ALGEBRAIC GEOMETRY NOTES 1. Conventions and Notation Fix A

E. FRIEDLANDER J. WARNER

1. Conventions and Notation Fix a ﬁeld k. At times we will require k to be algebraically closed, have a certain characteristic or cardinality, or some combination of these. AN and PN are aﬃne and projective spaces in N variables over k. That is, AN is the set of N-tuples of elements of k, and PN N+1 is the set of equivalence classes of A − 0 under the relation (a0, . . . , aN ) ∼ (b0, . . . , bN ) if and only if there exists c ∈ k with cai = bi for i = 0,...,N.

2. Monday, January 14th Theorem 2.1 (Bezout). Let X and Y be curves in P2 of degrees d and e respectively. Let {P1,...,Pn} be the points of intersection of X and Y . Then we have: n X (1) i(X,Y ; Pi) = de i=1 3. Wednesday, January 16th - Projective Geometry, Algebraic Varieties, and Regular Functions We ﬁrst introduce the notion of projective space required to make Bezout’s Theorem work. Notice in A2 we can have two lines that don’t intersect, violating (1). The idea is that these two lines intersect at a ”point at inﬁnity”. Projective space includes these points.

3.1. The Projective Plane. Define the projective plane, P2, as a set, as equivalence classes 3 of A under the relation (a0, a1, a2) ∼ (b0, b1, b2) if and only if there exists c ∈ k with cai = bi for i = 0, 1, 2. It is standard to write [a0 : a1 : a2] for the equivalence class represented by 3 (a0, a1, a2) ∈ A . As a set, the projective plane can be written in many ways as the union of three copies of 2 the affine plane, A . Here is the standard way of doing so. Let Ui = {[a0 : a1 : a2] | ai 6= 0} for i = 0, 1, 2, and notice the following bijections of sets 2 2 2 A → U0 A → U1 A → U2 (a, b) 7→ [1 : a : b](a, b) 7→ [a : 1 : b](a, b) 7→ [a : b : 1] 3 The above maps are bijections because each element of Ui has a unique representative in A whose ith coordinate is 1. Also, they cover P2 because every element of P2 has some nonzero 2 coordinate. The points of P − U0 are all of the form [0 : a : b], and we consider these to be the points at infinity for the affine plane U0. Notice that these points are in one-to-one correspondence with P1, the projective line. P1 in turn can be covered by two copies of A1, Date: Spring 2013. 1 the affine line, each of whose complement in P1 is just a single point at infinity P0 = A0. Hence, we can decompose the projective plane as follows: 2 2 1 2 1 0 2 1 0 P = A t P = A t A t P = A t A t A

3.2. Algebraic Varieties. Let f1, . . . , fm ∈ k[x1, . . . , xN ], and denote by Z(f1, . . . , fm) the N N set of common zeroes of the fi in A . That is, Z(f1, . . . , fm) = {a ∈ A | fi(a) = 0 ∀i}. Such a set is called an affine algebraic variety. Notice that if all fi vanish at a, then so does any linear combination of the fi so that any element in the ideal generated by the fi also vanishes at a. The Hilbert Basis Theorem says that k[x1, . . . , xN ] is Noetherian, so all ideals are finitely generated. Thus, there is no loss of generality in starting with finitely many polynomial equations. That is, any collection of polynomial equations generates an ideal that is also generated by finitely many polynomial equations, and the common zeroes of each will be the same.

Example 3.1. Consider the polynomial f(x, y) = x − y2 ∈ C[x, y]. Z(f) = {(a, b) ∈ C2 | a = b2}. The real points of the variety can be drawn as a sideways parabola. To each affine algebraic variety we can assign a ring, called the coordinate ring. To do so, let X be an affine algebraic variety, and define the ideal I(X) = {f ∈ k[x1, . . . , xN ] | f(P ) = 0 ∀P ∈ X}. Then the coordinate ring is the quotient k[x1, . . . , xN ]/I(X). We would like to define a similar concept for projective space. A problem arises due to the fact that a point in projective space is represented by many points in affine space, that is, the zeroes of a polynomial aren’t well defined in projective space. For example, consider the polynomial 1 + x + y ∈ C[x, y], and notice that it’s values at the points (1, −2) and (2, −4) are 0 and −1 respectively. But [1 : −2] = [2 : −4] in P1 so that the polynomial doesn’t well-define a function to C. We can solve this problem by only considering homogeneous polynomials, that is, polynomials all of whose monomials have the same degree. Homogeneous polynomials of degree 0 are just constants. Consider the following proposition. Proposition 3.2. In projective space, the zero locus of a polynomial is well-defined if and only if the polynomial is homogeneous. Proof. Let f be a polynomial. What we mean by a well-defined zero locus in projective space is if any representative of a point is zero in f, then all such representatives are zero. Suppose f is homogeneous of degree d. Then the equation f(ca) = cdf(a) for all c 6= 0, a ∈ AN+1, shows that the zero locus is well-defined. Next, suppose f is not homogeneous. Write f as the sum of its homogeneous components, P f = fi. Choose some a with f(a) = 0 but fi(a) 6= 0 for some i (if no such a existed, then Z(f) ⊂ Z(fi) for all i. Applying the Nullstellenzats shows then that f must be homogeneous). Then notice that f(λa) is a non-constant polynomial in λ, and thus is not zero for all λ. Hence, the zero locus of f is not well-defined if f is not homogeneous. The above proposition shows that the zero set of a homogeneous polynomial is well-defined N in P . Now let F1,...,Fm ∈ k[x0, . . . , xN ] be a collection of homogeneous polynomials, not necessarily of the same degree, and denote by Z(F1,...,Fm) the set of common zeroes of N N the Fi in P . That is, Z(F1,...,Fm) = {a ∈ P | Fi(a) = 0 ∀i}. Such a set is called a 2 projective algebraic variety. Again by Hilbert’s Basis theorem, we only require finitely many polynomial equations to define varieties. Similar to the case of affine varieties, each projective variety has a homogeneous coordinate ring. Let Y be a projective variety, and define the ideal I(Y ) generated by all homogeneous polynomials vanishing at all points of Y . Since I(Y ) is generated by homogeneous elements, it is a homogeneous ideal. The desired homogeneous coordinate ring is the quotient k[x0, . . . , xN ]/I(Y ). 3.3. Regular Functions. The concept of a regular functions on a variety differs slightly depending on if the variety is projective or affine. We describe both here. Definition 3.3. Let X ⊂ AN be an affine variety, and let P be a point on X. A function f : X → k is said to be regular at P if there is a Zariski open neighborhood U ⊂ X containing P , and polynomials g, h ∈ k[x1, . . . , xn] with h nowhere zero on U such that f = g/h on U. If f is regular at all points of some open subset U, we say f is a regular function on U. The regular functions on any open set U form a ring under pointwise multiplication and addition. Denote this ring by OX (U). Example 3.4. Consider the variety X = Z(f) from example 3.1. Any polynomial g ∈ C[x, y] determines a regular function on X, and another polynomial g0 determines the same regular function if and only if g0 − g ∈ I(X). This shows that the coordinate ring of X embeds into the ring of regular functions. In fact, in the affine case, the map is surjective and the coordinate ring of X is isomorphic to the ring of regular functions. That is, any function that is locally a rational function everywhere on X is actually a polynomial. Notice we have yet to use anything specific about our variety X. Consider the rational function f(x) = 1/(x − 1). This defines a regular function at all points P on X except when x = 1. This corresponds to the two real points (1, 1) and (1, −1) on X. To see this, notice that the Zariski open set U = X T Z(x − 1)C contains all such points, and the function f is a ratio of two polynomials, of which the denominator doesn’t vanish on X. Above we constructed a regular function on an open set U by taking a rational function and discarding the points at which the denominator vanishes. In this way, rational functions yield regular functions on open sets U ⊂ X, but sometimes different rational functions yield the same regular function, as the following example illustrates. 4 Example 3.5. Consider the hypersurface X = Z(x1x2 − x3x4) inside A . Let U be the Zariski open set of X given by the complement of the equation x2x3 = 0. Then the rational functions x1/x3 and x4/x2 determine the same regular function on U. It is also true that some regular functions are not equal to any rational function. Let U 0 be the Zariski open set of X defined by the complement of the closed set Z(x2, x3). Then on 0 U , x2 and x3 are never simultaneously 0. Hence we have a regular function f which equals x1/x3 when x3 6= 0 and x4/x2 when x2 6= 0. Notice f is well-defined because when both x3 and x2 are non-zero, the two expressions are equal. There is a case in which the regular functions on an open set of an affine variety are given exactly by rational functions defined on the open set. Theorem 3.6. Let X be an affine variety, and A(X) its coordinate ring. If A(X) is a unique factorization domain, then for any open U ⊂ X, the regular functions on U are exactly all rational functions g/h with h nowhere vanishing on U. 3 Notice that the coordinate ring in example 3.5 is not a unique factorization domain, as the element x1x2 = x3x4 has two distinct factorizations into irreducible elements. Next let’s define regular functions on projective varieties.

Definition 3.7. Let Y ⊂ PN be a projective variety, and let P be a point on Y . A function f : Y → k is said to be regular at P if there is a Zariski open neighborhood U ⊂ Y containing P , and homogeneous polynomials g, h ∈ k[x0, . . . , xn] of the same degree with h nowhere zero on U such that f = g/h on U. Notice that since g and h have the same degree, the function g/h is well defined on projective space. If f is regular at all points of some open subset U, we say f is a regular function on U. As with affine varieties, the regular functions on an open neighborhood U ⊂ Y form a ring under pointwise addition and multiplication. We denote this ring OY (U). The following is a characterization of regular functions on complements of hypersurfaces of projective space.

Proposition 3.8. Let F be a homogeneous polynomial in k[x0, . . . , xN ]. Consider the open set U = PN − Z(F ). Then 1 G N OP (U) = k x0, . . . , xN , = i | G homogeneous, deg(G) = ideg(F ) F 0 F The above proposition allows us to prove the following two exercises.

Exercise 3.9. Prove that the ring of regular functions on PN is k. Proof. PN is an open set deﬁned by the complement of Z(1) = ∅. Hence, the above proposition gives: N G O N ( ) = | G homogeneous, deg(G) = 0 = k P P 1i because the homogeneous polynomials of degree 0 are the constants. N x1 xN Exercise 3.10. Prove that the ring of regular functions of U0 ⊂ P is k ,..., . x0 x0 N Proof. U0 is the complement in P to the hypersurface Z(x0). A homogeneous polynomial of degree i has the form

X i0 iN G(x0, . . . , xN ) = ai0...iN x0 . . . xN i0+...+iN =i i Dividing G by x0 and dividing the power of x0 among the variables according to their powers yields the expression:

i0 i1 iN i1 iN G(x0, . . . , xN ) X x0 x1 xN X x1 xN = ai ...i ... = ai ...i ... i 0 N i0 i1 iN 0 N x0 x x x x0 x0 i0+...+iN =i 0 0 0 i0+...+iN =i x1 xN Any element of k ,..., can be written in this form. x0 x0

Notice that the regular functions on U0 in exercise 3.10 are isomorphic as a ring to the N N coordinate ring of A , as is suggested by our identification of U0 with A . 4 4. Friday, January 18th - Hilbert’s Basis Theorem, Zariski Topology, and Ideals and Subsets We would like to establish a one-to-one correspondence between certain ideals of our ring N A = k[x1, . . . , xN ] and certain subsets of A . We begin by establishing a theorem of Hilbert concerning finite generation, then we discuss a natural topology to place on AN , and finally we discuss how to move between ideals and subsets. The Hilbert Basis Theorem allows us to work in complete generality and only consider finitely generated ideals of A = k[x1, . . . , xN ]. Definition 4.1. A commutative ring R is called Noetherian if any of the three following equivalent conditions are satisfied. (1) Any ascending chain of ideals stabilizes. That is if

I0 ⊂ I1 ⊂ ... ⊂ In ⊂ ...

is a chain of ideals, then there exists some m such that Im = Im+j for all j = 1, 2,.... (2) Any nonempty collection of ideals contains a maximal element. That is, if {Iα}α∈J is a collection of ideals for some nonempty indexing set J, then there exists α0 such that Iα ⊂ Iα0 for all α ∈ J. (3) All ideals in R are finitely generated. That is for any ideal I, there exists elements {r1, . . . , rn} such that I = (r1, . . . , rn). Before stating the theorem, let’s discuss some typical examples of Noetherian rings. Example 4.2. (1) Any PID is Noetherian by the third condition above. (2) Any field is a PID because fields contain no nontrivial ideals. (3) If R is Noetherian, and I is any ideal, then R/I is Noetherian by the correspondence theorem. Theorem 4.3 (Hilbert’s Basis Theorem). If R is Noetherian, then R[x] is also Noetherian.

Proof. Let a ⊂ R[x] be an ideal, and suppose it is not finitely generated. Let f0 be a polynomial of smallest degree in a. Inductively define bn := (f0, . . . , fn) and choose fn+1 in a − bn of smallest degree. Notice since all bn are finitely generated, and a is not, this process does not end. Consider the infinite series of polynomials f0, f1, f2,... and let r0, r1, r2,... be the series in R of leading terms. Since R is Noetherian, there must exist some m with rm ∈ (r0, . . . , rm−1 for if not, the chain of ideals

(r0) ⊂ (r0, r1) ⊂ (r0, r1, r2) ⊂ ... P would never stabilize. Hence, let ui ∈ R such that rm = uiri, and deﬁne the polynomial P deg(fm)−deg(fi) g = fm − uifix . Notice by construction, deg(g) < deg(fm), contradicting minimality as g 6∈ bm. Hence, a must be ﬁnitely generated.

