Course Notes: Algebraic Geometry – Schemes 2

Course Notes | Amnon Yekutieli | 22 May 2019

Spring 2018-19

Amnon Yekutieli

Contents 1. Review of Prior Material 3 2. Aﬃne Schemes 10 3. Schemes 21 4. Fibered Products of Schemes 22 5. Closed Subspaces 30 References 41

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Course Notes | Amnon Yekutieli | 22 May 2019

1. Review of Prior Material

Lecture 1, 27 Feb 2019

We ﬁx a nonzero commutative base ring . All rings will be commutative -rings by default. Let X be a topological space. Recall that a presheaf of -modules on X is a functor M : Open(X)op → Mod , where Open(X) is the category of open sets of X, and Mod is the category of -modules. More concretely, the presheaf M is the data of a -module Γ(U, M) for every open set U ⊆ X, and a module homomorphism

restV/U : Γ(U, M) → Γ(V, M) for every inclusion V ⊆ U. The conditions are that

restW/U = restW/V ◦ restV/U for every double inclusion W ⊆ V ⊆ U, and that restU/U = id for every U. We often use the abbreviation m|V := restV/U (m) ∈ Γ(V, M) for a section m ∈ Γ(U, M). The presehaves of -modules on X form a category. The morphisms are the obvious pre ones. We denote it by Mod X . Here X is the constant sheaf on X with values in (but we didn’t deﬁne sheaves yet...) Given a presheaf M and a point x ∈ X, we have the stalk Mx of M at x. This is a -module. Recall the formula:

Mx = lim Γ(U, M), U→ where the direct limit is on the open neighborhoods U of x. Taking the stalk at x is a functor pre Mod X → Mod . A presheaf M is a sheaf if it satisfies the two sheaf axioms. These can be encoded as Ð follows: for every open set U ⊆ X and every open covering U = i∈I Ui, the sequence of -modules Ö δ0−δ1 Ö 0 → Γ(U, M) −→ Γ(Ui0 , M) −−−−−→ Γ(Ui0 ∩ Ui1 , M) i0 ∈I i0,i1 ∈I is exact. Here the homomorphisms , δi are induced by the restrictions. For more details see [Ye4, Sec 3 and Prop 7.10]. The sheaves of -modules on X, also called X -modules, form a full subcategory of pre Mod X , that we denote by Mod X . The sheafification functor assigns to each presheaf M a sheaf Sh(M), and a homomorphism of presheaves τM : M → Sh(M), which is universal for homomorphisms into sheaves. Namely: if N is a sheaf and φ : M → N is a homomorphism of presehaves, then there is a unique homomorphism of sheaves 0 0 φ : Sh(M) → N such that φ = φ ◦ τM. Exercise 1.1. Read the proof of the sheafication, including an understanding of the Gode- ment sheaf GSh(M). This is [Ye4, Thm 6.1]. We will need this next week when we define the structure sheaf of an affine scheme.

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Exercise 1.2. State the categorical property of the functor Sh, as an adjoint (from which pre side?) to the inclusion functor Mod X → Mod X . The sheaﬁcation does not change the stalks: for every x ∈ X the homomorphism: ' τM,x : Mx −→ Sh(M)x is bijective. A ringed space over is a pair (X, OX ), consisting of a topological space X, and a sheaf of -rings OX on X. Let (X, OX ) be such a ringed space. A sheaf of OX -modules, also called an OX -module, is a sheaf of -modules on X, together with a structure of a Γ(U, OX )-module for every open set U ⊆ X, which respects restrictions to open subsets. The category of OX -modules is denoted by Mod OX . The morphisms are the obvious ones. pre The sheaﬁfcation functor respects the OX -module structure: if M ∈ Mod OX then Sh(M) ∈ Mod OX , and τM is OX -linear. Let φ : M → N be a homomorphism in Mod OX . Its kernel is the OX -module Ker(φ) such that Γ(U, Ker(φ)) = Ker Γ(U, φ) : Γ(U, M) → Γ(U, N ) .

The image of φ is the OX -module Im(φ) := Sh(Impre(φ)), where Impre(φ) is the presheaf deﬁned by Γ(U, Impre(φ)) = Im Γ(U, φ) : Γ(U, M) → Γ(U, N ) . Note that Ker(φ) is a subsheaf of M and Im(φ) is a subsheaf of N . A sequence of homomorphisms φi φi+1 S = ··· Mi −−→ Mi+1 −−−→ Mi+1 ··· in Mod OX is called exact if for every point x ∈ X the sequence of homomorphisms i i+1 i φx i+1 φx i+1 ··· Mx −−→ Mx −−−→ Mx ··· in Mod OX,x is exact. We know that S is exact iﬀ for every i there is equality Im(φi−1) = Ker(φi) of these subsheaves of Mi. Given an open set U ⊆ X we write OU := OX |U . Thus (U, OU ) is also a ringed space. There is a restriction functor

Mod OX → Mod OU , M 7→ M|U . Sheaves, and homomorphisms between sheaves, can be glued. Ð Notation: given an open covering X = i∈I Ui, and indices i, j,... ∈ I, we often write

(1.3) Ui, j,... := Ui ∩ Uj ∩ · · · .

Theorem 1.4 (Gluing Sheaf Homomorphisms). Let M and N be OX -modules, let X = Ð i∈I Ui be an open covering, and let

φi : M|Ui → N |Ui be homomorphisms of OUi -modules satisfying the condition

φi |Ui, j = φj |Ui, j : M|Ui, j → N |Ui, j . (This is the 0-cocycle condition.)

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Then there is a unique homomorphism of OX -modules φ : M → N such that

φ|Ui = φi : M|Ui → N |Ui for all i. Ð Theorem 1.5 (Gluing Sheaves). Suppose X = i∈I Ui is an open covering. For every i let

Mi be an OUi -module, and for every i, j let ' φi, j : Mi |Ui, j −→ Mj |Ui, j be an isomorphism of OUi, j -modules. The condition is that

Moreover, the OX -module M, with the collection of isomorphisms {φi }, is unique up to a unique isomorphism. Exercise 1.6. Read and make sure you understand these last two thms. Proofs can be found here: [Ye4, Thm 7.3] and [Ye4, Thm 7.4].

Let M, N ∈ Mod OX . The OX -module HomOX (M, N ) is deﬁned by

Γ(U, HomOX (M, N )) := HomMod OX |U (M|U , N |U ).

The OX -module M ⊗OX N is the sheaf associated to the presheaf

(1.7) U 7→ Γ(U, M) ⊗Γ(U,OX ) Γ(U, N ). Exercise 1.8. Find an example of M and N such that the presheaf tensor product (1.7) is not a sheaf. (Hint: line bundles on P1)

Let f : Y → X be a map of topological spaces. For a Y -module N , its pushforward, or direct image, is the X -module f∗(N ) deﬁned by −1 Γ(U, f∗(N )) := Γ( f (U), N ) for open sets U ⊆ X. We get a functor

(1.9) f∗ : Mod Y → Mod X . −1 For a X -module M, the pullback, or inverse image, is the Y -module f (M) deﬁned by Γ(V, f −1(M)) := lim Γ(U, M), U→ where V ⊆ Y is open, and U runs over the open sets in X that contain f (V). We get a functor −1 (1.10) f : Mod X → Mod Y . There is adjunction: an isomorphism of -modules −1 (1.11) HomMod X (M, f∗(N )) HomMod Y ( f (M), N )

5 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019 which is functorial in M and N . Exercise 1.12. Prove the adjunction formula (1.11). Exercise 1.13. Prove that the functor f −1 in (1.10) is exact.

Exercise 1.14. Prove that the functor f∗ in (1.9) is left exact. Find and example showing that it is not exact. (Hint: line bundles on P1)

Let (X, OX ) and (Y, OY ) be ringed spaces. A map of ringed spaces

(1.15) ( f , ψ) : (Y, OY ) → (X, OX ) consists of a map of topological spaces f : Y → X, together with a homomorphism of X -rings

(1.16) ψ : OX → f∗(OY ). We shall often use the notation ( f , f ∗) instead of ( f , ψ). The notation ( f , ψ) is common in most textbooks, but it is a bit redundant, so we will only use it when it is need to clarify matters. Note that in many cases (see examples below) the homomorphism f ∗ is literally pullback of functions; and this makes the notation ( f , f ∗) good heuristically. On the other hand, we sometimes omit all mention of the structure sheaves, and just talk about a map f : Y → X of locally ringed spaces.

Lecture 2, 6 Mar 2019

Given a maps (1.15) of ringed spaces, the direct image is a functor

f∗ : Mod OY → Mod OX . ∗ There is another kind of inverse image here: for M ∈ Mod OX we deﬁne f (M) ∈ Mod OY by ∗ −1 ( ) ⊗ −1 ( ) f M := OY f (OX ) f M . Again there is adjunction: ∗ HomMod OX (M, f∗(N )) HomMod OY ( f (M), N ).

