mR. We get the following commutative square: ] γt 0 ...... OX,x0 Ot0 = R ...... ] ..... γt 1 ...... OX,x1 Ot1 = K

Assume that x0 ∈ {x1}. Then deﬁne γ : Spec R → Spec OX,x0 → X. Recall: A domain R is a valuation ring of K if K = K(R) and ∃ valuation v : K× → G such that R = {a ∈ K|v(a) ≥ 0} ∪ {0}. Note: R is a local ring, mR = {a ∈ K|v(a) > 0}. Fact: Every local ring R0 ⊆ K is dominated by some valuation ring R of K 0 0 (that is, R ⊆ R with mR0 ⊆ mR). R ⊆ K a local subring is a valuation ring of K iﬀ R0 is not domainated by strictly larger R ⊆ K.

Theorem 6.13. X N¨otherian. f : X → Y a morphism, ∆ : X → X ×Y X = P . Then ∆(X) ⊆ P is closed iﬀ for all valuation rings R with morphisms α, β with αi = fβ there exists at most one γ sch that fγ = α and γi = β.

Proof. ⇒: Let γ1, γ2 : Spec R → X be given, with γji = β, fγj = α. Define ϕ : Spec R → P by πjϕ = γj. Then γji = β imply that ϕ(t1) = ϕ(i(0)) ∈ ∆(X) and ϕ(t0) ∈ ϕ(t1) ⊆ ∆(X) = ∆(X). Thus, π1(ϕ(t0)) = γ1(t0) = γ2(t0) = π2(ϕ(t0)), so γ1 = γ2 by the lemma. ⇐: Let z1 ∈ ∆(X) and z0 ∈ {z1} ⊆ P . Claim: z0 ∈ ∆(X). 0 Set K = k(z1), R = Im(OP,z0 → OP,z1 → K) ⊆ K. The fact implies that there is a valuation ring R of K such that OP,z0 → R ⊆ K is a local ring homomorphism. β : Spec K → P , πk : P → X. Exercise: z1 ∈ ∆(X) implies that π1β = π2β : Spec K → X. Now the lemma implies that there is ϕ : Spec R → P such that β = ϕi, ϕ(t0) = z0. Set γj = πjϕ : Spec R → X. Then fγ1 = fπ1ϕ = fπ2ϕ = fγ2. So the algebraic assumption says that γ1 = γ2, so ϕ = ∆π1ϕ, thus z0 = ϕ(t0) ∈ ∆(X), so the claim holds. So now X Nötherianimplies that X = X1 ∪ ... ∪ Xn where Xi is closed and irreducible. So Xi = {xi}, xi ∈ X, we set zi = ∆(xi) ∈ P . So X ⊆ {x1, . . . , xn}, thus ∆(X) ⊆ {z1, . . . , zn} = {z1} ∪ ... ∪ {zn} ⊆ ∆(X) by the claim, and so the theorem holds. Definition 6.23 (Properness). f : X → Y is proper if f is separated, of finite type, and f is universally closed, which means that any base extension of f is a closed map. (takes closed sets to closed sets.) Definition 6.24 (Redefinition of a Variety). Let k be a field. A prevariety over k is a reduced scheme of finite type over k. A variety is a separated pre-variety X, that is, the structure morphism X → Spec(k) is separated.

74 Definition 6.25 (Complete). A variety is complete if X → Spec(k) is proper. ie, X ×k Y → Y is closed for all Y . Theorem 6.14. X Nötherian, f : X → Y of finite type. Then f is proper iff ∀ valuation rings R with a commutative diagram f ... X ...... Y ...... β ...... α ...... ∃!γ ...... i ... Spec K(R) ...... Spec R there ∃!γ such that fγ = α, γi = β Corollary 6.15. X a complete variety, C a curve, P ∈ C a nonsingular point, then f : C \{P } → X is a morphism of varieties, then we can extend f to C → X.

This follows by setting Y = Spec(k), R = OC,P a DVR, and K(R) = K(C) the function field of C. Corollary 6.16. 1. Closed immersions are proper. 2. Compositions of proper morphisms are proper. 3. Base extensions of proper maps are proper 4. Properness is determined locally. That is, f : X → Y a morphism it is proper iff for all V ⊆ Y open subscheme, f : f −1(Y ) → Y is proper. Exercise: Assume that P is a property of morphisms such that closed em- beddings are P , compositions of P -morphisms are P and base extensions of P f g morphisms are P . Then products of P -morphisms are P , X → Y → Z, if gf is P and g is separated then f is P , and if f : X → Y is P then fred : Xred → Yred is P . Projective Morphisms n = Proj A[x , . . . , x ] = n × Spec A. PA 0 n PZ Spec Z Definition 6.26 (Projective Space over a Scheme). For any scheme Y , set n = n × Y PY PZ Definition 6.27 (Projective Morphism). f : X → Y is projective if there exists n a closed immersion i : X → PY s.t. f = πY ◦ i. f is quasi-projective if there is an open immersion j : X ⊂ X0 and a projective morphism f 0 : X0 → Y s.y. f = f 0 ◦ j. Example: A variety X is quasi-projective iff X → Spec k is quasiprojective, and likewise projective. Example: S is a graded ring, gneerated bny finitely many elements of S1 as an S0-algebra. Then Proj(S) → Spec(S0) is projective.

75 Theorem 6.17. A projective morphism of N¨otherianschemes is proper. Proof. Enough to show that n → Spec is proper. PZ Z Separated: Show that ∆ : n → n× n is a closed immbedding. D (x x ) → PZ P P + i j D+(xi)×D+(xj) gives Z[x0, . . . , xn](xixj ) ← Z[x0, . . . , xn](xi)⊗ZZ[x0, . . . , xn](xj ). Properness; R a valuation ring, K = K(R). We want γ such that πγ = α n and γi = β. Let t0 = mR, t1 = 0 ∈ Spec(R). Set p1 = β(t1) ∈ P . If p1 ∈ V (xi), n−1 n then induction imples that there is a γ : Spec R → V (xi) ' P ⊆ P . WLOG: p1 ∈ D+(x0x1 . . . xn) ⇒ xi/xj ∈ OP,p1 for all i, j. ] ] β : k(p1) → K, set fij = β (xi/xj) ∈ K. A valuation v : K → G, R = {v ≥ 0}. Choose m such that v(fm0) is minimal. v(fim) = v(fi0/fm0) = v(fi0) − v(fm0) ≥ 0, so fim ∈ R for all i. So we have a ring homomorphism Z[x0/xm, . . . , xn/xm] → R by xi/xm 7→ fim. γ n Thus, Spec R → D+(xm) ⊆ P is as desired. Exercise: 1) There exists a closed Segre Embedding n × m → nm+n+m. PZ PZ PZ 2) Composition of projective morphisms is projective. affine Example: X → Y → Spec k, X a prevariety, Y a variety, then X is a variety. OX modules X is a scheme, F , G are sheaves of OX -modules.

homOX (F , G ) = {homomorphisms of OX -modules F → G }.

We deﬁne a sheaf of OX -modules H omOX (F , G ) by U 7→ homOX |U (F |U , G |U ).

F ⊗OX G by the sheafification of U 7→ F (U) ⊗OX (U) G (U). Let f : X → Y be morphisms of schemes, H an OY -module. f∗F is an ] f∗OX -module, f : OY → f∗OY a ring homomorphism implies that f∗F is an OY -module. −1 −1 The adjoint property: homOY (H , f∗F ) ⇐⇒ homf OY (f H , F ). ∗ −1 −1 Definition 6.28 (Pullback). f H = f H ⊗f OY OX , this is an OX -module. Continued...see Sheaves from last semester

Proposition 6.18. M˜ P = MP , M˜ (D(a)) = Ma. Corollary 6.19. Γ(X, M˜ ) = M and M 7→ M˜ is an exact, fully faithful functor ˜ ˜ from A-mod to OX -mod with inverse the global section functor, ⊕iMi = ⊕iMi ˜ ˜ ˜ and M ⊗A N = M ⊗OX N We have done this several times before, and so will not prove them again. Proposition 6.20. f : Spec A = X → Spec B = Y a morphism, M an A- ∗ module and N a B-module. Then f∗M˜ = M˜ B, f N˜ = N ⊗˜B A where MB is M as a B-module using f ] : B → A. −1 ] Proof. Let b ∈ B, Γ(D(b), f∗M˜ ) = Γ(f (D(b)), M˜ ) = Γ(D(f (b)), M˜ ) = Mf ](b) = (MB)b. ] −1 ∗ ˜ ˜ Let P ∈ X. Q = (f ) (P ) = f(P ) ∈ Y .(f N)P = Nf(P ) ⊗OY,f(P ) OX,P =

NQ ⊗BQ AP = (N ⊗B A)P .

76 Deﬁnition 6.29 (Quasicoherent and Coherent Sheaves). Let X be a scheme. An OX -module F is quasi-coherent if there exists an open cover X = ∪Ui, ˜ Ui = Spec Ai and Ai modules Mi such that F |Ui ' Mi. F is coherent if the Mi are ginitely generated.

Example: i : Y → X a closed subscheme implies that i∗OY is a coherent OX - −1 ˜ ˜ module. U = Spec A ⊆ X open, i (U) = Spec(A/I), i∗OY = i∗(A/I) = A/I.

Lemma 6.21. X = Spec A, f ∈ A. F a quasi-coherent OX -module.

1. If s ∈ Γ(X, F ) and s|D(f) = 0 ∈ Γ(D(f), F ), then ∃n > 0 such that f ns = 0 ∈ Γ(X, F )

n 2. If t ∈ Γ(D(f), F ) then ∃s ∈ Γ(X, F ) and m > 0 so that s|D(f) = f t.

Proof. P ∈ X, there exists an open nbhd P ∈ U = Spec(B) ⊆ X and B-module ˜ M such that F |U ' M. So there exists g ∈ A such that P ∈ D(g) ⊆ U. ˜ ˜ F |D(g) = M|D(g) = M ⊗B Ag. Thus, we can write X = D(g1) ∪ ... ∪ D(gm) ˜ such that F |D(gi) = Mi, with Mi a Agi -module.

Corollary 6.22. Γ(D(f), F ) = Γ(X, F )f .

Proposition 6.23. X a scheme, F an OX -module.

1. F is quasi-coherent iff for all open Spec A = U ⊆ X, there exists an ˜ A-module M such that F |U = M 2. X Nötherian: F coherent iff the same is true but M finitely generated.

Proof. 1. Assume that F is quasicoherent, U = Spec A ⊆ X is open. Then F |U is a quasicoherent OX -module. Write U = D(gi) ∪ ... ∪ D(gr), ˜ gi ∈ A. F |D(gi) ' Mi. Set M = F (U). M 7→ Γ(U, F ) gives an OU - ˜ homomorphism α : M → F . The lemma implies that Mi = F (D(gi)) =

Mgi . Therefore, α is an isomorphism over D(gi) implies that it is one over U.

77 2. Assume X N¨otherian, F is coherent. Then A is N¨otherian,so Mi = Mgi

is a ﬁnitely generated module over Agi for all i. We must show that M

is a ﬁnitely generated A-module: take m1, . . . , mN ∈ M such that Mgi is gneerated by {mj/1} for all i. Then M is generated by {m1, . . . , mN } as an A-module.

Corollary 6.24. X = Spec A. Then there is an equivalence M 7→ M˜ and F → Γ(X, F ) between A-modules and quasi-coherent OX -modules. 0 00 0 00 Recall: 0 → F → F → F → 0 is exact iﬀ 0 → Fp → Fp → Fp → 0 is exact for all p ∈ X. We only get automatically that 0 → Γ(U, F 0) → Γ(U, F ) → Γ(U, F 00) is exact. Proposition 6.25. X is aﬃne scheme, 0 → F 0 → F → F 00 → 0 exact sequence, if F 0 is quasi-coherent, then Γ(U, F ) → Γ(U, F 00) is surjective, so Γ(U, −) becomes an exact functor.

Proposition 6.26. Any kernel, cokernel or image of an OX -homomorphism of quasicoherent OX -modules is quasicoherent. An extension of quasi-coherent OX -modules is quasi-coherent. If X is Nötherian,then the same holds for coherent modules. Proof. WLOG, X = Spec A. ϕ : M˜ → N˜. This gives 0 → K → M → N → C → 0 an exact sequence, as −˜ is exact, 0 → K˜ → M˜ → N˜ → C˜ → 0 is exact. So the kernel and cokernel must be quasicoherent. The image is coker(K˜ → M˜ ), and so is also quasicoherent. We now assume that 0 → M˜ → F → N˜ → 0 is a short exact sequence ˜ ˜ with M, N quasicoherent and F an OX -module. Define F = Γ(X, F ). The proposition states that 0 → M → F → N → 0 is a short exact sequence of A-modules. Then we obtain ...... 0 ...... M˜ ...... F˜ ...... N˜ ...... 0 ...... ' ...... ' ...... 0 ...... M˜ ...... F ...... N˜ ...... 0 So by the five lemma, F˜ ' F . Definition 6.30 (Ideal Sheaf). Let X be a scheme, i : Y → X a closed immer- i] sion. Then the ideal sheaf of Y is IY = ker(OX → i∗OY ) ⊆ OX . Proposition 6.27. {closed subschemes of X} correspond to {quasicoherent sheaves of ideals ⊆ OX }.

Proof. Assume that i : Y → X is a closed immersion, then we have 0IY → i] OX → i∗OY → 0. i∗OY is quasicoherent (we will prove this next, it is true because i is separated and quasicompact). Then IY is quasicoherent as well. Assume that I ⊆ OX is quasicoherent. For p ∈ X, Ip ⊆ OX,p. Deﬁne Y = {P ∈ X : IP ⊆ mP ( OX,P } = Supp(OX /I ) = {P ∈ X|(OX /I )P 6= 0}. Let i : Y → X be the inclusion.

78 −1 Define OY = i (OX /I ). Then (Y, OY ) is a locally ringed space. OX,P = −1 id (OX /I )P = OX,P /IP . Then i (OX /I ) → OY , which corresponds to a map ϕ OX /I → i∗OY by the adjoint property. ] ϕ ] Then we define i : OX → OX /I → i∗OY , so (i, i ): Y → X is a morphism of locally ringed spaces. Claim: This is a closed immersion. ˜ WLOG X = Spec A. Then I = I ⊆ OX , with I ⊆ A an ideal. Then Y = V (I) ⊆ X is closed, and i : Y → X is a homeomorphism onto its image. We must check that Y = Spec A/I. OY,P = (OX /I )P = AP /IP = (A/I)P = OSpec A/I,P . Corollary 6.28. All closed subschemes of Spec A are in correspondence with ideals of A. All closed subschemes of an affine scheme are affine.

Corollary 6.29. If U1,U2 ⊆ X open aﬃne subschemes, and if X is separated, then U1 ∩ U2 is aﬃne.

∆ ... X ...... X × X ...... ⊆ ...... U1 ∩ U2 ...... U1 × U2 Proof. As ∆ is a closed embedding, the bottom map must be as well. The product of affines over an affine is affine, and so then U1 ∩ U2 is affine, as it is a subscheme of U1 × U2. Proposition 6.30. Let f : X → Y be a morphism.

∗ 1. G a quasi-coherent OY -module, then f G is a quasi-coherent OX -module. 2. X,Y both Nötherian,same for coherent. 3. Assume X is Nötherianor that f is separated and quasicompact, then if F is quasicoherent OX -module, then f∗F is a quasicoherent OY -module Proof. 1. Let P ∈ X, Let Spec B = V ⊆ Y open affine such that f(P ) ∈ V . −1 ˜ Take U = Spec A ⊆ X open such that P ∈ U ⊆ f (V ). G |V = M with ∗ ¯∗ ¯∗ ˜ ˜ M a B-module. Then (f G )|U = f (G |V ) = f (M) = M ⊗B A. 2. Similar, but noting finite generation everywhere. 3. WLOG, Y is affine. Last time, we showed that (c) is true when X is affine. Assumptions imply that there exists a finite open affine cover X = ∪Ui such that Ui ∩ Uj has a finite open affine cover Ui ∩ Uj = ∪Uijk finite. As F is a sheaf, we obtain an exact sequence of OY -modules

0 → f∗F → ⊕if∗F |Ui → ⊕i,j,kf∗F |Uijk . So f∗F is the kernel of a map of quasicoherent OY -modules.

79 Previously, we’ve shown that the following properties are all equivalent

1. Closed subschemes of aﬃne schemes are aﬃne

2. If X is separated and U1,U2 ⊆ X open aﬃne, then U1 ∩ U2 aﬃne. 3. Pushforward of a quasicoherent sheaf is quasicoherent

4. The ideal sheaf of a closed subscheme is quasi-coherent.

We now prove number 1:

Proof. If i : Y → X is a homeomorphism onto i(Y ), and i(Y ) is closed, then (i∗OY )P is OY,P if P ∈ Y and 0 else. For f ∈ Γ(X, OX ) write Xf = D(f) = {P ∈ X : f∈ / mP ⊆ OX,P }. Assume that i : Y → X = Spec A is a closed subscheme. Let P ∈ Y , −1 then there exists open affine V ⊆ Y . Note: {Yf = i (Xf )} is a basis for the topology on Y . Thus, there exists f ∈ A such that P ∈ Yf ⊆ V , Yf = Vf is affine. So we can write X = Xf1 ∪ ... ∪ Xfr such that Yfi affine for all i.(f1, . . . , fr) = (1) ⊆ A, ] ] ] thus (i (f1), i (f2), . . . , i (fr)) = (1) ⊆ Γ(Y, OY ). By exercise 2.17b in Hartshorne, Y is affine iff ∃f1, . . . , fr ∈ Γ(Y, OY ) such that Yfi is affine for all i and (f1, . . . , fr) = 1 ⊆ Γ(Y, OY ).

S = ⊕d≥0Sd a graded ring, X = Proj(S), and M = ⊕d∈ZMd a graded S- module. Sd · Me ⊆ Me+d for P ∈ X deﬁn an S(P )-module M(P ) = {m/f : m ∈ M, f ∈ S homogeneous of the same degree}. ˜ ˜ ` For U ⊆ X open, deﬁne OX -module M by Γ(U, M) = {s : U → P ∈U M(P )|s(P ) ∈ M(P )∀P ∈ U and s is locally a quotient}. ˜ Proposition 6.31. 1. MP = M(P ) ˜ ˜ 2. M|D+(f) = M(f) for f ∈ S+, deg(f) > 0.

