MATH 8253 WEEK 9

CIHAN˙ BAHRAN

Preliminary

Lemma 1. Let f : X → Y be a continuous map of topological spaces. If A ⊆ X is irreducible , then so is f(A) ⊆ Y .

Proof. Suppose f(A) ⊆ C ∪ D where C,D are closed subsets of Y . Then A ⊆ f −1(C ∪ D) = f −1(C) ∪ f −1(D) , and since A is irreducible , either A ⊆ f −1(C) or A ⊆ f −1(D). Thus either f(A) ⊆ C or f(A) ⊆ D. This shows that f(A) is irreducible. 

Solutions

2.4.1. Let k be a field and P ∈ k[T1,...,Tn]. Show that Spec(k[T1,...,Tn]/(P )) is reduced (resp. irreducible; resp. integral) if and only if P has no square factor (resp. admits only one irreducible factor; resp. is irreducible)

Write A = k[T1,...,Tn]. We may assume (P ) is a proper since By Proposition 2.4.2, Spec(A/(P )) is reduced if and only if A/(P ) is a reduced ring and that holds if and only if (P ) is a radical ideal of A.

e1 en Note that A is a UFD, so we can write P = p1 ··· pn where pj’s are mutually non- + associate irreducibles in A and ej ∈ Z . Then the element f = p1 ··· pn lies in (P ). So if (P ) is radical, then P divides f and this forces ej = 1 for all j. Conversely, if P has no square factor, then f = P . So if g ∈ (P ), then P | gn for some n; and hence n pj | g for all j and that yields pj | g because pj is irreducible. Thus P = f | g, that is, g ∈ (P ). » By Proposition 2.4.7, Spec(A/(P )) is irreducible if and only if A/(P ) has a unique » minimal . Assume that there is a unique prime ideal p of A which is minimal among the primes containing P . As p is prime, by reordering we may assume p1 ∈ p. So (p1) ⊆ p. But (p1) is also a prime ideal containing P ; hence p = (p1). If n were larger than 1, we would have (p1) = p ⊆ (p2) so the irreducibility of p1 forces (p1) = (p2) contradicting the fact that p1 and p2 are non-associate. Thus n = 1, that is, P admits only one irreducible factor. Conversely, assume P = pn where p is an irreducible. Then every prime ideal containing P also contains p and hence (p) is the unique prime ideal which is minimal among the primes containing P . Finally, Spec(A/(P )) is an integral iff A/(P ) is an integral domain iff (P ) is prime iff P is irreducible. 2.4.2. Let X be a scheme and x ∈ X. Show that the image of the morphism Spec OX,x → X is the set of points of X that specialize to x.

1 MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 2

Let y be in the image. Then y lies in an affine open subset U ∼= Spec A which contains x. So we can write x = p and y = q where p, q ∈ Spec A. So OX,x gets identified with the localization Ap and writing ϕ : A → Ap for the localization map, we have q ∈ im(Spec ϕ); hence q ⊆ p. By Lemma 4.11 (b) y specializes to x. And also every such point lies in the image because im(Spec ϕ) = {q ∈ Spec A : q ⊆ p}.

2.4.3. Let OK be a discrete valuation ring with field of fractions K and uniformizing parameter t (i.e., a generator of the maximal ideal). Show that Spec K[T ] can be identified with an open subscheme of Spec OK [T ]. Determine the set of closed points of Spec K[T ] which specialize to the point coresponding to the maximal ideal (T, t) of OK [T ]. See also Proposition 10.1.40(c).

2.4.4. We say that a scheme X is connected if the underlying is connected. For example, an irreducible scheme is connected. Let X be a scheme having only finitely many irreducible components {Xi}i. Show that X is connected if and only if for any pair i, j, there exists indices i0 = i, i1, . . . ir = j such that

Xil ∩ Xil+1 6= ∅ for every l < r. Also show that X is integral if and only if X is connected, and integral at every point x ∈ X.

We may assume X 6= ∅. Say {Xi : i ∈ I} are the irreducible components of X. For i, j ∈ I declare i ∼ j if and only if Xi ∩ Xj 6= ∅. This defines a reflexive and symmetric relation on I so its transitive is an equivalence relation for which we also write ∼. So we can restate the first part of the question as “X is connected iff ∼ yields a single equivalence class on I.” S For every equivalence class E ⊆ I, write XE := i∈E Xi. As E is a finite set and each Xi is closed, XE is closed. Also by construction, if F is another equivalence class under ∼, then XE ∩ XF = ∅. But G [ XE = Xi = X, E∈I/∼ i∈I so each XE is also open. Moreover, because of the following standard point-set lemma and noting that irreducibility is stronger than being connected, we can conclude that XE is connected.

Lemma 2. Let {Xi : i ∈ E} be a finite collection of connected subspaces of a topological space X, where |E| > 1. If for every i ∈ E there exists j ∈ Er{i} such that Xi∩Xj 6= ∅, S then i∈E Xi is connected.