Deﬁnition 1.17. A locally ringed -space is a ringed -space (X, OX ), such that for every point x ∈ X the stalk OX,x is a local ring. The maximal ideal of OX,x is denoted by mx, and the residue ﬁeld is denoted by k(x).

Deﬁnition 1.18. If (Y, OY ) is another locally ringed space, then a map of locally ringed spaces ( f , ψ) : (Y, OY ) → (X, OX ) is a map of ringed spaces, such that for every point y ∈ Y, with image x := f (y) ∈ X, the induced -ring homomorphism

ψx : OX,x → OY,y is a local homomorphism, namely ψx(mx) ⊆ my. The category of locally ringed -spaces is denoted by LRSp/. Here are three types of locally ringed spaces. Example 1.19. The category Top of topological spaces. The base ring is = . A space X ∈ Top is made into a ringed space by putting on it the sheaf OX of continous -valued functions (for the metric topology of ). Then (X, OX ) belongs to LRSp/. Moreover, the functor Top → LRSp/ is fully faithful.

6 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Example 1.20. The category Mfld of differentiable (of type C∞) real manifolds. The base ring is = . A manifold X ∈ Mfld is made into a ringed space by putting on it the sheaf OX of differentiable -valued functions. Then (X, OX ) belongs to LRSp/. Moreover, the functor Mfld → LRSp/ is fully faithful.

Example 1.21. Let be an algebraically closed ﬁeld, and let Var be the category of algebraic varieties over . Here OX is the sheaf of algebraic -valued functions. Then (X, OX ) belongs to LRSp/. Moreover, the functor Var → LRSp/ is fully faithful.

Given a locally ringed space (X, OX ) and an open subset U ⊆ X, the pair (U, OX |U ) is a locally ringed space. We call it an open subspace of X. The inclusion map

(U, OX |U ) → (X, OX ) is called an open embedding. Just like sheaves, locally ringed spaces and maps between them can be glued.

Theorem 1.22 (Gluing Maps of LR Spaces). Let (X, OX ) and (Y, OY ) be objects of Ð LRSp/, let Y = i∈I Vi be an open covering, and for every i let

( fi, ψi) : (Vi, OY |Vi ) → (X, OX ) be a map in LRSp/. We assume that this condition holds: for every i, j ∈ I there is equality

( fi, ψi)|Vi, j = ( fj , ψj )|Vi, j of maps

(Vi, j , OY |Vi, j ) → (X, OX ) in LRSp/. Then there is a unique map

( f , ψ) : (Y, OY ) → (X, OX ) in LRSp/ such that

( f , ψ)|Vi = ( fi, ψi) for every i. See picture in ﬁgure 1.

Proof. The existence and uniqueness of a map of topological spaces f : Y → X satisfying

f |Vi = fi is clear. We need to produce the homomorphism of sheaves of rings

ψ : OX → f∗(OY ) on X. By the adjunction formula (1.11), this amounts to producing a homomorphism of sheaves of rings −1 (1.23) ψ : f (OX ) → OY on Y. Now we are given homomorphisms −1 ψi : f (OX )|Vi → OY |Vi that agree on double intersections. According to Theorem 1.4 these can be glued uniquely to a homomorphism of sheaves of -modules ψ as in (1.23), such that ψ|Vi = ψi. This is a homomorphism of sheaves of rings, because this propeorty can be checked locally. Also on stalks it is a local homomorphism. So ( f , ψ) is a map of locally ringed spaces.

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Figure 1. Gluing maps of spaces

( ) Theorem 1.24 (Gluing LR Spaces). Let Ui, OUi i∈I be a collection of objects of LRSp/. For every i, j ∈ I there is an open subset Ui, j ⊆ Ui, and an isomorphism ' ( fi, j , ψi, j ) : (Ui, j , OUi |Ui, j ) −→(Uj,i, OUj |Uj,i ) in LRSp/. These conditions hold:

(a) For every i there are equalities Ui,i = Ui and ( fi,i, ψi,i) = id. (b) For every i, j, k there are equalities

fi, j (Ui, j ∩ Ui,k ) = Uj,i ∩ Uj,k

of subsets of Ui, and

( fj,k , ψj,k ) ◦ ( fi, j , ψi, j ) = ( fi,k , ψi,k ) of isomorphisms

(Ui, j ∩ Ui,k , OUi |Ui, j ∩Ui,k ) → (Uk,i ∩ Uk, j , OUk |Uk,i ∩Uk, j ) in LRSp/.

Then there is an object (X, OX ) in LRSp/, with open embeddings

( fi, ψi) : (Ui, OUi ) → (X, OX ), such that ( fj , ψj ) ◦ ( fi, j , ψi, j ) = ( fi, ψi) as morphisms

(Ui, j , OUi |Ui, j ) → (X, OX ), and such that Ø X = fi(Ui). i∈I Moreover, the space (X, OX ), with the collection of morphisms {( fi, ψi)}i∈I , are unique up to a unique isomorphism.

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Figure 2. Gluing spaces: the input

Figure 3. Gluing spaces: the output

See ﬁgures 2 and 3 for an illustration. The data ( ) ( ) (1.25) Ui, OUi i∈I , fi, j , ψi, j i, j ∈I is called gluing data or descent data..

Proof. This is done in a few steps. Ý Step 1. We deﬁne the set X. Consider the disjoint union U := i∈I Ui. We deﬁne a relation ∼ on this set as follows: two points x, y ∈ U are in the relation x ∼ y if there are i, j ∈ I

9 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019 such that x ∈ Ui, j ⊆ Ui, y ∈ Uj,i ⊆ Uj , and fi, j (x) = y. This is an equivalence relation (exercise). We let X := U/∼, the quotient set. The canonical surjection is π : U → X. For each i there a map fi : Ui → X, gotten by composing the inclusion Ui ⊆ U with π. The map fi is injective (exercise). We identify Ui with its image fi(Ui) ⊆ X. With this identiﬁcation, there is equality Ui, j = Uj,i of subsets of X. Step 2. We put a topology on X. The disjoint union U gets the disjoint union topology. Then we put on X the quotient topology relative to the surjection π : U → X. Ð Each Ui is an open set of X, and so X = i Ui is an open covering (exercise).

Step 3. Now we construct the sheaf of rings OX . On each open set Ui ⊆ X we have a sheaf of rings OUi . On double intersections we have isomorphisms of sheaves of rings ' ψi, j : OUi |Ui, j −→ OUj |Ui, j , and these agree on triple intersections. By Theorem 1.5 we get a sheaf OX on X, with isomorphisms of sheaves ' ψi : OUi −→ OX |Ui such that ψj ◦ ψi, j = ψi on Ui, j . We see that OX is a sheaf of rings, and the stalks OX,x are local rings. So (X, OX ) is a locally ringed -space. By construction there are open embeddings

( fi, ψi) : (Ui, OUi ) → (X, OX ) satisfying the required compatibility. Step 4. The uniqueness is due to Theorem 1.22. Exercise 1.26. Finish the proof of the theorem. Exercise 1.27. Find an example of a collection of spaces as in the theorem, such that all the spaces Ui are separated topological spaces (i.e. Hausdorff), yet X is not separetd. Exercise 1.28. Suppose that the spaces and the gluing data are in Mfld (Example 1.20), the indexing set I is countable, and the topological space X is Hausdorff. Prove that (X, OX ) is an object of Mfld. Remark 1.29. The gluing in Thm 1.24 will be used to construct fiber products of schemes. Then the fiber products will allow us to define the notion of a separated map of schemes. This is the relative and generalized variant of the Hausdorff condition.

2. Affine Schemes Let A be a -ring. The prime spectrum of A is the set Spec(A) of prime ideals of A. As we all know, this is a topological space with the Zariski topology. By deﬁnition the closed sets of Spec(A) are the sets Z(a) := {p ∈ Spec(A) | a ⊆ p}, where a is some ideal in A. We sometimes refer to Z(a) as the zero locus of the ideal a. Here is the reason: to a prime ideal p we associate the local ring Ap and the residue ﬁeld

k(p) := Ap/pp. An element a ∈ A has a class a(p) ∈ k(p), coming from the canonical ring homomorphism A → k(p). Then (2.1) Z(a) = {p ∈ Spec(A) | a(p) = 0 for all a ∈ a}. Exercise 2.2. Prove the formula above.