Corollary 6.32. M˜ is quasicoherent. If S is N¨otherianand M ﬁnitely generated S-module, then M˜ is coherent.

Definition 6.31 (Twisted Modules). Define M(n) = M as an S-module, but with grading M(n)e = Me+n. ˜ If X = Proj(S), define OX (n) = S(n). For any OX -module, F , define

F (n) = F ⊗OX OX (n). Example: ˜ 1. OX (0) = S = OX .

m 2. S = A[x0, . . . , xm], X = Proj(S) = PA . Then Γ(D+(xi), OX (n)) = r S(n)xi = {f/xi |f ∈ Sr+n} = {elements of degree n} ⊆ Sxi

80 Claim: Γ(X, OX (n)) = Sn. We have a map Sn = S(n)0, Let t ∈ Γ(X, OX (n)), set ti = t|D+(xi) ∈ Sxi element of degree n, if ti = tj on D+(xixj), then ti = tj ∈ Sxixj , not that Sxi ⊆ Sxixj , Sxi ∩ Sxj = S ⊆ Sxixj , thus ti = tj ∈ Sn for all i, j.

Proposition 6.33. Let X = Proj(S), S generated by S1 as an S0-algebra.

1. OX (n) is an invertible OX -module. ˜ ˜ ˜ 2. M ⊗OX N = M ⊗S N.

·f −n Proof. 1. Let f ∈ S1. Γ(D+(f), OX (U)) = S(n)(f) → S(f), which says that ˜ ˜ OX (n)|D+(f) = S(n)(f) ' S(f) = OD+(f).

2. M ⊗S N is a graded S-module, by (M ⊗S N)e =submodule generated by m ⊗ n where m ∈ Mr, n ∈ Nt with r + t = e. Let P ∈ X. Then we have

a map M(P ) ⊗S(P ) N(P ) → (M ⊗S N)(P ) by m/f ⊗ n/g 7→ m ⊗ n/fg. S is generated by S1, so this is an isomorphism. Why? Because P 6⊇ S+ ⇒ P 6⊇ S1 ⇒ ∃h ∈ S1 such that h∈ / P . Let m ⊗ n/f ∈ (M ⊗ N)(P ), go to r r m/h ⊗ h n/f ∈ M(P ) ⊗ N(P ) where m ∈ Mr, n ∈ Nt and s ∈ Sr+t. ˜ ˜ ˜ ˜ ˜ Construct an isomorphism ϕ : M ⊗OX N → M ⊗ N by s ∈ Γ(U, M ⊗N) 7→ ˜ ˜ [P 7→ sp ∈ MP ⊗OX,P NP = (M ⊗ N)(P )], which is an isomorphism.

˜ ˜ ˜ ˜ ˜ Corollary 6.34. M(n) = M ⊗OX OX (n) = M ⊗ S(n) = (M ⊗S S(n))˜= M(n).

This says that OX (m) ⊗ OX (n) = OX (m + n).

Deﬁnition 6.32. Let X = Proj(S), F an OX -module. Deﬁne Γ∗(F ) =

⊕n∈ZΓ(X, F (n)). This is a graded S-module.

Let s ∈ Sd, t ∈ Γ(X, F (n)) gives s ∈ Γ(X, OX (d)), so t ⊗ s ∈ Γ(X, F (n)) ⊗ OX (d) = Γ(X, F (n + d)) Note: X = Proj S where S = A[x0, . . . , xn]. Γ∗(OX ) = S. This is NOT always true! (Side Note: Look at the operation Γ∗(Proj S). Is it a functor? What properties does it have? Filling in lower degrees?) Let X be a scheme, I an invertible LX -module, and f ∈ Γ(X, L ).

Deﬁnition 6.33. Xf = {P ∈ X|fP ∈/ mP LX }. Xf ⊆ X is open.

Lemma 6.35. Let X be a quasi-compact scheme, L an invertible OX -module, f ∈ Γ(X, L ) and F a quasi-coherent OX -module. n ⊗n 1. Let s ∈ Γ(X, F ), s|Xf = 0, then s ⊗ f = 0 ∈ Γ(X, F ⊗ L ).

81 X a schemem L an invertible OX -module.

−1 Deﬁnition 6.34. L = H om(L , OX ).

−1 If L |U ' OU then L |U ' H om(OU , OU ) ' OU . −1 Note: Global OX -hom: L ⊗L → OX by s⊗α 7→ α(s) is an isomorphism on U. −1 Let f ∈ Γ(X, L ), Xf = {P ∈ X : fP ∈/ mP LP } ⊆ X open. L |Xf →

OXf , α 7→ α(f). −1 −1 Deﬁne f ∈ Γ(Xf , L ) to be the inverse image of 1 by this isomorphism. −1 −1 So f ⊗ f = 1, adn L ⊗ L ' OX .

Lemma 6.36. Let X be a quasi-compact scheme, L an invertible OX -module, f ∈ Γ(X, L ) and F a quasicoherent OX -module.

n ⊗n 1. s ∈ Γ(X, F ), s|Xf = 0 ⇒ s ⊗ f = 0 ∈ Γ(X, F ⊗ L )

2. Assume X = U1 ∪ ... ∪ Ur with Ui open aﬃne and L |Ui ' OUi for all i, ⊗n and Ui ∩ Uj quasicompact. t ∈ Γ(Xf , F ) ⇒ ∃s ∈ Γ(X, F ⊗ L ) such n that s|Xf = t ⊗ f .

⊗d Corollary 6.37. S = ⊕d≥0Γ(X, L ) is a graded ring, f ∈ S1. Γ∗F = ⊗e ⊕e∈ZΓ(X, F ⊗ L ) is a graded Smodule. Then (Γ∗F )(f) = Γ(Xf , F ) d ⊗d Proof. (Γ∗F )(f) = {s/f |s ∈ Γ(X, F ⊗ L )} −d ⊗−d d −d f ∈ Γ(Xf , L ). (Γ∗F )(f) → Γ(Xf , F ) by s/f 7→ s ⊗ f . This is injective and surfective.

Let S = ⊕d≥0Sd be a graded ring, X = Proj S. ˜ Recall, OX (n) = S(n). f ∈ Sd implies f ∈ S(d)0 ⇒ f ∈ Γ(X, OX (d)). F ˜ an OX -module, Γ∗F = ⊕e∈ Γ(X, F (e)). Construct an OX -hom β : Γ∗F → F ˜ Z by β|Xf : Γ∗F(f) → F |Xf , which corresponds to a module homomorphism m −m (Γ∗F )(f) → Γ(Xf , F ) by s/f 7→ s ⊗ f .

Proposition 6.38. If F is quasi-coherent, and S is ﬁnitely generated by S1 as ˜ an S0-algebra, then β : Γ∗F → F is an isomorphism.

Proof. Assume S0[f1, . . . , fr] → S is surjective, fi ∈ S. Ui = Xfi aﬃne. X = ˜ ˜ U1 ∪ ... ∪ Ur. O(1)|Ui = S(1)(xi) = S(xi) = OUi , Ui ∩ Uj = Xfifj aﬃne, so the lemma implies that β| is an isomorphism for all i. Xfi Let I ⊆ S be a homogeneous ideal. Natural inclusion i : Y = Proj(S/I) → ˜ X = Proj(S) i(Y ) = V (I), i∗OY = S/I (exercise) ˜ ˜ So OX → S → S/I = i∗OY , so Y ⊆ X is a closed subscheme. Note that ˜ IY = I ⊆ OX .

r r Corollary 6.39. 1. Assume Y ⊆ PA a closed subscheme with PA = Proj S, S = A[x0, . . . , xr]. Then there exists a homogeneous ideal I ⊆ S such that Y = Proj(S/I).

82 2. A morphism ϕ : Y → Spec A is projective iﬀ Y = Proj(S),S0 = A and S is ﬁnitely generated by S1.

Proof. 1. ⊆ r a quasicoherent subsheaf, ⊗ (d) ⊆ (d) implies IY OPA IY O O that I = Γ ⊆ Γ r = S a homogeneous ideal. ∗IY ∗OPA r Proj(S/I) ⊆ has ideal sheaf I˜ = Γ ˜ = ⊆ r . Thus, Y = PA ∗IY IY OPA Proj(S/I).

r 2. ϕ is projective iﬀ it factors through PA as a closed immersion for some r iﬀ Y = Proj S, with S = A[x0, . . . , xr]/I.

Deﬁnition 6.35 (Twisting Sheaf). Let Y be any scheme. The twisting sheaf r r π r ∗ of = × Y → is (1) = π r (1). PY PZ PZ O OP Deﬁnition 6.36 (Immersion). A morphism i : X → Z is an immersion if we can factor it as i : X → Z1 → Z with X → Z1 an open immersion and Z1 → Z a closed immersion. Exercise: A composition of immersions is an immersion.

Deﬁnition 6.37 (Very Ample). Let X be a scheme over Y and L an invertible r OX -module. L is very ample relative to Y if ∃ an immersion i : X → PY such that L = i∗O(1). Note: If ϕ : X → Y is projective, then ϕ is proper, and X has a very ample invertible sheaf relative to Y . Suppose that ϕ : X → Y is proper and X has a very ample invertible L r relative to Y . We have an immersion i : X → PY , ϕ proper implies that i is r proper, so i(X) ⊆ PY is closed, and so i is a closed immersion. Deﬁnition 6.38 (Generated by global sections). Let X be a scheme, F an α OX -module, F is generated by globa section if ∃ ⊕i∈I OX → F a surjective OX -hom. IE, ∃si ∈ Γ(X, F ) such that FP is generayed by {(si)P } as an OX,P -module for all P ∈ X. Examples:

1. X = Spec A, F = M˜ .

2. X = Proj S, S generated by S1 as an S0-module. Then OX (1) is generated by global sections.

Definition 6.39 (Ample Sheaf). An invertible OX -module L on a Nötherian ⊗n scheme X is ample if ∀ coherent OX -modules F , ∃n0 > 0 such that F ⊗ L generatd by global sections for all n ≥ n0. Later: If X is of finite type over Spec A, A Nötherian,then L is ample iff L ⊗m is very ample relative to A for some m > 0.

83 Theorem 6.40. Let X be projective over Spec A, A N¨otherian.If O(1) is very ample relative to A, then O(1) is ample. r ∗ r Proof. i : X → PA is a closed immersion, O(1) = i OP (1). F a coherent r OX -module. Then i∗F is coherent on PA. For P ∈ X,(i∗F )P = FP . ⊗m r Exercise: Γ(X, F ⊗ O(1) ) = Γ(PA, (i∗F )(m)). r WLOG: X = PA = Proj A[x0, . . . , xr]. ˜ F coherent implies that F |D+(xi) ' Mi, with Mi a ﬁnitely generated Bi module, where Bi = A[x0/xi, . . . , xr/xi]. Let si1, . . . , siN generate Mi as a Bi- module. Then the lemma implies that there exists tij ∈ Γ(X, F ⊗ OX (n)) such n that tij|D+(xi) = sij ⊗ xi . Take n large enough for all i, j. Claim: F (n) = F ⊗ OX (n) is generated by {tij} ⊂ Γ(X, F (n)). ' n Why? Because F |D+(xi) → F (n)|D+(xi) by s 7→ s ⊗ xi .

Corollary 6.41. X projective over Nötherian A, F a coherent OX -module. Then there exists ⊕OX (ni) → F surjective with a finite sum. Proof. The theorem says that F ⊗O(n) = F (n) is generated by global sections. ⊕N Thus, OX → F (U) → 0 exact, tensor with O(−n), and we get the result. Definition 6.40 (*-scheme). X is a *-scheme if it is Nötherian,separated, integral and OX,P is a DVR whenever dim OX,P = 1. Examples: nonsing alg variety X normal, Nötherian,separated and integral. Definition 6.41 (Prime Divisor). Assume X is a *-scheme, a prime divisor is a closed integral subscheme of codimension 1.

∗ Note that OX,Y is a DVR, we have a valuation vY : k(X) → Z by OX,Y = {f ∈ k(X)|vY (f) ≥ 0}. We call vY (f) the order of vanishing of f along Y . v(fg) = vY (f) + vY (g) and vY (f + g) ≥ min(vY (f), vY (g)). We set Div(X) to be the free abelian group on the prime divisors. A principal P ∗ 0 divisor is one of the form (f) = Y vY (f)[Y ] with f ∈ k(X) . For D,D ∈ Div(X) write D ∼ D0 ⇐⇒ D − D0 is principle. Define the divisor class group C`(X) = Div(X)/{principal divisors}. n Example: X = Pk , a prime divisor is Y = V (f) and deg(Y ) = deg(f). n P P There is an ismorphism C`(P ) → Z by ni[Yi] 7→ ni deg(Yi). 1. U ⊆ X open implies there exists a surjective group hom C`(X) → C`(U) by [Y ] 7→ [Y ∩ U] if Y ∩ U 6= ∅ and 0 otherwise. So if Z = X \ U and codim(Z; X) ≥ 2, then C`(X) ' C`(U). If Z ⊆ X is a prime divisor, then have Z → C`(X) → C`(U) → 0. 2. A is a Nötheriandomain. Then A is a UFD iff U = Spec A is normal and C`(U) = 0.

3. π : X × Am → X gives an isomorphism π∗ :C`(X) → C`(X × Am) by [Y ] 7→ [Y × Am].

84 Example: X = Pn × Pm, p : X → Pn and q : X → Pm, then p∗ :C`(Pn) → C`(X) is injective, by [Y ] 7→ [Y × Pm]. Let H ⊆ Pm be a hyperplane, U = X \ Pn × H = Pn × Am. Then we have an isomorphism C`(Pn) → C`(X) → C`(Pn × An). Similarly q∗ :C`(Pm) → C`(X) is injective. In fact, the image of q∗ is Z[Pn × H]. So we get 0 → C`(Pm) = Z → C`(X) → C`(Pn × Am) = Z → 0, but the last map has a section, so we get C`(X) = Z[Pn × H] ⊕ Z[H0 × Pm]. Carier Divisors Let X be any scheme. For U ⊆ X open, let S(U) ⊆ OX (U) be the set of non-zerodivisors. We define K to be the sheafification of pre − K , which has −1 pre − K (U) = S(U) OX (U). This is a sheaf of rings, OX -module, OX → K . Example: if X is an integral scheme, S(U) = OX (U)\{0}. Then for U affine, Γ(U, pre − K ) = K(OX (U)) = k(U) and Γ(U, K ) = k(X) for any nonempty U ⊆ X open. Note that OX ⊆ pre − K is a sub-presheaf, so pre − K is a decent presheaf. Recall that a decent presheaf satisfies the first sheaf axiom, and OX ⊂ pre−K ⊂ K . Define K ∗(U) to be the set of invertible elements of K (U). This is a sheaf ∗ ∗ of abelian groups. So OX ⊆ K . Definition 6.42 (Cartier Divisor). A Cartier Divisor is an element D ∈ Γ(X, K ∗/O∗). It is principleif it is in the image of Γ(X, K ∗) → Γ(X, K ∗/O∗). NOTE: This is not surjective! There is a sheafification involved. Convention: We use additive notation. Let D be a Cartier Divisor. Then there exists an open cover X = ∪Ui ∗ and fi ∈ Γ(Ui, K ) such that D|Ui is the image of fi and on Ui ∩ Uj, fi/fj ∈ ∗ Γ(Ui ∩ Uj, OX ). Whenever {fi} satisfy this condition, they define a Cartier divisor D. Definition 6.43. We define CaC`(X) = {cartier divisors}/{principal cartier divisors}. Write D ∼ D0 iff D − D0 = 0 ∈ CaC`(X). Assume that X is a *-scheme. Then D is a Cartier, Y ⊆ X a prime divisor, we write vY (D) = vY (fi) where Y ∩ Ui 6= ∅ and D|Ui =image of di. This gives ∗ ∗ P us a group homomorphism Γ(X, K /OX ) → Div(X) by D 7→ Y vY (D)[Y ]. This induces a map CaC`(X) → C`(X), and is injective when X is normal.

Deﬁnition 6.44 (Locally Factorial). X is locally factorial if OX,P is a UFD for all P ∈ X. Proposition 6.42. X is locally factorial *-scheme, then CaC`(X) ' C`(X). Deﬁnition 6.45 (Picard Group). The set of all isomorphism classes of invertible OX -modules under tensor product. ∗ ∗ ∗ Let D ∈ Γ(X, K /OX ) cartier. Then D|Ui is the image of fi ∈ Γ(U, K ).

Deﬁnition 6.46. L (D) ⊆ K is the sub OX -module such that L (D)|Ui is −1 generated by fi .

85 L (D) is an invertible OX -mod as OUi → L (D)|Ui given by multiplication −1 by fi . ∗ ∗ Proposition 6.43. 1. Γ(X, K /OX ) ↔ {invertible subsheafs of L }. −1 2. L (D1) ⊗ L (D2) ' L (D1 − D2).

3. D1 ∼ D2 iﬀ L (D1) ' L (D2).

Proof. 1. Let L ⊆ K be any invertible OX -module. There exists an open

cover X = ∪Ui such that L |Ui ' OUi , so Γ(Ui, L ) ' Γ(Ui, OUi ). De-

ﬁne fi to be the section corresponding to 1 ∈ Γ(Ui, OUi ) by the chosen ∗ −1 isomorphism. fi/fj ∈ Γ(Ui ∩ Uj, OX ) ⇒ {fi } deﬁne a Cartier divisor D. −1 2. L (D1) ⊗ L (D2) → L (D1 − D2) by h1 ⊗ h2 7→ h1h2.

3. Assume L (D1) ' L (D2) as OX -modules. By the last part, K ⊂ L (D1− D2) ' OX . Let 1 ∈ Γ(X, OX ) which corresponds to f ∈ Γ(X, K ). f ∈ ∗ −1 Γ(X, K ), so D1 − D2 =image of f .

Corollary 6.44. If X is any scheme, then the map CaC`(X) → Pic(X) is injective. Proposition 6.45. If X is integral, then CaC`(X) ' Pic(X).