Proof. Omitted. 

Thus {XE : E ∈ I/∼} is the set of connected components of X. Therefore X is connected if and only if I/∼ is a singleton. For the second part, the “only if” part is clear. For the “if” part, assume X is connected and integral at every point x ∈ X. So X is reduced at every x ∈ X, which means by definition that X is reduced. Also by Proposition 2.4.12(b), every point x has a unique irreducible component passing through it. This means that the irreducible components of X are mutually disjoint, that is, Xi ∩ Xj = ∅ whenever i, j ∈ I with i 6= j. So ∼ is just equality and therefore by above {Xi : i ∈ I} are the connected components of X. But X is connected, so |I| = 1 and X is irreducible. MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 3

2.4.5. Let X be a scheme. Show that every irreducible component of X is contained in a connected component (in the topological sense). Show that if X is locally Noetherian, then the connected components are open. If X is Noetherian, then there are only finitely many connected components.

We may assume X 6= ∅. Let X0 be an irreducible component of X. Pick x ∈ X0. Then the connected component, say C, containing x is the union of all connected subsets of X containing x. Since X0 is one of these, X0 ⊆ C. Suppose X is a Noetherian scheme. Then X has finitely many irreducible components. So Exercise 2.4.4 is applicable, and we have shown in its solution that X has finitely many connected components. Note that this implies that the connected components of X are open. Suppose X is locally Noetherian. To show that the connected components of X are open, it suffices to show that every x ∈ X has a connected neighborhood. We know that x has a Noetherian neighborhood U, so we may pick the connected component of U containing x. 2.4.6. Let A be a ring. We say that e ∈ A is idempotent if e 6= 0 and e2 = e. We will say that e is indecomposable if it cannot be written as a sum of two idempotent elements. Let X be a scheme. (a) Show the equivalence of the following properties: (i) X is connected; (ii) OX (X) has no other idempotent elements than 1; (iii) Spec OX (X) is connected. (b) Show that any local scheme (i.e. the spectrum of a local ring) is connected. (c) Let us suppose that the connected components of X are open (e.g., X locally Noetherian). Let U be a connected component of X. Show that there exists a unique idempotent element e of OX (X) such that e|U = 1 and e|XrU = 0. Show that this establishes a canonical bijection from the set of connected components onto the set of indecomposable idempotent elements of OX (X), the converse map being given by e 7→ V (1 − e).

(a) (iii) ⇒ (ii): We show the contrapositive. Write A = Spec OX (X). Suppose A has an idempotent e 6= 1. Then 1 − e is also an idempotent and Ae, A(1 − e) are nonzero rings such that A ∼= Ae × A(1 − e) via a 7→ (ae, a(1 − e)). Therefore the nontrivial decomposition

Spec A ∼= Spec Ae t Spec A(1 − e) shows that Spec A is disconnected. (ii) ⇒ (i): We show the contrapositive. Suppose X is disconnected. Then X is the disjoint union of two nonempty open subsets, say U and V . Then we have a ring isomorphism ∼ OX (X) = OX (U) × OX (V ) .

But OX (U) and OX (V ) are nonzero rings, so OX (X) has nontrivial idempotents. (i) ⇒ (iii): BILEMEDIM MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 4

(b) Let A be a local ring and e ∈ A be an idempotent. Since e2 = e, if e 6= 1, then e is a non-unit therefore lies in the unique maximal ideal, which is the (Jacobson) radical of A. Therefore 1 − e is a unit, but 1 − e is also an idempotent so 1 − e = 1 and so e = 0, a contradiction. Thus A has no nontrivial idempotents and therefore by (a) Spec A is connected. (c) Note that connected components are always closed, so Xr U is open. Hence

OX (X) → OX (U) × OX (Xr U)

s 7→ (s|U , s|XrU ) is a ring isomorphism and therefore there is a unique e ∈ OX (X) such that e|U = 1 and e|XrU = 0. Since (1, 0) is an idempotent in the product ring OX (U) × OX (X r U), so is e in OX (X). Moreover, e is indecomposable. Suppose not, so e = f + g where f, g are idempotents 2 in A := OX (X). Since e = e = fe + ge, we may assume that f, g ∈ Ae. But then f, g ∼ ∼ are nontrivial idempotents in the ring Ae and therefore since Ae = OX (U) = OU (U), by (a) U is not connected; a contradiction. So we may write ε(U) for e and get a well-defined map

ε : {connected components of X} → {indecomposable idempotents of OX (X)} .

Now let e be an indecomposable idempotent of OX (X). Recall that any subset S of OX (X) gives rise to the closed subset

V (S) = {x ∈ X : sx = 0 for all s ∈ S} of X. We claim that V (e) is a connected component of X. First, since 1 = e + (1 − e) we have ∅ = V (e, 1 − e) = V (e) ∩ V (1 − e) . Second, if x ∈ X ????????????????????????????