Corollary 4.4. The ring A = k[x1, . . . , xn] is Noetherian. Proof. Apply the theorem n times. Let’s see why this result is useful. First a proposition. 5 Proposition 4.5. Let T ⊂ A be a collection of polynomials, and let I = (T ). Then Z({T }) = Z(I). Hence, we may always assume that a variety is defined by an ideal. Proof. Clearly if a point disappears on a set of polynomials, it disappears on any subset, so that we have the inclusion Z({T }) ⊃ Z(I). Now, since I consists of all R-linear combinations of elements of T , if a point disappears on all elements of T it also disappears on all elements of I. Hilbert’s theorem thus allows us to assume that any variety is always defined by finitely many equations. That is, suppose T is a collection of polynomials, and let I = (T ). Since I is finitely generated, let T 0 be a finite generating set. Then Z({T }) = Z(I) = Z({T 0}). Now let us concern ourselves with establishing a topology on AN . S Lemma 4.6. Let I1,I2 and Iα for α ∈ J be ideals in A. Then Z(I1) Z(I2) = Z(I1I2) = T Z(I1 I2) and ! ! \ [ X Z(Iα) = Z Iα = Z Iα α∈J α∈J α∈J T Proof. Let’s consider the first set of equalities. By the inclusions I1I2 ⊂ I1 I2 ⊂ I1 and S T the observation in the proof of proposition 4.5, it follows that Z(I1) Z(I2) ⊂ Z(I1 I2) ⊂ Z(I1I2). Next, suppose A disappears on I1I2, but not on I1. Let f be any polynomial in I2 and g be any in I1 with g(a) 6= 0. Then since f(a)g(a) = 0 is must follow that a disappears S on f, so that a ∈ Z(I2). This provides the containment Z(I1I2) ⊂ Z(I1) Z(I2) which finishes the proof of the first set of equalities. For the second set of equalities, the last equality follows from proposition 4.5 because the sum of ideals is the ideal generated by the union. The other equality follows by definition. The lemma shows that sets of the form Z(I) for some ideal I satisfy the properties of closed sets of a topology.

Definition 4.7. The Zariski topology on AN is the unique topology whose closed sets are the zero sets of a collection of polynomials in A. Example 4.8. (1) Since any nonzero polynomial in k[x] has only finitely many roots, the closed sets of A1 are empty, finite, or all of A1. This is also known as the cofinite topology. (2) Since plane curves always intersect in finitely many points, the closed sets of A2 are empty, finite collections of points, plane curves along with finitely many points, or all of A2. We can define a Zariski topology on PN in much the same way, taking the closed sets to be zero loci of finitely many homogeneous polynomial equations. We end this section with a discussion of the certain ideals in A and closed subsets of AN .

Definition√ 4.9. Let R be a commutative ring, with I ⊂ R an ideal. Define the radical of I, n + denoted I, to be the collection√ of all elements in r ∈ R such that r√∈ I for some n ∈ Z . Notice that by commutativity, I is an ideal. We say I is radical if I = I. 6 4.1. Ideals and Subsets. Let S ⊂ AN be any subset. Define I(S) := {f ∈ A | f(s) = 0, ∀s ∈ S}. Notice that I(S) is a radical ideal (this follows because fields have no nontrivial nilpotent elements). With this definition, we can now move back and forth between subsets of AN and ideals in A, as illustrated below.

AN A

S −→ I(S)

Z(T ) ←− T The following exercise shows that if S happens to be closed, then S is recovered by Z(I(S)). This establishes half of the desired correspondence between radical ideals and closed subsets. Exercise 4.10. Let T be any subset of A. Then Z(I(Z(T ))) = Z(T ). Proof. Let f ∈ T , and let a ∈ Z(T ). By deﬁnition, f(a) = 0, so that f ∈ I(Z(T )). Thus T ⊂ I(Z(T )), so that by the observation in the proof of proposition 4.5, we have Z(I(Z(T ))) ⊂ Z(T ). Next, suppose that a ∈ Z(T ), ie, that f(a) = 0 for all f ∈ T , and let g ∈ I(Z(T )). By deﬁnition, g(a) = 0, so that a ∈ Z(I(Z(T ))) completing the other inclusion.

5. Wednesday, January 23rd - Hilbert’s Nullstellenzats Equipped with a topology, it is often convenient to have the following more general deﬁ- nition of an aﬃne algebraic variety.

Definition 5.1. An affine algebraic variety over k is a locally closed subset of some AN . A locally closed subset is the intersection of an open set and a closed set. In other words, an affine algebraic variety is an open set of a closed set with the subspace topology. Definition 5.2. An irreducible closed subset Y inside a topological space X is one which satisfies the following condition: if Y = Y1 ∪ Y2 with Y1,Y2 ⊂ Y closed, then either Y = Y1 or Y = Y2. It turns out that under the correspondence of radical ideals and closed subsets mentioned above, that irreducible subsets correspond to prime ideals. Proposition 5.3. Y is irreducible if and only if I(Y ) is prime. Proof. Suppose Y is irreducible, and let fg ∈ I(Y ). Then Y ⊂ Z(fg) = Z(f) ∪ Z(g). Then we have Y = (Z(f) ∩ Y ) ∪ (Z(g) ∩ Y ). By irreducibility, it follows that either Y ⊂ Z(f) in which case f ∈ I(Y ) or Y ⊂ Z(g) in which case g ∈ I(Y ). Next, suppose I(Y ) is prime, and let Y = Y1 ∪ Y2. So I(Y ) = I(Y1 ∪ Y2) = I(Y1) ∩ I(Y2). Suppose I(Y ) 6= I(Y1) and choose f ∈ I(Y1) − I(Y ). Then for any g ∈ I(Y2), we must have fg ∈ I(Y ) so that g ∈ I(Y ) so that I(Y ) = I(Y2). Then it follows that Y = Y2 and Y is irreducible. 7 Definition 5.4. The dimension of a closed set Y inside AN or PN is the length,d, of the longest chain of irreducible closed subsets in Y

Y0 ⊂ Y1 ⊂ ... ⊂ Yd The Krull dimension of a commutative ring is the length of the longest ascending chain of prime ideals. For example, the chain of closed sets {point ⊂ line ⊂ plane} shows that planes have dimension 2, as is expected. If Y = Z(I) ⊂ AN has coordinate ring A(Y ), the correspondence between prime ideals and irreducible closed subsets along with the correspondence theorem shows that the dimension of Y is the Krull dimension of A(Y ). 5.1. Hilbert’s Nullstellensatz. We present the Hilbert’s Nullstellensatz in three separate theorems. Theorem 5.5 (Theorem 1). Let k be a field and let K/k be a finitely generated commutative k-algebra. If K is also a field, then K is algebraic over k. In particular, K is finite dimensional over k. Proof. We will use the fact that a finitely generated, algebraic field extension is finite. To see this, let K = ha1, . . . , ani as a k-algebra. Since a1 is algebraic, k[a1] is a finite field extension of degree equal to that of the minimal polynomial of a1, say r1. Then, a2 is algebraic over k[a1], so k[a1, a2] is finite over k[a1] say of degree r2. In general, ai is algebraic over k[a1, . . . , ai−1], so k[a1, . . . , ai−1, ai] is finite over k[a1, . . . , ai−1] of some degree ri. It follows that K = k[a1, . . . , an] has degree r1r2 . . . rn over k. Now let’s begin with the proof. Suppose that K is not algebraic, and initially, let’s suppose the transcendence degree of K is 1. This means there is some x ∈ K with K/k(x) algebraic. By the previous discussion, this means K/k(x) is finite. Let {e1, . . . , el} be a basis of K over k(x), and write out the multiplication table of K: X aijk eiej = ek bijk k where the a’s and b’s are in k[x]. Now, let f1, . . . , fm be any elements of K, and let A be the k-algebra they generate. Expand each fi in terms of the ej: X cij f = e i d j j ij

and notice that every element in A is some linear combination of the ei over k(x), where the coefficients have denominators that are products of the a’s, b’s, c’s, and d’s. Since k[x] has infinitely many irreducibles, there must exist some element in K that is not in A. Hence K is not finitely generated, a contradiction. Now, suppose the transcendence degree of K over k is greater than 1. Since k(a1, . . . , an) = 0 0 K, there must exist some k = k(x1, . . . , xr) with K/k of transcendence degree 1. Then we can appeal to the first case. Corollary 5.6. If k is algebraically closed, then there does not exist a non-trivial finitely generated field extensions of k. 8 Proof. Suppose K/k is a finitely generated field extension of k. Then by the theorem, K is algebraic over k. Let a ∈ K and let f ∈ k[x] be the minimal polynomial of a. Since k is algebraically closed, f is linear, so that a ∈ k. Hence K = k. Theorem 5.7 (Theorem 2 - Weak Nullstellensatz). If k is algebraically closed, then each N maximal ideal of k[x1, . . . , xN ] has the form (x1−a1, . . . , xN −aN ) for some (a1, . . . , aN ) ∈ A .

Proof. Let R = k[x1, . . . , xN ] and let m be a maximal ideal of R. The quotient map R → R/m is a surjective map onto a ﬁeld, with kernel equal to m. Since R is ﬁnitely generated over k, so is R/m so that R/m ∼= k by the corollary. Any map from R into k is evaluation at some (a1, . . . , aN ), and the result follows. 6. Friday, January 25th - Hilbert’s Nullstellensatz and the Segre Embedding The following is a corollary to the Weak Nullstellensatz.

Corollary 6.1. Let S ⊂ k[x1, . . . , xN ]. If Z(S) = ∅, then hSi = k[x1, . . . , xN ].

Proof. Suppose hSi= 6 k[x1, . . . , xN ]. Then there exists some maximal ideal ma containing N hSi for some a ∈ A . This is a contradiction however, because then a ∈ Z(S) = ∅. Here is the usual statement of Hilbert’s Nullstellensatz.

Theorem 6.2 (Theorem 3 - Hilbert’s Nullstellensatz)√. Assume k is algebraically closed and d let I ⊂ k[x1, . . . , xN ]. If g vanishes√ on Z(I), then g ∈ I, ie, there exists d such that g ∈ I. In other words, I(Z(I)) = I

Proof. Let I = (f1, . . . , fm), and deﬁne J = (f1, . . . , fm, xN+1g − 1) ⊂ k[x1, . . . , xN+1]. It N+1 follows that Z(J) = ∅ in A , because if (a1, . . . , aN ) ∈ Z(I), then aN+1g(a1, . . . , aN+1−1 = 1 for all aN+1. By the above corollary, we know that k[x1, . . . , xN+1] = J. Next consider the map

φ : k[x1, . . . , xN+1] → k[x1, . . . , xN , frac1g] ( xi for i = 1,...,N xi 7→ 1 g for i = N + 1 P Now write 1 = (pifi) + pm+1(xN+1g − 1) ∈ k[x1, . . . , xN+1] and apply φ to obtain m X 1 1 1 = p (x , . . . , x , f ∈ k[x , . . . , x , i 1 N g i 1 N g i=1 Clearing denominators gives the desired result. Let’s use Hilbert’s Nullstellensatz to show that the nilradical of a finitely generated commutative k algebra over an algebraically closed field is equal to the intersection of all maximal ideals. That is, the Jacobsen radical is equal to the nilradical in the case of finite generation.

Proposition 6.3. Let I ⊂ k[x1, . . . , xN ] be any ideal. Then \ √ \ p = I = m I⊂p I⊂m p prime m maximal 9 Proof. Since maximal ideals are prime, we have: \ \ p ⊂ m I⊂p I⊂m p prime m maximal

Also, if f ∈ m for all maximal√m containing I, then f vanishes at all points of Z(I). Hilbert’s nullstullensatz then says f ∈ I. To finish the proof, we need to see that if f r ∈ I for some r, then f ∈ p for all prime ideals r p containing I. Since f ∈ I ∈ p, and since p is radical, it follows that f ∈ p. Corollary 6.4. Let R be a finitely generated commutative k-algebra, with k an algebraically ∼ closed field, that is, R = k[x1, . . . , xN ]/I. Then the nilradical of R, the intersection of all prime ideals, is equal to the Jacobson radical of R, the intersection of all maximal ideals.

Proof. Prime (maximal) ideals in R correspond to prime (maximal) ideals in k[x1, . . . , xN ] √containing I. By the proposition, both the nilradical and the Jacobson radical equal the ideal I/I in R.

For any commutative ring with identity R, define Specm(R) = {maximal ideals in R} and Spec(R) = {prime ideals in R}. Notice that Specm(R) ⊂ Spec (R). We can place a topology on Spec(R) by requiring closed sets to be of the form Z(I) = {p prime | I ⊂ p} for some ideal I ⊂ R. Corollary 6.5. Let R be a finitely generated commutative k-algebra over, k an algebraically closed field. Then Specm(R) is dense in Spec(R). T Proof. Suppose Specm(R) ⊂ Z(I) for some I. This implies that I ⊂ m for all maximal T ideals m. By the corollary, we have that I ⊂ p, so that Z(I) = Spec(R). 6.1. The Segre Embedding. The Segre embedding allows us to view the product of two projective spaces as a projective variety in some suitably large projective space. Define a map ψ : Ps × Pt → Pst+s+t by

(a0, . . . , as) × (b0, . . . , bt) 7→ (. . . , aibj,...) where we order the coordinates on the right hand side lexicographically. This map is well deﬁned on projective space because scalar multiples on the left correspond to scalar multiples on the right. Now, let k[zij] where i = 1, . . . , s and j = 1, . . . , t, and deﬁne

I = ker{ϕ : k[zij] → k[x0, . . . , xs, y0, . . . , yt]}

where ϕ(zij) = xiyj. Then the Im(ψ) = Z(I). To see this, note the equality

f ◦ ψ[(a0, . . . , as) × (b0, . . . , bt)] = ϕ(f)(a0, . . . , as, b0, . . . , bs)

for any f ∈ k[zij]. In particular, if f ∈ I, then the above equation shows that Im(φ) ⊂ Z(I). Next, suppose (. . . , cij,...) ∈ Z(I). Not all homogeneous coordinates can be zero, so we may assume that c00 6= 0. Choose any a0, b0 such that a0b0 = c00. Next, for i = 1, . . . , s and j = 1, . . . , t, deﬁne ci0 c0j ai := bj := b0 a0 10 Then since the cij satisfy the polynomial zi0z0j − z00zij we have ci0 c0j ci0c0j c00cij aibj = = = = cij b0 a0 c00 c00

so that ψ[(a0, . . . , as) × (b0, . . . , bt)] = (. . . , cij,...) and Z(I) ⊂ Im(ψ).

7. Monday, January 28th - The Category of Affine Varieties We would like to discuss the category of affine varieties free from any particular embedding in affine space. Concentrating on the ring of regular functions allows us to do so. This is made possible by the fact that if two affine varieties over an algebraically closed over k are ’isomorphic’, a concept to be defined soon, then the their rings of regular functions are also isomorphic as k-algebras. Hence, we have the following categorical definition of affine varieties. Definition 7.1. Define the category of affine varieties over k to be one in which the objects are finitely generated, reduced, commutative k-algebras, and morphisms are k-algebra homomorphisms.

Notice that an object in this category A is isomorphic to k[x1, . . . , xN ]/I for some N and some I, but we may also have an isomorphism of A with k[y1, . . . , yM ]/J. Refusing to pin down such a representation of A, we have extracted ourselves from committing to a certain embedding of the aﬃne variety whose ring of regular functions is A.