Proof. Must show that any invertible L is a submodule of K . Let L |U ' OU . Then 1 ∈ Γ(U, OU ) corresponds to f ∈ Γ(U, L ). Assume that ∅= 6 V ⊆ X open, h ∈ L (V ), then U ∩ V 6= ∅. h ⊗ f −1 ∈ Γ(U ∩ V, L ⊗ −1 L ) = Γ(U ∩ V, OX ) ⊆ Γ(U ∩ V, K ) = Γ(V, K ). −1 Deﬁne an OX -hom L → K by h 7→ h ⊗ f . This is injective, as fp generator for Lp for all p ∈ U. Corollary 6.46. X N¨otherian,Integral, Separated, Locally Factorial implies C`(X) = CaC`(X) = Pic(X)

n Corollary 6.47. Pic(Pk ) = {O(m)|m ∈ Z}. n Proof. X = Pk . We know that C`(X) = Z[H] ' Pic(X). Must show that [H] ↔ O(1). H = V (h) where h ∈ k[x0, . . . , xn] is a linear form. Set fi = h/xi ∈ OX (D+(xi)). fi/fj = xj/xi is a unit on D+(xixj), so {fi} deﬁnes a Cartier Divisor. D ∈ CaCl(X) thar corresponds to [H] since vY (D) = 0 if Y 6= H and 1 if Y = H. So we get OX (1) → L (D) an isomorphism, by s 7→ s/h.

Definition 6.47 (Effective Divisor). A Cartier Divisor D is effective if ∃ an open cover X = ∪Ui and nonzerodivisors fi ∈ OX (Ui) such that D|Ui is the image of fi.

86 D is effective gives us a closed subscheme of X by ID generated by fi on U. This gives us a correspondence between effective Cartier Divisors and closed subschemes that are locally generated by one nonzerodivisor. Note: If X is a *-scheme and D an effectve Cartier Divisor, then vY (D) = P vY (fi) ≥ 0 so Y vY (D)[Y ] is effective. If X is a locally factorial *-scheme, then effective Cartier divisors correspond to effective Weil divisors. Commutative Algebra Fact: If A is a normal NötherianDomain, then A =

∩ht(P )=1AP ⊆ K(A). So fi ∈ k(Ui) gives fi ∈ OUi (Ui) ⇐⇒ (fi) ∈ Div(Ui) eﬀective.

Proposition 6.48. D ⊆ X an eﬀective Cartier divisor implies that ID = L (−D) ⊆ K .

This is because OD ⊆ OX ⊆ K is locally generated by fi. −1 Let ϕ : X → Y , f ∈ OY (Y ) then ϕ (Yf ) = Xϕ]f . In fact, we can do this for any line bundle L . Let L be an OY -module. Then ∗ −1 ∗ −1 ∗ −1 ϕ = ϕ L ⊗ϕ OY OX . For s ∈ Γ(Y, L ) deﬁne ϕ s = ϕ s ⊗ 1 ∈ Γ(X, ϕ L ). ∗ ∗ −1 Xϕ∗s = {P ∈ X|(ϕ s)P ∈/ mP (ϕ L )P } = ϕ (Ys). Morphisms to Pn. n If ϕ : X → PA is a morphism, then 1. X is a scheme over A

∗ 2. L = ϕ O(1) is an invertible OX -module.

∗ 3. Generated by si = ϕ xi ∈ Γ(X, L ) for 0 ≤ i ≤ n where O(1) is generated n by x0, . . . , xn ∈ Γ(P , O(1)).

n Claim: 1+2+3 determines a unique ϕ : X → PA. −1 −1 −1 −1 Note: si ∈ Γ(Xsi , L ). sj/si = sj ⊗ si ∈ Γ(Xsi , L ⊗ L ) =

Γ(Xsi , OX ). −1 We must have Xsi = ϕ (D+(xi)). This defines ϕ : Xsi → D+(xi) by A- algebra homomorphism A[x0/xi, . . . , xn/xi] → Γ(Xsi , OX ) by xj/xi 7→ sj/si. Theorem 6.49. Let X be a scheme over A. Then there is a correspondence n {ϕ : X → PA over A} to {(L , s0, . . . , sn)|L generated by s0, . . . , sn}/ '. n n n−1 n n Example: if k is a field, then Pk = Ak ∪ Pk , A = D+(x0) and P ∈ P . Then P = 0 ∈ An. So π : An \{0} → Pn−1, id : Pn−1 → Pn−1 defines the projection from a point Pn \{P } → Pn−1. n n Let x1, . . . , xn ∈ Γ(P , O(1)). These generate O(1)|P \{P }. This defines the projection from P . n ∗ ∗ Note: Assume ϕ : X → PA is given by si = ϕ xi ∈ L = ϕ O(1). ϕ is a closed immersion iff ϕ : Xsi → D+(xi) is a closed immersion for all i iff Xsi affine and A[x0/xi, . . . , xn/xi] → OX (Xsi ) is surjective. Recall the definition of an ample line bundle. Example: If X is projective over a Nötherianring, L = O(1) is very ample. If X is affine, and L is any invertible sheaf.

87 Proposition 6.50. X a N¨otherianscheme, L an invertible OX -module. Then TFAE 1. L ample 2. L ⊗m ample for all m ≥ 1 3. L ⊗m ample for some m ≥ 1. Proof. 1 ⇒ 2 ⇒ 3: trivial ⊗m ⊗k Assume that L is ample, F a coherent OX -module. F ⊗ L is coher- ⊗k ⊗m ⊗n ent. For 0 ≤ k ≤ m − 1 choose nk > 0 such that (F ⊗ L ) ⊗ (L ) is generatedby global sections for all n ≥ nk. ⊗n N = max{k + nmk} implies that F ⊗ L is generated by global sections for all n ≥ N.

Theorem 6.51. X is of finite type over Nötherian A. Then L is ample iff L ⊗m isvery amples relative to Spec A for some m > 0. n n Pk : O(m) is ample iff m ≥ 1. If m < 0 then Γ(P , O(m)) = 0. O(0) gen by global sections. Remark: ϕ : X → Y a morphism, (X NötherianOR ϕ separated and ] quasi-compact) Let Z = ϕ(X) ⊆ Y . IZ = ker(ϕ : OY → ϕ∗OX ). IZ is quasicoherent implies that (Z, OX /IZ ) ⊆ Y is a closed subscheme called the scheme-theoretic image of X. Exercise: X is reduced implies that Z = ϕ(X) is reduced. Example: Y = Spec k[x, y]/(xy, y2), X = D(x) = Spec k[x, x−1] ⊆ Y . j : 1 X ⊆ Y . So j(X) = Yred = A . f g Application: X → Y → Z an immersion. Then X ⊆ gf(X) → Z is an immersion. Exercise: Check This! Hint: Assume f is closed and g is open.

Lemma 6.52. X N¨otherianScheme, U ⊆ X open, F a coherent OU -module. 0 0 Then there exists a coherent OX -module F such that F |U ' F .

Note: i : U → X, so i∗F is quasicoherent has (i∗F )|U ' F . Proof. If not, let U ⊆ X be a maximal open set such that the lemma is false, and WLOG, F is a counterexample. Let Y ⊆ X be open affine, Y 6⊆ U. Then j : U ∩ Y ⊆ Y . G = j∗(F |U∩V ) is a quasicoherent OY -module, and G |U∩Y = F |U∩Y . Let M = Γ(Y, G ) = Γ(U ∩ Y, F ) be a module over A = Γ(Y, OX ). = M˜ . Now, U ∩Y = Y ∪...∪Y , f ∈ A. Then M˜ = | = | is G f1 fn i fi G Yfi F Yfi coherent. Thus Mfi is a finitely generated Afi -module. Choose m1, . . . , mN ∈ 0 0 ˜ 0 M generated Mfi for all i. Set M = (m1, . . . , mN ) ⊆ M. G = M is a coherent 0 0 OY -module. G ⊆ G and G |U∩Y = G |U∩Y = F |U∩Y . 0 0 0 Set X = U ∪ Y . Define an OX0 -module F by F (V ) = {(a, b)|a ∈ F (U ∩ 0 V ), b ∈ G (Y ∩ V ) with a|U∩Y ∩V = b|U∩Y ∩V ∈ F (U ∩ Y ∩ V )}. 0 0 0 0 F |U = F , and F |Y = G , so F is a coherent OX0 -module. The contradicts that F is a maximal counterexample.

88 Theorem 6.53. X is a scheme of finite type over a Nötherianring A. L ⊗m an invertible OX -module. Then L is ample iff L is very ample relative to Spec A for some m > 0.

Proof. Assume that L ⊗m is very ample over A. Then there exists an immersion ¯ n X ⊆ X → PA where the first is open and the second is closed, such that ⊗m L ' OX (1). Let F be coherent OX -module. We know that OX¯ (1) is ample, so the ¯ ¯ lemma implies that there exists a coherent OX¯ -module F such that F |X = F . ¯ F ⊗ OX¯ (N) generated by global sections for all N >> 0, so F ⊗ OX (N) is ⊗m generated by global sections for N >> 0. Thus OX (1) = L is ample, so L is ample. Now we assume that L is ample. Let P ∈ X. There exists an open affine neighborhood P ∈ U ⊆ X such that L |U ' OU . Y = (X \ U)red → X is a ⊗n closed subscheme, IY ⊆ OX is coherent implies htat IY ⊗ L is gneerated ⊗n by global sections for some n. So there exists s ∈ Γ(X, IY ⊗ L ) such that ⊗n sP ∈/ mP (IY ⊗ L )P . ⊗n ⊗n ⊗n So IY ⊆ OX , so I ⊗ L ⊆ L , thus s ∈ Γ(X, L ). P ∈ Xs ⊆ U, so (IY )P = OX,P ,(IY )Q ⊆ mQ for all Q ∈ Y , so L |Y ' OU . s|U corresponds to f ∈ Γ(U, OX ). Xs = Us = Uf is affine. Therefore for all P ∈ X, there exists n > 0 ⊗n and s ∈ Γ(X, L ) such that P ∈ Xs and Xs affine. As X is Nötherian, ⊗n Q X = Xs1 ∪...∪Xsk , where s1, . . . , sk ∈ Γ(X, L ) and Xsi are affine. n = ni, n/ni ⊗n so we can replace si with si ∈ Γ(L ). ⊗n WLOG, ni = n for all i. We replace L with L . So WLOG, there exist s1, . . . , sk ∈ Γ(X, L ) such that Xsi is affine and X = Xs1 ∪ ... ∪ Xsk .

Set Bi = Γ(Xsi , OX ). As X is of finite type over Spec A, we know that Bi is a finitely gneerated A-algebra generated by bi1, . . . , biN ∈ Γ(Xsi , OX ). So there exists n > 0 and c ∈ Γ(X < ⊗ ⊗n) such that c | = b sn. ij OX L ij Xsi ij i ⊗n Now, L is an invertible OX -module generated by the global setions n {si , cij}. k(N+1)−1 We define a morphism over A, ϕ : X → PA = Proj A[xij : 1 ≤ i ≤ ∗ ⊗n ∗ ∗ n n, 0 ≤ j ≤ N]. Then ϕ O(1) = L , ϕ (xij) = cij and ϕ (xi0) = si . Note that −1 Xsi = ϕ (D+(xi0)) → D+(xi0) is a closed immersion, so Bi ← O(D+(xi0)) is n surjecitve, mapping xij/xi0 to cij/si = bij. k k(N+1)−1 ⊗n Thus, ϕ : X → ∪i=1D+(xi0) ⊆ PA is an immersion, so L ' ϕ∗O(1) is very ample. Remark: Y a scheme, X ⊆ Y an integral closed subscheme. D ⊆ Y is an effective Cartier divisor. Assume that X 6⊆ D, then D|X = X ∩ D ⊆ X is an ∗ effective Cartier divisor, and L (D)|X = i L (D) = L (D ∩ X). So ID|X = ID∩X . 2 2 3 2 2 Example: Pk = Proj k[x, y, z]. X = V (zy − x + xz ) ⊆ P , and P0 = (0 : 1 : 0) corresponds to (x, z) ⊆ k[x, y, z]. L (P0) is ample, but not very ample.

89 2 Claim: OX (1) ' L (3P0). Let L = V (z) ⊆ P , this is just the line at infinity. 2 2 2 C`(P ) = Z[L] and OP (1) = L ([L]). So OX (1) = OP (1)|X = L ([L])|X = L (L ∩ X). Show that L ∩ X = 3P0 ∈ Div(X). Notice that X = Xy ∪ Xz IL∩X ⊆ OX is generated by z/y on Xy and 1 on Xz. 3 3 OX,P0 = k[x/y, z/y](x/y,z/y)/(z/y − (x/y) + (x/y)(z/y) ) and mP0 = (x/y). 3 Then vP0 (L ∩ X) = vP0 (z/y) = 3 as z/y = (x/y) times a unit, therefore L ∩ X = 3P0 ∈ Div(X) so L (P0) is ample. Claim: L (P0) is not very ample. Otherise there would exist s ∈ Γ(X, L (P0)) such that sP0 ∈/ mP0 L (P0)P0 .(s)0 ⊆ X is an effective cartier divisor. If L (P0)|U = OU then s|U corresponds to f ∈ Γ(U, OU ), and (s)0 ∩U = V (f) ⊆ U is a closed subscheme. (s)0 ∼ P0 so deg((s)0) = deg P0 = 1, so (s0) = Q ∈ Div(X), Q 6= P0. So P0 ∼ Q and P0 6= Q on a nonsingular projective curve X, so X is rational and X ' P1, contradiction. Let f : Y → X be an affine morphism, A = f∗OY is a quasi-coherent sheaf ] of OX algebras. f : OX → f∗OY = A . If U ⊆ X open affine, then V = f −1(U) ⊆ Y is open affine, and A (U) = −1 OY (f (U)) = OY (V ), thus, f : V → U given by OX (U) → A (U). Let X be any scheme, A a quasi-coherent OX -algebra. Want: affine f : Y → X such that A = f∗OY . Functor of Points Let X be a scheme.

Deﬁnition 6.48 (Functor of Points). We deﬁne the functor of points to be a contravariant functor FX :schemes to sets with FX (Y ) = homsch(Y,X). If 0 ∗ 0 h : Y → Y is a morphism, then F (h) = h : FX (Y ) → FX (Y ) by g 7→ gh. ¯ Example: X a variety over k = k. Then FX/k(Y ) = homk(Y,X), then FX/k(Spec k) = {the set of points of X}. Proposition 6.54. The set of morphisms ϕ : X → X0 are in correspondence with the natural transformations α : FX → FX0 .

0 Proof. Given ϕ : X → X , then αY : FX (Y ) → FX0 (Y ) can be deﬁned by g 7→ ϕg. 0 Given α : FX → FX0 , αX : FX (X) → FX0 (X) set ϕ = αX (id) : X → X . Let Y be a scheme, and g ∈ FX (Y ). As we have a natural transformation, ∗ ∗ we know that αY g = g αX . Thus, if we take id ∈ FX (X), it is mapped to g ∈ FX (Y ) then to αY (g) in FX (Y ). But also it is mapped to ϕ in FX0 (X) and then to ϕg in FX (Y ), so they must be equal, and so αY (g) = ϕg.

0 Corollary 6.55. X ' X ⇐⇒ FX ' FX0 .

Example: Let X ×S Y be the ﬁbered product over f : X → S and g : Y → S.

Then X ×S Y is uniquely determined by FX×S Y (Z) = {(p, q)|p : Z → X, q :

Z → Y, fp = gq} = FX (Z) ×FS (Z) FY (Z). X is a scheme, A a quasi-coherent OX -algebra, γ : OX → A .

90 ˜ ˜ We deﬁne FA :schemes→sets by FA (Y ) = {(f, f)|f : Y → X, F : A → ˜ ] γ f 0 f∗OY an OX -alg hom, and f : OX → A → f∗OY } on objects and if h : Y → Y ∗ 0 ˜ ] ˜ is a morphism, then we deﬁne h : FA (Y ) → FA (Y ) by (f, f) 7→ (fg, f∗(h )◦f).

Deﬁnition 6.49 (SpecA ). SpecA is the unique scheme represented by FA if it exists. FSpecA = FA .

Note: FSpecA (SpecA ) = FA (SpecA ) so id maps to (π, π˜) the natural pro- ] γ π˜ jection π : SpecA → X, π : OX → A → π∗O with O = OSpecA . ] Example: f : Y → X an affine morphism, A = f∗OY , γ = f : OX → A , then SpecA = Y ] ] Let Z be any scheme, g : Z → Y a morphism, g : OY → g∗OZ , so f∗(g ): ] f∗OY = A → (gf)∗OZ define FY (Z) → FA (Z) by g 7→ (gf : Z → X, f∗(g )). ] ] f ϕ˜ Assume (ϕ, ϕ˜) ∈ FA (Z). ϕ : Z → X and ϕ : OX → A → ϕ∗OZ . Let U ⊆ X be open affine, V = f −1(U) ⊆ Y affine. −1 ϕ˜ : A (U) = OY (V ) → ϕ∗OZ (U) = IZ (ϕ (U)) defines a morphism ϕ−1(U) → V .

Proposition 6.56. Let X be a scheme, A be a quasi-coherent OX -algebra. Then SpecA exists, π : SpecA → X is aﬃne, and π˜ : A → π∗O is an isomorphism.

Proof. If U ⊆ X open affine, then SpecA |U = Spec(A (U)), because π : ˜ ˜ Spec A (U) → X has π∗OSpecA = π∗A (U) = A (U) = A |U . 0 −1 0 Assume U ⊆ U open subset, U affine, then SpecA |U 0 = πU (U ) We πU 0 have Spec A |U 0 = Spec A |U ×U 0 U ⊆ Spec A |U → U ⊇ U , so we glue {Spec A (U)|U ⊆ X affine } together to Spec A . U1,U2 ⊆ X open affine, and Spec (U ) ⊇ π−1(U ∩U ) = Spec | = π−1(U ∩U ) ⊆ Spec (U ). A 1 U1 1 2 A U1∩U2 U2 1 2 A 2

Let X be a NötherianScheme, S = ⊕d≥0Sd a graded quasi-coherent OY - 0 algebra. For U ⊆ X open affine, πU : Proj S (U) → U. If U ⊆ U a smaller open affine, then S (U) → S (U 0) define Proj S (U 0) → Proj S (U). So we have a fiber square πU ... Proj S (U) ...... U ...... inc . inc ...... π 0 0 ...... U ...... 0 Proj S (U ) .... U 0 0 S is quasi coherent implies that S (U ) = S (U) ⊗OX (U) OX (U ). −1 Define FS (Y ) = homsch(Y, Proj S ) = {(f, gU )|f : Y → X, gU : f (U) → Proj S (U) compatible}.

Proposition 6.57. There exists a unique scheme Proj S representing FS.

πU 0 Proof. If U ⊆ X open aﬃne, then Proj(S|U ) = Proj(S(U)) → U. If U ⊆ U −1 0 any open subset, then Proj(S|U 0 ) = πU (U ), now glue.