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For an element s ∈ A we deﬁne (2.3) D(s) = {p ∈ Spec(A) | s < p}. This is an open set: it it the complement of the closed set Z(a), where a := (s), the principal ideal generated by s. We call such an open set a principal open set. Analogously to (2.1) we have (2.4) D(s) = {p ∈ Spec(A) | s(p) , 0}. Proposition 2.5. The principal open sets are a basis of the topology of Spec(A). Namely Ð every open set set U is a union U = i D(si) for a suitable collection {si } of elements of A. Exercise 2.6. Prove the proposition above. Deﬁnition 2.7. Let A be a ring, and write X := Spec(A) for this topological space. For an open set U ⊆ X we let S(U) ⊆ A be the multiplicatively closed set S(U) := {s ∈ A | s(p) , 0 for all p ∈ U}.

Let AS(U) be the localization of A w.r.t. S(U). Lemma 2.8. The assignment U 7→ S(U) is a presheaf of rings on X = Spec(A), that we denote by Opre. Proof. This is easy: if V ⊆ U is a smaller open set, then S(U) ⊆ S(V), so by the universal property of localization there is a unique A-ring homomorphism AS(U) → AS(V). Deﬁnition 2.9. Let A be a ring. The structure sheaf of Spec(A) is the sheaf of rings pre OSpec(A) := Sh(O ).

Lecture 3, 13 Mar 2019

pre By construction, O is a presheafs of A-rings on X := Spec(A), and OX is a sheaf of A-rings on X. Proposition 2.10. Let A be a ring and write X := Spec(A). For every point x = p ∈ X the stalk of OX at x is OX,x = Ap, the local ring at p. More precisely, there is a unique A-ring isomorphism OX,x Ap.

Proof. Let S(x) := A − p, the complement of p. By deﬁnition we have Ap = AS(x). The universal property of localization says that there is at most one A-ring isomorphism ' Ap −→ OX,x. We will produce it. Let us denote by Ux the set of principal open neighborhoods of x. These are the open sets U = D(s) for some s ∈ S(x). By Proposition 2.5, Ux is a basis of open neighborhoods of x. Because the stalks of the presheaf Opre and its associated sheaf O are the same, we have pre (2.11) OX,x = lim Γ(U, O ) = lim AS(U) U→ U→ where U runs over Ux. For each U = D(s) ∈ Ux we have S(U) ⊆ S(x), so there is a unique A-ring homomorphism AS(U) → AS(x) = Ap. Going to the limit in U we get an A-ring homomorphism

ψ : OX,x → Ap.

In the other direction, every element s ∈ S(x) belongs to S(U) for U := D(s) ∈ Ux, and hence s is invertible in AS(U). This means that every s ∈ S(x) is invertible in the A-ring OX,x. Hence there is a unique A-ring homomorphism

φ : Ap = AS(x) → OX,x .

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The homomorphism φ is surjective, because every f ∈ OX,x is the image of some fraction a/s ∈ AS(U), see (2.11). But s ∈ S(x), so a/s ∈ Ap and f = φ(a/s). Finally consider the A-ring homomorphism ψ ◦ φ from Ap to itself. By uniqueness there is equality ψ ◦ φ = idAp . This shows that φ is injective. In conclusion, φ is an isomorphism of A-rings. Corollary 2.12. Spec(A), OSpec(A) is a locally ringed space.

Deﬁnition 2.13. An aﬃne -scheme is a locally ringed space (X, OX ) ∈ LRSp/ which is isomorphic to Spec(A), OSpec(A) for some -ring A.

We now provide a more explicit description of the structur sheaf OSpec(A) in terms of its Godement sheaf from [Ye4, Sec 6]. Recall that for a presheaf of -modules M on a space X, its Godement sheaf GSh(M) is deﬁned by Ö Γ(U, GSh(M)) := Mx x ∈U for an open set U ⊆ X. Then Sh(M) ⊆ GSh(M) is the subsheaf of geometric sections. Here is what this means. There is a canonical homomorphism of presehaves M → GSh(M). For an open set U ⊆ X, a section m ∈ Γ(U, GSh(M)) is called geometric if for every point x ∈ U there is an open set V such that x ∈ V ⊆ U, and a section m0 ∈ Γ(V, M), such that 0 m|V = m ∈ Γ(V, GSh(M)). Specializing to our case it says that following: Proposition 2.14. Let (X, OX ) := Spec(A), OSpec(A) . Then OX is the subsheaf of GSh(Opre) consisting of the geometric sections. Speciﬁcally, let U ⊆ X be an open set, and let pre Ö f ∈ Γ(U, GSh(O )) = Ap. p∈U

The section f belongs to Γ(U, OX ) iff for every point x = p ∈ U there is an open set V such that x ∈ V ⊆ U, and elements a ∈ A and s ∈ S(V), such that pre Ö f |V = a/s ∈ Γ(V, GSh(O )) = Aq. q∈V See Figure 4. Here are some example of affine schemes. Example 2.15. Consider the ring A = . The affine scheme X := Spec() has these points: for every (positive) prime number p there is a maximal ideal m := (p). These are closed points of X, since Z(p) = {m}. The local ring is

(p) = {a/s | s < (p)} ⊆ .

The residue field is ¿p. The zero ideal p := (0) is also prime. Is is the generic point of X; namely its topological closure is X. The local ring and the residue field at p are . Exercise 2.16. Analyze the affine scheme Spec(A) for the ring A := [t], the polynomial ring in one variable over a field .

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Figure 4. Picture for Prop 2.14

Example 2.17. Suppose is a nonzero ring. For every n ≥ 0 there is an affine scheme n A := Spec([t1,..., tn]) called the n dimensional affine space over . Recall that for an ideal a ⊆ A, its zero locus is Z(a) = {p ∈ Spec(A) | a ⊆ p}. The radical of a is the ideal √ a := {a ∈ A | ai ∈ a for some i > 0} ⊆ A. Lemma 2.18. Let a ⊆ A be an ideal. Then √ Ù a = p p∈Z(a) and √ Z(a) = Z( a). Exercise 2.19. Prove the lemma. (Hint: do not use the Nullstellensatz.) Lemma 2.20. For ideals a, b ⊆ A the following are equivalent: √ √ (i) a = b. (ii) Z(a) = Z(b). Proof. Clear from Lemma 2.18. Definition 2.21. A topological space X is called quasi-compact if every open covering of X has a finite subcovering. Remark 2.22. The term “compact” is usually reserved for spaces that are Hausdorff and quasi-compact. Schemes are almost never Hausdorff. There is an analogous notion of separation, that we will study later. Cf. Exercise 1.27 and Rem 1.29.

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Proposition 2.23. Let A be a ring. The topological space X := Spec(A) is quasi-compact. Ð Proof. Let X = i∈I Ui be an open covering. For each i there is an ideal ai such that Í Ui = X − Z(ai). Write a := i ai. Then there is equality Ù Z(A) = = Z(ai) = Z(a) i∈I √ √ √ of subsets of X. By Lemma 2.20 we know that A = a. Since 1 ∈ A, we see that 1 ∈ a, Í 0 and hence 1 ∈ a. This says that we can express 1 as a ﬁnite sum: 1 = i∈I0 ai with I ⊆ I Í a ﬁnite subset and ai ∈ ai. We see that A = i∈I0 ai, and therefore Ù = Z(A) = Z(ai) i∈I0 Ð and X = i∈I0 Ui.

Lemma 2.24. Let s1,..., sm ∈ A. TFAE: Ð (i) Spec(A) = i D(si). Í (ii) There exists a1,..., am ∈ A s.t. 1A = i ai ·si. Proof. Í (i) ⇒ (ii): As in the proof of the proposition, A = i ai, where ai := (si). So the element Í 1A is a linear combination 1A = i ai ·si for some ai ∈ A. Í Í (ii) ⇒ (i): Here 1A ∈ i ai, so A = i ai, and, as in the proof of the proposition, Ð Spec(A) = i D(si).

By construction, for every s ∈ A the element s is invertible in the rings Γ(D(s), Opre) and Γ(D(s), OX ). Lemma 2.25. Let (X, OX ) := Spec(A), OSpec(A) . Suppose U ⊆ X is an open set, f ∈ Γ(U, OX ), and x = p ∈ U. Then there are elements a, s ∈ A such that x ∈ D(s) ⊆ U and

f |D(s) = a/s ∈ Γ(D(s), OX ). Proof. By Proposition 2.14 the point x = p ∈ U has an open neighborhood V ⊆ U, and elements b, t ∈ A, such that t ∈ S(V) and f |V = b/t ∈ Γ(V, OX ). According to Proposition 2.5 we can replace V with a smaller open neighborhood D(r) for some r ∈ A, i.e. x ∈ D(r) ⊆ V. Deﬁne the elements a := b·r ∈ A and s := t ·r ∈ A, and the open set W := D(s). Since D(r) ⊆ V ⊆ D(t), it follows that D(r) = D(t ·r) = D(s) = W. Also

a/s = (b·r)/(t ·r) = f |W ∈ Γ(W, OX ). Lemma 2.26. Let (X, OX ) := Spec(A), OSpec(A) . SupposeV ⊆ X is a quasi-compact open set, and f ∈ Γ(V, OX ). Then there are ﬁnitely many elements a1,..., am, s1,..., sm ∈ A Ðm such that V = i=1 D(si), and

f |D(si ) = ai/si ∈ Γ(D(si), OX ) for every i.