2.4.7. We say that a topological space X verifies the separation axiom T0 if for every pair of points x 6= y, there exists an open subset which contains one of the points and not the other one. Show that the underlying topological space of a scheme verifies T0.

Let x, y be distinct points in X. Let U ∼= Spec A be an affine neighborhood of x. If y∈ / U then we are done. If y ∈ U, we can make the identifications x = p and y = q where p and q are distinct prime ideals of A. Since p 6= q, either p * q or q * p. WLOG assume p * q; so there exists f ∈ p − q. Then q lies in the principal D(f) but p doesn’t. 2.4.8. Let X be a quasi-compact scheme (Exercise 3.14). Show that X contains a closed point. See also Exercise 3.3.26 for a counterexample when X is not quasi- compact.

Let’s define a topological space to be cool if it is nonempty and every nonempty closed subset contains a closed point. Lemma 3. Let X be a topological space which has a finite open covering of cool subsets. Then X contains a closed point. MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 5

Proof. We show by induction on n that if X can be written as a union of n cool and open subsets then X has a closed point. The basis case is easy since then X is cool and hence the closed subset X contains a closed point. Sn Now assume that the claim is true for n − 1. Suppose X = j=1 Uj where each Uj is Sn cool and open. By induction hypothesis, the subspace X1 = j=2 Uj contains a closed point x. Note that this means {x} ∩ X1 = {x}. Consider the closed subset {x} ∩ U1 of U1. There are two cases:

• {x} ∩ U1 = ∅. Then

{x} = ({x} ∩ X1) ∪ ({x} ∩ U1) = {x} hence x is a closed point of X. • {x} ∩ U1 6= ∅. Then by the coolness of U1 its closed subset {x} ∩ U1 contains a closed point x1 of U1. So we have x1 ∈ {x} and {x1} ∩ U1 = {x1}. We want to show that x1 is a closed point of X. Observe that

{x1} ∩ X1 ⊆ {x} ∩ X1 = {x} .

Suppose x ∈ {x1}. Then either x = x1 or x∈ / U1. In the former case x = x1 is closed in both X1 and U1, hence in X and we are done. So assume x∈ / U1. Then X r U1 is a closed subset of X containing x. This forces x1 ∈ X r U1; a contradiction. Thus x∈ / {x1} and hence {x1} ∩ X1 = ∅. Thus

{x1} = ({x1} ∩ X1) ∪ ({x1} ∩ U1) = {x1}

which means x1 is a closed point in X.  Note that affine schemes are cool topological spaces because every nonempty closed subset V (I) in Spec A contains a maximal ideal. Thus quasi-compact schemes, which can be covered by finitely many affine schemes contain a closed point. 2.4.9. Let X be a Noetherian scheme. Show that the set of points x ∈ X such that OX,x is reduced (resp. is an integral domain) is open.

Let’s call a scheme X nice if {x ∈ X : OX,x is reduced} is open in X. Suppose {Ui}i is an open covering of X such that each Ui is nice. Then [ {x ∈ X : OX,x is reduced} = {x ∈ Ui : OX,x is reduced} i [ = {x ∈ Ui : OUi,x is reduced} i is open, hence X is nice. Therefore we may assume that X = Spec A where A is a Noetherian ring. Suppose Ap is reduced for some p ∈ Spec A. It suffices to find f ∈ Ar p such that Af is reduced. Indeed localizations of reduced rings are reduced, so for every q ∈ D(f) the ring Aq would be reduced. Consider the ideal I = {a ∈ A : sa = 0 for some s ∈ Ar p} of A, which is the kernel of the localization A → Ap. As Ap is reduced, I is a radical ideal of A. And since A is Noetherian, I has a finite generating set, say {a1, . . . , an}. So there exists si ∈ Ar p such that siai = 0 for each i. MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 6

Qn n Let f = i=1 si ∈ Ar p. We claim that Af is reduced. So suppose (a/1) = 0 in Af . Then f kan = 0 for some k ∈ Z+. So an ∈ I and hence a ∈ I as I is radical. So n X a = ciai i=1 for some ci ∈ A. Therefore fa = 0, thus a/1 = 0 in Af . Since every element of Af is of the form a/1 times a unit, this implies that Af is reduced.

For the integral domain part, it suffices to show that if Ap is a domain then so is Af (the argument reducing to this is the same as above). Suppose b/1 · c/1 = 0 in Af . k + Then f bc = 0 for some k ∈ Z . Hence bc ∈ I, but since Ap is a domain I is a prime ideal, hence WLOG b ∈ I. Therefore

n X b = diai i=1 for some di ∈ A. Thus fb = 0, so b/1 = 0 in Af . 2.4.10. Let f : X → Spec A be a quasi-compact morphism (Exercise 3.17). Let # I be an ideal of A. Show that f(X) ⊆ V (I) if and only if fSpec A(I) ⊆ OX (X) is nilpotent.