91 n Example: If X is a scheme, and S = OX [τ0, . . . , τn] then Proj(S) = PX = Pn × X. 1 1 Example: P = Proj(k[x, y]), S = ⊕d≥0Sd. Let a ∈ Z, then Sd = OP ⊕ 0 O(a) ⊕ O(2a) ⊕ ... ⊕ O(da). So if f ∈ Sd, f ∈ Sd0 , then f ∈ O(ia) and f 0 ∈ O(ja), so ff 0 ∈ O((i + j)a). 1 −1 a So we get π : Proj S → P with π (D+(x)) = Proj S(D+(x)) = Proj k[y/x][s, x t] = 1 1 −1 a 1 1 A × P , and π (D+(y)) = Proj(k[x/y][s, y t]) = A × P . Set λ = y/x, then this is k[λ−1][s, xaλat], so we glue along the (A1 \ 0) × P1’s vis (λ, (s : t)) 7→ (λ−1, (s : λat)). π 1 Then F−a = Proj(S) → P is the Hirzebruch Surface. X is a scheme, S = ⊕d≥0Sd graded OX -algebra. Then ProjS is the unique scheme such that hom(Y, ProjS ) = FS (Y ). −1 0 Note: Have OU (1) on π (U), compatible: U ⊆ U a smaller open aﬃne, 0 then Proj S (U ) ⊆ Proj S (U), so OU 0 (1) =pullback of OU (1). Glue to get −1 O(1) on ProjS : Γ(V, O(1)) = {(σU )|σU ∈ Γ(V ∩ π (U)), OU (1) which are compatible. Remark: S = ⊕d≥0Sd is a graded ring, u ∈ S0 a unit. Then deﬁne θd : Sd → d Sd by θd(s) = u s. This gives an isomorphism of graded S0-algebras θ : S → S. Note: h ∈ S+ homogeneous implies that θ = id : S(h) → S(h), so it induces id : Proj S → Proj S.

Deﬁnition 6.50. Let S = ⊕d≥0Sd graded OX -algebra. L an invertible OX - 0 0 0 ⊗d module. Then S ∗ L = S = ⊕d≥0Sd where Sd = Sd ⊗ L . Let π : P = ProjS → X and π0 : P 0 = ProjS 0 → X. Lemma 6.58. We have an natural isomorphism ϕ : P 0 → P over X and ∗ 0 ∗ OP 0 (1) = ϕ OP (1) ⊗ (π ) L .

Proof. Let U ⊂ X open affine, with OU → L |U an isomorphism with 1 corresponding to f ∈ γ(U, L ). This defines an isomorphism S (U) → S 0(U), with 0 ⊗d d S (U)d → S (U)d = S (U)d ⊗ Γ(U, L ) by s 7→ s ⊗ f . This defines an isomorphism ϕ : Proj S 0(U) → Proj S (U). Remark implies that ϕ is independent of f ∈ Γ(U, L )∗. So we can glue to an isomorphism ϕ : P 0 → P . 0 The sections of OP 0 (1) correspond to elements in Γ(U, S1) = Γ(U, S1) ⊗ ∗ Γ(U, L ) correspond to sections of OP (1) ⊗ π L .

Definition 6.51. Assume X is Nötherian. S = ⊕d≥0Sd satisfies (+) if

1. S0 = OX

2. S1 is a coherent OX -module

3. S is generated (locally) by S1.

⊗d Note: If U ⊆ X is a small open aﬃne, then ∃S1 |U → Sd|U → 0 implies that Sd is coherent.

92 Lemma 6.59. Assume that S satisﬁes (+). Let π : P = Proj S → X. Then π is prokper and if there exists an ample invertible OX -module L on X, then ⊗n π is projective, and OP (1) ⊗ L is very ample relative to X for some n > 0.

Proof. Let U ⊂ X open affine such that S (U) generated by S1(U) as A-algebra, with A = OX (U). Then S1-coherent implies that S1(U) is a finitely gneerated A-module, so ∃ a graded A-algebra homomorphism A[x0, . . . , xN ] → S (U) close N π surjective. Thus, Proj S (U) → PU → U. Thus π|U is projective, so it is proper. Thus, π : Proj S → X is proper. ⊗n If L is ample, then S1 ⊗ L is generated by global sections. As X is ⊗n Nötherian, S1 ⊗L is coherent, so it is generated by finitely many global sections. Thus, there exists a surjection of graded OX -algberas OX [T0,...,TN ] → ⊗n 0 ⊗n n S ∗ (L ), and so P = Proj(S ∗ L ) → Proj OX [Ti] = PX is closed, and so ∗ ⊗n OP 0 (1) = ϕ6 ∗ (OP (1) ⊗ π (L )) is very ample. Definition 6.52 (Tensor Algebra). Let A be a ring and M be an A-module. d d Then T M = M ⊗ ... ⊗ M, d-times. T (M) = ⊕d≥0T M is called the tensor d algebra. S(M) = ⊕d≥0S (M) = T (M)/(x ⊗ y − y ⊗ x) is the symmetric algbera. ⊕r If M = A , then S(M) = A[T1,...,Tr].

If X is a Nötherianscheme, E a locally free OX -module of rank r and ∨ ∨ E = H om(E , OX ) the dual sheaf is also locally free of rank r. Then S(E ) = ∨ + [U 7→ S(Γ(U, E )) (sheafification) is a graded OX -algebra. ∨ ∨ ∨ S(E )0 = OX , and S(E )1 = E , ... this satisfies (+). Set π : Y = Spec S(E ∨) → X. ⊕r ∨ If U = Spec A ⊆ X open, E |U ' OU , then S(E )|U ' OU [T1,...,Tr], so −1 r π (U) = Spec OU [T1,...,Tr] = U × A . Thus π : Y → X affine bundle (in −1 fact, a vector bundle), so assume f : U → π (U) ⊆ Y is a section, (πf = idU ), then ] ∨ ∨ f Γ(U, E ) → Γ(U, S(E )) → Γ(U, OX ) ∨ gives an OX -homomorphism f : E → OX over U, ie, a section f ∈ Γ(U, E ). THus E is the sheaf of sections of π : Spec S(E ∨) → X.

Deﬁnition 6.53. P(E ) = ProjS(E ∨). π : P(E ) → X, OE (1). ⊗r −1 r−1 r−1 If E |U ' OU , then π (U) = U × P = PU . Example: L an invertible OX -module, π : P(L ) → X an isomorphism, ∗ −1 OL (1) = π (L ).

m ∨ Proposition 6.60. 1. If rank(E ) ≥ 2, then π∗OE = S (E ) for m ≥ 0 and 0 for m < 0.

∗ ∨ 2. Have surjection π E → OE (1).

m ∨ Proof. Have global S (E ) → π∗OE (m), and in U ⊆ X open aﬃne, then m ∨ −1 ∨ f ∈ Γ(U, S (E )) gives a section of OE (m) over π (U) = Proj Γ(U, S(E ))

93 ⊗r −1 r−1 isomorphism locally when E |U ' OU , π (U) = PU , already computed r−1 Γ(PU , O(m)). ∨ For the secton part, we have an OX -homomorphism E → π∗OE (1) which ∗ ∨ gives an OP(E )-homomorphism π E → OE (1), and can check that it is surjecitve locally. Universal Property

FP(E )(Y ) = {(f, L , θ)|f : Y → X, L an invertible OY -module, and θ : f ∗E ∨ → L surjective}. Left to reader. 2 ¯ 2 3 3 Exercise: Pk, k = k. Then P = {L ⊂ k | dim L = 3} = {E2 ⊆ k | dim E2 = 2}. ⊕3 2 0 → E → O 2 → OP (1) → 0 with E locally free of rank 2, then P(E ) = 3 P 3 F `(k ) = {(E1,E2)|E1 ⊂ E2 ⊂ k , dim(Ei) = i}. 2 ∗ ⊕3 π : (E ) → . Flag of locally free OX -module, O (−1) ⊆ π E ⊆ O cor- P P E P(E ) 3 respond to ﬂags of vector bundles B1 ⊆ B2 ⊆ P(E )×k , Bi = {((E1,E2),~v)|~v ∈ Ei}.

7 Schemes II

Differentials Let R be a ring, S a commutative R-algebra and M an S-module. Definition 7.1 (R-derivation). D : S → M is an R-derivation if D(fg) = fD(g) + gD(f), D(f + g) = D(f) + D(g) for all f, g ∈ S and D(f) = 0 if f ∈ R (iff D is R-linear) Universal Derivation: Let F be the free S-module with basis {df|f ∈ S}, and F 0 ⊂ F the submodule generated by d(fg)−fdg−gdf, d(f +g)−df −dg and 0 df for all f ∈ R. Then defime ΩS/R = F/F and d = dS = dS/R : S → ΩS/R by d(F ) = df + F 0. This satisfies the universal property that if D : S → M is any ˜ ˜ R-derivation, then ∃!S-homomorphism D :ΩS/R → M such that D = D ◦ dS/R. Note: P (x1, . . . , xn) ∈ R[x1, . . . , xn] and f1, . . . , fn ∈ S. Then D(P (f1, . . . , fn)) = Pn ∂P (f1, . . . , fn)D(fi). i=1 ∂xi ⊕n n Proposition 7.1. S = R[x1, . . . , xn], then ΩS/R ' S = ⊕i=1S · dxi and ⊕n d : S → S is given by df = (df/dx1, . . . , df/dxn).

Proposition 7.2. R → S → T ring homomorphisms, then ΩS/R ⊗S T → ΩT/R → ΩT/S → 0 is exact as T -modules.

2 δ Proposition 7.3. If S is an R-algebra and T = S/I, then I/I → ΩS/R⊗S T → 2 ΩT/R → 0 wotj δ taking f + I to df ⊗ 1.

0 0 0 Proposition 7.4. Let R and S be R-algebras and S = S⊗RR . Then ΩS0/R0 = 0 ΩS/R ⊗S S

−1 Proposition 7.5. U ⊆ S a multiplicative subset, then U ΩS/R = ΩU −1S/R.

94 Corollary 7.6. If S is a localization of a ﬁnitely generated R-algebra, then ΩS/R is a ﬁnitely generated S-module.

−1 0 0 Proof. S = U S , S = R[s1, . . . , sn]. Then ΩS/R is generated by ds1, . . . , dsn 0 −1 as an S -module, so ΩS/R = U ΩS0/R is generated by ds1, . . . , dsn as an S- module. Sheaves of Differentials Let X be a topological space, R, S sheaves of rings, and R → S a sheaf homomorphism. Define pre − Ω = pre − ΩS /R = [U 7→ ΩS(U)/R(U)]. If V ⊆ U is open, then we get a diagram S (U) → S (V ) → pre − Ω(V ), but also S (U) → pre − Ω(U), which goes to pre − Ω(V ) by restriction, and this is an R(U)-derivation. Then pre − Ω is a presheaf. + Define ΩS /R = (pre − Ω) . ] −1 Let ϕ : X → Y be a morphism of schemes, ϕ : ϕ OY → OX . −1 Define ΩX/Y = ΩOX /ϕ OY , and if we have X → Spec(k) is a scheme over k, then ΩX = ΩX/k = ΩX/ Spec(k). We call this the relative cotangent sheaf and the cotangent sheaf.

Proposition 7.7. If ϕ : X → Y is a morphism of aﬃne schemes, X = Spec(S) ˜ and Y = Spec(R), then ΩX/Y ' ΩS/R.

Corollary 7.8. ΩX/Y is always quasi-coherent. If ϕ : X → Y is locally of finite type, then ΩX/Y is coherent. Proof. P ∈ X, take open affine neighborhoods ϕ(P ) ∈ V ⊆ Y adn P ∈ U ⊆ −1 ϕ (V ) ⊆ X.ΩX/Y |U = ΩU/V is quasicoherent. If ϕ is lcoally of finite type, then we can take U, V such that OX (U) finitely generated OY (V )-algebra. Proposition 7.9. Take a fiber square: ... X ...... Y ...... 0 . g . g ...... f . 0 ...... 0 X .... Y 0∗ Then g ΩX/Y = ΩX0/Y 0

f g ∗ Proposition 7.10. If X → Y → Z morphisms then f ΩY/Z → ΩX/Z → ΩX/Y → 0 is exact.

2 Proposition 7.11. If g : Y → Z, X ⊆ Y a closed subscheme, then IX /IX → ΩY/Z ⊗ OX → ΩX/Z → 0 is exact.

n n Theorem 7.12. Let Y be a scheme. X = PY = P × Y . Then ∃ an exact ⊕n+1 Z sequence 0 → ΩX/Y → OX (−1) → OX → 0.

95 ... n ...... X PZ ...... Y ...... Spec Z Proof. WLOG, Y = Spec(Z) and X = Proj(S) for S = Z[x0, . . . , xn]. Set ⊕n+1 E = S(−1) . We have a map E → S by ei 7→ xi, it has kernel M, so we get an exact sequence 0 → M → E → S. This gives us an exact sequence ˜ ⊕n+1 0 → M → O(−1) → OX → 0 Pn ∂f Notice: f ∈ (x0, . . . , xn) homogeneous of degree d, then xi = df. Q i=0 ∂xi Pn ∂f Deﬁne d : X → E˜ by d(f) = ei. Note that d( X ) ⊆ M˜ , ie d : X → O i=0 ∂xi O O ˜ ˜ ˜ M is a derivation. This induces d :ΩX/Z → M, we will check that this is an isomorphism locally on D+(xi) = Spec Z[x0/xi, . . . , xn/xi]. ˜ Enough to check that Γ(D+(xi), M) is a free S(xi)-module with basis {d(xj/xi)|j 6= i}.

0 → M(xi) → E(xi) → S(xi) → 0 which takes ej → xj but there’s a map from 1 xj S(xi) → E(xi) taking 1 → ei/xi. And therefore, d(xj/xi) = ej − 2 ei. xi xi Singular Varieties Let k = k¯ Definition 7.2 (Nonsingular Variety). A variety X over k is nonsingular at P ∈ X if OX,P is a regular local ring. X is nonsingular if all points are nonsingular. Theorem 7.13. X is an irreducible separated scheme of finite type over k = k¯. Set n = dim(X). Then ΩX/k is locally free of rank n iff X is a nonsingular variety. Recall that F ⊂ K is a field extension, and each a ∈ K has a minimal polynomial pa(T ) ∈ F [T ] such that pa(a) = 0 ∈ K. K is separable over F if pa(T ) does not have multiple roots for all a ∈ K. 0 Exercise: If K/F is separable, then ΩK/L = 0. (0 = dK (pa(a)) = pa(a)dK (a) 0 with pa(a) 6= 0) Corollary 7.14. X is a variety over k implies that a dense open subset of X is nonsingular.

Proof. K = k(X). FACT: k is a perfect ﬁeld implies that any ﬁnitely generated extenseion k ⊂ K is separaly generated. IE, there exists a transcendence basis x1, . . . , xn ∈ K suchj that k(x1, . . . , xn) ⊆ K is separable. K is separable over F = k(x1, . . . , xn) for n = dim(X). Let S = k[x1, . . . , xn], ⊕n F = S0 ⇒ ΩF/k = (ΩS/k)0 = F . α So ΩF/k ⊗F K → ΩK/k → ΩK/F = 0.

96 As any k-derivation D : F → K can be extended to D : K → K we can conclude that α is injective. n n Thus, ΩK/k ' K . Spec(R) ⊆ X open. K = R0 ⇒ (ΩR/k)0 = ΩK/k = K . n Thus ∃0 6= f ∈ R : (ΩR/k)f = ΩRf /k = (Rf ) .

Lemma 7.15. If (R, m) is a local ring, k = R/m, and k ⊆ R. Then δ : m/m2 →' ΩR/k ⊗R k is an isomorphism. Theorem 7.16. X nonsing variety over k = k¯ and Y ⊆ X irreducible closed 2 subscheme, then Y is nonsingular iﬀ ΩY/k is locally free and 0 → IY /IY → ΩX ⊗ OY → ΩY → 0 is exact.

Proof. Assume the latter conditions, set q = rank(ΩY ), n = rank(ΩX ) = dim(X). It is enough to show that q = dim(Y ). The second condition causes 2 IY /IY to be locally free of rank n − q, and by Nakayama, IY is locally generated by n − q elements. The Principle Ideal Theorem says that dim(Y ) ≥ q. Let P ∈ Y , mP ⊆ 2 OY,P , then the lemma says that mP /mP = ΩY,P ⊗ OY,P /mP = ΩY,P ⊗ k. So 2 dimk(mP /mP ) = q. Thus dim(Y ) = dim OY,P ≤ q, so dim(Y ) = q, so Y is nonsingular. Assume that Y is nonsingular. ΩY/k is then locally free of rank q = dim(Y ). 2 ϕ We know that IY /IY → ΩX ⊗ OY → ΩY → 0 is exact, so all that remains is 2 showing that δ : IY /IY → ΩX ⊗ OY is injective. Let P ∈ Y be any closed point. Localize at P . Then ker ϕP is a free OY,P - module of rank r = n−q, so there exist x1, . . . , xr ∈ IY,P such that dx1, . . . , dxr 0 form a basis for ker ϕP . Let I ⊆ IY be a subideal generated by x1, . . . , xr and Y 0 = Z(I0) ⊆ X. Thus, dx1, . . . , dxr must generate a free subsheaf of rank r in ΩX ⊗ OY 0 . So 0 02 δ 0 we get that I /I → ΩX ⊗ OY 0 → ΩY 0 → 0 and δ must be injective. So the ﬁrst part tells us that Y 0 is a nonsingular variety and dim(Y 0) = 0 0 rank ΩY 0 = n − r = q, and Y ⊆ Y ⊆ X closed subschemes, Y,Y both have dimension q, so Y 0 = Y and δ0 = δ. Exercise: (R, m) is a regular local ring, f ∈ m. Then R/(f) is regular local 2 2 iﬀ f∈ / m . Use: Minimum number of gens of m = dimk m/m , k = R/m.

n Theorem 7.17 (Bertini’s Theorem). Let X ⊆ Pk a closed irreducible nonsingular subvariety, k algebraically closed. Then there exists a hyperplane H ⊆ Pn such that X 6⊂ H and X ∩ H is nonsingular.

n Proof. V = Γ(P , O(1)) = kx0 ⊕ ... ⊕ kxn. Then points of P(V ) correspond to hyperplanes by f 7→ Z(f). Let P ∈ X be a closed point, BP = {f ∈ P(V )|X ⊆ Z(f) or P ∈ X ∩ Z(f) is a singular point}. Will show: ∪P ∈X BP (P(V ) is proper closed. Set r = dim(X). Claim: BP ⊆ P(V ) is a linear subspace of dimension n n − r − 1. Check that f0 ∈ V such that P/∈ Z(f0) ⊆ P . For f ∈ V , we have

97 f/f0 ∈ OX,P . Then OX∩Z(f),P = OX,P /(f/f0). f/f0 ∈ mP ⇐⇒ P ∈ Z(f), 2 f/f0 = 0 ∈ OX,P ⇐⇒ X ⊆ Z(f), f/f0 ∈ mP /mP ⇐⇒ f ∈ BP . 2 ¯ Note: ϕP : V → OX,P /mP by f 7→ f/f0 is surjective because k = k so mP is generated by linear forms. 2 dimk V = n + 1, dimk(OX,P /mP ) = r + 1. So dim ker ϕP = n − r, so BP = P(ker ϕP ) so dim BP = n − r − 1. Now we deﬁne B = {(P, f) ∈ X × P(V )|f ∈ BP } Check: This is actually a closed subvariety of X × P(V ). π1 : B → X is a −1 surjective morphism and π1 (P ) = BP has dim BP = n−r −1 and dim(X) = r and so dim B = dim(X) + dim BP = n − 1 and ∪P ∈X BP = π2(B) ⊆ P(V ). Then π2 : X × P(V ) → P(V ) is a proper map, so π2(B) is closed of dimension at most n − 1.