14 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Proof. By Lemma 2.25, for every point x = p ∈ V there are elements ax, sx ∈ A such that x ∈ D(sx) ⊆ V and

f |D(sx ) = ax/sx ∈ Γ(D(sx), OX ). Ð We have an open covering V = x ∈V D(sx). Because of quasi-compactness we can pass to a ﬁnite subcovering, that is to a ﬁnite subset {x1,..., xm} of V. Finally, by letting ai := axi and si := sxi we are done. A ring homomorphism ψ : A → B induces a map of sets (2.27) Spec(ψ) : Spec(B) → Spec(A) with formula Spec(ψ)(q) := ψ−1(q). Lemma 2.28. Let ψ : A → B be a ring homomorphism. (1) The map Spec(ψ) is continuous. (2) Suppose B = As for some s ∈ A. Then the image of Spec(ψ) is D(s), and

Spec(ψ) : Spec(As) → D(s) is a homeomorphism. Exercise 2.29. Prove the last lemma.

Lecture 4, 27 Mar 2019

Theorem 2.30. Let A be a ring and s ∈ A an element. Consider the aﬃne scheme (X, OX ) := Spec(A), OSpec(A) . There is a unique A-ring isomorphism

As Γ(D(s), OX ). Proof.

Step 1. Since the element s is invertible in the ring Γ(D(s), OX ), there is a unique A-ring homomorphism

(2.31) φ : As → Γ(D(s), OX ). We need to prove that φ is bijective.

Step 2. In this step we we will prove that φ is injective. Let’s write U := D(s). Since OX is a subsheaf of the Godement sheaf GSh(OX ), there is an embedding of A-rings Ö Γ(U, OX ) Γ U, GSh(OX ) = Ap. p∈U It thus suﬃces to prove that the homomorphism 0 Ö (2.32) φ : As → Ap p∈U is injective. Let’s write B := As. By Lemma 2.28 we know that Spec(B) = U as topological subspaces of X. For every p ∈ U the element s is invertible in Ap, and hence the 0 Î homomorphism Ap → Bp is bijective. So we can rewrite (2.32) as φ : B → p∈U Bp, and we need to prove it is injective. 0 0 Suppose b ∈ B is such that φ (b) = 0. Then φp(b) = 0 in Bp for all p ∈ U. This means that there is some element tp ∈ B such that p ∈ D(tp) and tp ·b = 0 in B. Now Ð U = Spec(B) = p∈U D(tp).

15 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

By quasi-compactness of U we can pass to a ﬁnite subcovering, indexed by {p1,..., pm}. Ðm Let ti := tpi ∈ B. Then ti ·b = 0 in B, and Spec(B) = i=1 D(ti). According to Lemma Í 2.24 as in the proof of Prop 2.23, there are elements c1,..., cm ∈ B such that 1B = i ci ·ti. Í Then b = 1B ·b = i ci ·ti ·b = 0. Step 3. We now prove that the homomorphism φ from (2.31) is surjective. As before let U := D(s). Take an element f ∈ Γ(U, OX ). We know that U = Spec(B) is a quasi-compact topological space. By Lem 2.26 there are ﬁnitely many elements ai, si ∈ A, 1 ≤ i ≤ m, Ðm such that U = i=1 D(si), and f |D(si ) = ai/si ∈ Γ(D(si), OX ) for every i. Step 2, when applied to the element si ·sj ·s ∈ A instead of to s, shows that for every i, j the A-ring homomorphism φi, j : Asi ·sj ·s → Γ D(si ·sj ·s), OX is injective. Recall that B = As, so that Asi ·sj ·s = Bsi ·sj . We know that −1 −1 f |D(si ·sj ·s) = ai ·si = aj ·sj ∈ Γ D(si ·sj ·s), OX . · −1 · −1 → Therefore ai si = aj sj in Bsi ·sj . The kernel of the localization homomorphism B

Bsi ·sj is known: there is a positive integer li, j such that

li, j (si ·sj ) ·(ai ·sj − aj ·si) = 0 in B. Taking l := max({li, j }) we obtain l (2.33) (si ·sj ) ·(ai ·sj − aj ·si) = 0 in B. l l+1 Deﬁne bi := ai ·si and ti := si . Then D(ti) = D(si) and −1 −1 (2.34) f |D(ti ) = ai ·si = bi ·ti ∈ Γ(D(ti), OX ). Also, from (2.33) we have

(2.35) ti ·bj = tj ·bi in B. Since m Ø (2.36) Spec(B) = D(ti), i=1 Í by Lem 2.24 we can ﬁnd elements c1,..., cm ∈ B such that 1B = i ci ·ti. Let Õ b := ci ·bi ∈ B. i For every i we have – using (2.35) and (2.34) –

Õ −1 Õ −1 Õ −1 φ(b)|D(ti ) = cj ·bj = ti · cj ·ti ·bj = ti · cj ·tj ·bi = ti ·bi = f |D(ti ) j j j in Γ(D(ti), OX ). But by (2.36) and the sheaf axioms this implies that φ(b) = f .

Lecture 5, 3 Apr 2019

Convention 2.37. From now on the expression Spec(A) refers to the locally ringed space Spec(A), OSpec(A) .

Corollary 2.38. For a ring A, with (X, OX ) := Spec(A), the canonical ring homomorphism A → Γ(X, OX ) is bijective.

16 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Proof. Take s = 1 in Thm 2.30.

Definition 2.39. The category of affine -schemes, denoted by Schaff/, is the full subcategory of LRSp/ on the affine schemes (see Definition 2.13).

Recall that Rng/ is the category of (commutative) -rings.

Proposition 2.40. The assignment that sends a locally ringed space (X, OX ) to the ring Γ(X, OX ), and a map of locally ringed spaces ( f , φ˜) : (X, OX ) → (Y, OY ) to the ring homomorphism

Γ(X, φ) : Γ(Y, OY ) → Γ(X, OX ), is a functor Γ : (LRSp/)op → Rng/.

Exercise 2.41. Prove proposition 2.40.

Before continuing, here is a general useful notion.

Deﬁnition 2.42. A morphism f : C → D in a category C is called an epimorphism if it has this cancellation property: for every morphisms g0, g1 : D → E in C, if g0 ◦ f = g1 ◦ f then g0 = g1.

Proposition 2.43. Let f : A → B be a morphism in the category Rng of commutative rings. If f is either surjective or a localization, then f is an epimorphism in Rng.

Exercise 2.44. Prove proposition 2.43.

Let (Y, OY ) be a LRSp and s ∈ Γ(Y, OY ). For a point y ∈ Y we denote by s(y) the image of the element s in the residue ﬁeld k(y). And we write

D(s) := {y ∈ Y | s(y) , 0}.

Lemma 2.45. Let (Y, OY ) be a LRSp and s ∈ Γ(Y, OY ). Then: (1) The set D(s) is open in Y. (2) The element s is invertible in the ring Γ(D(s), OY ).

Proof. (1) Take a point y ∈ D(s). Since s(y) , 0, it is an invertible element of the ﬁeld k(y). Because the stalk OY,y is a local ring, it follows that s is an invertible element of OY,y. Let t ∈ OY,y be the inverse of s, so s·t = 1 in OY,y. There is an open neighborhood V of y such 0 0 that t ∈ Γ(V, OY ). There is a smaller open neighborhood V of y s.t. s·t = 1 in Γ(V , OY ). Then s(y0) , 0 for all y0 ∈ V 0, so V 0 ⊆ D(s).

(2) Let V := D(s). As we saw above, every point y ∈ V has an open neighborhood Vy ⊆ V −1 and an element ty ∈ Γ(Vy, OY ) such that s·ty = 1. This means that ty = s in Γ(Vy, OY ), a fact that makes it unique. We conclude that ty = ty0 in Γ(Vy ∩ Vy0 , OY ). By the sheaf property we get t ∈ Γ(V, OY ), and it satisﬁes s·t = 1.

Theorem 2.46. Let A ∈ Rng/ and let (Y, OY ) ∈ LRSp/. Write B := Γ(Y, OY ) and (X, OX ) := Spec(A). Given a -ring homomorphism φ : A → B, there is a unique map

( f , φ˜) : (Y, OY ) → (X, OX ) in LRSp/, such that Γ(X, φ˜) = φ : A → B.