8. Wednesday, January 30th - Varieties, Regular Functions, and Contravariance Below are the four concepts of ”variety” discussed so far. (1) Affine - a closed set in some AN . (2) Quasi-affine - a locally closed set in some AN , ie, an open set of an affine variety. (3) Projective - a closed set in some PN . (4) Quasi-projective - a locally closed set in some PN , ie, an open set of a projective variety. The concept of a quasi-projective variety is the most general of the four. In fact, it en- compasses the other three. A projective variety is an open set of itself, and is thus quasi- N N projective. Suppose X ⊂ A is an affine variety, view X ⊂ U0 ⊂ P as the image of N the homeomorphism ϕ0, and consider its projective closure, X in P . Then I claim that X = X ∩ U0, so that X is quasi-projective. It is clear that X ⊂ X ∩ U0. Now, suppose x ∈ X ∩ U0, and write x = [1 : x1 : ... : xN ]. Let f ∈ I where X = Z(I) ∈ k[y1, . . . , yN ]. Then since 0 = β(f)(1, x1, . . . , xN ) = f(x1, . . . , xN ) (cf. Hartshorne, Exercise I.2.9), it follows that x ∈ X. To show that any quasi-affine variety is a quasi-projective variety, consider the following lemma. Lemma 8.1. Let X be a topological space, U ⊂ X open, and Y ⊂ U locally closed in U. Then Y is locally closed in X. Proof. Since Y is locally closed in U, there exist open V and closed E in X such that Y = (U ∩ V ) ∩ E. It follows that Y is locally closed in X. 11 N In the above lemma, let X = P ,U = U0, and Y be the quasi-affine variety in question. Then we see that Y is a quasi-projective variety. We would now like to establish a contravariance between varieties and their rings of regular functions. We being with the affine case. First we define a morphism of varieties. Definition 8.2. Let X and Y be quasi-projective varieties. A map ϕ : X → Y is a morphism if it is continuous, and for any open V ⊂ Y and any regular function f : V → k, the map f ◦ ϕ : ϕ−1(V ) → k is regular.

Proposition 8.3. Let X and Y be aﬃne varieties, with A(X) = k[x1, . . . , xN ]/I and A(Y ) = ∗ k[y1, . . . , yM ]/J. Then any map of k-algebras f : A(Y ) → A(X) induces a morphism of varieties f : X → Y

Proof. We begin by defining f. Let a ∈ X, and well-define a map evala : A(X) → k by ∗ evaluation at a. Consider the map of k-algebras evala ◦ f : A(Y ) → k which must be evaluation at some point b ∈ Y . What is b in terms of f ∗ and a? Well, let g ∈ A(Y ). Since ∗ ∗ ∗ ∗ f is a map of k-algebras, we have f (g(y1, . . . , yM )) = g(f (y1), . . . , f (ym)). Hence ∗ ∗ ∗ evala ◦ f (g(y1, . . . , yM )) = g(f (y1)(a), . . . , f (ym)(a)) ∗ ∗ so we see that b = (f (y1)(a), . . . , f (ym)(a)). From this reasoning, we also have the formula (2) f ∗ ◦ g = g ◦ f as functions X → k. To see this is a morphism, we must verify that f is continuous, and that it pulls back regular functions. First, suppose that V ⊂ Y is closed. That is, V = Z(J 0) for some J 0 ⊃ J. Let I0 = hf ∗(J 0)i. Then f −1(V ) = Z(I0) is closed. To see this, note that the equation f ∗ ◦ g(a) = g ◦ f(a), which holds for all g ∈ A(Y ) implies that f(a) ∈ V if and only if a ∈ Z(I0). Before we show that f preserves regular functions, we introduce the notion of a basic open set. Definition 8.4. Let X be any affine variety, with ring of regular functions A(X). For any element f ∈ A(X), let D(f) = {x ∈ X | f(x) 6= 0} = Z(f)c ∩ X. Such open sets in X are called basic open sets. c Notice that basic open sets form a basis for the Zariski topology, as if V = Z(f1, . . . , fr) c c is any open set, then V = Z(f1) ∪ ... ∪ Z(fr) . The ring of regular functions on D(f) ⊂ X 2 is A(X)f , the localization of A(X) at the multiplicative set {1, f, f ,...}. Now, let V ⊂ Y be open, let g : V → k be regular, and take any x ∈ X with f(x) ∈ V . Since the basic open sets form a basis for the Zariski topology, we can find a basic open set 0 0 D(q) such that f(x) ∈ D(q) ⊂ V . Then g|V 0 is regular on V , so on V we have the equality p g = q for p, q ∈ A(Y ). Then on f −1(V ) we have p ◦ f f ∗(p) g ◦ f = = q ◦ f f ∗(q) Taking any representative of f ∗(p), f ∗(q) in A(Y ) yields the desired local form of g ◦ f as a rational function in a neighborhood of x. 12 9. Friday, February 1st - Presheaves and Sheaves Given any topological space X, we can construct a category whose objects are open sets, and whose morphisms are inclusions. Call this category Top(X). Definition 9.1. Let X be a topological space. A presheaf of sets P on X is a contravariant functor P : Top(X) → Sets The elements of P (U) are called the sections of U. The definition thus assigns to every open set U a set P (U) and to every inclusion V ⊂ U a set map ρUV : P (U) → P (V ). We can similarly define the notion of a presheaf of groups, rings, or k-algebras by allowing the target category to be groups, rings, or k-algebras, respectively. Notice then that the maps ρUV must then respect the structure of the desired category. The maps ρUV are called restriction maps.

Example 9.2. Let X be a variety, and deﬁne a presheaf of k-algebras OX by OX (U) := {regular functions on U} and ρUV (f) = f|V for all regular f : U → k. Motivated by this example, we often write f|V for ρUV (f) for any presheaf P . Also motivated by this example, we often think of presheaves as functions on open sets.

Example 9.3. Let X be a variety, and define a presheaf of k-algebras RX by RX (U) := {rational functions on U} and ρUV (f) = f|V for all rational f : U → k. This example, although similar to the previous example, differs in a very significant way in that the definition of a regular function is a local notion. We’ll see the implications of this difference when we discuss sheaves. Suppose P is a presheaf in some category, and suppose we have f ∈ P (U) and g ∈ P (V ) such that f|U∩V = g|U∩V . Does there necessarily exist some element in P (U ∪ V ) that restricts to f and g appropriately, that is, since f and g agree on the intersection of U and V , do they necessarily patch together to give a function on U ∪ V ? If we require nothing of our functions except that they are set maps, then the answer is yes. The following example shows that arbitrary presheafs don’t necessarily patch together like this.

Example 9.4. Let X be the affine variety from example 3.5, and consider the presheaf RX from example 9.3. Then, the rational functions x1/x3 and x4/x2 agree on the open set where neither denominator vanishes, but there is no rational function defined on the larger open set where at least one denominator doesn’t vanish that restricts to both. Hence, we see that the presheaf RX doesn’t patch together. A presheaf that patches together in a unique way is called a sheaf. Let’s make this notion more precise. Definition 9.5. A sheaf of sets (groups, rings, k-algebras) on a topological space X, F , is a presheaf of sets (groups, rings, k-algebras) on X that satisfies the following ”Sheaf axiom:” For any open U ⊂ X, and any open covering Ui ⊂ U of U, the following sequence is exact Q Q ρ ρU U ∩U −ρU U ∩U i UUi Y i,j i i j j i j Y 0 → F (U) −−−−→ F (Ui) −−−−−−−−−−−−−−→ F (Ui ∩ Uj) i i,j where in the case of sets, what we mean by the 0 is that the first map is injective. 13 Let’s examine what the exactness of the above sequence entails.

(1) First, by exactness in the middle term we see that if we have fi ∈ F (Ui) such that

fi|Ui∩Uj = fj|Ui∩Uj for all i, j, then there exists some f ∈ F (U) such that f|Ui = fi for all i. This is exactly what we mean by patching together. (2) Second, the injectivity of the ﬁrst arrow says that if f and g are in F (U) such that

f|Ui = g|Ui for all i, then f = g. This is the uniqueness of patching, that is, if functions fi patch together, then they do so uniquely.

The local nature of the deﬁnition of a regular function shows that OX is a sheaf of k- algebras for any variety X. Thus, for any variety X, we often consider the object (X, OX ) of the variety paired with its sheaf of regular functions. Revisiting the concept of a morphism of varieties, we now see why we required the condition that the morphism, in some sense, ’preserve’ regular functions. Suppose f : X → Y is a morphism of varieties. The condition on regular functions says that f induces a map −1 0 OY (V ) → OX (f (V )) for all V given by g → g ◦ f. For any open sets V ⊂ V ⊂ Y , and any regular function g : V → k, consider the following commutative diagram:

f g f −1(V ) / V / k O O

? f ? f −1(V 0) / V 0 The commutativity of the above diagram implies the commutativity of the following diagram:

−1 OY (V ) / OX (f (V ))

ρVV 0 ρf−1(V )f−1(V 0) 0 −1 0 OY (V ) / OX (f (V )) This leads us to the following deﬁnition of a morphism of presheaves. Deﬁnition 9.6. Let P and P 0 be two presheaves on a topological space X.A morphism 0 0 of presheaves ϕ : P → P is a map ϕU : P (U) → P (U) such that for any open V ⊂ U the following diagram commutes.

ϕ P (U) U / P 0(U)

ρUV ρUV ϕ P (V ) V / P 0(V ) A morphism of sheaves ϕ : F → F 0 is a morphism of F and F 0 considered as presheaves.

10. Monday, February 4th - Irreducibility and Stalks Assuming irreducibility of a variety leads to nice results. Before describing the benefits of considering irreducible varieties, let’s look at an example of a reducible variety. 14 Example 10.1. Let X = Z(xy) ⊂ A2. Notice that (xy) is a radical ideal, but not prime, so that X is reducible, with ring of regular functions A(X) ∼= k[x, y]/(xy). X = Z(x) ∪ Z(y) is a decomposition of X as the union of two closed sets. Let’s consider the basic open sets D(x) and D(y). A basic open set of an affine variety is again affine, with coordinate ∼ ∼ ring given by localization. Hence we have OX (D(x)) = (k[x, y]/(xy))x = k[x, 1/x], and ∼ ∼ OX (D(y)) = (k[x, y]/(xy))y = k[y, 1/y]. What are the restriction maps ρXD(x) and ρXD(y)? Answer for D(x):

ρXD(x) : k[x, y]/(xy) → k[x, 1/x] x 7→ x y 7→ 0 The answer for D(Y ) is similar. This is intuitive, as the regular function y on X restricts to 0 on D(x) because y = 0 when x 6= 0. Deﬁnition 10.2. Let F be a presheaf on a topological space X. For any p ∈ X deﬁne the stalk of F at p to be: F := lim F (U) p −→ p∈U

Fp inherits the structure of the category of the sheaf F . Hence an element in Fp is represented by some section s ∈ F (U), and is equal to another representative t ∈ F (V ) if there is some open W ⊂ U ∩ V with s|W = t|W . An element of Fp is called a germ. Let’s consider the concept of a stalk applied to the sheaf of regular functions of an aﬃne variety.

Proposition 10.3. Let X be an aﬃne variety, with sheaf of regular functions OX , and ring ∼ of regular functions A(X). Let p ∈ X. Then OX,p = A(X)mp . If A(X) is a domain, ie, if X is irreducible, then ∼ \ ∼ \ A(X) = OX,p = A(X)mp p∈X mp where the intersection is considered inside of Frac(A(X)).

Proof. Since mp is the kernel of the map A(X) → k which takes a polynomial in A(X) and evaluates at p, it follows that g ∈ mp if and only if g(p) = 0. Any element in A(X)mp can be represented by f/g where g(p) 6= 0, and thus defines an element in OX,p (f/g ∈ OX (D(g)), and p ∈ D(g)). We thus have a map of k-algebras A(X)mp → OX,p. The map is surjective because any regular function h in a neighborhood U of p is equal to a rational function f/g in a smaller neighborhood U 0 of p, with g not vanishing on U 0 and so not on p. Thus the element in OX,p represented by h is the image of f/g ∈ A(X)mp . To see that the map is injective, suppose that f/g and f 0/g0 determine the same element in 0 0 OX,p. This means there is some open W on which f/g = f /g . By shrinking W if necessary, we may assume W = D(q) is a basic open set. Then notice that q(fg0 − f 0g) = 0 on X so 0 0 that f/g = f /g ∈ A(X)mp (Notice that q vanishes outside of W , and q∈ / mp). The final statement follows from the fact that a domain is equal to the intersection of its localizations at maximal ideals inside of its field of fractions. Let R be a domain. Since T R ⊂ Rm for all m it follows that R ⊂ Rm. Next, suppose that f/g ∈ Rm for all m, ie, that −1 g∈ / m for all m. Then g must be a unit, so that f/g = fg ∈ R. 15 The final statement of the above proposition shows why irreducible varieties are often easier to work with. Below are two more reasons why irreducible varieties are nice. (1) Any two non-empty open sets of an irreducible topological space intersect. In fact, any non-empty open set is dense.

(2) If U2 ⊂ U1 ⊂ X with U2 non-empty, then the restriction map ρU1U2 : OX (U1) → OX (U2) is injective. This follows from the continuity of regular functions and (1). Condition (1) listed above allows us to form a directed set out of the collection of nonempty open sets of an irreducible variety. Hence, we can take a direct limit over all open sets, not just those containing a particular point p as in the definition of a stalk. This leads us to the following proposition. Proposition 10.4. Let X be an irreducible variety. Then lim O (U) ∼ Frac(A(X)) −→ X = U Proof. Notice that irreducibility implies that A(X) is a domain, so the fraction field is well- defined. The proof of this proposition is similar to that of proposition 10.3. For the sake of notation, let the direct limit in the statement be K(X). Given f/g ∈ Frac(A(X)), we can define the element of K(X) represented by f/g ∈ OX (D(g)). This defines a map of k-algebras. Surjectivity follows from the local nature of a regular function as in proposition 10.3. For injectivity, notice that if f/g = f 0/g0 on some nonempty open U, then fg0 = f 0g on U so that by (2) above we have fg0 = f 0g on all of X. Thus f/g = f 0/g0 as elements in Frac(A(X)). 11. Wednesday, February 6th - Linear Algebraic Groups and Grassmannians

11.1. Linear Algebraic Groups. We’re used to thinking of GLn as the set of invertible n × n matrices. GLn has more structure. It is an aﬃne variety with a group structure which turns its ring of regular functions into a Hopf algebra. Let’s investigate this below. n2 First, notice that GLn is an open subset inside of A deﬁned by the complement of the polynomial equation det(xij) = 0. Thus, GLn is actually the basic open set D(det(xij)). Hence, its ring of regular functions is given by localizing at det(xij): ∼ 1 ∼ k[GLn] = k[xij][ ] = k[xij, t]/(det(xij)t − 1) det(xij)

The second isomorphism actually shows that we can consider GLn as a closed subset of n2+1 A : it is defined by the vanishing of the equation det(xij)t − 1. Next, to see that the group law on GLn is actually a map of affine varieties, we need to define the product of two affine varieties.