# Assume f(X) ⊆ V (I). Let a ∈ I. We want to show that the section s = fSpec A(a) ∈ OX (X) is nilpotent. Fix x ∈ X and write p = f(x). Via the local homomorphism Ap → OX,x which maps a/1 to sx, we obtain sx ∈ mx since a ∈ I ⊆ p.

So if we pick an affine open subset U = Spec B of X and consider t = s|U ∈ OX (U) = B, then for every q ∈ Spec B the element t/1 ∈ Bq lies in the maximal ideal qBq, that is, t ∈ q. Thus t is nilpotent. Now being quasi-compact, X has a finite open covering by affine open subsets, say

U1,...,Un. By above, s|Ui ∈ OX (Ui) is nilpotent for every i = 1, . . . , n. So picking m m m m > 0 such that 0 = (s|Ui ) = (s )|Ui for every i, we get s = 0 since OX is a sheaf. Conversely, assume f(X) * V (I), so there exists x ∈ X such that p = f(x) does not # contain I. So pick a ∈ Ir p. As a/1 is a unit in Ap, writing s = fSpec A(a), we get that sx is a unit in OX,x. As OX,x 6= 0, sx is not nilpotent, therefore s cannot be nilpotent.

2.4.11. Let f : X → Y be a morphism of irreducible schemes with respective generic points ξX , ξY . We say that f is dominant if f(X) is dense in Y . Let us suppose that X,Y are integral. Show that the following properties are equivalent: (i) f is dominant; # (ii) f : OY → f∗OX is injective; (iii) for every open subset V of Y and every nonempty open subset U ⊆ f −1(V ), the map OY (V ) → OX (U) is injective; (iv) f(ξX ) = ξY ; (v) ξY ∈ f(X).

(v) ⇒ (iv): Since {ξX } = X and f is continuous, we have

ξY ∈ f(X) = f({ξX }) ⊆ {f(ξX )} . MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 7

Therefore f(ξX ) specializes to ξY . But since ξY is a (actually the) of Y , by definition this implies that f(ξX ) = ξY . (iv) ⇒ (iii): We may assume that V is nonempty as the statement is vacuously true for V = ∅. Since ξY is the generic point of Y , we have ξY ∈ V and similarly ξX ∈ U. Then since f(ξX ) = ξY we have a commutative diagram a OY (V ) / OX (U)

b d  c  OY,ξY / OX,ξx of rings. By Proposition 2.4.18, OY,ξY is a field (which immediately implies that c is injective) and b is injective. So d ◦ a = c ◦ b is injective and hence a is injective. −1 # −1 (iii) ⇒ (ii): If f (V ) 6= ∅, then by assumption fV : OY (V ) → OX (f (V )) = f∗OX (V ) is injective for every open V in Y . SIKINTI?????????????????? Thus f # is injective as a sheaf map. (ii) ⇒ (i): We show the contrapositive. Suppose that f is not dominant. So there exists a nonempty open subset V of Y such that V ∩ f(X) = ∅, or equivalently f −1(V ) = ∅. ??????????????????????????

(i) ⇒ (v): The proof is almost the same with (v) ⇒ (iv). Since f(X) ⊆ {f(ξX )}, the assumption forces {f(ξX )} = Y . So f(ξX ) specializes to ξY and hence f(ξX ) = ξY . 2.4.12. Let B be a graded ring. Let Y be a reduced closed subscheme of Proj B. Show that there exists a homogenous ideal I of B such that Y ∼= Proj B.

As a topological subspace Y can be identified with V+(I) for a homogenous ideal I of B because that’s what the closed subspaces of Proj B look like. √ Lemma 4. With the notation above, I is also a homogenous ideal.

Proof. Suppose that p is a prime ideal containing I. Then for every d ≥ 0 h I = ⊕d≥0(I ∩ Bd) ⊆ ⊕d≥0(p ∩ Bd) = p . and we know by Lemma 3.35 that ph is prime. So we have shown that p ∈ V (I) implies h p ∈ V+(I). Hence \ p ⊇ \ ph ⊇ \ q

p∈V (I) p∈V (I) q∈V+(I) but since V+(I) ⊆ V (I) the reverse containment is trivial; hence √ I = \ = \ q

p∈V (I) q∈V+(I) is homogenous.  √ √ Now we have√ that I is homogenous and moreover V+(I) = V+( I). Thus we may assume I = I. √ The ring B/I is reduced since its nilradical is I/I = 0. Thus the scheme Spec B/I is reduced. So in particular for every homogenous f ∈ A := B/I, OSpec A(D(f)) = Af MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 8 is a reduced ring. This implies that the subring A(f) of Af is also reduced. Since A(f) = OProj A(D+(f)) and {D+(f): f homogenous} forms an affine open covering of Proj A, by Proposition 4.2 part (b) Proj A = Proj B/I is a reduced scheme which can be identified as a closed subscheme of Proj B whose underlying topological space is V+(I). But by Proposition 4.2 part (d), there is a unique structure of a reduced closed sub- scheme on V+(I). Since Proj B/I and Y are both such subschemes, we get Y ' Proj B/I.