17 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Proof. Step 1. We prove uniqueness of f . Take a point y ∈ Y, and let x = p := f (y) ∈ X. Because φ˜y : OX,x → OY,y is a local homomorphism, and Γ(X, φ˜) = φ, we have a commutative diagram of rings

φ (2.47) A / B

φ˜y OX,x / OY,y

k(x) / / k(y)

The homomorphism k(x) → k(y) is injective. Comparing the two paths in this diagram we see that φ KerA −→ B → k(y) = KerA → k(p) = p. This formula determines the value f (y) = p. Step 2. Now we prove that the homomorphism of sheaves of rings

φ˜ : OX → f∗(OY ) is unique. Since the principal open sets U = D(s) ⊆ X, for s ∈ A, are a basis for the topology, it is enough to prove the uniqueness of the ring homomorphism

Γ(U, φ˜) : Γ(U, OX ) → Γ(V, OY ), for U := D(s) and V := f −1(U) ⊆ Y. Let’s examine this commutative diagram of rings

φ=Γ(X,φ˜) (2.48) A / B

Γ(U,φ˜) As = Γ(U, OX ) / Γ(V, OY ) Because the left vertical arrow is an epimorphism (see Prop 2.43), it follows that the ring homomorphism Γ(U, φ˜) is unique. Step 3. Here we start with the existence. We deﬁne the function f : Y → X by the formula from step 1, namely a point y ∈ Y is sent to the prime ideal

φ (2.49) p := KerA −→ B → k(y) ∈ X. We need to prove that f is continuous. It suﬃces to show that for a principal open set U = D(s) ⊆ X, its preimage f −1(U) is open. A little calculation using formula (2.49) shows that f −1(U) = D(φ(s)), and this is open in Y by Lemma 2.45. Step 4. Now we construct the homomorphism of sheaves of rings

(2.50) φ˜ : OX → f∗(OY ) on X. Let U ⊆ X be an open set. We need to specify the ring homomorphism

(2.51) φ˜U : Γ(U, OX ) → Γ(V, OY ) where V := f −1(U) ⊆ Y.

18 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

First we consider U = D(s) for s ∈ A. Then V = D(φ(s)). We know from Thm 2.30 that Γ(U, OX ) = As. The element φ(s) is invertible in Γ(V, OY ) by Lemma 2.45. So there is a unique A-ring homomorphism

(2.52) φ˜s : As → Γ(V, OY ). In particular, for s = 1, we get Γ(X, φ˜) = φ. By the uniqueness, if t ∈ A is another element, then the diagram of A-rings

(2.53) As / As ·t

φ˜s φ˜s ·t Γ(D(φ(s)), OY ) / Γ(D(φ(s·t)), OY ) is commutative. Now we consider an arbitrary open set U ⊆ X. We can cover U by principal open sets: Ð Ð U = i∈I D(si). Then V = i∈I D(φ(si)). There are exact sequences Ö Ö (2.54) 0 → Γ(U, OX ) → Γ(D(si), OX ) → Γ(D(si ·sj ), OX ) i i, j and Ö Ö (2.55) 0 → Γ(V, OY ) → Γ(D(φ(si)), OY ) → Γ(D(φ(si ·sj )), OY ). i i, j By the previous paragraph there are A-ring homomorphisms ˜ φsi : Γ(D(si), OX ) → Γ(D(φ(si)), OY ) and these agree on double intersections by diagram (2.53). So we obtain the ring homomorphism φ˜U in (2.51). As the open set U ⊆ X changes, these become a homomorphism of sheaves of rings φ˜ : OX → f∗(OY ).

Step 5. It remains to prove that ( f , φ˜) is local, i.e. φ˜y : OX,x → OY,y is a local homomorphism for every y ∈ Y and x := f (y). Let’s return to the construction of the map f . For a point y ∈ Y with image x = p = f (y) ∈ X we have this solid commutative diagram of rings:

(2.56) A / A/p / / k(p) !

φ ! Γ(Y, OY ) / k(y)

Because the slanted arrow is an injection, it extends to the ﬁeld of fractions k(p), i.e. the dashed arrow exists. On the other hand, all our ring constructions are over A, so we have the next commutative diagram

(2.57) A / Ap = OX,x

φ φ˜y Γ(Y, OY ) / OY,y

19 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Merging (2.56) and (2.57) we get this diagram:

(2.58) A / Ap = OX,x / k(p)

φ φ˜y Γ(Y, OY ) / OY,y / k(y) The left square is the commutative diagram (2.57), and the outer rectangle is the commutative diagram (2.56). Because A → Ap is an epimorphism, it follows that the right square is commutative. But this implies that φ˜y is a local homomorphism. Corollary 2.59. The assignment that sends a ring A to the aﬃne scheme Spec(A) is a functor op Spec : (Rng/) → Schaﬀ/

Proof. Say φ : A → B is a homomorphism in Rng/. Let (X, OX ) := Spec(A) and (Y, OY ) := Spec(B). By Thm 2.46 there is a map

Spec(φ) := ( f , φ˜) : (Y, OY ) → (X, OX ) in LRSp/. If ψ : B → C is another homomorphism in Rng/, with (Z, OZ ) := Spec(C), then there is a map Spec(ψ) := (g, ψ˜) : (Z, OZ ) → (Y, OY ) in LRSp/. Now Γ(Y, ψ˜) ◦ Γ(X, φ˜) = ψ ◦ φ, so the uniqueness clause in Thm 2.46 says that Spec(φ) ◦ Spec(ψ) = Spec(ψ ◦ φ).

Likewise Spec(idA) = idSpec(A). Corollary 2.60. The functor Spec : (Rng/)op → LRS/ is adjoint to the functor Γ : (LRS/)op → Rng/. Proof. Thm 2.46 produces a bijection ' (2.61) HomRng/ A, Γ(Y, OY ) −→ HomLRSp/ (Y, OY ), Spec(A) . We need to prove that this is bifunctorial. This is an exercise.

Exercise 2.62. Prove that the bijection (2.61) is functorial in A and in (Y, OY ). Corollary 2.63. The functor op Spec : (Rng/) → Schaﬀ/ is an equivalence of categories, with quasi-inverse Γ.

Proof. By deﬁnition, Schaﬀ/ is the essential image in LRSp/ of the functor Spec. We need to prove that Spec is fully faithful. Since there is adjunction (Cor 2.60), it is enough to prove that the unit of the adjunction

ηA : A → Γ(X, OX ), where (X, OX ) := Spec(A), is an isomorphism in Rng/ for every A. But this is Cor 2.38.

20 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Next week we will talk about schemes and ﬁber products.

Lecture 6, 10 Apr 2019

Deﬁnition 2.64. Let (X, OX ) be a LRSp and let U ⊆ X be an open subset. The LRSp (U, OX |U ) is called an open subspace of (X, OX ).

It is clear that (U, OX |U ) is a LRSp, and that the inclusion (U, OX |U ) → (X, OX ) is a map in LRSp/.

Proposition 2.65. Let A be a ring and s ∈ A. Let (X, OX ) := Spec(A). Consider the principal open set U := D(s) ⊆ X and the open subspace (U, OX |U ) of (X, OX ). Let φ : A → As be the canonical ring homomorphism. Then the map

Spec(φ) : Spec(As) → Spec(A) induces an isomorphism ' Spec(As) −→(U, OX |U ) in LRSp/.

Proof. Let’s write (Y, OY ) := Spec(As). We know that the image of the set Y is the subset U ⊆ X, and moreover that Y → U is a homeomorphism of topological spaces. It remains to prove that the sheaf homomorphism φ˜ : OX |U → OY on the topological space Y = U is an isomorphism. It is enough to check that it induces bijections on all stalks. But the stalks are OX,y = OY,y = Ap for y = p ∈ U, and the homomorphism induced by φ is the identity.

3. Schemes

Recall that an aﬃne -scheme is a locally ringed space (U, OU ) ∈ LRSp/ which is isomorphic to Spec(A) for some -ring A.

Definition 3.1. A -scheme is a locally ringed -space (X, OX ) that admits an open Ð covering by affine -schemes. Namely there is an open covering X = i∈I Ui, such that each open subspace (Ui, OX |Ui ) is an affine -scheme.

Deﬁnition 3.2. Let (X, OX ) and (Y, OY ) be -schemes. A map of -schemes

( f , ψ) : (Y, OY ) → (X, OX ) is, by deﬁnition, a maps of LR -spaces. Thus the category Sch/ of -schemes is the full subcategory of LRSp/ on the -schemes.