Definition 11.1. Suppose X ⊂ As and Y ⊂ At are two affine varieties with ideals I ⊂ k[x1, . . . , xs] and J ⊂ k[y1, . . . , yt]. Then define the product of X and Y to be the variety s+t X × Y ⊂ A given by the set of points (a1, . . . , as+t) such that (a1, . . . , as) ∈ X and (as+1, . . . , as+t) ∈ Y . Notice this a variety defined by the vanishing of the I and J viewed as subsets of k[x1, . . . , xs, y1, . . . , yt]. ∼ Proposition 11.2. k[X × Y ] = k[X]⊗kk[Y ] 16 Proof. A regular function f on X×Y is represented by a polynomial in k[x1, . . . , xs, y1, . . . , yt] P and thus we can write f = higi where hi ∈ k[x1, . . . , xs] and gi ∈ k[y1, . . . , yt] for all i. P The desired isomorphism is then ϕ : f 7→ fi ⊗ gi. To see this map is independent of the choice of representing polynomial, suppose f = f 0 0 on X ×Y , that is, suppose that f −f = ai+bj for some i ∈ I, j ∈ J and a, b ∈ k[x1, . . . , yt]. Applying ϕ to both sides of the equation yields

X X 0 0 gkhk − glhl = ϕ(a)(i ⊗ 1) + ϕ(b)(1 ⊗ j) = 0 k l so that ϕ is well-deﬁned.

Next, consider the map f : GLn × GLn → GLn given by matrix multiplication. Is this a morphism of varieties? Instead of checking continuity and the condition on regular functions, let’s see if we can deﬁne the corresponding map on regular functions, f ∗. Consider the map:

∗ f : k[xij, t]/(det(xij)t − 1) / k[xij, t]/(det(xij)t − 1) ⊗ k[xij, t]/(det(xij)t − 1)

/ P xij k xik ⊗ xkj

t / t ⊗ t This map is well-defined by the property that the determinant of a product of matrices is the product of the determinants. Now, with this f ∗ in hand, lets determine the induced map as defined in proposition 8.3. Let (aij) and (bij) be two invertible matrices. Then ∗ X ∼ X f(aij, bij) = (f (xij)(aij, bij)) = ( aik ⊗ bkj) = ( aikbkj) k k so that f ∗ induces matrix multiplication. Hence, we see that matrix multiplication is a map of affine varieties.

Exercise 11.3. Show that the inverse map ι : GLn → GLn is also a map of affine varieties. Proof. The proof is similar to the argument for the multiplication map. Since the inverse of a given matrix can be defined at each entry by a cofactor divided by the determinant, we can give the following map ι∗ on global regular functions: ∗ ι : k[GLn] → k[GLn] xij 7→ tcji t 7→ det(xij) where cji is the cofactor of xji in the matrix (xij). The map is well-defined because the determinant of the inverse is the inverse of the determinant.

Notice that the product on GLn gives a map ∆ : k[GLn] → k[GLn]⊗k[GLn] and the inverse gives a map S : k[GLn] → k[GLn]. These maps (coproduct and antipode, respectively) give k[GLn] the structure of a Hopf algebra. The maps satisfy the duals of commutative diagrams satisifed by a group. If we deﬁne a group variety to be a group object in the category of varieties, ie, a variety G with morphisms G × G → G, G → G, and ? → G (here ? is any 17 singleton) satisfying the usual diagrams deﬁning a group, then the ring of regular functions k[G] is a Hopf algebra. We thus have an antiequivalence of categories

Group varieties Finitely generated, reduced, commutative ↔ over k Hopf algebras over k Under this equivalence, if the multiplication map on the group variety is commutative, then the Hopf algebra is cocommutative, ie, we can twist the tensor product in the coproduct without changing its value. Hence, under the above equivalence we also have

Abelian group Finitely generated, reduced, commutative, ↔ varieties over k cocommutative Hopf algebras over k Let’s see this antiequivalence with another example.

Example 11.4. Consider U3 ⊂ GL3, the collection of unipotent, upper-triangular matrices. Every matrix in U3 has the form:  1 a b   0 1 c  0 0 1

This is a subgroup of GL3 because it is closed under multiplication and inverses, and it contains the identity. It is also a closed subvariety as it is defined by the extra equations (x11 − 1, x22 − 1, x33 − 1, x21, x31, x32). Notice that since there are no conditions on x12, x13, 3 and x23, as an affine variety U3 is isomorphic to A . Hence, their rings of regular functions are ∼ also isomorphic as k-algebras. Thus k[U3] = k[x, y, z]. What is the Hopf algebra structure of k[U3] that corresponds to matrix multiplication? It could be calculated by defining a map on k[U3] to be the same as that on k[GL3] defined above, where k[U3] is considered a quotient of k[GL3]. This map will be well defined because U3 is closed under multiplication. More explicitly, the group structure on U3 is given by  1 a b   1 a0 b0   1 a + a0 b + b0 + ac0   0 1 c   0 1 c0  =  0 1 c + c0  0 0 1 0 0 1 0 0 1 The coproduct corresponding to this multiplication is

∆ : k[x12, x13, x23] → k[x12, x13, x23] ⊗ k[x12, x13, x23] x12 7→ 1 ⊗ x12 + x12 ⊗ 1 x13 7→ 1 ⊗ x13 + x13 ⊗ 1 + x12 ⊗ x23 x23 7→ 1 ⊗ x23 + x23 ⊗ 1 Notice this coproduct is not cocommutative, and it arises because the group is not abelian. Another classical example.

1 Example 11.5. Let Ga be the affine line A with group structure given by addition. Then under this group operation, the coproduct on k[Ga] = k[x] sends x to 1 ⊗ x + x ⊗ 1. This can be extended to any number of copies of the affine line. In the case of A3, we see that the Hopf algebra obtained in this way is different than the one in the previous case, although the varieties are isomorphic (the difference is that they are not isomorphic as group varieties). 18 11.2. Grassmanians. Let V = k4 be a four dimensional vector space with a fixed basis. Consider Grass(2,V ), the set of all subspaces W of V of dimension 3. Any such W is spanned by two linearly independent vectors, which form a 2 × 4 matrix of rank 2. Map this matrix to P5 by taking the determinants of all possible 2 × 2 minors. For example, map the matrix:   v1 w1  v2 w2     v3 w3  v4 w4 to the homogeneous coordinates:

[v1w2 − w1v2 : v1w3 − w1v3 : v1w4 − w1v4 : v2w3 − w2v3 : v2w4 − w2v4 : v3w4 − w3v4] Notice we have ordered the minors lexicographically. The coordinates obtained in this way are called Plücker coordinates. This is a well-defined embedding because two matrices give the same homogeneous coordinates if and only if they differ by a 2 × 2 invertible matrix, which gives a linear automorphism of W . It can be shown that the image of this embedding 5 is equal to the closed set Z(x1,2x3,4 − x1,4x2,3 + x1,3x2,4) in P where we have labeled the coordinates according to the rows defining the minor. Hence, Grass(2, 4) ,→ P5 is a closed embedding as a projective variety. Notice that the Grass(2, 4)∩Ui is the open set of Grass(2, 4) where the determinant of the ith minor doesn’t vanish. Hence we may multiply by a 2×2 invertible matrix to assume that this minor is the 2 × 2 identity. There will then be four remaining entries, and any choice ∼ 4 of these entries defines a distinct W in V , so that Grass(2, 4) ∩ Ui = A . This is illustrated below for U0  1 0   0 1     v3 w3  v4 w4 Hence, we’ve shown that the Grassmannian can be covered by open sets of dimension 4. If we consider the general case of r-dimensional spaces in a vector space V of dimension n, then the argument shows that Grass(r, V ) can be covered by open sets of dimension r(n − r). If we assume that the Grassmannian is irreducible, then its dimension is r(n − r), because the dimension of an irreducible variety is equal to the dimension of any nonempty open subset.

12. Friday, February 8th - Grassmannians, Projections, and Planar Curves 12.1. Grassmannians and Flag Varieties. The Pl¨ucker coordinates of a Grassmannian depend on the choice of basis for V and the choice of order for the minors. Here we discuss a choice free embedding of the Grassmannian into projective space, given by exterior powers. Given a vector space V of dimension n, and a subspace W of dimension r, consider the r r r r subspace Λ W ⊂ Λ V . Notice that the dimension of Λ W is r = 1, so that we have a map: Grass(r, V ) / P(ΛrV ) W / ΛrW The map is injective, and depends on no choices. 19 Suppose we consider the product of two Grassmanians, Grass(r, V )×Grass(r0,V 0). Choose a basis for V and V 0 as well as an ordering of minors to obtain embeddings:

n n0 ( )−1 0 0 ( 0 )−1 Grass(r, V ) ,→ P r Grass(r ,V ) ,→ P r The Segre embedding then shows we have a map:

n n0 n n0 0 0 ( )−1 ( 0 )−1 ( )( 0 )−1 Grass(r, V ) × Grass(r ,V ) ,→ P r × P r ,→ P r r If we give the product in the middle the product topology, then the ﬁrst map is a closed embedding, so the composition is also (the Segre embedding is a closed map: it’s a homeomorphism onto its image). Exercise 12.1. Let V be an n-dimensional vector space. Let r ≤ r0 ≤ n. Show that the subset: {(U, W ) | U ⊂ W } ⊂ Grass(r, V ) × Grass(r0,V ) is a closed subvariety.

Proof. The result of the exercise holds more generally. That is, suppose V is of dimension n and choose 0 < i1 < i2 < . . . < it < n. Then the subset

{(U1,U2,...,Ut) | U1 ⊂ U2 ⊂ ... ⊂ Ut} ⊂ Grass(i1,V ) × Grass(i2,V ) × ... × Grass(it,V ) is a closed subvariety. Varieties of this form are called ﬂag varieties.

N 12.2. Projections. Suppose N > m and consider L = Z(x0, . . . , xm) ⊂ P . Notice that ∼ N−m−1 L = P , so that dim(L) = N − m − 1. Deﬁne the following projection: PN − L / Pm

[a0 : a1 : ... : aN ] / [a0 : a1 : ... : am] The map is well defined because we’ve ensured the first m coordinates are not all zero. This map is an example of a projection. Notice it is surjective, but not injective (infinitely many points map to each point in the image). L is called the center of the projection. The above projection is a morphism of varieties, where PN − L is a quasi-projective variety (needs proof). In general, let X ⊂ PN be any subvariety of dimension m. We’d like to, in some sense, ’embed’ X into Pm. To do so, find linear L ⊂ PN of dimension N − (m + 1) disjoint from X (Why is this possible?), and consider the composition: X / PN − L / Pm

12.3. Planar Curves. A planar curve of degree d is a hypersurface in P2 deﬁned by a homogeneous polynomial of degree d.

2 2 2 Example 12.2. X = Z(x0 + x1 + x2) is a planar curve of degree 2. The twisted cubic curve in P3 is not a planar curve (needs proof). We will build up to the proof of the following theorem: 20 Theorem 12.3. There is a natural 1-1 correspondence between isomorphism classes of smooth projective curves and ﬁeld extensions of k which are ﬁnitely generated of transcendence degree 1.

Under this correspondence, P1 will correspond to the field of rational functions of 1 vari- able, k(t). Definition 12.4. Let X and Y be irreducible varieties. A rational map f : X → Y is an equivalence class of pairs hU, fU i of nonempty open sets U ⊂ X and morphisms fU : U → Y , where we identify two equivalence classes if fU = fV on U ∩ V . We say the rational map is dominant if fU (U) is dense in Y for some (and hence any (needs proof)) U. We can think of a rational map as the germ of a morphism. The relation defined is an equivalence relation because if two morphisms agree on some nonempty open set, they agree on the whole variety (this is used to show transitivity of the relation). Proposition 12.5. There is a bijection between dominant rational maps f : X → Y and k-algebra homomorphisms k(Y ) → k(X).