8 Cohomology

Let R be a ring. Deﬁnition 8.1 (Injective). An R-modules I is injective if for every R-module M and submodule M 0 and R-homomorphism ϕ : M 0 → I, there exists an extension to M.

Deﬁnition 8.2 (Divisible Abelian Group). A Z-module T is divisible if for all n ∈ Z the map T → T by multiplication by n is surjective. Examples: Q, Q/Z. Lemma 8.1. Any divisible Z-module is injective. Proof. M 0 ⊂ M, ϕ0 : M 0 → T . Consider {(N, ψ)|M 0 ⊂ N ⊂ M, ψ : N → T extends ϕ0}. Zorn’s Lemma says that there is a maximal element (N, ψ). Claim: N = M. Else take x ∈ M \ N. 0 → N → (N, x) → Z/mZ → 0. Either m = 0 in which case (N, x) = N ⊕ Zx and we can extend ϕ(x) = 0 or m 6= 0 then mn ∈ N. So there exists y ∈ T with ϕ0(mx) = my, so ψ(x) = y.

Deﬁnition 8.3. If M is a Z-module, Mˆ = homZ(M, Q/Z). Exercise: M → Mˆˆ, x 7→ [f 7→ f(x)] is injective.

Corollary 8.2. Every Z-module M is contained in a divisible module. Proof. Take F free, F → Mˆ → 0. So M → Mˆˆ → F ˆ =direct product of Q/Zs, with all the maps inclusions.

Note: T a Z-module, M an R-module, then homZ(R,T ) is an R-module. homZ(M,T ) ' homR(M, homZ(R,T )) by f 7→ [x 7→ [r 7→ f(rx)]].

Lemma 8.3. T a divisible Z-module implies that I = homZ(R,T ) is an injective R-module.

98 Proof. M 0 ⊆ M are R-modules. Want: surj ... 0 homR(M,I) ...... homR(M ,I) ...... ' . ' ...... surj . .. 0 ...... homZ(M,T ) . homZ(M ,T ) Which is true. Theorem 8.4. Every R-module M is a submodule of an injective R-module.

Proof. Lemma implies that there exists f : M → T injective Z-hom when T is divisible. We deﬁne M → I = homZ(R,T ) by x 7→ fx with fx(r) = f(xr) Note that M ⊆ I: this is because f was injective to begin with.

Deﬁnition 8.4 (Abelian Category). A category C is abelian if hom(A, B) is an abelian group for all A, B ∈ obj(C ), hom(B,C) × hom(A, B) → hom(A, C) is a group homomorphism, ﬁnite direct products exist, kernels exist, cokernels exist, etc. See Weibel.

For any I an object of C , we have a contravariant functor hom(−,I): C → Ab. It is left exact, ie 0 → A0 → A → A00 → 0 exact implies 0 → hom(A00,I) → hom(A, I) → hom(A0,I) exact. Deﬁnition 8.5 (Enough Injectives). I is injective if hom(−,I) is exact. A category C has enough injectives if every object is a subobject of an injective object.

Corollary 8.5. If (X, OX ) is a ringed space, then the category of OX -modules has enough injectives.

Proof. Let F be an OX -module. For p ∈ X, there exists an inclusion FP → IP Q an injective OX,P -module. Set I(U) = p∈U IP . Claim: I is an injective OX -module. For any OX -module, G , we have Q homOX (G , I) = p∈X homOX homOX,P (GP , IP ). So if G 0 ⊆ G submodule, then hom(G , I) → hom(G 0, I ) is surjective. Corollary 8.6. Let X be a topological space, then the category Ab(X) of sheaves of abelian groups on X has enough injectives.

Proof. Set OX to be the constant sheaf Z, the sheaf of locally constant functions to Z. Injective Resolution Let M be an object of C , then if C has enough injectives, there exists an exacct sequence 0 → M → I0 → I1 → ... with Ij injective, that is, M has an injective resolution. Let F : C → D be a left exact functor. Then we have a complex F (I·) = 0 → F (I0) → F (I1) → ....

99 Deﬁnition 8.6. RjF (M) = Hj(F (I·)). Example: Left Exact implies that 0 → F (M) → F (I0) → .... We throw away M so that H0(I·) = R0F (M) = F (M).

Deﬁnition 8.7. Let X be a topological space, F a sheaf of abelian groups, and Γ: Ab(X) → Ab by Γ(F ) = Γ(X, F ) is left exact. Then set Hj(X, F ) = RjΓ(F ). Let A·,B· be complexes in C , a morphism f : A· → B· is a collection of n n fn : A → B such that the appropriate squares all commute. This induces a map H(f): Hn(A·) → Hn(B·). · · n f, g : A → B are homotopic (f ∼ g) iﬀ there exists morphisms hn : A → n−1 n−1 n B with fn − gn = dB hn + hn+1dA. This implies that H(f) = H(g): Hn(A·) → Hn(B·). Lemma 8.7. Consider two chain complexes A· and I·, 0 → M → A· and 0 → M 0 → I· an injective resolution, with ϕ : M → M 0. Then there exists a morphism f :(M → A·) → (M 0 → I·) and any other is homotopic to f. Proof. See Weibel.

0 0 0 0 0 −1 Note: (f1 − g1 − dBh1)dA = dB(f0 − g0 − h1dA) = dBdB h0 = 0. Corollary 8.8. RjF (M) is independent of the choice of injective resolution.

0 j Corollary 8.9. A morphism ϕ : M → M induces a unique ϕ∗ : R F (M) → RjF (M 0) Lemma 8.10 (Horseshoe Lemma). Let 0 → M 0 → M → M 00 → 0 be a short exact sequence in a category with enough injectives. Then there exists...

Theorem 8.11. Let C be abelian with enough injectives and F : C → C 0 a left exact functor.

1. RnF : C → C 0 is additive, ie, RnF (M ⊕ M 0) = RnF (M) ⊕ RnF (M 0) 2. R0F (M) = F (M) and F nF (I) = 0 if I is injective and n ≥ 1. 3. 0 → M 0 → M → M 00 → 0 short exact gives a long exact sequence 0 → F (M 0) → F (M) → F (M 00) → R1F (M 0) → ...

4. The δ morphisms from RnF (M 00) → Rn+1F (M 0) are natural.

Deﬁnition 8.8 (F-acyclic). J ∈ obC is F -acyclic if RnF (J) = 0 for all n ≥ 1. Example: J injective implies that J is F -acyclic for all left exact functors F . Lemma 8.12. 0 → Y 0 → Y 1 → ... exact with Y i F -acyclic implies that 0 → F (Y 0) → F (Y 1) → ... is exact.

100 Theorem 8.13. 0 → M → J 0 → ... a resolution by F -acyclic objects J n. Then RnF (M) = Hn(F (J ·)) Definition 8.9 (δ-functor). If C and C 0 are abelian categories, then a δ-functor 0 n 0 from C → C is a sequence of functors T = (T : C → C )n≥0 together with a morphism δn : T n(M 00) → T n+1(M 0) for each short exactr sequence 0 → M 0 → M → M 00 → 0 such that 1. Long exact sequence 2. Naturality. Example: T n = RnF . Definition 8.10 (Universal δ-functor). A δ-functor T is universal if, given any 0 0 δ functor U and natural transformation f0 : T → U , then there are unique i i fi : T → U such that the appropriate diagram commutes. Definition 8.11 (Erasable). An additive functor F : C → C 0 is erasable if for all M ∈ ob(C ) there exists a monomorphism P : M → J such that F (P ) = 0 : F (M) → F (J). Example: RnF is erasable for n ≥ 1, M → I a monomorphism and I injective, then RnF (M) → RnF (I) = 0. Theorem 8.14. Let T = (T n) be a δ-functor. If every T n is erasable for all n ≥ 1 then T is a universal δ-functor. n Proof. Assume that U = (U ) is a δ-functor with f0 : T0 → U0 a natural transformation. Let M ∈ ob(C ). p q 0 → M → J → X → 0 such that T 1(p) = 0. Then we take the long exact sequences, horizontally, and f0 vertically. We 0 0 1 1 use that ker δT = Im(q∗) ⊆ ker δU ◦f0. Then ∃!f1(M): T (M) → U (M). Must check that f1 is a natural transformation. Check: f1 commutes with δ. 2 To get f2, we use 0 → M → J → X → 0 such that T (p) = 0. Proceed by induction using the long exact sequence.

Definition 8.12. Let X be a topological space, and F a sheaf of abelian groups (hereafter referred to as an abelian sheaf). Let Γ = Γ(X, −). Then Hn(X, F ) is defined to be RnΓ(F ). Definition 8.13 (Flasque). A sheaf F is flasque if all the restriction maps are surjective.

Lemma 8.15. (X, OX ) a ringed space. F injective as an OX -module. Then F is ﬂasque.

Proof. For U ⊆ X open, Set OU = j!(OX |U ) by j : U → X the inclusion and for W ⊆ X open, OU (W ) = OX (W ) if W ⊆ U and 0 else. Note, hom(OU , F ) = F (U). If V ⊆ U, then 0 → OV → OU implies that hom(OU , F ) → hom(OV , F ) is surjective.

101 Lemma 8.16. Assume that 0 → F 0 → F → F 00 → 0 exact.

1. F 0 is flaque implies that 0 → F 0(X) → F (X) → F 00(X) → 0 is exact 2. F 0 and F flasque implies that F 00 is flasque.

Proof. Let γ ∈ Γ(X, F 0). Consider {(U, β)|U ⊆ X open and β ∈ Γ(U, F ) with 00 β 7→ γ|U ∈ F (U)}. 0 0 0 0 Then (U , β ) ≤ (U, β) iﬀ U ⊆ U and β = β|U 0 . Zorn’s lemma gives us a maximal element (U, β). Claim: U = X. Assume not, then there exists (V, σ) such that V 6⊆ U and 0 σ ∈ Γ(V, F ), σ 7→ γ|V . Then β|U∩V − σ|U∩V ∈ F (U ∩ V ). Then ﬂasque 0 implies that ∃α ∈ F (V ) with α|U∩V = β − σ. Set β1 = σ + α ∈ F (V ). Then ˜ ˜ β|U∩V = β1|U∩V , so glue β, β1 to β ∈ Γ(U ∪ V, F ), then (U ∪ V, β) ≥ (U, β), contradiction.

Proposition 8.17. If F is flasque, then Hn(X, F ) = 0 for all n ≥ 1. Proof. 0 → F → I → G → 0 is a short exact sequence, with I injective and G flasque. So we get a long exact sequence 0 → F (X) → I(X) → G (X) → H1(X, F ) → .... But Hi(X,I) = 0, so we get H1(X, F ) = 0, H2(X, F ) = H1(X, G ) = 0, etc. Corollary 8.18. If 0 → M → F 0 → ... is a flasque resolution, then Hn(X, M ) = Hn(Γ(X, F ∗)).

Proposition 8.19. (X, OX ) a ringed space. The right derived functos fo Γ: n Mod(X) → Mod(OX (X)) are given as M 7→ H (X, M ). Proof. Injective resolution in Mod(X) 0 → M → I0 → ..., we know that In is flasque for all n, so the corollary implies this. Definition 8.14 (Direct Limit Sheaf). lim = [U 7→ lim (U)]+. −→ Fα −→ Fα Lemma 8.20. X Nötherianimplies that Γ(U, lim ) = lim Γ(U, ). −→ Fα −→ Fα Proof. Check sheaf axioms using finite open covers. Lemma 8.21. X Nötherianand flasque, then lim is flasque. Fα −→ Fα Proof. V ⊆ U ⊆ X open, (U) → (V ) surjective implies that lim (U) → Fα Fα −→ Fα lim (V ) is surjective. −→ Fα Proposition 8.22. X Nötherian, { } directed system, then Hn(X, lim ) = Fα −→ Fα lim Hn(X, ). −→ Fα

Proof. Let I = indA(Ab(X)) be the category of all directed systems of abelian sheaves. T n : I → Ab by T n({ }) = lim Hn(X, ) and U n : I → Ab by Fα −→ Fα U n({ }) = Hn(X, lim ). Fα −→ Fα Note that lim is exact, so T n and U n form δ-functors. −→

102 And T 0({ }) = lim Γ(X, ) = Γ(X, lim ) = U 0({ }). Fα −→ Fα −→ Fα Fα By theorem from before, ETS that T n and U n are universal δ-functors, because they agree on the 0th term. ˜ Q Given F ∈ Ab(X), set F (U) = p∈U Fp, the sheaf of discontinuous sections. 0 → F → F˜ exact and F˜ is ﬂasque. ˜ ˜ {Fα} ∈ ob(I), and {Fα} ∈ ob(I), and we have 0 → {Fα} → {Fα}. So now sett that T n({ ˜ }) = lim Hn(X, ˜ ) = 0 and simialry for U n. Fα −→ Fα Lemma 8.23. Y ⊆ X a closed subspace, j : Y → X the inclusion, and F an n n abelian sheaf on Y . Then H (Y, F ) = H (X, j∗F ).

Proof. 0 → F → I0 → I1 → ... an injective resolution on Y . In is flasque, so n 0 j∗(I ) is flasque. So 0 → j∗F → j∗I → ... is a flasque resolution (exactness follows from Y being closed). n n ∗ n ∗ n So H (Y, F ) = H (Γ(Y,I )) = H (Γ(X, j∗I )) = H (X, j∗F )

Remark: Y ⊆ X closed, U = X \Y . Then F an abelian sheaf on X, we have j : Y → X and i : U → X inclusions. Then we can set FU = i!(F |U ) = 0 unless V ⊆ U, and if V ⊆ U, then F (V ) and also FY = j∗(F |Y ) where FY (U) = 0 −1 if V ⊆ U and equals j F (Y ∩ V ) if V 6⊆ U. Then FY,P = FP if P ∈ Y and zero else, as FU = Fp iﬀ p ∈ U and is 0 otherwise. So we have 0 → FU → F → FY → 0 is exact.

Theorem 8.24. Let X be a Nötheriantopological space, and F an abelian sheaf on X. Then Hn(X, F ) = 0 for all n > dim(X). Proof. Step 1: Reduce to X irreducible: If X is reducible, let Y be a maximal component in X. Take U = X \Y , then we have the above short exact sequence. n n This gives us a long exact sequence on cohomology, H (X, FU ) → H (X, F ) → n n H (X, FY ) = H (Y, F |Y ). If the theorem is true for irreducible spaces, then n n the last one is zero, and so we have isomorphisms H (X, FU ) → H (X, F ), so by induction, it is enough to show that the theorem holds when X is irreducible. Step 2: dim X = 0, X irreducible implies that X has open sets ∅,X and nothing else. Then Ab(X) = Ab, so Γ : Ab(X) → Ab is the identity, and so it is exact, so Hn(X, F ) = 0 for n ≥ 1. Step 3: Assume that X is irreducible of dimension > 0. If σ ∈ F (U), then get ZU → F by 1 7→ σ. Let α = {σ1, . . . , σm} with σi ∈ F (Ui). Define ` Fα = Im(⊕ZUi → F ) the subsheaf generated by α. Set B = U⊆X F (U). A = {α ⊆ B|α finite}. A is a directed poset, so {Fα} is a directed system. lim = . And ten, as Hn(X, ) = Hn(X, lim ) = lim Hn(X, ), it −→ Fα F F −→ Fα −→ Fα suffices to prove this for F = Fα finitely generated. 0 If α ⊆ α, then we have 0 → Fα0 → Fα → G → 0, with G generated by the number of sections in α \ α0. So WLOG, F is generated by one section. So we have 0 → K → ZU → F → 0 exact. Thus, it is enough to show vanishing for F ⊆ ZU a subsheaf.

103 For p ∈ U, Fp ⊆ ZU,p = Z. Deﬁne d = min{e ∈ Z+|Fp = eZ for some p}. Choose p such that Fp = dZ, d = σp, σ ∈ F (V ) ⊆ ZU (V ), p ∈ V . Must have F |V = dZU |V = dZV . ·d So we have 0 → ZV → F → F /ZV → 0. Supp(F /ZV ) ⊆ U \ V , so n dim U \ V < dim(X). So induction on dimension gives us that H (X, F /ZV ) = 0 for n > dim(X). Set Y = X \ U, 0 → ZU → ZX → ZY → 0 is ses, dim Y < dim X, so n n−1 n n H (X, ZY ) = 0 for n ≥ dim(X). H (ZY ) → H (ZU ) → H (ZX ). So WLOG, F = ZX . n If X is irreducible, then ZX is ﬂasque. So H (X, ZX ) = 0 for all n ≥ 1.