Proposition 3.3. Let (X, OX ) be a -scheme and let U ⊆ X be an open set. Then the open subspace (U, OX |U ) is a -scheme. Proof. Take a point x ∈ U. We need to find an open neighborhood V of x in U such that the open subspace (V, (OX |U )|V ) is an affine -scheme. We know (by Definition 3.1) that there is an open neighborhood W of x in X such that the open subspace (W, OX |W ) is an affine -scheme; say (W, OX |W ) = Spec(A). Because U ∩W is open in W, we can find an element s ∈ A such that x ∈ D(s) ⊆ U ∩ W. Write V := D(s). Now (V, (OX |U )|V ) = (V, OX |V ), and by Prop 2.65 we know that (V, OX |V ) Spec(As) as LRSp’s. This proposition makes the next definition sensible. comment: [(190425) next def new ]

Deﬁnition 3.4. Let (X, OX ) be a scheme.

21 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

(1) An open subscheme of (X, OX ) is a scheme (U, OU ) such that U ⊆ X is an open subset, and OU = OX |U . (2) An affine open subscheme of (X, OX ) is an open subscheme (U, OU ) which is an affine scheme, i.e. (U, OU ) Spec(A) for some ring A. (3) An affine open set of (X, OX ) is an open subset U ⊆ X such that the open subscheme (U, OX |U ) is an affine scheme.

In item (2) above the ring A is Γ(U, OU ), by Corollary 2.38. Here are two examples of schemes which are not aﬃne.

Example 3.5. Let be a nonzero ring. (If you prefer, you can assume that is a field, or even an algebraically closed field.) We define 1-dimensional projective space to be the 1 1 following scheme P. Let U0 := Spec([t1]) and U1 := Spec([t0]). So U0 U1 A, the 1-dimensional affine space over , see Exa 2.17. Inside U0 we have the affine open −1 subscheme U0,1 := Spec([t1, t1 ]), and inside U1 we have the affine open subscheme −1 −1 ' −1 −1 U1,0 := Spec([t0, t0 ]). The ring isomorphism [t0, t0 ] −→ [t1, t1 ], t0 7→ t1 , induces ' an isomorphism of schemes φ0,1 : U0,1 −→ U1,0. By Thm 1.24 we can glue U0 and U1 along 1 φ0,1. The resulting LRSp is the scheme P.

1 1 Exercise 3.6. Show that P is not an aﬃne scheme. (Hint: write (X, OX ) := P. By direct calculation show that the ring Γ(X, OX ) = . Then use Cor 2.63.)

Example 3.7. Let be a nonzero ring. (If you prefer, you can assume that is a field, 2 or even an algebraically closed field.) Consider the affine plane A = Spec([t1, t2]). The 2 “origin” is the closed subset Z := Z(t1, t2) ⊆ A. If is a field, then Z contains only one 2 point; but in general it is the topological space Spec(), a closed subset of A. 2 2 Define the “punctured plane” U := A − Z. This is an open subscheme (U, OU ) of A.

Exercise 3.8. Show that the punctured plane (U, OU ) is not an aﬃne scheme. (Hint: deﬁne

−1 2 Ui := Spec([t1, t2, ti ] ⊆ A, an aﬃne open subscheme. Show that U = U1 ∪ U2. Use this covering to calculate the ring Γ(U, OU ). Show that Γ(U, OU ) = [t1, t2]. Then use Cor 2.63.)

Exercise 3.9. Let U be the punctured plane from Example 3.7. Show that there is a 1 "canonical" surjective map of schemes g : U → P. Show that when is a ﬁeld, then set-theoretically this is the usual way we get the projective line.

4. Fibered Products of Schemes Fibered products are very important in the theory of schemes. To be clear what we mean, here’s a deﬁnition

Deﬁnition 4.1. Let f : Y → X and g : Z → X be morphisms in a category C.A ﬁbered product of Y and Z over X is an object W ∈ C, with morphisms p : W → Y and q : W → Z, satisfying these two conditions: (i) f ◦ p = g ◦ q. (ii) Given an object W 0, and morphisms p0 : W 0 → Y and q0 : W 0 → Z, such that f ◦ p0 = g ◦ q0, there exists a unique morphism r : W 0 → W such that p0 = p ◦ r and q0 = q ◦ r.

22 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Figure 5. Fibered product

This is illustrated in the next commutative diagram. 0 W q0 r

q W / Z p0 p g f Y / X It is clear that the fibered product (W, p, q) is unique, up to a unique isomorphism. The notation for the fibered product is Y ×X Z, leaving the morphisms implicit. See Figure 5. Example 4.2. In the category Set the fiber product is

Y ×X Z = {(y, z) | f (y) = g(z)} ⊆ Y × Z, with the obvious projections p, q. Here is another name for the same categorical construct. Deﬁnition 4.3. A diagram q W / Z

p g f Y / X in the category C is called cartesian if W = Y ×X Z. We can also describe the ﬁbered product in terms of representable functors. Given f : Y → X and g : Z → X as in Deﬁnition 4.1, let F : Cop → Set be this functor:

F(U) := HomC(U,Y) ×HomC(U,X) HomC(U, Z) ∈ Set .

23 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

I.e. the elements of F(U) are pairs (p0, q0) of morphisms p0 : U → Y and q0 : U → Z such 0 0 that f ◦ p = g ◦ q . The ﬁbered product W = Y ×X Z is an object of C that represents F:

F HomC(−, W) as functors Cop → Set. The pair of projections (p, q) ∈ F(W) is the universal pair. The next theorem gives a suﬃcient condition for the existence of the ﬁbered product in LRSp/. Theorem 4.4. Let f : Y → X and g : Z → X be maps in LRSp/. Assume that there is a collection of open subspaces {Xi }i∈I of X, a collection of open subspaces {Yi }i∈I of Y, and a collection of open subspaces {Zi }i∈I of Z, such that these three conditions hold:

(a) For every i ∈ I there are inclusions f (Yi) ⊆ Xi and g(Zi) ⊆ Xi.

(b) For every i ∈ I the ﬁbered product Yi ×Xi Zi exists. (c) For every pair of points (y, z) ∈ Y × Z such that f (y) = g(z), there is an index i ∈ I such that y ∈ Yi and z ∈ Zi.

Then the ﬁbered product Y ×X Z exists. Moreover, for every i the canonical map

Yi ×Xi Zi → Y ×X Z is an open embedding, and Ø Y ×X Z = Yi ×Xi Zi . i∈I The proof requires two lemmas. Lemma 4.5. Let f : X 0 → X be an open embedding in LRSp/. Then f is a monomor- phism in the category LRSp/. Lemma 4.6. Let f : Y → X and g : Z → X be maps in LRSp/. Let X 0 ⊆ X, Y 0 ⊆ Y and Z 0 ⊆ Z be open subspaces, such that f (Y 0) ⊆ X 0 and g(Z 0) ⊆ X 0. Assume the ﬁber 0 product Y ×X Z exists, with projections p, q. Let W be the open subspace 0 −1 0 −1 0 W := p (Y ) ∩ q (Z ) ⊆ Y ×X Z, 0 0 0 0 0 0 0 with induced projections p : W → Y and q : W → Z . Then W = Y ×X0 Z . See Figure 6 for an illustration of this lemma. Exercise 4.7. Prove Lemmas 4.5 and 4.6.

Proof of Thm 4.4. Let’s write Wi := Yi ×Xi Zi. The projections are pi : Wi → Yi and qi : Wi → Zi. See Figure 7 for an illustration. For a pair of indices i, j deﬁne Xi, j := Xi ∩ Xj , Yi, j := Yi ∩ Yj and Zi, j := Zi ∩ Zj . Let −1 −1 Wi, j := pi (Yi, j ) ∩ qi (Zi, j ) ⊆ Wi . By Lemma 4.6 we know that

Wi, j Yi, j ×Xi, j Zi, j . ' But Yi, j = Yj,i, etc., so there is an isomorphism hi, j : Wi, j −→ Wj,i, and this isomorphism is determined by the properties of fibered products, namely by its compatibility with the projections to Y and to Z. This implies that the conditions of Thm 1.24 are satisfied: hj,k ◦ hi, j = hi,k . In other words, these three isomorphism between three incarnations of the fibered product Yi, j,k ×Xi, j, k Zi, j,k are compatible. According to Thm 1.24 we get a Ð locally ringed space W, with a covering W = i∈I Wi by open subspaces. The projections pi : Wi → Y and pj : Wj → Y agree on Wi, j = Wj,i, now seen as open subspaces of W. Therefore they glue to a map p : W → Y. Likewise the projections −1 −1 qi : Wi → Z glue to a map q : W → Z. By Lemma 4.6 we know that Wi = p (Yi)∩q (Zi) as subspaces of W.

24 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Figure 6. For Lem 4.6

Figure 7.