13. Monday, February 11th - Sheafification, Etale´ Space and Adjunction For this lecture, if F is a presheaf of sets on some topological space X, and U ⊂ X is open with some section s ∈ F(U), then for p ∈ U, let sp be the image of s in the stalk of Fp. Let’s consider a motivating example of a sheaf. Example 13.1. Let f : Y → X be continuous map of topological spaces. Deﬁne a presheaf of sets on X as follows: if U ⊂ X is open, let:

F(U) := {continuous sections of f over U} = {s : U → Y | s is continuous, f ◦ s = idU }

If V ⊂ U, define the restriction map ρUV to just be restriction of functions. It follows that F is a presheaf. Is it a sheaf? First, let Ui be a covering of U, and suppose we have sections si ∈ F(Ui) for all i that agree on all intersections. We can then glue these functions to define a function s : U → Y . We must show that s is a continuous section. To see that s is continuous, let V be open in Y , and consider some element p such that −1 −1 −1 s(p) ∈ V . p ∈ Ui for some i, so p ∈ si (V ) ∈ s (V ). It follows that s (V ) is open, so s is continuous. Also, f ◦ s(p) = f ◦ si(p) = p, so that s is a section. To see that the gluing is unique, we only need to say that functions are determined by their values on points. 13.1. Sheafification. We would like to define a functor from the category of presheaves of sets on a topological space X to the category of sheaves of sets on X. Suppose F is a presheaf of sets on X and define for any open set U in X:     1. s(p) ∈ Fp, ∀p ∈ U  +  G   F (U) = s : U → Fp | 2. ∀p ∈ U, ∃V and t ∈ F(V ) s.t.  p∈X    p ∈ V ⊂ U and s(q) = tq ∀q ∈ V  Again, we define restriction maps to be restriction of functions. The definition is very messy, so let’s try to make some sense of it. The intuitive idea is that elements of F +(U) look like 21 elements of F(U) locally. That is, if s ∈ F +(U), then for any p ∈ U, we can find some element in the presheaf t ∈ F(U) whose image in the stalk of F is equal to the ’image’ of s in the stalk of F for all points in some small neighborhood of p. + Is F a sheaf? Suppose Ui covers U, and consider sections si ∈ F(Ui) that agree on intersections. Glue the functions to obtain a map s → tFp. We must show that s satisfies the two conditions listed above. Let p ∈ U, so that p ∈ Ui for some i. Then s(p) = si(p) ∈ Fp so that the first condition is satisfied. Next, find V ⊂ Ui containing p and t ∈ F(V ) satisfying the second condition. Then, since V ⊂ U, and s = si on V , it follows that s satisfies the second condition as well. Uniqueness of gluing again follows from the fact that functions are defined pointwise. Notice we have a natural map of presheaves θ : F → F + which takes a section s on U to the function s defined by s(p) = sp for all p ∈ U. Suppose that G is another sheaf of sets on X, and that ϕ : F → G is a map of presheaves. I claim there is a unique morphism of sheaves ψ : F + → G such that F θ / F +

ψ ϕ G + p commutes. Indeed let s ∈ F (U), and for each p ∈ U, find Vp and t ∈ F(Vp) agreeing with p 0 p p0 s on stalks, ie, s(q) = tq for all q ∈ Vp. Consider two such points p and p so that tq = tq for p p0 all q ∈ Vp ∩Vp0 . Hence t and t restrict to the same section on Vp ∩Vp0 . Consider the images p rp = ϕ(t ) ∈ G(Vp). Since ϕ is a morphism of presheaves, it commutes with restriction, so that the rp agree on the intersections of the open cover Vp of U. Since G is a sheaf, we can glue the rp together uniquely to obtain r ∈ G(U). Define ψ(s) = r. Is ψ the desired map? First, if we start with s ∈ F(U), then considering s ∈ F +(U), in p the notation above we can choose Vp = U and t = s for all p. Then ψ ◦ θ(s) = ψ(s) = ϕ(s) so that the diagram commutes. The map is unique because for any section s on U, there is an open covering of U such that s looks like t on each open piece. Then, since ψ commutes with restrictions, the value of s is forced because commutativity forces the values of each t. Notice that F + is unique by a standard universal diagram argument. Also, with use of the universal property, we can see that if F is a sheaf, then F + ∼= F. Consider the following pathological example Example 13.2. Let X be any nonempty topological space. Define a presheaf of abelian groups on X by F(X) = Z, F(U) = 0 for all proper open U, and restriction maps are forced to be trivial. Then notice that all stalks are also trivial, so that a nonzero global section of X is zero pointwise. The example shows why it is nice to work with sheaves: sections of sheaves behave like functions; they are determined by their pointwise behavior.

13.2. Etale´ Space of a Presheaf. Let F be a presheaf of sets on a topological space X. Deﬁne sp´e(F) as a set to be the disjoint union of the stalks of F. Notice we have a surjection

π : spé(F) X s ∈ Fp 7→ p 22 For any open U ⊂ X, and s ∈ F(U) we can define a section of π over U by s : U → spé(F) p 7→ sp Give spé(F) the strongest topology making s continuous for all s ∈ F(U), ie, throw in as many open sets as possible while maintaining the continuity of these maps. The topology can be defined by letting V ⊂spé(F) be open if and only if s−1(V ) is open for all open W ⊂ U and all s ∈ F(W ). Lemma 13.3. Let W be open in U, and let s ∈ F(W ). Then s : W → spé(F) is a homeomorphism onto its image. Proof. Since we are considering disjoint unions of stalks, the map s is one-to-one. By definition of the topology on spé(F), s is continuous. We want to show it is also an open map. Suppose that V ⊂ W is open. To see that s(V ) is open, consider any W 0 and any t ∈ F(W 0). Then we have: −1 0 t (s(V )) = {p ∈ W ∩ V | tp = sp} 0 If this set is empty, then we are fine. If not, take any p ∈ W ∩ V with tp = sp. This means there is an open neighborhood W 00 ∈ W 0 ∩ V containing p where t = s on W 00. Hence we have p ∈ W 00 ⊂ t−1(s(V )), so that s(V ) is open. This lemma will help solve the following exercise. Exercise 13.4. Show that π is a continuous map in this topology, and that F + is isomorphic to the sheaf of continuous sections of π as defined in example 13.1 Proof. Let U be open in X. Then −1 G π (U) = Fp p∈U Thus, for any W open in X and any s ∈ F(W ), we have s−1(π−1(U)) = W ∩ U is open, so that π is continuous. Next, we wish to show that for any open U in X, we have F +(U) = {continuous sections of π over U}

Notice that s is a section over U if and only if s(p) ∈ Fp for all p, which is the first condition in defining the sheaf associated with a presheaf. Next, we will show that the second condition of sheafification is equivalent to continuity of s. Suppose that the second condition is satisfied. Let V be any open set in spé(F), and choose p ∈ U such that s(p) ∈ V . Find W and t satisfying the condition. Since V is open, it follows that t−1(V ) is open. It contains p and its image under s is in V . This shows that s is continuous. Next, suppose that s is continuous. Consider p ∈ U, and s(p) ∈ Fp. Find open W containing p and t ∈ W with tp = s(p). By the lemma, t(W ) is open in spé(F), so that −1 V = s (t(W )) is open in U. Notice that if q ∈ V if and only if s(q) = tq. Hence, s satisfies the second condition of sheafification. 23 13.3. Adjunction. Sheafification defines a functor, α, from the category of presheaves on X to the category of sheaves on X. The functor on morphisms of presheaves is given by the universal property. The forgetful functor, β, defines a functor from the category of sheaves on X to the category of presheaves on X. The forgetful functor on morphisms is identity, because morphisms of sheaves are morphisms of presheaves. Notice that α ◦ β is the identity on Shvs(X) because the sheafification of a sheaf is itself. The universal property shows there is a set bijection

HomPreshvs(F, β(G)) ↔ HomShvs(α(F), G) for all presheaves F and sheaves G. This set bijection is functorial in the sense that if we have a map f : F → F 0 of presheaves and a map g : G → G0 of sheaves, then the following diagram commutes:

0 0 HomP reshvs(F , β(G)) o / HomShvs(α(F ), G)

0 0 HomP reshvs(F, β(G )) o / HomShvs(α(F), G ) Let θ : F → α(F) and θ0 : F 0 → α(F 0) be the natural maps in the universal property. Let ϕ : F 0 → β(G) be a morphism of presheaves, and let ψ : α(F 0) → G be the corresponding map of sheaves. The commutativity of the above diagram is equivalent to the equality: g ◦ ψ ◦ α(f) ◦ θ = α(β(g)) ◦ ϕ ◦ f This equality follows by the fact that α(β(g)) = g and the following two commutative diagrams: 0 F θ / α(F) F 0 θ / α(F 0)

f α(f) ψ ϕ " F 0 / α(F 0) β(G) θ0 The above discussion shows that the functors α and β are adjoint. Let’s examine adjunction in the case of morphisms between the categories of sheaves on topological spaces X and Y induced by a continuous map f : X → Y .

−1 13.4. Adjunction of f and f∗. Let X and Y be topological spaces, and let f : X → Y be a continuous map. Then f deﬁnes two functors on the categories of sheaves of X and Y as deﬁned below. Let F, F 0 ∈ Shvs(X), and G, G0 ∈ Shvs(Y). Let U ⊂ X and V ⊂ Y be open subsets.

−1 −1 Definition 13.5. Define f∗F ∈ Shvs(Y) by f∗F(V ) = F(f (V )) and f G ∈ Shvs(X) as the sheaf associated to the presheaf: U → lim G(V ) −→ V ⊃f(U) 0 0 Suppose α : F → F is a map of sheaves on X. Then define f∗(α): f∗F → f∗F by −1 0 −1 f∗(α)V :F(f (V )) →F (f (V ))

s 7→αf −1(V )(s) 24 Also, if β : G → G0 is a map of sheaves on Y . Then deﬁne f −1(β): f −1G → f −1G0 on U by lim G(V ) / lim G0(V ) −→V ⊃f(U) −→V ⊃f(U) ht, V i / hβ(t),V i

−1 This completes the deﬁnition of the functors f and f∗. Notice that U ⊂ f −1(f(U)) and f(f −1(V )) ⊂ V , so that the following natural morphisms of presheaves are well-deﬁned:

−1 ϕ ψ −1 f f∗F(U) / F(U) G(V ) / f∗f G(V )

−1 ht, f (V )i / t|U s / hs, V i

Next, suppose that α : f −1G → F is a morphism of sheaves on X. Notice then that f∗(α) ◦ ψ : G → f∗F is a morphism of sheaves on Y . Conversely, if β : G → f∗F is a morphism of sheaves on Y , then ϕ ◦ f −1(β): f −1G → F is a morphism of sheaves on X. These two operations are inverses of one another (needs to be shown), so we have a set bijection: −1 HomShvs(X)(f G, F) ↔ HomShvs(Y)(G, f∗F) −1 Furthermore, as above, the bijection is functorial (needs to be shown), so that f and f∗ are adjoint functors.

14. Wednesday, February 13th - Rational Maps and Blow Ups Suppose ϕ : X → Y and ψ : Y → Z are two dominant rational maps. Let’s define their composition ψ ◦ ϕ : X → Z. Choose any representatives hU, ϕU i and hV, ψV i, and let −1 their composition be the rational map represented by hϕU (V ), ψV ◦ ϕU i. To see that this 0 0 is well-defined, suppose hU , ϕU 0 i and hV , ψV 0 i are two other representatives, and consider −1 0 −1 −1 0 0 0 hϕU 0 (V ), ψV 0 ◦ ϕU 0 i. Notice that ϕU (V ) ∩ ϕU 0 (V ) ⊂ U ∩ U and ϕU = ϕU 0 on U ∩ U . −1 −1 0 −1 0 −1 0 0 Hence, ϕU (V ) ∩ ϕU 0 (V ) = ϕU (V ∩ V ) = ϕU 0 (V ∩ V ). Then, since ψV = ψV 0 on V ∩ V , −1 −1 0 it follows that ψV ◦ ϕU = ψV 0 ◦ ϕU 0 on ϕU (V ) ∩ ϕU 0 (V ). (Why is the image dense?) Thus, we can consider the category of varieties and dominant rational maps. An isomorphism in this category is called a birational map. If a birational map exists between two varieties, we say they are birationally equivalent. We state the following theorem without proof: Theorem 14.1. X and Y are birationally equivalent if and only if k(X) ∼= k(Y ) as k- algebras if and only if there is some U ⊂ X and some V ⊂ Y that are isomorphic as quasi-affine varieties.

Let’s now deﬁne the blow-up of An at the origin, O = (0, 0,..., 0). Consider the variety n n−1 A × P , with aﬃne coordinates (x1, . . . , xn) and homogeneous coordinates (y1, . . . , yn). Consider the closed set:

X = {(x1, . . . , xn, y1, . . . , yn) | xiyj = xjyi, i, j = 1, 2, . . . , n} 25 There is a natural morphism ϕ : X → An by restricting the projection: X / An × Pn−1

ϕ % An The pair (X, ϕ) is the blowup of An at O. Let’s consider some properties of (X, ϕ). (1) First, the ﬁber of ϕ above a point P consists of only one element if P 6= O, and −1 ∼ n thus, ϕ gives an isomorphism of varieties X − ϕ (O) = A − O. To see this, let n P = (a1, . . . , an) be nonzero in A , so assume ai 6= 0. Next, suppose that P × −1 (b1, . . . , bn) ∈ ϕ (P ) so that for each j we have:

ajbi bi bj = =⇒ (b1, . . . , bn) = (a1, . . . , an) ai ai

Notice bi 6= 0 else all bj = 0 so we may assume bi = ai and thus bj = aj for all j. This shows that ϕ−1(P ) consists of the singleton P × P . We can then well-define the inverse map ϕ−1(P ) = P × P . Both ϕ and ϕ−1 are algebraic maps, so they are morphisms of varieties. (2) The fiber of ϕ above O is a copy of Pn−1. Indeed, ϕ−1(O) consists of all points of the form O × Q with Q ∈ Pn−1 having no restriction. (3) Notice that (2) gives a 1-1 correspondence between points in ϕ−1(O) and Pn−1. Here we make this correspondence more explicit. Any line, L in An through the origin can 1 be parametrized by the equations xi = ait, t ∈ A and not all ai are zero. Consider 0 −1 −1 0 L = ϕ (L − O) ⊂ X − ϕ (O). By (1), L is parametrized by xi = ait and yi = ait, 0 t 6= 0. Since the yi are homogeneous, we take yi = ai. Now we can close L in X by −1 n−1 allowing t to be zero. This closure meets ϕ (O) at the point Q = (a1, . . . , an) ∈ P . (4) X is irreducible. To see this, notice that it contains the subset X − ϕ−1(O) which is irreducible by (1) and dense by (3) (it contains all possible L0 whose closures contain all of ϕ−1(O). Since the closure of an irreducible subset is irreducible, it follows that X is irreducible. The copy of Pn replacing O in X is called the exceptional curve, and is denoted E. In some sense, what we’ve done is replace the origin with a point for each line tangent to An that passes through the origin (which is every line that passes through the origin). Now let’s define the blow-up of any affine variety Y at O.

Definition 14.2. Let Y ⊂ An be a closed subset containing O. Define the blowup of Y at ˜ ˜ −1 n O as the pair (Y , ϕY˜ ) where Y = ϕ (Y − O). Through a linear change of variables of A we can define the blow-up of Y at any point P . Notice that Y and Y˜ are birationally equivalent by theorem 14.1 because ϕ induces an isomorphism between the open sets Y − O and Y˜ − ϕ−1(O). It seems that Y˜ depends on the particular embedding of Y into An, but in fact it does not. Let’s work with a specific example to get a better sense of what is happening.

Example 14.3. Let Y = Z(y2 − x2(x + 1)) ⊂ A2, and let’s blow up Y at O. 26 Let [t : u] be homogeneous coordinates in P1. Then X is defined in A2 × P1 by the single equation xu = ty. Let’s compute ϕ−1(Y ): −1 2 1 2 2 ϕ (Y ) = {(x, y, t, u) ∈ A × P | xu = ty, y = x (x + 1)} First suppose t 6= 0, so that we can assume t = 1 and treat u like an affine coordinate. This leads to the equation: 0 = x2u2 − x2(x + 1) = x2(u2 − x − 1) This leads to two families of points: (0, 0, 1, u) whose closure is E, and (u2 −1, u3 −u, 1, u) which is Y˜ . Supposing t = 0 requires that x = 0 and y = 0, so these points add nothing to Y˜ . Notice that the points in Y˜ lying above the origin are (0, 0, 1, 1) and (0, 0, 1, −1) which correspond to the slopes of the curve defining Y at the origin. In some sense then, we have replaced O with all of the slopes of lines passing through it that are tangent to Y .