2.5.1. Let X be a topological space. Let {Xi}i be a covering of X by closed subsets Xi. We assume that it is a locally finite covering, that is to say that every point x ∈ X admits an open neighborhood U which meets only a finite number of Xi. Show that dim X = supi dim Xi.

First, we observe that we can reduce to the case where the covering is finite. So assume that the claim holds for finite coverings by closed subsets. By assumption, every x ∈ X has a neighborhood Ux such that Ux intersect only finitely many Xi. So {Xi ∩ Ux}i is a finite covering of Ux by closed subsets of Ux. Therefore by our assumption we have

dim Ux = sup dim(Xi ∩ Ux) ≤ sup dim Xi . i i

But {Ux : x ∈ X} is an open covering of X, so

dim X = sup{dim Ux : x ∈ X} ≤ sup dim Xi . i therefore X = supi dim Xi.

Thus we may assume {Xi}i = {X1,...,Xn} is a finite covering of X. It suffices to prove the case n = 2, the rest follow by induction (the basis case n = 1 is trivial) because X1 and X2 ∪ · · · ∪ Xn is also a covering of X by closed subsets. So we may write C and D to be closed subsets of X such that C ∪ D = X where we want to prove that dim X ≤ max{dim C, dim D}. If X0 is an irreducible component of X, then X0 = (C ∩ X0) ∩ (D ∩ X0) so X0 = C ∩ X0 or X0 = D ∩ X0, that is, X0 ⊆ C or X0 ⊆ D. In either case, we have dim X0 ≤ max{dim C, dim D}. By Proposition 5.5(c), we have dim X ≤ max{dim C, dim D}. 2.5.2. Let X be a scheme and Z be a closed subset of X. Show that for all x ∈ X, we have codim({x},X) = dim OX,x and codim(Z,X) = minz∈Z dim OX,z.

We first work out some basic facts about irreducible sets and dimension which are not (or just stated without proof) in Liu’s book. If Y is an irreducible subset of X, we define codim(Y,X) in the same way as in Definition 2.5.7, observing that it is equal to codim(Y,X). Lemma 5. Let X be a topological space. Let Y be an irreducible and let U be an open subset of X such that Y ∩ U 6= ∅. (a) Y ∩ U is irreducible. (b) If Z is an irreducible closed subset of Y such that Z ( Y , then Z ∩ U ( Y ∩ U. (c) codim(Y ∩ U, U) ≥ codim(Y,X). (d) If Y ⊆ U, then codim(Y,U) = codim(Y,X). MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 9

Proof. (a) Suppose Y ∩ U ⊆ A ∪ B where A, B are closed in X. Then Y ⊆ A ∪ B ∪ (Xr U) and since Y * XrU by assumption, by the irreducibility of Y either Y ⊆ A or Y ⊆ B. In particular, either Y ∩ U ⊆ A or Y ∩ U ⊆ B. (b) We show the contrapositive: assume Z ∩ U = Y ∩ U. So Y = (Y ∩ U) ∪ (Y r U) = (Z ∩ U) ∪ (Y r U) . As Z ⊆ Y , we get Y = Z ∪(Y rU). But Y is irreducible and Y 6= Y rU by assumption, so Y ⊆ Z. (c) Note that it makes sense to talk about codim(Y ∩U, U) because Y ∩U is irreducible by (a). Now if

Y ⊆ Y0 ( Y1 ( ··· ( Yn is a chain of irreducible closed subsets of X containing Y , then by (a) and (b)

Y ∩ U ⊆ Y0 ∩ U ( Y1 ∩ U ( ··· ( Yn ∩ U is a chain of irreducible closed subsets of U containing Y ∩ U. The assertion follows from this observation. (d) “≥” follows from (c). To see “≤”, let

Y ⊆ Y0 ( Y1 ( ··· ( Yn be a chain of irreducible closed subsets of U containing Y . Then if we let Zj to be the closure of Yj in X for each j, then Yj = Zj ∩ U and Zj is irreducible for each j. Hence

Y ⊆ Z0 ( Z1 ( ··· ( Zn is a chain of irreducible closed subsets of X containing Y .  Note that Lemma 5(d) says that the codimension can be calculated locally. Now we solve the first part of the question. Let U ∼= Spec A to be an affine open neighborhood of x in X where x corresponds to p ∈ Spec A. Then by Lemma 5(d) and Proposition 2.5.8(b), we have codim({x},X) = codim({x},U) = codim({p}, Spec A) = codim({p}, Spec A) = codim(V (p), Spec A)

= dim Ap = dim OX,x .