NOTE: Above, FU is not a sheaf, to correct things, sheaﬁfy

n n Lemma 8.25. Z ⊆ X closed, G a sheaf on Z, then H (Z, G ) = H (X, j∗G ). ¯ n Note: Supp(FU ) = U ⊆ U, then FU = j∗(FU |U¯ ), so H (X, FU ) = n ¯ H (U, FU |U¯ ). n Next: If X = Spec(A) and F is a quasi-coherent OX -module, then H (X, F ) = 0 for n ≥ 1. Exercise: If M is an A-module, assume homA(A, M) → homA(I,M) is surjective for all ideals I ⊆ A. Then M is injective. Hint: N 0 ⊆ N, ϕ : N 0 → M. Let M 0 ⊆ M an A-module, I ⊆ A an ideal, Ip(M 0 ∩InM) ⊆ M 0 ∩In+pM for all p, n ≥ 0. The Artin-Rees lemma says that if A is N¨otherianand M ﬁnitey generated, then for some n > 0 we have equality for all p ≥ 0.

n Deﬁnition 8.15. ΓI (M) = {m ∈ M|I m = 0 for some n > 0}.

Lemma 8.26. A N¨otherian, M injective implies that ΓI (M) is injective.

Proof. Let J ⊆ A be an ideal. ϕ : J → ΓI (M) an A-homomorphism. If a ∈ J, then Inϕ(a) = 0. J ﬁnitely generated implies that ∃p > 0 such that ϕ(IpJ) = 0. By Artin- Rees, there exists n > 0 such that Ip(J ∩ InA) = J ∩ Ip+nA. So we have ϕ(J ∩ In+p) = 0. So the map from J → A gives a map from n+p n+p n+p J/J ∩ I → A/I . As we have J/J ∩ I → ΓI (M) ⊆ M, and M n+p injective, we get a map A/I → M. The image is contained in ΓI (M), as n+p n+p I ψ(A/I ) = 0, and so the inclusion really extends to a map to ΓI (M). Lemma 8.27. M an injective A-module, A N¨otherian, f ∈ A. Then θ : M → Mf is surjective.

2 r r+1 Proof. Ann(f) ⊆ Ann(f ) ⊆ ... ⊆ Ann(f ) = Ann(f ) = .... Let x ∈ Mf , then x = θ(y)/f n for some y ∈ M. So we have a ses 0 → Ann(f r) → A → (f n+r) → 0. Then A → M by f ry and (f n+r) ⊆ A, so we get a map (f n+r) → M by f n+r 7→ f ry. This map can be extended to A, call the image of 1 in this map z. Then f n+rz = f ry, then θ(z) = θ(y)/f n = x.

F a sheaf on X, σ ∈ F (U). Supp(σ) = {p ∈ U|σp 6= 0 ∈ Fp} ⊆ U relatively closed. For Z ⊆ X closed, ΓZ (U, F ) = {σ ∈ F (U)| Supp(σ) ⊆ Z}

104 0 0 Deﬁnition 8.16. Sheaf HZ (F ) by Γ(U, HZ (F )) = ΓZ (U, F ). 0 Let j : X \ Z → X, then 0 → HZ (F ) → F → j∗(F |X\Z ), which becomes short exact if F is ﬂasque. Assume X = Spec(A) and F = M˜ . For m ∈ M = Γ(X, M˜ ), then Supp(m) = V (Ann(m)), p ∈ Supp(m) ⇐⇒ m/1 6= 0 ∈ Mp ⇐⇒ Ann(m) ⊆ P .

0 ˜ Lemma 8.28. X = Spec(A), N¨otherian, Z = V (I) ⊆ X. Then HZ (M) =

Γ^I (M)

0 0 Proof. 0 → HZ (F ) → F → j∗(F |U\Z ) tells us that HZ is the kernel of a map of quasi-coherent sheaves. 0 ˜ ˜ Enough to show that Γ(X, HZ (M)) = ΓI√(M), m ∈ ΓZ (X, M) ⇐⇒ Supp(m) ⊆ Z iff V (Ann(m)) ⊆ V (I), iff pAnn(m) ⊇ I iff In ⊆ Ann(m) (by Nötherian n property), and this is iff I m = 0, iff m ∈ ΓI (M). Proposition 8.29. A a Nötherianring, X = Spec(A). Then M an injective ˜ A-module implies that M is a flasque OX -module.

Proof. N¨otherianinduction on Y = Supp M˜ . On Y = {point}, clear. Let U ⊆ X open. Show Γ(X, M˜ ) → Γ(U, M˜ ) surjective. WLOG, Y ∩ U 6= ∅. Choose f ∈ A such that Xf ⊆ U and Xf ∩ Y 6= ∅. Set Z = V (f) = X \ Xf . Set I = (f) ⊆ A. ˜ n We know that ΓI (M) is injective, and Supp(ΓI (M)) ⊆ Y ∩ Z.(f m = 0 for all m ∈ ΓI (M), so f∈ / P implies ΓI (M)P = 0)

Induction implies that Γ^I (M) is ﬂasque, and M = Γ(X, M˜ ) → Γ(U, M˜ ) → ˜ ˜ Γ(Xf , M˜ ) = Mf gives us Γ(X, ΓI (M)) = ΓZ (X, M˜ ) → ΓZ (U, M˜ ) = Γ(U, ΓI (M)). Recall that M → Mf is surjective. 0 Let σ ∈ Γ(U, M˜ ). Then ∃τ ∈ Γ(X, M˜ ) with the same image in Γ(Xf , M˜ ). 0 ˜ Set τ = τ |U . Then (σ − τ)|Xf = 0, so σ − τ ∈ ΓZ (U, M). 0 0 0 0 So ∃α ∈ ΓZ (X, M˜ ) with α 7→ σ − τ. Therefore, α + τ ∈ Γ(X, M˜ ) maps to σ ∈ Γ(U, M˜ ).

Theorem 8.30. X affine Nötherianscheme, F a quasi-coherent OX -module, then Hn(X, F ) = 0 for all n > 0. Proof. X = Spec(A), F = M˜ . A resolution of M by injective A modules gives us a resolution of F by flasque OX -modules. The global section functor gives back the original sequence, so Hn(X, M˜ ) = Hn(I·) which is M if n = 0 and 0 otherwise.

Corollary 8.31. X N¨otherianscheme, F a quasi-coherent OX -module, then there exists a monomorphism 0 → F → G where G is a ﬂasque quasicoherent OX -module.

105 Proof. X = U1 ∪ U2 ∪ ... ∪ Un where Ui = Spec Ai where Ai is Nötherian. ˜ F |Ui = Mi, Mi an Ai-module. There exists Mi ⊆ Ii for some injective Ii. ˜ ˜ ˜ FUi → Ii injective, so F → j∗(Ii) is injective over Ui. As Ii is injective, Ii is flasque, and j∗(I˜i) is flasque. n ˜ 0 → F → ⊕i=1j∗(Ii) is flasque and quasicoherent.

Exercise: X be a scheme, f1, . . . , fr ∈ Γ(X, OX ), then if Xfi is aﬃne for all i, and (f1, . . . , fr) = (1) ⊆ Γ(X, OX ), then X is aﬃne. The ideas is that A = Γ(X, OX ), and we have ϕ : X → Spec(A). Then show that Γ(Xfi , OX ) = Afi , and so we get isomorphisms ϕ : Xfi → Spec(Afi ). So

X = ∪Xfi and so ϕ is a global isomorphism. Theorem 8.32 (Serre’s Criterion). Let X be a N¨otherianscheme. TFAE

1. X is aﬃne

n 2. H (X, F ) = 0 for all quasicoherent OX -modules F and n > 0.

1 3. H (X, I ) = 0 for all coherent sheaves of ideals I ⊆ OX .

Proof. 1 implies 2 implies 3 are done already. Assume 3. Let P ∈ X be a closed point. Let P ∈ U be an open aﬃne neighborhood of P . Let Y = X \ U. We get 0 → IY ∪P → IY → k(P ) → 0, 1 Γ(X, IY ) → Γ(X, k(P )) → H (X, IY ∪P ) = 0. ∃f ∈ Γ(X, IY ) ⊆ A = Γ(X, OX ) such that Y ⊆ V (f) and f 7→ 1 ∈ k(P ), so P ∈ Xf ⊆ U, Xf = Uf is aﬃne. Choose f1, . . . , fr ∈ A such that X =

Xf1 ∪ ... ∪ Xfr and Xfi is aﬃne. It remains to show that (f1, . . . , fr) = (1). r 1 Claim: F ⊆ OX is a coherent subsheaf implies that H (X, F ) = 0. For r−1 r r = 1 this is three. For r > 1, we have 0 → OX → OX → OX → 0 giving r−1 1 us 0 → F ∩ OX → F → I → 0, and so H (I ) = 0 and by induction, 1 r−1 H (X, F ∩ OX ) = 0, so the claim is proved. r P Take OX → OX → 0 by (g1, . . . , gr) 7→ gifi. This map is surjective, and so we take the kernel and call it F , and get a short exact sequence. r 1 Γ(X, OX ) → Γ(X, OX ) → H (X, F ) = 0 exact so (f1, . . . , fr) = (1). Cech Cohomology Let X be a topological space and X = ∪i∈I Ui. U = (Ui)i∈I , for i0, . . . , ip ∈ I, set Ui0,...,ip = Ui0 ∩ ... ∩ Uip . Let F be an abelian sheaf on X. Deﬁnition 8.17 (Alternating Function). A function α : Ip+1 → {sections of

F } is called alternating if α(i0, . . . , ip) ∈ F (Ui0,...,ip ) and if α(i0, . . . , it, it+1, . . . , ip) = −α(i0, . . . , it+1, it, . . . , tp) and if α(i1, i1, i2, . . . , ip) = 0.

Deﬁnition 8.18 (Cech Complex). Cp(U, F ) = {alternating α : Ip+1 → sections of } and maps d : Cp → Cp+1 by dα(i , . . . , i ) = Pp+1(−1)kα(i ,...,ˆi , . . . , i )| . F 0 p k=0 0 k p+1 Ui0,...,ip+1 Check: d2α = 0. Then we have Hˇ p(U, F ) = Hp(C∗(U, F )).

106 Lemma 8.33. Hˇ 0(U, F ) = Γ(X, F ). 0 Q 0 Proof. C (U, F ) = i∈I F (Ui). If α ∈ C (U, F ), then dα(i, j) = α(j)|Ui,j −

α(i)|Ui,j , so dα = 0 iﬀ the sections are all compatible which is true iﬀ ∃σ ∈ F (X) such that σ|Ui = α(i). Therefore, Hˇ 0(U, F ) = ker(C0 → C1) = F (X).

1 1 Example: X = P = Proj k[x, y]. F = ΩP = Ω. U = {U, V } with U = −1 −1 D+(x) = Spec k[t], t = y/x and V = D+(y) = k[t ]. U ∩ V = Spec k[t, t ]. Then we denote Cp = Cp(U, Ω). C0 = Γ(U, Ω) ⊕ Γ(V, Ω) = k[t]dt ⊕ k[t−1]d(t−1), C1 = Γ(U ∩ V, Ω) = k[t, t−1]dt. C2 = 0. d : C0 → C1 by dt 7→ −dt and d(t−1) 7→ −t−2dt. f(t)dt ⊕ g(t−1)d(t−1) ∈ ker d iff −f(t) − t−2g(t−1) = 0 iff f = g = 0, so Hˇ 0(U, Ω) = Ω(X) = 0. −2 −1 −1 P i Im(d) = {(−f(t) − t g(t ))dt} ⊆ k[t, t ]dt. This is { ait |a−1 = 0}, so Hˇ 1(U, Ω) ' k generated by t−1dt. p Let X = ∪i∈I Ui, and U = (Ui)i∈I . Then we have defined C (U, F ).

Deﬁnition 8.19. Deﬁne an abelian sheaf C p(U, F ) = C p given by Γ(V, C p(U, F )) = p C (U, j∗(F |V )) where j is the inclusion V → X.

0 d 1 0 0 → F → C → C → .... σ ∈ F (V ), then σ ∈ C (U, j∗F |V ) and

σ(i) = σ|Ui∩V . Lemma 8.34. This is an exact sequence of sheaves.

0 1 Proof. Exact at p = 0. 0 → Γ(X, j∗(F |V )) → C → C → .... Let p ≥ 1. p p x ∈ X, choose k ∈ I with x ∈ Uk. Let αx ∈ Cx . Then α ∈ C (V ) for x ∈ V . WLOG, V ⊆ Uk. p−1 Deﬁne hα ∈ C by hα(i0, . . . , ip−1) = α(k, i0, . . . , ip−1) ∈ F (Uk,i0,...,ip−1 ∩

V ) = F (Ui0,...,ip ∩ V ). p p−1 Now define h : Cx → Cx by αx 7→ (hα)x. Check that (dh + hd)α = α, so p ∗ ∗ dh + hd = id − 0 so id ∼ 0 and so H (Cx ) = 0, and so C is exact. Proposition 8.35. If F is flasque, then Hˇ p(U, F ) = 0 for p > 0. Proof. Let C p = C p(U, F ). This is flasque for all p. So 0 → F → C 0 → C 1 → ... is an exact sequence of flasque sheaves. Taking the global section functo we get Cp, but it is still exact because it is a flasque sequence. So Hˇ p(U, F ) = Hp(C∗) = 0.

Note: F is an abelian sheaf, 0 → F → I 0 → ... an inejctive resolution, then we get a map Hˇ p(U, F ) → Hp(X, F ).

Theorem 8.36. If X is N¨otherianseparated scheme, and U = (Ui)i∈I is an ˇ p open aﬃne covering, and F is quasi-coherent OX -module, then H (U, F ) = Hp(X, F ).

107 Proof. 0 → F → G → R → 0 with G ﬂasque and quasicoherent, then R is quasicoherent. If X is seprated, then Ui0,...,ip is aﬃne, so we get 0 → F (Ui0,...,ip ) →

G (Ui0,...,ip ) → R(Ui0,...,ip ) → 0. So we get the long exact sequence 0 → Hˇ 0(F ) → Hˇ 0(G ) → Hˇ 0(R) → Hˇ 1(F ) → 0 and an exact sequence H0(F ) → H0(G ) → H0(R) → H1(F ) → 0, and so taking the vertical maps, for p = 1, this proves it. We can proceed by induction obtaining 0 → Hˇ p(R) → Hˇ p+1(F ) → 0 and 0 → Hp(R) → Hp(F ) → 0 and looking at the vertical maps, whcih must be isomorphisms.

r Let A be a N¨otherianring, S = A[x0, . . . , xr], and X = Proj S = PA. Then let F be any OX module. Recall that Γ∗(F ) = ⊕n∈ZΓ(X, F (n)). Note: f ∈ Sm = Γ(X, O(m)) gives f : F (n) → F (n + m) by σ 7→ σ ⊗ f. p p Thus, f : H (X, F (n)) → H (X, F (n + m)), so Γ∗(F ) is a graded S- module.

Theorem 8.37. 1. S → Γ∗(OX ) is an isomorphism of graded S-modules.

p 2. H (X, OX (n)) = 0 for 0 < p < r any n.

r 3. H (X, OX (−r − 1)) ' A

0 r r 4. Perfect Pairing: H (X, OX (n))×H (X, OX (−n−r−1)) → H (X, OX (−r− 1)) = A.

p p Proof. F = ⊕n∈ZOX (n) is quasicoherent. Then H (X, F ) = ⊕n∈ZH (X, OX (n)).

Let U = {U0,...,Ur} with Ui = D+(xi) ⊆ X. Then Ui0,...,ip = D+(xi0 . . . xip ), so we look at Γ(U , (n)) = S(n) , and so Γ(U , ) = i0,...,ip OX (xi0 ...xip ) i0,...,ip F S . xi0 ...xip C∗(U, ): ⊕ S → ⊕ S → ... → ⊕r S → S F i0 xi0 i0

1. ker(ﬁrst map) = ∩Sxi = S. 2. Cp(U, ) = Q S = Q ⊕ Γ(U ∩U , (n)). F xr i0<...

FH → 0, and FH = ⊕n∈ZOH (n), with the ﬁrst map being multiplication p p by xr. By induction on r, we get that H (X, FH ) = H (H, FH ) = 0 for 0 < p < r − 1. 0 0 1 So we get a long exact sequence H (X, F ) → H (X, FH ) → H (X, F (−1)) → 1 1 1 1 H (X, F ) → H (X, FH ), so we get that H (X, F ) ' H (X, F (−1)).

108 p−1 p For 1 < p < r − 1, we have H (X, FH ) = 0 → H (X, F (−1)) ' p p H (X, F ) → H (X, FH ) = 0.

r−2 0 r−1 xr r−1 So we now have H (X, FH ) → H (X, F (−1)) → H (X, F ) → r−1 δ r xr r H (X, FH ) → H (X, F (−1)) → H (X, F ) → 0. Note that if r = 2, r−2 we don’t get H (X, FH ) = 0, but the ﬁrst map is zero.

r `0 `r−1 −1 r−1 Ann(xr) ⊆ H (X, F (−1)) has basis {x0 . . . xr−1 xr |`i < 0} and H (X, FH ) `0 `r−1 has basis {x0 . . . xr−1 |`i < 0}. T his tells us that δ is injective, and so the map before it is zero and we have an isomorphism Hr−1(X, F (−1)) = Hr−1(X, F ).

`0 `r 3. Image(last map)= spanA{x0 . . . xr |ì ∈ Z with some ì ≥ 0}, so we get r `0 `r r H (X, F ) = spanA{x0 . . . xr |ì < 0∀i}. And so H (X, OX (−r − 1)) = −1 A(x0x1 . . . xr) ' A.

r `0 `r P 4. H (X, OX (−n−r−1)) = spanA{x0 . . . xr | ì = −n−r−1, ì < 0}. So 0 r r look at H (X, OX (n)) × H (X, OX (−n − r − 1)) → H (X, OX (−r − 1)). m0 mr `0 `r `0+m0 `r +mr map (x0 . . . xr ) × (x0 . . . xr ) 7→ x0 . . . xr . r The right hand side 6= 0 ∈ H (X, OX (−r − 1)) iff mi + ì = −1 for all i, so ì = −mi − 1 for all i.