15. Friday, February 15th - Smoothness n Suppose Y ⊂ A is an affine variety of dimension r whose ideal is generated by f1, . . . , ft. Let P be a point in Y . We say that Y is nonsingular at P is the rank of the matrix: ∂f i (P ) ∂xj is n − r. Y is nonsingular if it is nonsingular at all points P ∈ Y . The matrix of partial derivatives is called the Jacobian. Let’s give an algebraic definition of the tangent space of an affine variety X at a point x ∈ X. Consider the ideal mx of regular functions which vanish at x. Let’s identify to 2 functions if they agree up to first order, that is, if there difference is in mx. Thus, we are 2 considering the vector space mx/mx. This is called the Zariski cotangent space. It consists of the first order germs of regular functions vanishing at x. 2 ∗ Next we define the Zariski tangent space to be the dual, (mx/mx) . Roughly, the idea is that given a tangent direction on our variety X, we can define a linear functional on 1st order germs of functions vanishing at X by sending them to their derivative in the given direction.

16. Wednesday, February 20th - Smoothness and Dedekind Domains We would like to find the algebraic equivalent to the geometric notion of nonsingular. The correct idea is regularity, which is defined now. Definition 16.1. Let A be a commutative, noetherian, local ring, with maximal ideal m, ∼ 2 and residue field k = A/m. A is said to be regular if dim(A) = dimk(m/m ), where the left dimension is the Krull dimension of the ring A, and the right dimension is as a vector space. The next theorem connects the two notions of regularity and nonsingularity.

Theorem 16.2 (Zariski). Let Y ⊂ An be an affine variety. Then Y is nonsingular at P ∈ Y if and only if OY,P is regular. Before proving the theorem, we will need two lemmas. 27 Lemma 16.3. Let R be a commutative local ring, with maximal ideal m, and let k = R/m be the residue field. Then there is a k-linear isomorphism 2 2 ϕ : m/m → mRm/m Rm a [a] 7→ 1 Proof. ϕ is well-defined because if a − b ∈ m2, then a b a − b − = ∈ m2R 1 1 1 m ϕ is k-linear because for any c + m ∈ k we have

ca + b ca b ϕ((c + m)[a] + [b]) = ϕ([ca] + [b]) = ϕ([ca + b]) = = + 1 1 1 hai b = (c + m) + = (c + m)ϕ([a]) + ϕ([b]) 1 1 ϕ is injective because hai ∈ m2R ⇒ a ∈ m2 1 m Finally, ϕ is surjective because if b 6∈ m implies that b is a unit so we have ab−1 hai ϕ([ab−1]) = = 1 b Lemma 16.4. Let R be a commutative ring. Let m ⊂ R be a maximal ideal, let k = R/m be the residue field, and let I ⊂ m be an ideal. Let m be the maximal ideal in R/I given by m = {a + I | a ∈ m}. Then there is a k-linear isomorphism ϕ : m/(I + m2) → m/m2 [a] 7→ [a + I] Proof. ϕ is well defined and injective because f − g ∈ (I + m2) if and only if f − g = i + m for i ∈ I, m ∈ m2 if and only if f − g + I = m + I if and only if f − g + I ∈ m2. Surjectivity is immediate, and linear is shown as in the previous lemma. We can now prove the theorem. Proof. Let J be the Jacobian of Y at P . We would like to show the following equality: 2 dimk(m/m ) + rank(J) = n 2 With this equality, notice then that if Y is nonsingular at P , then dimk(m/m ) = dim(Y ) = dim(OY,P ), so that OY,P is regular. Similarly if OY,P is regular, then we have 2 n = dimk(m/m ) + rank(J) = dim(OY,P ) + rank(J) = dim(Y ) + rank(J) so that Y is nonsingular at P . n To show the desired equality, first define a k-linear map θ : k[x1, . . . , xn] → k by ∂f ∂f θ(f) = (P ),..., (P ) ∂x1 ∂xn 28 Notice that θ satisfies the formula (?) θ(fg) = θ(f)g(P ) + θ(g)f(P )

Let aP = (x1 − a1, . . . , xn = an) be the maximal ideal associated to P = (a1, . . . , an), and restrict θ to aP . Notice that the restriction is still surjective, as θ(xi − ai) = ei. Also, we will show that ker(θ) = a2 . First, if f ∈ a2 , then (?) shows that θ(f) = 0. Next, write P P P f = hi(xi − ai) ∈ aP and suppose θ(f) = 0. Then for each j = 1, . . . , n we have: X ∂hi 0 = (a − a ) + δ h (P ) = h (P ) ∂x i i ij j j i j 2 2 so that hj ∈ aP for each j and thus f ∈ aP . Hence ker(θ) = aP . We thus have a vector space isomorphism: 0 2 n θ : aP /aP → k [f] 7→ θ(f)

Next, let b be the ideal of Y , and choose generators f1, . . . , ft for b. Notice that the span of the rows of J is contained in θ(b). For the reverse inclusion, use (?). We see then that 0 2 2 0 rank(J) = dim(θ(b)). Restricting θ to the subspace (b + aP )/aP we see that the image of θ 2 2 is the subspace θ(b) so that dim(θ(b)) = dim((b + aP )/aP ). Next, by the lemmas above, we have the following isomorphisms of vector spaces: 2 ∼ 2 ∼ 2 m/m = aP /aP = aP /(b + aP ) Finally, by counting dimensions of vector spaces (subtracting dimensions of quotients) we have: 2 2 2 2 rank(J) = dim(θ(b)) = dim((b + aP )/aP ) = dim(aP /aP ) − dim(aP /(b + aP )) 2 2 2 = dim(aP /aP ) − dim(m/m ) = n − dim(m/m ) 16.1. Dedekind Domains. We continue our study of smooth curves by studying the local rings at their points. These are Dedekind domains. Definition 16.5. Let A be a commutative domain, and let K be the fraction field of A. An element b ∈ K is said to be integral over A if there is a monic polynomial f(t) ∈ A[t] such that f(b) = 0. The collection of all elements in K that are integral over A is the integral K K closure of A in K, denoted A . If A = A , we say that A is integrally closed. Definition 16.6. A commutative ring A is called a Dedekind domain if it is an integrally closed noetherian domain of krull dimension 1.

17. Friday, Febrary 22nd - Discrete Valuations and the p-adic Numbers Definition 17.1. Let K/k be a field extension. A discrete valuation on K/k is a map ν : K → Z ∪ {∞} satisfying (i) ν(x) = ∞ ⇔ x = 0 (ii) ν(xy) = ν(x) + ν(y) (iii) ν(x + y) ≥ min{ν(x), ν(y)} (iv) ν(x) = 0 if x ∈ k∗ A discrete valuation on K is a map as above without property (iv). 29 Notice that (ii) and (iv) above imply 1 (3) ν = −ν(x) x for all x ∈ K∗. Hence, if we have a ring R whose fraction field is K, we only need to specify the values of ν on R. This will uniquely define a discrete valuation on K. Also notice that the image of the valuation is an ideal in Z, so that we may divide by some prime number to assume that the valuation is surjective.

Definition 17.2. An absolute value on a field K is a map |·| : K → R≥0 satisfying (i) ν(x) = 0 ⇔ x = 0 (ii) |xy| = |x||y| (iii) |x + y| ≤ |x| + |y| If in addition, |·| satisfies (iv) |x + y| ≤ max{|x|, |y|} then we say the absolute value is non-Archimedean. The next proposition shows the relationship between discrete valuations and absolute values. We take the convention that c∞ = 0 for 0 < c < 1. Proposition 17.3. Let ν be a discrete valuation on K/k, and choose any real number c ν(a) such that 0 < c < 1. Then the map |·|ν defined by |a|ν = c is a non-Archimedean absolute value on K. n Proof. Since c ≥ 0 for all n ∈ Z ∪ {∞}, |·|ν maps K into the nonnegative reals. Also, n since c = 0 if and only if n = ∞, it follows that |a|ν = 0 if and only if a = 0. Property (ii) of absolute values follows from the laws of exponents. Notice that the non-Archimedean property implies roperty (iii) of absolute values, so we only need to show that the function defined is non-Archimedean. This follows from property (iii) of discrete valuations. Let’s consider some examples of discrete valuations. Example 17.4. Let p be prime, and consider Z and its fraction field Q. For any n ∈ Z, e write n = p m, with p - m, and define νp(n) = e. Assigning νp(0) = ∞ we obtain a function on Z satsifying the axioms of a discrete valuation. This map then uniquely extends to a discrete valuation on Q as ( a m pm | a, pm+1 a 0 6= 7→ - b −m pm | b, pm+1 - b where the fraction is in lowest terms, and m ≥ 0. Example 17.5. We consider an analagous example to that of example 17.4. Let k be a field, a ∈ k, and let K be the fraction field of k[t]. For any p(t) ∈ k[t], write p(t) = (t − a)eq(t), with (t − a) - q(t), and define νa(p(t)) = e. Assigning νp(0) = ∞ we obtain a function on k[t] satsifying the axioms of a discrete valuation. This map then uniquely extends to a discrete valuation on K as ( p(t) m (t − a)m | p(t), (t − a)m+1 p(t) 0 6= 7→ - q(t) −m (t − a)m | q(t), (t − a)m+1 - q(t) 30 where the fraction is in lowest terms, and m ≥ 0. We can also define ν∞ := ν0, ie, so that −1 ν∞(p(t)) is the highest power of t dividing p(t) (which will always be negative). As above, this extends to k(t). Any discrete valuation on a field K has an associated local ring.

Proposition 17.6. Let ν : K → Z be a discrete valuation on K. Then the set of elements in K with non-negative valuation form a local ring, with maximal ideal equal to the set of elements with positive valuation. Proof. Properties (ii) and (iii) of a discrete valuation show that elements with nonnegative valuation are closed under multiplication and addition, so they form a ring. The same two properties show that the set of elements with positive valuation is an ideal. Next, let Rν be the ring associated to ν, with m the ideal of elements with positive valuation. Then notice that r ∈ Rν is a unit if and only if ν(r) = 0 (this follows from (3)). Hence, m is the unique maximal ideal in Rν.

The ring Rν is the discrete valuation ring associated to ν. Let’s consider the valuation of example 17.4.

Example 17.7. Let p be a prime and consider the valuation νp from example 17.4. Notice that a ν ≥ 0 ⇔ p b p b - Thus, the discrete valuation ring of νp is just the localization of Z at the prime ideal (p). This is indeed local, with maximal ideal given by (p) · Z(p).

We can complete Z and Q with respect to the absolute value associated to νp to obtain the p-adic integers and the p-adic numbers.

18. Nonsingular Curves We continue to work towards the correspondence between ﬁnitely generated ﬁeld extensions of k of transcendence degree 1 and nonsingular projective curves. Let A be a Dedekind domain, and suppose m ⊂ A is a maximal ideal. If we localize at m, we still obtain a Noetherian, local, integrally closed domain of dimension 1, ie, a local Dedekind domain. The following result says, in particular, that this is a discrete valuation ring. Proposition 18.1. The following are equivalent for R, a Noetherian local domain of dimension 1. (i) R is a discrete valuation ring (ii) R is integrally closed (iii) R is a regular local ring (iv) m ⊂ R is principal

n Proof. (i)⇒(ii). Let s ∈ K be integral over Rν, ie, find some monic polynomial f(x) = x + P i n P i aix ∈ Rν[x] such that f(s) = 0. This means that s = −( aix ). Using the properties of a discrete valuation, we see that this leads to the conclusion that nν(s) ≥ min{iν(s)}. Notice this inequality implies that ν(s) must be positive. 31 (i)⇒(iv). Take any element in r ∈ Rν such that ν(r) = 1. Clearly (r) ⊂ m. Next, let s ∈ m be nonzero. If ν(s) = n, it follows that s ν = 0 rn n ∗ Thus s = αr for some unit α ∈ Rν. Hence s ∈ (r), so that Rν has a principal maximal ideal. (iv)⇒(i). Given a generator r, define a valuation on K, the fraction field of R as follows:  m m is the largest integer such that a ∈ (rm) a  ν = −m m is the largest integer such that b ∈ (rm) b 0 a∈ / (r), b∈ / (r) It follows that the discrete valuation ring of this valuation is R. The rest of the implications are deferred for now. Let K be a finitely generated field extension of k of transcendence degree 1, and define CK a topological space as follows. The points of CK are discrete valuations of K/k, and the closed subsets are the whole set and the finite sets. The following lemma will prove helpful in establishing our correspondence. Lemma 18.2. Let A be a Dedekind domain. Suppose R is a discrete valuation ring with A ⊂ R. Then m ∩ A is maximal in A. Proof. If m ∩ A = (0), then A consists only of units in R and is thus a field. This contradicts that A is a Dedekind domain. So m ∩ A is nonzero. Since non-zero primes in Dedekind domains are maximal, we show that m ∩ A is prime. This follows because the product two elements with nonpositive valuation will also have a nonpositive valuation.