For the second part, if {Zλ : λ ∈ Λ} are the irreducible components of Z, then by definition

codim(Z,X) = min{codim(Zλ,X): λ ∈ Λ} and also

min dim OX,z = min{min dim OX,z : λ ∈ Λ} z∈Z z∈Zλ so we can reduce to the case where Z is irreducible. For every z ∈ Z, as {z} ⊆ Z, we have codim(Z,X) ≤ codim({z},X) . MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 10

Hence codim(Z,X) ≤ minz∈Z codim({z},X). On the other hand, since Z is irreducible, it has a generic point ξ ∈ Z, so in particular {ξ} = Z. Thus

codim(Z,X) = min codim({z},X) = min dim OX,z z∈Z z∈Z where the last equality is by the first part. 2.5.3. Show the following properties. (a) Let Z be a closed subset of a topological space X. Show that we have codim(Z,X) = 0 if and only if Z contains an irreducible component of X. Give an example (with X non-irreducible) where codim(Z,X) = 0 and dim Z < dim X. (b) Let X = Spec OK [T ] where OK is a discrete valuation ring, with uniformiz- ing parameter t 6= 0. Let f = tT − 1. Show that the ideal generated by f is maximal. Let x ∈ X be the corresponding point. Show that X is irreducible, dim OX,x = 1, and that codim({x},X) + dim{x} < dim X.

(a) Note that if Z0 is irreducible, codim(Z0,X) = 0 means that Z0 is maximal among irreducible closed subsets of X, that is, Z0 is an irreducible component of X. In general, if codim(Z,X) = 0, then (by the definition of codimension for general closed subsets) there exists an irreducible component Z0 of Z such that codim(Z0,X) = 0. Then by above Z0 must be an irreducible component of X. Conversely, if Z contains an irreducible component X0 of X, then X0 is also an irreducible component of Z. Therefore codim(Z,X) ≤ codim(X0,X) = 0, hence codim(Z,X) = 0. For the desired example, by above it suffices to find a scheme X who has two irreducible components of different dimensions; because we may then choose Z to be the irreducible component with smaller dimension, which ensures dim Z < dim X while codim(Z,X) = 0. So pick integral domains A, B such that dim A < dim B (for instance A = k[X] and B = k[X,Y ] where k is a field). Then let X = Spec(A × B) = Spec A t Spec B and Z = Spec A, Z0 = Spec B. Here Z and Z0 are irreducible because A and B are domains, and they are the irreducible components of X by Proposition 2.4.5(c) such that dim Z < dim Z0.

(b) Note that the quotient ring OK [T ]/(f) is isomorphic to the localization (OK )t.

Lemma 6. If OK is a discrete valuation ring with uniformizing parameter t, then every n nonzero proper ideal of OK is generated by t for some n ≥ 1.

Proof. By the book’s definition, OK is a local PID which is not a domain and (t) is the unique maximal ideal. So in particular OK is a UFD such that every irreducible element is associate with t. Then if (g) is an arbitrary nonzero proper ideal of OK , n then g is nonzero and non-unit so g = ut for some unit u ∈ OK and n ≥ 1. Thus n (g) = (t ).  n n Every nonzero element of (OK )t can be written of the form a/t where gcd(a, t ) = 1. By the above lemma, the ideal generated by a in OK cannot be a proper ideal, hence a n must be a unit. Thus a/t is a unit in (OK )t. This shows that (OK )t is a field, therefore m := (f) is maximal in OK [T ].

As OK [T ] is an integral domain, X = Spec OK [T ] is a fortiori irreducible. Note that

dim OX,x = dim (OK [T ])m = ht(m) . MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 11

Thanks to the chain 0 ( m, ht(m) ≥ 1. On the other hand, by Krull’s principal ideal theorem (which is applicable as OK [T ] is Noetherian and (f) = m is prime), we have ht(m) ≤ 1. Thus dim OX,x = 1. Since x is a closed point, by Exercise 2.5.2 we have codim({x},X) = codim({x},X) = dim OX,x = 1. And trivially dim{x} = 0. However, by Example 2.5.4 and Corollary 2.5.17, dim X = dim OK [T ] = 1 + 1 = 2.

2.5.4. Let A be a ring and p1,..., pr be prime ideals of A. Let I be an ideal of A contained in none of the pi. We want to show that I is not contained in p1 ∪· · ·∪pr. (a) Show that the property is true for r ≤ 2. (b) Assume that the property is true for r − 1 and that pr does not contain any pi, i ≤ r − 1. Let x ∈ I r (p1 ∪ · · · ∪ pr−1). Show that there exists a y ∈ (Ip1 ··· pr−1)r pr. (c) Show that either x or x + y is not in p1 ∪ · · · ∪ pr.