Theorem 8.38. X a projective scheme over Spec A, A N¨otherianand OX (1) is a very ample invertible sheaf relative to Spec(A). If F is a coherent OX -module, then

1. Hp(X, F ) is a ﬁnitely generated A-module for all p ≥ 0. 2. Hp(X, F (n)) = 0 for all p > 0 for n >> 0.

r ∗ r r Proof. Let f : X → P , OX (1) = f OP (1). f∗F is coherent on P . f∗(F (n)) = ∗ r r f∗(f OP (n) ⊗ F ) = OP (n) ⊗ f∗F = (f∗F )(n). p p r p r H (X, F (n)) = H (P , f∗F (n)) = H (P , (f∗F )(n)). r r WLOG, X = P . Note: Theorem is true for F = OP (n). The theorem is true for p > r, as all the cohomology vanishes. We proceed

r by decreasing induction on p. There exists E = ⊕OP (qi) with finitely many terms with 0 → R → E → F → 0 exact, so 0 → R(n) → E (n) → F (n) → 0 exact. Thus, Hp(Pr, E (n)) → Hp(Pr, F (n)) → Hp+1(Pr, R(n)). The outer modules are finitely generated by the previous theorem and inductions. As A is Nötherian,the middle module is finitely generated. If the outer modules are zero, then Hp(Pr, F (n)) = 0 for all n >> 0, p > 0.

Application: if X is a nonsingular curve, deﬁned genus to be g = dimk Γ(X, ΩX ) < 1 ∞. With Serre duality (to be shown), we see that g = dimk H (X, OX )

109 Euler Characteristic Let X be a projective scheme over k. Let F be a coherent OX -module and dim(X) = r.

Deﬁnition 8.20 (Euler Characteristic). The Euler characteristic of F is χ(F ) = P p p p≥0(−1) dimk H (X, F ).

Then 0 → F 0 → F → F 00 → 0 gives a long exact sequence 0 → H0(X, F 0) → H0(X, F ) → ... → Hr(X, F ) → Hr(X, F 00) → 0 an exact sequence of k-vector spaces. So χ(F ) = χ(F 0)+χ(F 00). In general, if we have 0 → F 0 → ... → F ` = 0, then P(−1)iχ(F i) = 0. Proposition 8.39. X a proper scheme over Nötherian Spec(A). L an invertible OX -module. Then L is ample iff ∀ coherent OX -modules F , we have Hp(X, F ⊗ L n) = 0 for all p > 0, n >> 0. Proof. ⇒: L ample, then L ⊗m is very ample relative to Spec A for some 0 r ⊗m m > 0. Thus, there exists an immersion X ⊆ X ⊆ PA with L = OX (1). 0 r p X is proper, so X = X is closed in PA. Choose n0 > 0 such that H (X, F ⊗ i L ⊗ OX (n)) = 0 for all p > 0 and for 0 ≤ i < m and n ≥ n0. Thus, p n H (X, F ⊗ L ) = 0 for n ≥ m + mn0. n ⇐: F a coherent OX -module. Show: F ⊗ L generated by global sections for all n >> 0. Let p ∈ X be a closed point. Then we get 0 → Ip → OX → k(P ) → 0 ses, so we have Ip ⊗ F → F → F ⊗ k(p) → 0, and to get a short exact sequence, we take 0 → IpF → F → F ⊗ k(P ) → 0. n n n So then we have 0 → IpF ⊗ L → F ⊗ L → F ⊗ L ⊗ k(p) → 0. For n n 1 n ≥ n0 = n0(p), we get Γ(X, F ⊗L ) → Γ(X, F ⊗L ⊗k(P )) → H (X, IpF ⊗ n n n L ) = 0, so the second term is (F ⊗ L )p ⊗ k(p). By Nakayama, (F ⊗ L )p is generated by global sections from Γ(X, F ⊗ L n). p ∈ U = Spec(B) ⊆ X, B Nötherian. So M = Γ(U, F ⊗ L n) is a finitely generated B-module. M 0 ⊆ M is the n 0 submodule generated by the imeage of Γ(X, F ⊗ L ). Mp = Fp = Mp. Then 0 there exists f ∈ B \ p suc that Mf = Mf . n Set Up,n = D(f) ⊆ U, then p ∈ Up,n and (F ⊗ L )|Up,n is generated by global sections. n1 Special case: may choose n1 and open Vp 3 p such that L |Vp is generated by global sections. Set Up = Vp ∩ Up,n0 ∩ ... ∩ Up,n0+n1−1. Then if n ≥ n0(p), 0 n n n1 m 0 we have F ⊗ L = (F ⊗ L ) ⊗ (L ) where n0 ≤ n < n0 + n1 with m ≥ 0. n n Then (F ⊗ L )|Up is generated by global sections from Γ(X, F ⊗ L ).

So X = Up1 ∪ Up2 ∪ ... ∪ Up` with n ≥ max(n0(p1), . . . , n0(p`)), and we get F ⊗ L n generated by global sections.

r Applications: A N¨otherian, X = PA = Proj(S) for S = A[x0, . . . , xr]. Let r ˜ M be a ﬁnitely generated graded S-module. We have maps Mn → Γ(P , M(n)), ˜ ˜ a homomorphism of graded S-modules ϕ : M → Γ∗(M) = ⊕n∈ZΓ(M(n)). r ˜ Claim: Mn ' Γ(P , M(n)) for all n >> 0

110 Injective: N = ker ϕ ⊆ M. Show that Nn = 0 for n >> 0. As N is ﬁnitely generated, it is enough to show that ∀m ∈ N and 0 ≤ i ≤ r, there exists p > 0 p ˜ so that xi m = 0. This is true because m = 0 in Γ(D+(xi), M(n)) = M(n)(xi) ⊆

Mxi . 0 Surjective: 0 → F → F → M → 0, F = ⊕S(qi) finite sum. So then 0 → F˜0(n) → F˜(n) → M˜ (n) → 0, and so for n >> 0, we get Γ(Pr, F˜(n)) → Γ(Pr, M˜ (n)) → H1(Pr, F˜0(n)) = 0, so we have a surjection, but this is really an r ˜ onto map Fn → Mn, and must also give a surejction Mn → Γ(P , M(n)). r X is projective over a field, k. X ⊆ Pk. Then there is OX (1) very ample. F is a coherent OX -module. Definition 8.21 (Hilbert Polynomial). We define the hilbert polynomial to be

PF (n) = χ(F (n)). In particular, PX (n) = POX (n).

Claim: PF (n) ∈ Q[n].

Proof. Nötherianinduction on Y = Supp F . Y = ∅ ⇒ F = 0 ⇒ PF (n) = 0. f· If Y 6= ∅, take f ∈ Γ(X, OX (1)) such that Y ∩ Xf 6= 0. 0 → R → F (−1) → F → M → 0 is an exact sequence of coherent OX -modules. Taking the Euler characteristics at n, we get PF (n)−PF (n−1) = PM (n)− PR(n). The closures of Supp R and Supp M ⊆ Y ∩ V (f) ( Y . By Nötherian induction, the functions on the right hand side are polynomials in Q[n], and therefore PF (n) ∈ Q[n]. r Note P = Proj S, S = k[x0, . . . , xr], M is a finitely generated graded S- r ˜ ˜ module. If n >> 0, then Mn = Γ(P , M(n)) = χ(M(n)). Note: PF (n) depends on OX (1) on OX (1), but PF (0) = χ(F ). Assume r dim X = r. Then the arithmetic genus if pa(X) = (−1) (χ(OX ) − 1) = r (−1) (PX (0) − 1), so if X is a connected curve, then pa(X) = 1 − χ(OX ) = 0 1 1 − dimk H (X, OX ) + dimk(H (X, OX )). So then, as X is projective, this is 1 dimk H (X, OX ). Ext Groups and Sheaves (X, OX ) a ringed space, F , G are OX -modules. Remember that hom(F , −) is a functor from Mod(X) → Ab is left exact.

Deﬁnition 8.22 (Ext). Extp(F , −) = Rp hom(F , −) from Mod(X) → Ab. E xtp(F , −) = RpH om(F , −).

0 p Proposition 8.40. 1. E xt (OX , G ) = G , E xt (OX , G ) = 0 for p > 0.

p p 2. Ext (OX , G ) = H (X, G )

Lemma 8.41. I ∈ Mod(X) injective, U ⊆ X open, then I |U ∈ Mod(U) injective. Proof. j : U → X inclusion.

111 ... j∗F ...... j!G ...... ⊂ ... j∗(I |U ) ...... I

p p Proposition 8.42. U ⊆ X open. Then E xtX (F , G )|U = E xtU (F |U , G |U ). Note: 0 → G 0 → G → G 00 → 0 gives a long exact sequence on Ext(F , −) Proposition 8.43. 0 → F 0 → F → F 00 → 0 gives a long exact sequence 0 → hom(F 00, G ) → hom(F , G ) → hom(F 00, G ) → Ext1(F 00, G ) → .... Proof. 0 → G → I 0 → .... Then as hom(−, I P ) is exact and contravariant, we get 0 → hom(F 00, I ∗) → hom(F , I ∗) → hom(F 0, I ∗) → 0, and so we get a long exact sequence in the ﬁrst variable.

Deﬁnition 8.23 (Locally Free Resolution). A locally free resolution of F is a resolution ... → E2 → E1 → E0 → F → 0 exact with Ep locally free of ﬁnite rank.

p Proposition 8.44. If E∗ → F → 0 is locally free resolution, then E xt (F , G ) = p H (H om(E∗, G )).

0 00 Proof. Both sides are δ-functors. RHS: 0 → G → G → G → 0, Ep is locally 00 free implies that H om(Ep, −) is exact, and so we get 0 → H om(E∗, G ) → 0 H om(E∗, G ) → H om(E∗, G ) → 0, so get long exact sequence. Agree for p = 0: E1 → E0 → F → 0, so 0 → H om(F , G ) → H om(E0, G ) → 0 H om(E1, G ) → ... is exact, and so H (H om(E∗, G )) = H om(F , H ) = E xt0(F , G ). Universality: Both sides vanish when G is injective.

∨ Note: E is locally free OX -module of ﬁnite rank, then E = H om(E , OX )is locally free. Exercise: H om(F , G ) ⊗ E ' H om(F , E ⊗ G ) ' H om(F ⊗ E ∨, G ) by ϕ ⊗ e 7→ [f 7→ e ⊗ ϕ(f)] and ψ 7→ [f ⊗ e∨ 7→ (e∨ ⊗ 1)(ψ(f))] ∨ ∨∨ Special Case: E = H om(O, E ) = H om(E , OX ) = E .

Lemma 8.45. I is an injective OX -module, E is a locally free OX -module, then I ⊗ E is injective.

Proof. hom(−, I ⊗ E ) = hom(− ⊗ E ∨, I ) is exact. Proposition 8.46. E locally free of ﬁnite rank, then Extp(F , E ⊗G ) = Extp(F ⊗ E ∨, G ) and E xtp(F , G ) ⊗ E = E xtp(F , E ⊗ G ) = E xtp(F ⊗ E ∨, G ). Proof. Everything is a δ-functor. They are the same for p = 0. They vanish for G injective.

112 p p Note: If A is a ring, M an A-module, then ExtA(M, −) = R homA(M, −). ({pt},A) is a ringed space, and modules on it and A-mod are the same thing.

If E is locally free of ﬁnite rank, then H om(E , F )x = homOX,x (Ex, Fx).

Proposition 8.47. X a N¨otherianScheme, F a coherent OX -module and G any -module. Then xtp( , ) = Extp ( , ) for all x ∈ X, p ≥ 0. OX E F G x OX,x Fx Gx

Proof. WLOG, X is aﬃne. There exists a locally free resolution E∗ → F → 0, so there exists a free resolution of OX,x-modules E∗,x → Fx → 0. p p p p E xt (F , G )x = H (H om(E∗, G ))x = H (H om(E∗, G )x) = H (hom(E∗,x, Gx)) = Extp ( , ). OX,x Fx Gx

Proposition 8.48. X a projective scheme over Nötherian Spec(A) with OX (1) p is very ample relative to A. F , G coherent OX -modules, p ≥ 0. Then Ext (F , G (n)) = Γ(X, E xtp(F , G (n))) for all n >> 0. Proof. p = 0: hom(F , G (n)) = Γ(X, H om(F , G (n))) is true for all n. p p ∨ F is lcoally free of finite rank. Ext (F , G (n)) = Ext (OX , F ⊗ G (n)) = p ∨ p p ∨ H (X, F ⊗ G (n)) = 0. E xt (F , G (n)) = E xt (OX , F ⊗ G (n)) = 0 for all p > 0 for all n, so it is ture for all n >> 0. F coherent: 0 → R → E → F → 0 where E is locally free. 0 → H om(F , G (n)) → H om(E , G (n)) → H om(R, G (n)) → E xt1(F , G (n)) → E xt1(E , G (n)) = 0 and we can take the twisting outside. So for n >> 0, we have 0 → hom(F , G (n)) → hom(E , G (n)) → hom(R, G (n)) → Γ(X, E xt1(F , G (n))) → 0. We also get 0 → hom(F , G (n)) → hom(E , G (n)) → hom(R, G (n)) → Ext1(F , G (n)) → Ext1(E , G (n)) = 0 for n >> 0. As the first three are the same, the last one must be. For p ≥ 2, we have long exact 0 → Extp−1(R, G (n)) → Extp(F , G (n)) → 0 for n >> 0. Also, for all n, 0 → E xtp−1(R, G (n)) → E xtp(F , G (n)) → 0. By induction, we get vertical maps, which prove the theorem. Let A be a ring, M an A-module, Vp M = M ⊗ ... ⊗ M/ < . . . x ⊗ x . . . >. 0 → M → N → R → 0 a ses of free modules of ranks m, n, r, then Vn N ' Vm Vr M ⊗A R is a natural isomorphism. This is important for sheafification. Vp Vp + Let F be an OX -module, F = (U 7→ F (U)) . Then if 0 → M → N → R → 0 a short exact sequence of locally free Vn Vm Vr OX -modules of ranks m, n, r, then N ' M ⊗ R. Let X be a nonsingular variety over a field k. Then ΩX is locally free of rank n.

Deﬁnition 8.24 (Canonical Sheaf of X). The canonical sheaf on X is ωX = Vn ΩX .

n ⊕n+1 n n n V n On Pk , we have 0 → ΩP → O(−1) → OP → 0. So ωP = ΩP = n 1 n+1 ⊕n+1 ⊗n+1 V n V n V ΩP ⊗ OP = (O(−1) ) = O(−1) = O(−n − 1). n Theorem 8.49 (Serre Duality, Version I). X = Pk , k a ﬁeld.

113 n 1. H (X, ωX ) ' k.

n n 2. F coherent OX -module implies that hom(F , ωX )×H (X, F ) → H (X, ωX ) = k is a perfect pairing.

p n−p ∗ 3. Ext (F , ωX ) ' H (X, F ) , a natural isomorphism of functors in F from Coherent Sheaves on X to k-modules. ∗ n (If V is a k-vector space, V = homk(V,H (X, ωX )))

n n Proof. 1. H (X, ωX ) = H (X, O(−n − 1)) ' k.

n 2. Let F = O(q). Then hom(F , ωX ) × H (X, F ) = Γ(X, O(−n − 1 − q)) × Hn(X, O(q)) → Hn(X, O(−n − 1)) = k is a perfect pairing. So now consider the case where F is coherent. Then ∃E1 → E0 → F → 0 exact, n Ei = ⊕O(qij) a ﬁnite sum. Note that H (X, −) is right exact (end of the n n n long exact sequence). So we get H (X, E1) → H (X, E0) → H (X, F ) → n ∗ n ∗ n ∗ 0, which gives us 0 → H (X, F ) → H (X, E0) → H (X, E1) . We also have 0 → hom(F , ωX ) → hom(E0, ωX ) → hom(E1, ωX ), with the last two n ∗ on each being equal. Thus, hom(F , ωX ) = H (X, F ) . 3. Both sides are contravariant δ-functors from Coh(X) → Mod(k), ie δ- functors Coh(X)op → Mod(k). They agree for p = 0. It remains to show that they are universal: we will show that both sides are erasable. That is, for all coherent F , there exists an epi u : E → F such that both functors vanish on u when p > 0.

So there is an epi E → F when E is a ﬁnite ⊕O(qi). WLOG, qi < ⊕n+1 p p 0 as O(q − 1) → O(q). So Ext (E , ωX ) = ⊕ Ext (O(qi), O(−n − p n−p 1)) = H (X, OX (−n − 1 − qi)) = 0 for p > 0. Also, H (X, E ) = n−p ⊕H (X, O(qi)) = 0 for p > 0.

Deﬁnition 8.25 (Dualizing Sheaf). Let X be a proper scheme of dimension n ◦ over a ﬁeld k. A dualizing sheaf for X is a coherent OX -module ωX together n ◦ with a k-linear trace map t : H (X, ωX ) → k such that forall coherent F we ◦ n n ◦ t get perfect pairings hom(F , ωX ) × H (X, F ) → H (X, ωX ) → k. Proposition 8.50. Let (ω, t) and (ω0, t0) be dualizing sheaves for X. Then there 0 0 n exists a unique isomorphism ϕ : ω → ω such that t = t ◦ ϕ∗ : H (X, ω) → Hn(X, ω0) → k.

0 Proof. The perfect pairing requirement gives hom(ω, ω0)×Hn(X, ω) → Hn(X, ω0) →t k, and so t : Hn(X, ω) → k is given by some ϕ ∈ hom(ω, ω0). That is, there 0 0 exists a unique ϕ : ω → ω with t = t ◦ ϕ∗. By symmetry, there exists a unique 0 0 ψ : ω → ω such that t = t ◦ ψ∗. ψϕ : ω → ω satisﬁes t = t ◦ (ψϕ)∗, and the identity does this, and so ψϕ = id and similarly for ϕψ.

N Assume that X is projective over k, choose a closed embedding X ⊆ Pk . Set r = codim(X, PN ).

114 ∗ r Deﬁnition 8.26. ω = xt ( ω N ). X E p OX P i Lemma 8.51. E xtP (OX , ωP ) = 0 for i < r. i i Proof. E xtP (OX , ωP ) is coherent as an OX -module. Thus, E xtP (OX , ωP )(q) is generated by global sections for all q >> 0. i N−i i Γ(P, E xtP (OX (−q), ωP )) ' H (P, OX (−q)) = ExtP (OX (−q), ωP ). This is zero if N − i > dim X iﬀ i < r. Note: Assume 0 → M 0 → M → M 00 → 0 ses. Then M 0 injective implies split exact, so hom(M,M 0) → hom(M 0,M 0) surjective. If M 0,M injective, then M 00 injective. And 0 → M 0 → M 1 → ... → M r → 0 with M i injective for i < r, then M r injective.

◦ r Lemma 8.52. There is a natural isomorphism homX (−, ωX ) ' ExtP (−, ωP ) of functors from Mod(X) → Mod(k).