Proposition 18.3. Let x ∈ K. Then the set {ν ∈ CK | x∈ / Rν} is finite. −1 Proof. If x = 0 then x ∈ Rν for all ν ∈ CK , so we may assume x 6= 0. Let y = x . Then x∈ / Rν if and only if y ∈ mν by (3). Hence we wish to show that {ν ∈ CK | y ∈ mν} is finite. If y ∈ k, then y∈ / mν for all ν so we may assume y∈ / k. Since k is algebraically closed, it K follows that y is transcendental over k. This shows that k[y] ,→ K. Let A = k[y] be the integral closure of k[y] in K, and C the corresponding smooth affine curve (theorem’s 6.3A and 3.9A in Hartshorne shows that A is a finitely generated Dedekind domain). As a regular function on C, y has only finitely many zeroes. We claim that there is a 1-1 correspondence between points on C and discrete valuations ν on K such that A ⊂ Rν. Given a point x ∈ C, its local ring is regular, and is thus a discrete valuation ring by proposition 18.1. Since the local ring is given by localization at mx, it clearly contains A. Also, points have distinct local rings inside of the fraction field. Next, given a valuation whose ring contains A, the ideal m ∩ A is maximal in A by the lemma, and thus corresponds to a point of C. 19. Nonsingular Curves We follow section I.6 in Hartshorne to show that in each birational equivalence class of curves, there is a unique non-singular projective curve. We will omit proof, and focus on the theory and bigger picture. First some results on birational equivalence we will use. 32 Theorem 19.1. There is an antiequivalence of categories between the category of varieties and dominant rational maps and the category of finitely generated field extensions of k given on objects by sending a variety X to its function field K(X). In particular, any finitely generated field extension of k is the function field of some variety X. Corollary 19.2. For any two varieties X and Y , the following are equivalent: (i) X and Y are birationally equivalent (ii) there are open sets U ⊂ X and V ⊂ Y with U ∼= V as varieties. (iii) K(X) ∼= K(Y ) as k-algebras In the case of curves then (varieties of dimension 1), it follows that each birational equivalence class is defined by a function field of transcendence degree 1 (the krull dimension of the coordinate ring of a variety is equal to the transcendence degree of its field of fractions over k). We will thus show that for a given function field of dimension 1 K, there is a unique non-singular projective curve CK whose function field is K. To motivate this construction, suppose Y is a non-singular curve, with P ∈ Y a point on Y . Then notice that OP is a regular local ring, and is thus a discrete valuation ring. This means that OP is the valuation ring of some discrete valuation ν : K → Z where K is the function field of Y . Notice that k ⊂ OP , and ν must vanish on k else some element of k would be in the maximal ideal of OP which is all elements with positive valuation. This is a contradiction as all elements of k are units. Hence, OP is actually a discrete valuation ring of K/k. Now, given a function field K of dimension 1, let the point of CK be all discrete valuation rings of K/k. Give CK the cofinite topology, that is, let the closed subsets be the whole space and finite sets. For any open U ⊂ CK , define the ring of regular functions on U to be \ O(U) = RP P ∈U where RP is the discrete valuation ring corresponding to the point P ∈ CK . The elements of OU can be thought of as functions on U by letting f(P ) be the residue f + mP where mP is the maximal ideal of RP . Is this residue in k? The following proposition answers this question affirmatively. Proposition 19.3. Any discrete valuation ring of K/k is isomorphic to the local ring of a point of some nonsingular affine curve. Since the residue field of local rings of affine curves is always k, our element f ∈ O(U) determines a map from U to k. Suppose f and g define the same function on U, are they then the same element? Again, the following proposition answers affirmatively.

Proposition 19.4. Let x ∈ K. Then the set {P ∈ CK | x∈ / RP } is finite. Suppose that f and g are elements of O(U) which determine the same function on U. −1 This means that f − g ∈ mP for infinitely many P . If f − g 6= 0, then let x = (f − g) in the proposition. It would follow that x∈ / RP for all P ∈ U, a contradiction, so that f = g. Finally, notice that for any f ∈ K, f must be a regular function on some open U because it is only missing from finitely many RP . This shows that the function field of CK is K.

Deﬁnition 19.5. An abstract nonsingular curve is an open subset U ⊂ CK , with the induced topology, and induced notions of regular functions 33 Proposition 19.6. Every nonsingular quasi-projective curve Y is isomorphic to an abstract nonsingular curve. Proof. We sketch the proof because it is enlightening. Let K be the function ﬁeld of Y . Each point P ∈ Y yields a local ring OP that is a discrete valuation ring of K/k. Furthermore, distinct points yield distinct rings (Lemma 6.4 Hartshorne). Let U ⊂ CK be all such discrete valuation rings. Then we have a bijection ϕ : Y → U given by ϕ(P ) = OP . There are more details to show (including that U is open), but this is the basic construction of the isomorphism.

Theorem 19.7. The abstract nonsingular curve CK is isomorphic to a nonsingular projective curve.

20. Monday, March 4th - Nonsingular Curves Given any affine algebraic variety over the complex numbers, we can consider it with its n ∼ 2n analytic topology in C = R . Since morphisms are given on coordinate algebras by polynomial functions, a morphism of varieties gives a continuous map in the analytic topology. A smooth projective planar curve C in this way corresponds to a compact, oriented, two- manifold. Such surfaces are classified up to there genus, and it is a theorem that the genus 1 of C is 2 (d − 1)(d − 2) where d is the degree of the homogeneous polynomial defining C. This shows that not all planar curves are isomorphic as varieties, so that not all function fields of dimension 1 are k(t).

21. Wednesday, March 6th - An Example of a non-rational curve Does there exist a function field that is not isomorphic to k(x), ie, is there a nonsingular curve not birational to P1? The answer is yes, and a proof is outlined in exercises I.6.1 and I.6.2 in Hartshorne, of which we complete below. Exercise 21.1 (I.6.1). Recall that a curve is rational if it is birationally equivalent to P1. Let Y be a nonsingular rational curve which is not isomorphic to P1. (a) Show that Y is isomorphic to an open subset of A1. (b) Show that Y is affine. (c) Show that A(Y ) is a unique factorization domain. ∼ 1 1 ∼ Proof. Let K = k(x) so that CK = P . Since Y is birational to P , K(Y ) = K, and the set of local rings of Y forms an open set U of CK which is isomorphic to Y . U cannot be all of P1 lest Y be isomorphic to P1. Hence U is missing at least one point, can thus be considered an open subset of A1. This proves (a). 1 ∼ Any closed set in A can be given as the zero set of one polynomial f, so that k[Y ] = k[t]f is affine. This proves (b). (c) is seen by showing that the localization of a unique factorization domain is still a unique factorization domain. Exercise 21.2 (I.6.2). An Elliptic Curve Let Y be the curve y2 = x3 − x in A2, and assume that the characteristic of the base field k 6= 2. In this exercise we will show that Y is not a rational curve, and hence K(Y ) is not a pure transcendental extension of k. (a) Show that Y is nonsingular, and deduce that A = A(Y ) ∼= k[x, y]/(y2 − x3 + x) is an integrally closed domain. 34 (b) Let k[x] be the subring of K = K(Y ) generated by the image of x in A. Show that k[x] is a polynomial ring, and that A is the integral closure of k[x] in K. (c) Show that there is an automorphism σ : A → A which sends y to −y and leaves x fixed. For any a ∈ A, defined the norm of a to be N(a) = a · σ(a). Show that N(a) ∈ k[x], N(1) = 1, and N(ab) = N(a) · N(b) for any a, b ∈ A. (d) Using the norm, show that the units in A are precisely the nonzero elements of k. Show that x and y are irreducible elements of A. Show that A is not a unique factorization domain. (e) Prove that Y is not a rational curve.

Proof. We would like the dimension of the Jacobian at all points of Y to be the dimension of the ambient space minus the dimension of Y . The Jacobian of Y is the following matrix 1 − 3x2 2y This matrix has rank 0 if and only if y = 0 and 1 − 3x2 = 0. Notice that on Y , this only happens if x = 0, 1, −1, so that 1 − 3x2 = 0 if and only if the characteristic of k is 2. Hence, the rank of the Jacobian at all points of Y is 1. Also, the dimension of Y is at least 1, but it can’t be 2 as all proper subsets of An have dimension less than 2. This shows that Y has no singular points. It follows that A(Y ) is integrally closed. This proves (a). We would like to show that x is transcendental over k. If not, then x satisﬁes some n n 2 3 polynomial ant + ... + a0, which means that anx + ... + a0 ∈ (y − x + x). Since n anx + ... + a0 has no y term, it follows that ai = 0 and that x is transcendental. Hence k[x] is a polynomial ring in K. To see that A is the integral closure of k[x], we only need to show that y is integrally closed over k[x] (because the product and sum of integral elements are integral). y satisﬁes the polynomial t2 − x3 + x in A[t]. This proves (b). The given σ is an automorphism because it is its own inverse. Any a ∈ A can be written as a = p(x)y + q(x), so that N(a) = −p2(x)y2 + q2(x) = (x − x3)p2(x) + q2(x) ∈ k[x]. The other equalities follow from the fact that σ is an automorphism of k-algebras. This proves (c). Let a ∈ A be a unit, with ab = 1. Then N(a)N(b) = 1 so that N(a) is a unit in k (the only units in k[x] are those in k). As above, N(a) = (x − x3)p2(x) + q2(x), so by a degree argument, we must have p(x) = 0 and q(x) a unit in k. This shows that a is a unit in k. The other inclusion is clear. Next, suppose that x = ab. Then N(x) = x2 = N(a)N(b), which shows that either a or b must be a unit (if not, N(a)N(b) would have a degree of at least 4). So x is irreducible. The same proof works to show that y is irreducible. Then notice that y2 = x(x2 − 1) in A(Y ) so that A(Y ) is not a unique factorization domain. This proves (d) Finally, by the previous exercise, we have shown that Y cannot be rational, else its ring of regular functions would be a unique factorization domain.

22. Friday, March 8th - Schemes Schemes are a generalization of varieties. The category of affine varieties over k is antiequivalent to the category of finitely generated commutative reduced k-algebras by sending a variety X to its ring of regular functions k[X]. Suppose we loosened the conditions on k[X], and just considered the category of commutative rings with identity. This is the motivation behind the definition of an affine scheme. 35 Definition 22.1. Let A be a commutative ring with identity. As a set, let Spec A consist of all prime ideals of A, and place a topology on A by letting the closed sets be of the form V (I) = {p ∈ Spec A | I ⊂ p}. If A is a finitely generated, commutative, reduced k-algebra, then notice that the points of the variety of A are in one-to-one correspondence with the maximal ideals. So Spec A is a larger set. In the case of varieties, maximal ideals correspond to points, and functions in a maximal ideal are exactly those regular functions which vanish at the point. Hence, an element in every maximal ideal must be the zero regular function. Here is an analagous result in the case of prime ideals. Proposition 22.2. If A is a commutative ring, then the intersection of all prime ideals in A is the ideal of nilpotent elements of A. Since A is the analog of a ring of regular functions on a variety, we should think of elements of A as functions on Spec A. To any a ∈ A and any mathfrakp ∈ Spec A, let a(p) = a ∈ k(p) where k(p) is the residue field of A at p, ie, the fraction field of A/p. Notice that the residue fields may be different for different prime ideals, so this ”function” maps into the disjoint union of the residue fields. Notice that a(p) = 0 if and only if a ∈ p. This shows that a function a is zero on Spec A if and only if it is in all prime ideals, if and only if it is nilpotent by the above proposition. Example 22.3. Let A = Spec Z. The prime ideals of Z are generated by prime numbers, and by 0. Each prime number represents a closed point, while (0) is the generic point of Spec Z, because it is contained in all points of Spec Z and therefore its closure is the whole space. Another property about a variety is that it carried with it a sheaf of rings, and the stalks of this sheaf were local rings (given by localizations of the ring of regular functions). In general, a topological space with a sheaf of rings whose stalks are local is called a locally ringed space. We’d like to give X =Spec A a sheaf of rings OX satisfying this property. To do so, let’s begin by defining the value of OX on some standard open sets. Let D(f) = {p ∈ Spec A | f∈ / p} for any f ∈ A. Tracing the analogy through, we see that in the case of varieties, D(f) would represent the complement to the hypersurface defined by f ∈ A. Thus we define OX (D(f)) to be the localization of A at f. Since any open set can be written as a union of sets of the form D(f), which we call basic open sets, by the sheaf axiom, we will see that this is enough to define the sheaf on all open sets.

23. Monday, March 11th - Faithfully Flat Descent Definition 23.1. A ring homomorphism f : R → A is flat if f gives A the structure of a flat R-module, ie, if the sequence

0 → M1 ⊗R A → M2 ⊗R A → M3 ⊗R A → 0 is exact whenever the sequence of R-modules

0 → M1 → M2 → M3 → 0 is exact. f is faithfully ﬂat if f gives A the structure of a faithfully ﬂat R-module, ie, if the sequence 0 → M1 ⊗R A → M2 ⊗R A → M3 ⊗R A → 0 36 is exact if and only if the sequence

0 → M1 → M2 → M3 → 0 is exact.

The natural map A → Af is flat for any f ∈ A, and if hf1, . . . , fni = A, then the map Q A → Afi is faithfully flat. S Lemma 23.2. hf1, . . . , fni = A if and only if Spec A = Spec Afi . −1 Proof. The proof relies on the fact that the prime ideals of Af are exactly the ideals S p where S = {1, f, f 2,...} and p is a prime ideal in A disjoint from S. Thus, in the equality S −1 Spec A = Spec Afi , we identify p with S p. If a prime ideal in A contained all the fi, it would contain all of A, so the forward direction follows by contradiction. If hf1, . . . , fni= 6 A, then the maximal ideal containing hf1, . . . , fni would be in Spec A but not Spec Afi for any i, so the reverse direction also follows by contradiction. Q 0 We now show that the map A → Afi = A is faithfully flat. That it is flat follows from the fact that the maps A → Afi are all flat. To see that it is faithfully flat, it suffices to show 0 that if M is an A-module, then M → M ⊗A A is injective. To see that this is sufficient, notice that the commutativity of the diagram / M _ N _

0 0 M ⊗A A / N ⊗A A would imply that M → N is injective, which is enough to show faithful ﬂatness because tensoring is always right-exact (ie, we only worry about the injectivity of exact sequences). 0 ∼ 0 0 Let M ⊗ Afi = Mfi , so that M ⊗A A = ⊕Mfi = M . Now suppose that m 7→ 0 ∈ M . This ci means that for each i, there is some ci such that fi m = 0. Now by the lemma, we know c1 cn P ci P ci that hf1 , . . . , fn i = A, so we can write 1 = αifi . This implies that m = αifi m = 0 so that M → M 0 is injective. Now let’s begin to see how this applies to deﬁning the structure sheaf of a scheme.

Proposition 23.3. If hf1, . . . , fni = A, then Y Y 0 → A → Afi ⇒ Afifj i i,j is exact. Proof. Injectivity follows from the discussion above, taking M = A. Now suppose that Q nij {ai/fi} ∈ i Afi is such that (aifj −ajfi)(fifj) = 0 for some nij. We can take N to be the N −N−1 maximum of the nij’s. Replacing ai/fi by aif f reduces us to the situation aifj = ajfi P Pi i for all i, j. Write 1 = αifi and deﬁne a = αiai ∈ A. Then notice that X X fja = αiaifj = αifiaj = aj so that a maps to our original element {ai/fi}. 37 In general, if R → A is faithfully ﬂat, then

0 → R → A ⇒ A ⊗R A is exact. Having defined the structure sheaf, OX of an affine scheme X =Spec A on standard open subsets D(f) = {p ⊂ A | f∈ / p} to be the localization Af , we can now use the above proposition to well define OX (U) for any open U ⊂ X. If we cover U with open sets of the form D(fi), then the following sequence is exact Y Y 0 → OX (U) → Afi ⇒ Afifj i i,j Q Q so we can define OX (U) to be the equalizer of i Afi ⇒ i,j Afifj , that is, the set of elements Q Q in Afi with equal image under both maps to Afifj . One must check that this definition is independent of the open cover chosen.