(a) Suppose I ⊆ p1 ∪ p2. Since I * p1, there exists x2 ∈ I r p1 and similarly there exists x1 ∈ Ir p2. Then necessarily x1 ∈ p1 and x2 ∈ p2. Consider x1 + x2 ∈ I. Since x1 ∈ p1 and (x1 + x2) − x1 = x2 ∈/ p1, we get x1 + x2 ∈/ p1. But similarly x1 + x2 ∈/ p2, so x1 + x2 ∈ Ir (p1 ∪ p2); a contradiction. Note that we didn’t need to use the fact that p1, p2 are prime here.

(b) Pick yi ∈ pir pr for each 1 ≤ i ≤ r − 1 and also y0 ∈ Ir pr. Then since pr is prime, Qr−1 y := i=0 yi ∈/ pr. Yet evidently y ∈ Ip1 ··· pr−1.

(c) Note that since y ∈ pi and x∈ / pi for every 1 ≤ i ≤ r − 1, we have x + y∈ / pi for every 1 ≤ i ≤ r − 1. Suppose x ∈ p1 ∪ · · · ∪ pr. Then by the choice of x, we have x ∈ pr. But y∈ / pr, so x + y∈ / pr. Thus x + y∈ / p1 ∪ · · · ∪ pr.

2.5.5. Let A be a graded ring, and let p1,..., pr be homogenous prime ideals of A. Let I be a homogenous ideal of A contained in none of the pi. We want to S show that there exists a homogenous element of I not contained in i pi. One can suppose that pr does not contain any pi, i < r.

(a) Show that there exists a homogenous element a ∈ Ip1 ··· pr−1r pr. S (b) Let b ∈ I be a homogenous element such that b∈ / i≤r−1 pi. Show that b or deg b deg a a + b is not in p1 ∪ · · · ∪ pr. Conclude.

(a) For each 1 ≤ i ≤ r − 1, since pi * pr and pi is a homogenous ideal, there exists a homogenous element ai ∈ pirpr. Similarly there exists a homogenous element a0 ∈ Irpr. Qr−1 Then a := i=0 ai is a homogenous element which lies in Ip1 ··· pr−1 but is not in pr since pr is prime. deg b deg a (b) Fix i ∈ {1, . . . , r − 1}. Note that a ∈ pi because a ∈ pi and b ∈/ pi because deg b deg a b∈ / pi and pi is prime. Then the homogenous element c := a + b lies outside pi. deg a Suppose b ∈ p1 ∪ · · · ∪ pr. Then by the choice of b, we have b ∈ pr, hence b ∈ pr. deg b But a∈ / pr, so a ∈/ pr as pr is prime. Thus c∈ / pr and so c∈ / p1 ∪ · · · ∪ pr.

2.5.8. Let OK be as in Exercise 5.3, and let K = Frac(OK ), k = OK /tOK be the residue field of OK . Let us set A = K×k and let ϕ : OK → A be the homomorphism induced by OK → K and OK → k. Show that Spec ϕ : Spec A → Spec OK is MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 12

surjective and that dim OK > dim A. Also show that A is a finitely generated OK -algebra (i.e., quotient of a over OK ).

We know that the only prime ideals of OK are 0 and m = (t). Since OK → K is injective, ϕ is injective, so ϕ−1(0) = 0. And ϕ−1(K ×0) = m. Thus Spec ϕ is surjective, as both primes of OK are realized as inverse images under ϕ. However, dim OK = 1 whereas being an artinian ring, dim A = 0. ∼ ∼ Also, K = OK [T ]/(tT − 1) by Exercise 2.5.3 and k = OK /m so A is a product of two finitely generated OK -algebras, so A itself is a finitely generated OK -algebra. 2.5.9. Let X be a scheme over a field k. Show that the points of X(k) are closed in X. If X is an over k, then x ∈ X is closed if and only if k(x) is algebraic over k.

We prove the initial statement first assuming X ∼= Spec A is affine. Then p ∈ X(k) means that structure morphism X → Spec k has a section Spec k → X with image p. Translating these morphisms between affine schemes to ring homomorphisms, this means that A is a k-algebra such that there is a k-algebra homomorphism ϕ : A → k with ϕ−1(0) = p. Then ϕ is surjective and has kernel p, therefore p is a maximal ideal, that is, a closed point in Spec A. Now let X be an arbitrary k-scheme with x ∈ X(k). Let y ∈ {x}. Then y has an affine open neighborhood U such that x ∈ U. But by Remark 3.31, x ∈ U(k) and by what we have shown above,

{x} ∩ U = {x} .

So y = x, and hence {x} = {x}. We follow a similar strategy for the second statement. Let A be a finitely generated k-algebra, and x be a point in X = Spec A, given by the prime ideal p. Then

k(x) = OX,x/mx = Ap/pAp is the field of fractions of the integral domain A/p. So if x is a closed point, then p is maximal and hence k(x) = A/p is a finite algebraic extension of k by Corollary 2.1.12. Conversely, if k(x) is an algebraic extension of k, then A/p must be a field by the following lemma; so p is maximal and x is a closed point.