It remains to prove that (W, p, q) is the ﬁbered product Y ×X Z. Take a space U and maps p0 : U → Y and q0 : U → Z such that f ◦ p0 = g ◦ q0. Deﬁne

0 −1 0 −1 Ui := p (Yi) ∩ q (Zi) ⊆ U.

25 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Ð Then U = i∈I Ui is an open covering. For every i there is a unique map ri : Ui → Wi that’s compatible with the projections, since Wi = Yi ×Xi Zi. By the arguments used above, the maps ri and rj agree on Ui, j := Ui ∩ Uj . So they glue to a map r : U → W. This r is unique because all its local pieces ri are unique. So W = Y ×X Z as claimed.

This is where lecture 6 ended. From here on it is self-reading during the vacation.

Before proving the existence of the fibered product of schemes, a little reminder on rings. The fibered coproduct is the dual (or opposite) notion of fibered product. To be a bit more precise, the fibered coproduct in a category C is the fibered product in the category Cop.

Lemma 4.8. Let f ∗ : A → B and g∗ : A → C be homomorphisms in Rng/. Then the ∗ ∗ ring D := B ⊗A C, with the canonical homomorphisms p : B → D and q : C → D, is the ﬁbered coproduct of f ∗ and g∗ in Rng/. Proof. We know from commutative algebra that for a ring E, the ring homomorphisms r∗ : D → E are in bijection with the pairs of A-ring homomorphisms p0 ∗ : B → E and q0 ∗ : C → E such that f ∗ ◦ p0 ∗ = g∗ ◦ q0 ∗.

The solid diagram g∗ A / C

f ∗ q∗ p∗ q0 ∗ B / D r∗

0 ∗ p . E in the category Rng/, as in the lemma, is called cocartesian. Theorem 4.9. Let f : Y → X and g : Z → X be maps in LRSp/, and assume these three spaces are schemes. Then the fibered product Y ×X Z in LRSp/ exists, and it is a scheme. Moreover, if X,Y, Z are all affine schemes, then the fibered product Y ×X Z is an affine scheme. Proof. Step 1. Suppose X = Spec(A), Y = Spec(B) and Z = Spec(C). There are ring homomor- ∗ ∗ phisms f : A → B and g : A → C. Define D := B ⊗A C ∈ Rng/ and W := Spec(D). The canonical ring homomorphisms B → D and C → D give rise to maps p : W → Y and q : W → Z. We claim that (W, p, q) = Y ×X Z. This is a consequence of Thm 2.46 and Lem 4.8. Indeed, given a space (U, OU ) ∈ LRSp/ with ring of functions E := (U, OU ), we know from Thm 2.46 that morphisms from U to affine schemes are in bijection with ring homomorphisms to E. By Lem 4.8, with arrows reversed, we see that maps U → W are in bijection with pairs of maps U → Y and U → Z whose composed maps to X aree equal. This is the property of the fibered product. Step 2. Take a pair of points (y, z) ∈ Y × Z such that f (y) = g(z), and let x := f (y). Choose some affine open neighborhood Xx of x in X. Then choose an affine open neighborhood Yy −1 −1 of y in f (Xx), and an affine open neighborhood Zz of z in g (Xx). We get a collection of open subspaces as in conditions (a) and (c) of Thm 4.4, with indexing set I consisting of these pairs (y, z), which is the fibered product of the underlying sets.

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Step 3. We have collections of affine open subschemes {Xi }i∈I of X, {Yi }i∈I of Y and {Zi }i∈I of Z, satisfying conditions (a) and (c) of Thm 4.4. These collections could be from the construction in step 2 above, but not necessarily so. According to step 1, for every index i ∈ I the fibered product Wi := Yi ×Xi Zi exists, and it is an affine scheme. Now Thm 4.4 applies. We get an object W = Y ×X Z in LRSp/, and its has an open covering by the affine schemes Wi. Hence W is a scheme. Fiber products will be used a lot. One operation they provide is "base change". Here are a few examples of this feature. But first a definition. Definition 4.10. Fix a scheme X (say in Sch/). By an X-scheme we mean a scheme Y together with a map πY : Y → X. The X-schemes form a category in the obvioous way, and we denote it by Sch/X.

If X = Spec(A) is affine, then the expressions Sch/X and Sch/A have the same meaning. Example 4.11. The projective line is "defined over ", or is "gotten by base change from 1 P". Here is what this means. In Example 3.5, taking := , we saw how to build the 1 1 -scheme P. But for any ring A we can also build PA. It is not hard (and maybe useful to try) to show that 1 1 PA P ×Spec() Spec(A) 1 in Sch/A. (By abuse of notation we sometimes write P × A instead of the longer expression above.) We can also take an arbitrary scheme X (not affine), and form the scheme 1 1 PX := P ×Spec() X. Example 4.12. Let X be a scheme. A group scheme over X is a scheme G ∈ Sch/X equipped with maps mult : G ×X G → G, inv : G → G and unit : X → G in Sch/X , that satisfy the group axioms.

Example 4.13. To be concrete, for n ≥ 1 we have the group scheme GLn,. It is an aﬃne group scheme: as a scheme over it is Spec(A), where A is the polynomial ring over in 2 the n variables ti, j , localized w.r.t. the determinant. You can try to write the formulas for the ring homomorphisms that yield the group operations. For an arbitrary scheme X we have the group scheme

GLn,X := GLn, ×Spec() X.

Example 4.14. We already know (from the last example) the -group scheme G := GL1,. As an -scheme it is G = Spec(A), where A := [t, t−1]. Now let B := [s, t]/(s2 +t2 −1). The aﬃne scheme H := Spec(B) is also a group scheme over . The multiplication is that of the circle. Try to guess the formulas. One can show that G and H are not isomorphic as group schemes over . (In fact, they are not isomorphic as -schemes.) However, after base change to ¼ they become isomorphic:

G ×Spec() Spec(¼) H ×Spec() Spec(¼) GL1,¼ as group schemes over ¼. Here is a quick discussion of “points of a scheme”, that could help understanding the examples on group schemes above. Let X be a -scheme. Given a -ring A, we deﬁne the set X(A) := HomSch/ Spec(A), X .

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The elements of X(A) are called the A-points of X. Key example: for the scheme n X = A = Spec([t1,..., tn]) we have, by virtue of Cor 2.63, n n A(A) = HomRng/ [t1,..., tn], A = HomSet {t1,..., tn}, A = A , the set of n-tuples of elements of the ring A. If G is a group scheme over , then the set G(A) is a group, in the ordinary sense of the word. For G = GLn,, the group G(A) = GLn,(A) is nothing but the group of invertible n × n matrices with entries in A. The notions of “points” and “base change” play the following game with each other. Suppose we have a homomorphism A → B in Rng/, and a scheme X ∈ Sch/. Deﬁne the A-scheme XA := X ×Spec() Spec(A). Since B is both in Rng/ and in Rng/A, we can consider the sets X(B) = HomSch/ Spec(B), X and XA(B) = HomSch/A Spec(B), XA . I leave it as an exercise to prove that these sets are equal; or more precisely, the canonical n function XA(B) → X(B) is bijective. For this reason we can write A (B) unambiguously – n n n it can mean either A(B) or AA(B), for B ∈ Rng/A, but these sets are equal (to the set B ). Likewise for the group GLn(A).

Example 4.15. Continuing Exa 4.14, the group scheme G = GL1, has the group G() = × GL1() = as its group of -points. For the group scheme H, the group of -points H() is the unit circle in the plane. This should help figure out the formulas for the operations of H: think of H() as a subgroup of ¼×, with s as the real part and t as the imaginary part. The two group schemes G and H are not isomorphic. One way to see this is to compare their groups of -points. In the group G() = R× the torsion subgroup is {±1}, whereas in the H() the torsion subgroup is the group of roots of 1 in ¼, that’s isomorphic to /. However, there is an isomorphism G¼ H¼ of group schemes over ¼, and hence the groups of ¼-points are: × H(¼) = H¼(¼) G¼(¼) = G(¼) = ¼ . comment: [(190426) More on these group schemes] The group scheme H is called the 1-dim nonsplit real torus, and the group scheme G = GL1, is called the 1-dim split real torus, and it is also called the mutiplicative group Gm,. The groups G and H are called the real forms of the group GL1,¼ = Gm,¼. Also H is called a twisted form of the group G. There is a way to classify these forms, and to show there are only two isoclasses of them – this is done by 1-st Galois cohomology, which the 0-dimensional version of étale cohomology. Here is a sketch. The classification is analogous to the classification of vector bundles in [Ye4, Thm 11.14] by 1-st nonabelian Čech cohomology, with values in the sheaf of automorphisms of the trivial vector bundle, i.e. in GLn(OX ). For vector bundles on a scheme (X, OX ) we do use the Zariski topology. However, for more complicated “bundles” the Zariski topology is not fine enough. (For the scheme Spec(), that has a single point, the Zariski topology is useless!)