24. Wednesday, March 13th - Another View of the Structure Sheaf Hartshorne provides another view of the structure sheaf of an affine scheme. Given a commutative ring A, and an open set U ⊂Spec A = X, define ( ( ) a 1. s(q) ∈ Aq O (U) := s : U → A | X p 2. s is locally given by a ratio of elements of A p∈U What we mean by the second condition is that given any q ∈ U, there is some open V ⊂ U a containing q, and elements a, b ∈ A, with b∈ / p for all p ∈ V such that s(p) = b ∈ Ap for all p ∈ V . We define restriction as restriction of functions so that OX is a presheaf. Also, functions satisfying the two properties listed will patch together (uniquely) to a function satisfying the two properties, so OX is a sheaf of rings. Using our old definition of the structure sheaf, it is easy to see the stalk at a point p ∈ Spec A. Using the fact that any open set contains a smaller open set of the form D(f) for some f ∈ A we have O = lim O (U) = lim O (D(f)) = lim A = A X,p −→ X −→ X −→ f p p∈U p∈D(f) f∈ /p It is also clear from Hartshorne’s definition that the stalk at a point is given by localization. ` This comes from the fact that any two maps s, t : U → p∈U Ap are equal in the stalk at p if and only if they agree on p (here we have to use the local nature of definition of the structure sheaf). Finally, to show that the two definitions of the structure sheaf are isomorphic, it is enough to construct a map of sheaves that induces an isomorphism on stalks. This map on a set of the form D(f) sends an element a ∈ Af to the map sa defined by s(p) = a ∈ Ap. We now make more explicit the fact that the category of affine schemes is antiequivalent to the category of commutative rings. Proposition 24.1. A map of rings ϕ : A → B induces a map of affine schemes Spec ϕ∗ : B → Spec A, which is just the inverse image of prime ideals on points.

The induced map on sheaves OSpec A → ϕ∗(OSpec B) is the same as the condition that regular functions map to regular functions for a morphism of varieties. 38 25. Monday, March 25th - Schemes

Deﬁnition 25.1. A ringed space is a topological space X together with a sheaf of rings OX . A locally ringed space is a ringed space with local stalks. An aﬃne scheme is a locally ringed space isomorphic to Spec A for some commutative ring A.A scheme is a locally ringed space ∼ which admits an open covering {Ui} such that Ui = Spec Ai for all i.

Definition 25.2. Let (X, OX ) and (Y, OY ) be locally ringed spaces. Then a map of locally # ringed spaces is a continuous map ϕ : X → Y together with a map of sheaves ϕ : OY → ϕ∗OX such that the induced map on stalks is a local homomorphism, i.e., the inverse image of the maximal ideal in the target is the maximal ideal in the domain. Let’s consider some examples of schemes that are not affine, but obtained by gluing affine schemes.

Example 25.3. Let X1 = Spec k[t] and X2 = Spec k[x] be affine schemes. The topological spaces of X1 and X2 are homeomorphic, whose points consist of the ideals (x−a) or (t−a) and the prime ideal (0). The ideal (x) is a closed point in X2, so let U2 be the open complement. Similarly for X1. Then map U1 to U2 by (t − a) 7→ (x − a), and form the topological space X by the disjoint union of X1 and X2 with points of U1 and U2 identified via the given map. The structure sheaf is defined on page 75 of Hartshorne. By our definition above, this is a locally ringed space which admits an affine open covering, so it is a scheme. If we had glued along the map (t − a) → (x − a−1) we would have obtained 1 a different scheme, which we will see is Pk. The following is a pathological example demonstrating the need for the local condition on stalks in our definition of a morphism of schemes. Example 25.4. Let R be a discrete valuation ring, and consider Spec R. Let K = Frac(R). Spec R consists of two points, the zero ideal (0) and the maximal ideal m. The kernel of any map R → K must be a prime ideal in R, but since this map cannot be onto, its kernel cannot be m, so it must be (0) and the map must be inclusion. This corresponds to the map of topological spaces Spec K → Spec R which sends (0) in K to (0) in R. This is a morphism of affine schemes because it corresponds to a morphism of rings. We could also define a continuous map of topological spaces Spec K → Spec R by sending (0) to m. To complete the data necessary for a morphism, we then need to define a map R → K (the map on global sections). Clearly this map must be inclusion. However, on stalks, this is not a map of local rings, and thus this does not define a map of locally ringed spaces.

26. Fiber Products To enlarge the examples of schemes we’ve been considering, we consider the notion of fiber product. Definition 26.1. Let X and Y be topological spaces, equipped with continuous maps f : X → Z and g : Y → Z to a topological space Z. Define the fiber product of X and Y over Z to be X ×Z Y = {(x, y)|f(x) = g(y)}, together with the natural projection maps to X and Y . 39 Notice that the notion of fiber product is a generalization of the fiber of a map over a point. Taking Y to be a single point y, we see that the fiber product of X and Y over Z is the fiber of f over g(y) ∈ Z. The fiber product of topological spaces satisfies the following universal property. Given any topological space T with maps into X and Y such that the diagram

8 X

T > Z

& Y

commutes, there is a unique map T → X ×Z Y such that the diagram:

:5 X

/ T X ×Z Y ? Z

$ ) Y commutes. Now let’s consider the fiber product in the category of schemes. Definition 26.2. Let X and Y be schemes over Z. The fiber product of X and Y over Z is a scheme X ×Z Y together with morphisms of schemes X ×Z Y → X and X ×Z Y → Y called projections satisfying the following universal property. Given any scheme T over Z and any maps T → X and T → Y such that

8 X

T > Z

& Y

commutes, there is a unique morphism of schemes T → X ×Z Y such that the diagram:

:5 X

/ T X ×Z Y ? Z

$ ) Y commutes. 40 In the case of affine schemes, the fiber product of Spec A and Spec B over R is Spec A ⊗R B, which is just a statement about tensor products in commutative ring theory. Let’s apply the notion of fiber product in an example.

Example 26.3. Consider Spec Z[x1, x2]. To get a handle on the prime ideals of Z[x1, x2], let’s consider the natural map from Z → Z[x1, x2], which gives a map of affine schemes Spec Z[x1, x2] → Spec Z. The fiber over (0) is given by the fiber product of Spec Q and Spec [x , x ] over Spec , which by the fact above, is Spec [x , x ] = 2 . The prime Z 1 2 Z Q 1 2 AQ ideals of Q[x1, x2] are maximal ideals, principal ideals generated by irreducible elements, and the zero ideal. Similarly, the fiber above (p) is given by the fiber product of Spec Fp and Spec [x , x ] over Spec , which by the fact above, is Spec [x , x ] = 2 . The prime Z 1 2 Z Fp 1 2 AQ ideals of Fp[x1, x2] are described similarly to those in Q[x1, x2]. 27. Proj We define the projective analogue to Spec. L Definition 27.1. Let S = Sd be a graded and commutative ring. The prime ideal S+ of elements with only positive degree homogeneous components is called the irrelevant ideal. As a set Proj S is all homogeneous prime ideals not containing S+. The closed sets of S are of the form VI = {p ∈ Proj S | I ⊂ p} for some homogeneous ideal I ⊂ S. The sheaf of rings on Proj S is defined on basic open sets as OProj S(Xf ) = (Sf )(0)

where (Sf )(0) is the degree zero elements in the localization of S at f. If S = A[x0, . . . , xn] n for some commutative ring A, then we write Proj S = PA. Example 27.2. Let’s combine examples 26.3 and 25.3 to calculate 1 . This is obtained by PZ glueing two 1 on their fibers along the map t 7→ x−1 as in example 25.3. The result is a AZ 1 over the ideals (p) in Spec . PFp Z 28. Varieties and Schemes Here we discuss how schemes are in fact a generalization of varieties. Theorem 28.1. There is an equivalence of categories between affine varieties over an algebraically closed field k, and integral, affine schemes of finite type over Spec k. Proof. Both are antiequivalent to the category of finitely generated integral domains over k. This can be extended to a more general result for the category of all varieties.

Deﬁnition 28.2. Let X be a scheme. An OX -module F is a sheaf of abelian groups on X such that for every open set U ⊂ X we have a module map

OX (U) ⊗Z F(U) → F(U) which commutes with the necessary restriction maps. The ﬁber of F at x ∈ X is Fx ⊗OX,x ⊕n k(x). The free OX -module of rank n is OX . We say an OX -module F is locally free of rank 41 ∼ ⊕n n if there exists an open cover Ui of X such that F|Ui = (OX |Ui ) . An invertible sheaf of OX -modules, L is a locally free module of rank 1. The motivation behind the term ’invertible’ is that we will be able to ﬁnd L0 such that 0 ∼ L ⊗OX L = OX .

29. OX -modules Suppose X is an aﬃne scheme, the spectrum of a commutative ring A. Let M be an ˜ ˜ A-modules. Then we can deﬁne M, an OX -module associated to M by letting M(Xf ) = Mf = M ⊗A Af and extending using the sheaf axiom.

Definition 29.1. Let X be a scheme and F be an OX -module. If there exists an affine open ∼ ˜ covering Ui = Spec Ai of X such that Fi|Ui = Mi for Ai-modules Mi, then we say that F is quasi-coherent. If all the Mi can be taken to be finitely generated, we say that F is coherent. Let R be a discrete valuation ring, and let K be its field of fractions. X =Spec R consists solely of two points. The three open sets of X are ∅, X, and {(0)}. Define an OX -module F by F(∅) = 0, F(X) = K, and F((0)) = K, with module maps given by the maps K,→ K and R,→ K. For some reason this should not be quasi-coherent.

30. Invertible Sheaves and Projective Space The trivial line bundle E over the circle S is the cylinder, and the sheaf of sections of π : E → S on any open set U ⊂ S is just real valued functions on U. The Mobius band M gives another line bundle on S, and on any proper open set U, the sheaf of sections of π0 : M → S is the real valued functions on U. The diﬀerence here is that the global sections have the restriction that f(0) = f(1) = 0, whereas for the cylinder, we only have the condition that f(0) = f(1).

Proposition 30.1. Let X be a topological space with an open covering Ui and sheaves of OX - modules Fi. Suppose we have the additional data of isomorphisms ϕij : Fi|Ui∩Uj → Fj|Ui∩Uj for all i, j such that ϕjk ◦ ϕij = ϕik on triple intersections. Then there is a unique sheaf of OX -modules F which restricts to Fi on Ui for all i.

n With this proposition, we will construct a family of invertible sheaves on P . Let Ui = n Spec k[t0i, t1i, . . . , tni] be the standard open set of P defined by complement of the closed n set defined by the homogeneous ideal (xi) in P = Projk[x0, . . . , xn]. Here we think of tij as xi the regular function on Uj. For any d ∈ , define a sheaf OX (d) on the standard open xj Z sets to be OX (d)(Ui) = Spec k[t0i, t1i, . . . , tni]. We have to give the data of the transition morphisms. Here we take the morphisms as follows: −1 −1 ϕij : k[t0i, t1i, . . . , tni, tji ] −→ k[t0j, t1j, . . . , tnj, tij ] d r 7−→ rtij

This is an isomorphism of modules because tij is invertible. Since tjktij = tik, the maps agree on triple intersections, so we have constructed an invertible sheaf on Pn. To see what the value is on arbitrary open sets, we use the sheaf axiom, noting that the restriction maps of our sheaf are defined with the transition morphisms ϕij. 42 31. Cartier Divisors Let X be an integral scheme, and let K be its function field (direct limit of the structure sheaf over all open sets). Then K defines a constant sheaf K on the irreducible scheme X whose value on each open set is K, and whose restriction maps are the identity. Definition 31.1. A Cartier Divisor D, is a global section of the sheaf of abelian groups ∗ ∗ K /OX . Equivalently, a Cartier divisor is the data of an open cover Ui of X, and an element ∗ ∗ fi ∈ K such that fi/fj ∈ OX (Ui ∩ Uj). A Cartier divisor is said to be principal if all of the fi are equal. Two Cartier divisors are said to be linearly equivalent iof their difference is principal. Given a Cartier divisor D, we can construct an invertible sheaf on X, L(D) as follows. On −1 −1 any Ui in the open cover defining D, let L(D)(Ui) = (fi ) ⊂ K, where (fi ) is the OX (Ui) −1 module generated by fi . Theorem 31.2. The map D → L(D) is an isomorphism of groups from linear equivalence classes of Cartier divisors to the Picard group of X.

32. Vector Bundles 32.1. Definitions. Definition 32.1. A topological real vector bundle of rank n ≥ 0 is a triple (E, B, π) where π : E B is a continuous surjection of topological spaces such that (i) for all p ∈ B, π−1(p) has the structure of an n-dimensional vector space over R. −1 (ii) for all p ∈ B, there is some open neighborhood U and a homeomorphism φU : f (U) → U × Rn which sends fibers to fibers linearly. Example 32.2. (1) (X × Rn, X, π) is the trivial vector bundle. (2) TM → M is the tangent vector bundle of a smooth manifold M. Definition 32.3. A map of vector bundles over a base space B is a continuous map f : E → E0 which commutes with projections, and sends fibers to fibers linearly. Definition 32.4. A section of a vector bundle is a continuous map s : B → E such that π ◦ s = idB. Claim: U 7→ {sections over U} defines a locally free sheaf on X. This follows because any local trivialization has n linearly independent sections, where n is the rank of the vector bundle.

32.2. Transition functions. Given π : E → X, we have transition functions φ ◦ φ−1 : Uj Ui n Uij × R where Uij = Ui ∩ Uj. Since fibers are mapped to fibers linearly, each point in Uij defines a linear automorphism of Rn, that is, the transition function gives a map

Tij : Uij → GLn(R)

Now suppose we have a point inside of a triple intersection Uijk. Then the transition functions must satisfy Tij · Tjk = Tik. In fact, if we only have an open cover, and transition functions satisfying the above condition, then a vector bundle is uniquely deﬁned on X. 43 32.3. Some Constructions. Moral: Most constructions of vector spaces carry over to vector bundles, ie, direct sum, tensor product, exterior powers, duals, homs 32.4. Equivalence of Categories. Theorem 32.5. There is an equivalence of categories between locally free sheaves of rank n over X and algebraic vector bundles of rank n over X. Any algebraic vector bundle of rank n over X is a topological vector bundle, but the reverse is not true. An example is OPn (−1) which has no global sections.