Lemma 7. Let k ⊆ K be an algebraic extension and let B be a subring of K that contains k. Then B is a field.

Proof. Let b ∈ Br {0}. Then b ∈ K is algebraic over k so the subring k[b] of B is a field; hence b has an inverse in B. 

Now let X be an arbitrary algebraic variety over k. Then X is a k-scheme that has Sn ∼ a finite open covering X = i=1 Xi such that Xi = Spec Ai where Ai is a finitely generated k-algebra. Let x ∈ X. If x is closed in X, then pick Xi such that x ∈ Xi.

Then x is closed in Xi and so by above k(x) = OX,x/mx = OXi,x/mx is algebraic over k. Conversely, if k(x) is algebraic over k, then for each Xi that contains x, by above x is closed in Xi. And if x∈ / Xi, then x lies in the XrXi and hence {x} ⊆ XrXi. MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 13

Thus, for every Xi, {x} ∩ Xi is either empty or equal to {x}, hence n [ {x} = ({x} ∩ Xi) = {x} , i=1 so x is closed in X. 2.5.11. (Schemes of dimension 0) (a) Let X be a scheme which is a (finite or not) disjoint union of open sub- ∼ Q schemes Xi. Show that OX (X) = i OX (Xi). (b) Show that any scheme of finite cardinal and dimension 0 is affine. (c) Let X = Spec A be a scheme of finite cardinal and dimension 0. Show that ∼ L every point x ∈ X is open. Deduce from this that A = p∈Spec A Ap. (d) Show that the statement (c) is false if we do not suppose that Spec A of dimension 0.

(a) Since OX is a sheaf and  Xi if i = j Xi ∩ Xj = ∅ if i 6= j , the diagram Y / Y O (X) / O (X ) O (X ) , X X i / X i i i where the parallel arrows are both the identity, is an equalizer diagram in the category ∼ Q of rings. Thus OX (X) = i OX (Xi).

(b) Since X has finite cardinal, it has finitely many irreducible components, say X1,...,Xn. Pick any x ∈ X1, Since x ⊆ X1 is a chain of irreducible closed subsets, because dim X = 0 we have x = X1. Hence every point of X1 is a generic point of X1, but then by Proposition 4.12 X1 must be a singleton. Similarly every Xi is a singleton, say {xi} and xi 6= xj if i 6= j. X = {x1, . . . , xn} and every point of X is closed, hence also open as |X| = n is finite. Therefore each {xi} must be affine, say isomorphic to Spec Ai where Ai is a ring with a unique prime ideal. Then n n ! ∼ G ∼ Y X = Spec Ai = Spec Ai . i=1 i=1 (c) This follows immediately from what we’ve shown above in (b).

(d) Let OK be a discrete valuation ring with field of fractions K. Then 0 and the unique maximal ideal are the only prime ideals of OK with respective localizations K and OK . However OK is not isomorphic to K × OK (the latter ring is not even a domain). 2.5.14. Let K ⊆ L be a finite field extension. Let x ∈ L. Then the multiplication by x is an endomorphism of L as a K-vector space. We let NormL/K (x) denote the determinant of this endomorphism. We also call it the norm of x over K.

(a) Show that NormL/K is a multiplicative map from L to K. (b) Let A ⊆ B be rings such that K = Frac(A), L = Frac(B), and that B is integral over A. Show that for any b ∈ B, NormL/K (b) is integral over A. (c) Let us moreover suppose that A is a polynomial ring over a field k. Show that NormL/K (B) ⊆ A. MATH 8253 ALGEBRAIC GEOMETRY WEEK 9 14

(a) Clearly NormL/K maps L to K. Given x ∈ L, let’s write mx for the endomorphism of L defined by “multiply by x”. Then observe that the map

θ : L → EndK (L)

x 7→ mx is a ring homomorphism, in particular multiplicative. And NormL/K is, by definition, the composition of θ with the multiplicative map det : EndK (L) → K. (b) Since L is a field, the map θ above is an embedding. Therefore the minimal polyno- mial f ∈ K[T ] of the K-linear endomorphism mb is precisely the minimal polynomial of the element b over K. So if we let g ∈ K[T ] to be the characteristic polynomial of mb, then by linear algebra we know that

• g(0) = ± det(mb) = NormL/K (b), • the irreducible factors of g and f are the same. r But f ∈ K[T ] is already irreducible, hence g = f for some r. Thus if b1, . . . , bn are the roots of f in a splitting field, then g(0) is a product of bi’s. But for each bi is the image of b under some field automorphism fixing K, hence is integral over B. Thus g(0), and hence NormL/K (b), is integral over B.

(c) By (b), NormL/K (B) lies in the integral closure of A in K. If A = k[T1,...,Tn] then A is a UFD, hence it is integrally closed. Thus NormL/K (B) ⊆ A.