∗ m m Proof. 0 → ωP → I an injective resolution in Mod(P ). Set J = H omP (OX , I ) ⊆ I m the subsheaf of sections killed by the ideal sheaf of X. m m Note: F an OX -module, ϕ : F → I an OP -hom, then Im(F ) ⊆ J . m m m Thus, J is an injective OX -module. So homX (−, J ) = homP (−, I ) is exact. i ∗ i H (J ) = E xtP (OX , ωP ) = 0 for i < r. So we have exact sequence 0 → 0 r−1 r r−1 r r J → ... → J → J , and we get J → J1 → 0 with J1 = r−1 r r Im d(d ) ⊆ J . The Note implies that J1 is injective, so we get 0 → r r r r r r J1 → J → J2 → 0 with J = J1 ⊕ J2 , so all are injective. ∗ r r+1 So the exact complex can be called J1 and J2 → J → ... can be ∗ called J2 . ◦ r r ∗ r r+1 ωX = E xtP (OX , ωP ) = H (J ) = ker(J2 → J ), so we get 0 → ◦ r r+1 ◦ ωX → J2 → J which gives, for F an OX -module, 0 → homX (F , ωX ) → r r+1 r r ∗ homX (F , J2 ) → homX (F , J ), so ExtP (F , ωP ) = H (homP (F , I )) = r ∗ r ∗ ◦ H (homX (F , J )) = H (homX (F , J2 )) = homX (F , ωX ). ◦ Proposition 8.53. ωX is a dualizing sheaf for X. ◦ Proof. F is any coherent OX -module. n = N − r = dim(X). homX (F , ωX ) × n r N−r N H (X, F ) = ExtP (F , ωP ) × H (P, F ) → H (P, ωP ) ' k is a perfect pairing, by Serre Duality for PN . ◦ ◦ ◦ N ◦ N Take F = ωX . Then homX (ωX , ωX )×H (X, ωX ) → H (P, ωP ) = k is per- ◦ ◦ n ◦ N fect pairing. id ∈ homX (ωX , ωX ) corresponds to t : H (X, ωX ) → H (P, ωP ), the trace map. Because everything has been natural, we can use t to factor the perfect pairing. Let X be a proper scheme over k of dimension n.

Deﬁnition 8.27 (Dualizing Sheaf). A dualizing sheaf for X is a coherent OX - ◦ n ◦ module ωX with a k-linear trace map t : H (X, ωX ) → k such that for all ◦ n coherent OX -mod F , we have a perfect pairing homX (F , ωX ) × H (F ) → n ◦ t H (X, ωX ) → k.

115 VN N Thus, it must be unique. ω N = Ω N is the dualizing sheaf for . IF P P P N ◦ r X ⊆ is a closed subscheme of codimension r, then ω = E xt N (OX , ω N ) P X P P dualizing for X. ◦ ◦ Notice: A choice of trace map t gives a natural isomorphism θ : homX (−, ωX ) → Hn(X, −)∗ of functors Coh(X)op → Mod(k).

i i ◦ n−i ∗ Corollary 8.54. ∃ natural transformations θ : ExtX (−, ωX ) → H (X, −) . Proof. We must show that both sides are δ-functors that agree for i = 0 and i ◦ op that ExtX (−, ωX ) is erasable in Coh(X) for i > 0. Choose OX (1) very ample. F coherent implies that there exists a surjection ⊕N i ⊕N ◦ i ◦ ⊕N O(−q) → F for q >> 0. ExtX (O(−q) , ωX ) = ExtX (OX (−q), ωX ) = i ◦ ⊕N i ◦ ⊕N Ext (OX , ωX (q)) = H (X, ωX (q)) = 0 for q >> 0, so done. Deﬁnition 8.28 (Regular Sequence). Let A be a local ring, m ⊆ A maximal ideal, and f1, . . . , fr ∈ m. Then f1, . . . , fr is a regular seqeunce if each fi is a nonzero divisor on A/(f1, . . . , fi−1). Deﬁnition 8.29 (Cohen-Macaulay Ring). A is Cohen-Macaulay (CM) if ∃ a regular sequence f1, . . . , fr ∈ m such that dim(A) = r. Facts:

1. A regular local ring is CM

2. If A is CM then f1, . . . , fr ∈ m is a regular sequence iﬀ dim A/(f1, . . . , fr) = dim A − r.

3. A is CM and f1, . . . , fr ∈ m a regular sequence, then A/(f1, . . . , fr) is CM.

4. A CM, f1, . . . , fr ∈ m a regular sequence, then I = (f1, . . . , fr) ⊆ A n n+1 ¯ 2 implies that (A/I)[t1, . . . , tr] → grI (A) = ⊕n≥0I /I by ti 7→ fi ∈ I/I is an isomorphism, so I/I2 ' (A/I)⊕r.

Koszul Complex ⊕r Vp ⊕r A a ring, f1, . . . , fr ∈ A and A has basis e1, . . . , er. Deﬁne Kp = (A ) for 0 ≤ p ≤ r. Basis {ei1 ∧ ... ∧ eip }.

d1 : K1 → K0 by ei 7→ fi. dp : Kp → Kp−1 by dp(ei1 ∧ ... ∧ eip ) = Pp j−1 j=1(−1) fij eij ∧ ... ∧ eˆij ∧ ... ∧ eip . So K∗ = K∗(f1, . . . , fr,A) is this complex.

Proposition 8.55. A a local ring, f1, . . . , fr a regular sequence for A. Then 0 → K∗ → A/(f1, . . . , fr) → 0 is exact.

Definition 8.30 (Cohen-Macaulay). A scheme X is Cohen-Macaulay if OX,P is CM for all P ∈ X. Definition 8.31 (Local Complete Intersection). If Y is a nonsingular variety over a field k and X ⊆ Y is a closed subscheme of codim r, and IX ⊆ OY the ideal sheaf of X. Then X is a local complete intersection in Y if IX,P is generated by r elements, ∀P ∈ X.

116 Note: X ⊆ Y a local complete intersection implies that X is Cohen-Macaulay 2 and IX /IX is a locally free OX -module of rank r, the ”Conormal Sheaf” of X in Y .

2 ∨ Deﬁnition 8.32 (Normal Sheaf). NX/Y = (IX /IX ) is the Normal Sheaf of X in Y .

2 Recall that X ⊆ Y nonsingular gives 0 → I /I → ΩY ⊗ OX → ΩX → 0 is exact, so dualizing we get 0 → TX → TY ⊗ OX → NX/Y → 0

N ◦ Theorem 8.56. X ⊆ Pk local complete intersection of codim r, then ωX ' Vr ω N ⊗ n . P NX/P Proof. Let U = Spec(A) ⊆ PN open affine such that I(X ∩ U) ⊆ A is generated by f1, . . . , fr ∈ A. Then let P ∈ U ∩ X. m = I(P ) ⊆ A is a maximal ideal. Then A = N is regular local, so CM. dim A /(f , . . . , f ) = dim = m OP ,P m 1 r OX,P dim X = N − r. So f1, . . . , fr is a regular sequence. So we look at 0 → K∗(f1, . . . , fr,Am) → Am/(f1, . . . , fr) → 0. Replace U with some Uh, so WLOG, 0 → K∗(f1, . . . , fr; A) → A/(f1, . . . , fr) → 0 is exact. ˜ Thus, 0 → K∗ → OX∩U → 0 is locally free resolution on U. ◦ r r r ˜ Therefore, ω |U = E xt N (OX , ω N ) = E xt (OX∩U , ωU ) = H (H omU (K∗, ωU )). X P P U ˜ ⊕R H om(K∗, ωU ) gives the sequence ... → ωU → ωU → 0 by (σ1, . . . , σr) 7→ P j−1 (−1) fjσj. ◦ Thus, ωX |U ' ωU /(f1, . . . , fr)ωU = ωU ⊗ OX . 2 Note: f1, . . . , fr form a basis of IX /IX , then f1 ∧ f2 ∧ ... ∧ fr is a basis of r 2 −1 r V V n (IX /IX ). Thus (f1 ∧ ... ∧ fr) is a basis of (NX/P ). ◦ We now define γ : ω | = ω /(f , . . . , f )ω → ω ⊗ N by taking X U U 1 r U U NX/P −1 σ¯ 7→ σ ⊗ (f1 ∧ ... ∧ fr) . Claim: γ is independent of choices, and so defines a global isomorphism ◦ ω → ω n ⊗ N . X P NX/P Let g1, . . . , gr ∈ I(X ∩ U) ⊆ A be another set of generators. Write gi = P ⊕r ⊕r P Vp cijfj with cij ∈ A. Define a map ϕ : A → A by ϕ(ei) = cijej. As is a functor, we get a map on exterior powers. So we get 0 → K∗(g1, . . . , gr; A) → A/I → 0 and 0 → K∗(f1, . . . , fr; A) → A/I → 0, with vertical maps given by exterior powers of ϕ and the identity from A/I to itself, with everything commuting. Vr ˜ As ϕ = det(cij): Kr(g) → Kr(f) and so det(cij): H omU (Kr(f), ωU ) → ˜ H omU (Kr(g), ωU ) gives a map ωU → ωU . det(cij ) Tus we get map ωU (f1, . . . , fr)ωU → ωU /(g1, . . . , gr)ωU and maps from Vr each to ω ⊗ N giving a commutative diagram. U NX/P ◦ Corollary 8.57. If X is a nonsingular projective variety then ωX = ωX = Vn ΩX for n = dim X.

N 2 Proof. X ⊆ closed subvariety. So 0 → / → Ω N ⊗ → Ω → 0. Pk IX IX P OX X VN Vr 2 Thus, ω N ⊗ OX = (ΩPP N ⊗ OX ) = IX /IX ⊗ ωX . P Vr Thus, ω = ω n ⊗ N X P NX/P

117 9 Curves

Let k = k¯ be an algebraically closed ﬁeld. Deﬁnition 9.1 (Curve). A curve is a complete connected non-singular variety of dimension 1 over k.

Fact: All curves are projective. P P Div(X) = {D = mpP } and deg(D) = mp. p∈X P L an invertible OX -module, s ∈ Γ(U, L ), U ⊆ X, then (s) = P ∈X vP (s)P gives Pic(X) ' C`(X) by L 7→ (s), s ∈ Γ(U, L ). The inverse map takes D to P OX (D) by Γ(U, OX (D)) = {f ∈ k(X)|vP (f) ≥ −mp} where D = mP P ∗ Fact: s ∈ k(X) , then s ∈ Γ(U, OX ), deg(s) = 0. So deg : C`(X) → Z is well-deﬁned. ωX = ΩX is the canonical sheaf. i −1 i 1−i ∗ Serre Duality: H (X, L ⊗ ωX ) = ExtX (L , ωX ) = H (X, L ) . 0 Deﬁnition 9.2 (Genus). The geometric genus is pg(X) = dimk H (X, ωX ). dim X The arithmetic genus is pa(X) = (−1) (χ(OX ) − 1).

1 When X is a curve, pa(X) = dimk H (X, OX ). By Serre Duality, we have:

Proposition 9.1. For a curve X, pa(X) = pg(X).

0 −1 1 ∗ Proof. Set i = 0 then Serre Duality says that H (X, L ⊗ ωX ) = H (X, L ) . 0 1 ∗ Then if L = OX , we get H (X, ωX ) = H (X, OX ) .

As ωX ∈ Pic(X), it corresponds to a divisor K = KX ∈ C`(X) the Canonical Divisor, with ωX = OX (KX ). 0 If D ∈ Div(X), we can deﬁne `(D) = dimk H (X, OX (D)). So, for example, `(KX ) = g, the genus of X. Lemma 9.2. 1. `(D) 6= 0 ⇒ deg(D) ≥ 0.

2. If `(D) 6= 0 and deg(D) = 0, then D ∼ 0 (ie, D = 0 in C`(X)) P Proof. 1. ∃0 6= s ∈ Γ(X, OX (D)), and (s) = P ∈X vP (s)P and so deg(D) = deg(s) ≥ 0. 2. If deg(D) = 0, then (s) = 0, so then D ∼ 0.

Theorem 9.3 (Riemann-Roch). Let D ∈ Div(X). Then

`(D) − `(K − D) = deg(D) + 1 − g

0 −1 1 Proof. `(K − D) = dimk H (X, OX (D) ⊗ ωX ) = dimk H (X, OX (D)). So 0 1 `(D) − `(K − D) = dimk H (X, OX (D)) − dim H (X, OX (D)) = χ(OX (D)). If D = 0, then `(0) − `(K) = deg(0) + 1 − g, that is, 1 − g = 1 − g. ETS that RR holds for D iﬀ RR true for D + P (P a point).

118 Let k(P ) = OX /IP and note that IP = OX (−P ). So we get 0 → OX (−P ) → OX → k(P ) → 0 ses. Tensor with OX (D + P ) and get 0 → OX (D) → OX (D + P ) → k(P ) → 0 ses. As Euler Characteristic is additive, χ(OX (D + P )) = χ(OX (D)) + 1. Examples: 1. g − 1 = `(K) − `(K − K) = deg(K) + 1 − g, so deg(K) = 2g − 2. 2. deg(D) > 2g − 2, then `(K − D) = 0, so `(D) = deg(D) + 1 − g. 3. Assume g(X) = 0. Then deg(K) = −2. Let P 6= Q ∈ X, D = P − Q ∈ Div(X). `(D) = `(D)−`(K−D) = 0+1−0 = 1. So then D ∼ 0, so P ∼ Q. Thus, X = P1 from last semester because f ∈ k(X) with (f) = P − Q gives a map f : X → P1, which turns out to be an isomorphism. Fact: A morphism f : X → Y of curves is ﬁnite or constant. Deﬁnition 9.3 (Degree of a Map). deg(f) = [k(X): k(Y )].

If P ∈ X and Q = f(P ) ∈ Y with Y a nonsingular curve, then mQ = (t) ⊆ ∗ OY,Q and f (t) ∈ OX,P . ∗ Define eP = vP (f (t)). This number is called the ramification index of f at P . If eP = 1, then f is unramified at P . If eP > 1, then f is ramified at P . In this case, Q is a branch point. A ramification point is called tame if char(k) = 0 or if char(k) 6 |eP . If it is not tame, then it is wild. ∗ ∗ P Recall: f : Div(Y ) → Div(X) by f (Q) = P ∈f −1(Q) eP P . This map in fact gives a map f ∗ :C`(Y ) → C`(X). This is becuase D ∈ Div(Y ), then ∗ ∗ f (OY (D)) ' OX (f D) Definition 9.4 (Separable Map). f : X → Y is separable if it is finite and f ∗ : k(Y ) ⊆ k(X) is a separable extension of fields. ∗ Proposition 9.4. f : X → Y is a separable map of curves, then 0 → f ΩY → ΩX → ΩX/Y → 0 is ses. ∗ Proof. We already know everything except that the map f ΩY → ΩX is injective. Note that these are both invertible OX -modules on an irreducible variety. ETS that the map is nonzero.

Let P0 ∈ X be the generic point. It is enough to show that (ΩX/Y )P0 = 0. This stalk is equal to Ωk[X]/k[Y ] localized at 0. Taking modules of diﬀeren- tials commutes with localization, do we get Ωk(X)/k(Y ) = 0 as the extension is separable. Let f : X → Y be ﬁnite morphism of curves. P ∈ X has image Q = f(P ) ∈ Y . mP = (u) ⊆ OX,P and similarly mQ = (t) ⊆ OY,Q. Then du ∈ ΩX,P is a ∗ ∗ generator, and similarly, dt ∈ ΩY,Q is a generator. f (dt) ∈ (f ΩY )P ⊆ ΩX,P . ∗ So ∃h ∈ OX,P with f (dt) = hdu. We will set the notation dt/du = h. This just means that f ∗(dt) = dt/du · du ∈ Ω(X,P ).

119 Proposition 9.5. Let f : X → Y be separable map of curves.

1. supp(ΩX/Y ) = {P ∈ X|eP > 1}, which is a ﬁnite set.

2. lengthOX,P (ΩX/Y,P ) = vP (dt/du).

3. If f is tamely ramiﬁed at P , then length(ΩX/Y,P ) = ep − 1. If f is wildly ramiﬁed, then length(ΩX/Y,P ) > ep − 1.

∗ ∗ Proof. 1. ΩX/Y,P = 0 iff ΩX,P generated by f (dt) iff f t generates mP ⊆ OX,P iff eP = 1.

∗ ∗ 2. (ΩX/Y )P ' Ω(X,P )/f ΩY,Q = ΩX,P /(f (dt)) ' OX,P /(dt/du).

∗ e 3. Set e = eP and t = f (t). Then t = au with a ∈ OX,P a unit. Then e e−1 dt = dau + aeu du. If the ramiﬁcation is tame, then e 6= 0 ∈ OX,P , so, as da = hdu for some h, vP (dt/du) = e − 1. e e If wild, then e = 0 ∈ OX,P , so dt = u da, so vP (dt/du) = vP (u h) >≥ e > e − 1.

Assume from now: f : X → Y separable finite morphism of curves. P Definition 9.5 (Ramification Divisor). R = P ∈X length(ΩX/Y,P )P ∈ Div(X). ∗ Proposition 9.6. KX ∼ f KY + R.

∗ −1 ∗ Proof. s ∈ Γ(V, ΩY ), then KY = (s) ∈ Div(Y ). f (s) ∈ Γ(f (V ), f ΩY ) ⊆ −1 ∗ Γ(f (V ), ΩX ). KX = (f s) ∈ Div(X). ∗ Let (KX ; P ) denote the coefficient of P in KX . Then (KX ; P ) = lengthOX,P (ΩX,P /(f s)) = ∗ ∗ ∗ length(f ΩY,P /(f s)) + length(ΩX/Y,P ) = (f KY ; P ) + (R; P ). Theorem 9.7 (Hurwitz Theorem). 2g(X) − 2 = n(2g(Y ) − 2) + deg(R) where n is deg(f). Example: f : X → Y any finite morphism of curves, then g(X) ≥ g(Y ). 1 Why? Because g(X) = g(Y ) + (n − 1)(g(Y ) − 1) + 2 deg(R). The ramification divisor has even degree. Theorem 9.8 (Lyroth’s Theorem). If k ⊆ L ⊆ k(t) and L is a field, then k = L or L = k(u) for some u ∈ L.

Proof. k 6= L, then k has transcendence degree 1, so let X = CL. Then L → k(t) gives a map f : P1 → X a ﬁnite map of curves. As g(P1) ≥ g(X) and g(P1) = 0, g(X) = 0, so X ' P1. Thus, L ' k(u) for some u ∈ L.

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