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Instead of the Zariski topology, we must use the étale topology of X, that we might not have time to study in this course. An étale covering f : Y → X is the algebraic geometry version of a covering space in algebraic topology. (The conditions are these: f is ﬁnite and étale.) The covering Y → X is called Galois if the group AutSch/X (Y) has the same size as the ﬁbers of f (assuming X is connected). The covering is universal if Y itself has no nontrivial étale coverings. This theory was discovered by Grothendieck. In the case of X = Spec(), the map (4.16) f : Spec(¼) → Spec() is a universal covering space, with Galois group ∆ = {id, σ}, where σ is complex conjuga- tion. We want to classify twisted forms of the -group-scheme Gm,. For that we need to knoe the group-sheaf of automorphisms of the group-scheme Gm, in the étale topology. (Think of the vector bundle analogy.) Let’s do it slowly. We start by identifying the group N() of automorphisms of Gm, as an -group- −1 scheme. Since Gm, = Spec([t, t ]), and we know the group structure, the conclusion is that the only automorphisms are id and γ : t 7→ t−1. So the group is N() = {id, γ}. We now guess that these are the -points of the -group-scheme N := Spec() t Spec() with two points, and with the obvious group structure. It turns out that the sheaf of automorphisms of Gm, in the étale topology of Spec() is indeed represented by the group scheme N. Most importantly, on the universal covering (4.16) the group of automorphisms of Gm,¼ as a ¼-group-scheme is N(¼) = {id, γ}. 1 The cohomology calculation we need now is of Het(Spec(), N), which is the étale 1 analogue of Hˇ (U, GLn(OX )) on a covering U. Since (4.16) is a Galois covering, the Čech cohomology is the same as the group cohomology. The action of the Galois group ∆ on the group N(¼) is trivial. Therefore the pointed set (by chance an abelian group) is 1 1 1 Het(Spec(), N) H (∆, N(¼)) Ext[∆](, N(¼)) N(¼) = {id, γ}.

So there are two twisted forms of Gm,, and we already know them. Details can be found in found in [Mil, Section 3.k].

Lecture 7, 1 May 2019

Recall that for a locally ringed space (X, OX ), a point x ∈ X and an element (a “function”) a ⊆ Γ(X, OX ), we let a(x) ∈ k(x) be the image of a in the residue ﬁeld k(x). New notation, to replace Z(a) and D(a) : comment: [ (190429) is this good notation? ]

Deﬁnition 4.17. Let (X, OX ) be a locally ringed space.

(1) Given an element a ⊆ Γ(X, OX ), its zero locus is the set

ZerX (a) := {x ∈ X | a(x) = 0} ⊆ X.

(2) Given an ideal a ⊆ Γ(X, OX ), its zero locus is the set Ù ZerX (a) := ZerX (a) ⊆ X. a∈a

(3) Given an element a ⊆ Γ(X, OX ), its non-zero locus is the set

NZerX (a) := {x ∈ X | a(x) , 0} ⊆ X.

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Proposition 4.18. Let (X, OX ) be a locally ringed space. The set NZerX (a) is open in X, and the sets ZerX (a) and ZerX (a) are closed in X.

Proof. According to Lemma 2.45 the set NZerX (a) is open. Then its complement ZerX (a) is closed. The claim about ZerX (a) is now clear.

5. Closed Subspaces comment: [(190425) new def 3.4 – open subscheme ] In Definition 2.64 we learned about open subspaces, and in Definition 3.4 we saw open subschemes. Closed subspaces and closed subschemes are much more complicated. Before talking about them we need to understand a few more basic things: supports and quotients of sheaves. Definition 5.1. Suppose (X, A) is a ringed space. (1) Let M be a sheaf of A-modules on X. The support of M is the set

Supp(M) := {x ∈ X | Mx , 0} ⊆ X. (2) Let M ∈ Mod A and let Y ⊆ X. We say that M is supported on Y if Supp(M) ⊆ Y. (3) Given a subset Y ⊆ X, we denote by ModY A the full subcategory of Mod A on the sheaves supported on Y.

Note that here we are looking at the stalks Mx, and (X, A) need not be a locally ringed space. Exercise 5.2. Let M N be a surjection in Mod A. Show that Supp(N ) ⊆ Supp(M). Definition 5.3. Let (X, A) be a ringed space and let M be an A-module. We say that M is locally finitely generated over A if there is an open covering {Ui }i∈I of X, and for each ⊕ei index i there is a surjection of A|Ui -modules (A|Ui ) M|Ui for some ei ∈ . Proposition 5.4. Let (X, A) be a ringed space and let M be an A-module. Assume that M is locally finitely generated. Then Supp(M) is a closed subset of X. Note that, unlike Proposition 4.18, here this need not be a LR space.

Proof. Being closed is a local question: it is enough to prove that Y ∩ Ui ⊆ Ui is closed for ⊕e all i. So, by replacing X with Ui, we may assume that there is a surjection A M of A-modules. Let m1,..., me ∈ Γ(X, M) be the corresponding generators on M (the images of the e copies of 1A). Then Ø Supp(M) = Supp(mi), i=1,...,e where for a global section m ∈ Γ(X, M) we deﬁne

Supp(m) := {x ∈ X | mx , 0 in Mx } ⊆ X. It now suﬃces to prove that Supp(m) is closed in X. We will prove that the complement V := X − Supp(m) is open in X. Take a point x ∈ V. Then mx = 0 in the stalk Mx. By the direct limit deﬁnition of the stalk, there is some open neighborhood W of x s.t. mx0 = 0 for all x0 ∈ W. This means that W ⊆ V. So V is open in X.

Exercise 5.5. Find an example of a scheme (X, OX ) and an OX -module M whose support is not closed. Try to characterize the sets Y ⊆ X that can occur as Supp(M) fore some OX -module M.

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−1 Recall that for a map f : Y → X in Top there is adjunction between f∗ and f , i.e. an isomorphism of -modules

−1 (5.6) HomMod X (M, f∗(N )) HomMod Y ( f (M), N ) which is functorial in M ∈ Mod X and N ∈ Mod Y .

Proposition 5.7. Suppose X is a topological space, and Y ⊆ X is a closed subset, with inclusion map f : Y → X. Then:

(1) Let N be a sheaf of -modules on Y. Then the sheaf f∗(N ) is supported on Y, and −1 the adjunction homomorphism N → f ( f∗(N )) is an isomorphism in Mod Y . (2) The functor

f∗ : Mod Y → ModY X is an equivalence.

Exercise 5.8. Prove this proposition.

The proposition tells us that when Y ⊆ X is a closed set, we can safely identify Y - modules with X -modules supported on Y. This is false for subsets that are not closed, as the next exercise shows.

Exercise 5.9. Find an example of a scheme (X, OX ), an open set U ⊆ X with inclusion g : U → X, and an OU -module M, such that the set Supp(g∗(M)) is bigger than U.

Given a sheaf M, we know what is a subsheaf N ⊆ M. But what’s a quotient sheaf of M ? Let X be a topological space, and let M ∈ Mod X .A quotient sheaf of M is a 0 X -module N together with a surjection π : M N in Mod X . Suppose π : M N 0 is another quotient sheaf of M.A homomorphism of quotients φ : N → N 0 is a homomorphism in Mod X such that π = φ ◦ π. In this way the quotients of M form a category. Note that if π : M N and π0 : M N 0 are quotients of M, then there is at most one homomorphism of quotients φ : N → N 0, and it is a surjection too. In case φ is an isomorphism, then of course there is an isomorphism of quotients φ−1 : N 0 → N , and these are unique. This means that we can identify the isomorphic quotients N and N 0 via the unique isomorphism φ between them. (Formally this means passing to isomorphism classes, and then choosing one representative from each isomorphism class.) Having done so, the set of (isoclasses of) quotients of M becomes partially ordered, by the relation of existence of morphism of quotients.

Exercise 5.10. This is a continuation of Exercise 5.2. Fix M ∈ Mod X . Show that the function Supp, from the set of quotients of M to the set of subsets of X, reverses the partial orders. (I.e. it is a contravariant functor.)

We now want to talk about quotients of sheaves of rings. Let (X, A) be a ringed space. A quotient ring of A is an A-ring B such that the structural homomorphism π : A → B is surjective. Suppose π : A B and π0 : A B0 are quotient rings of A.A homomorphism of quotient rings φ : B → B0 is an A-ring homomorphism. It is necessarily surjective, and there is at most one. It two quotient rings B and B0 of A are isomorphic, then this isomorphism is unique. Hence, as in the case of X -modules, we can identify isomorphic quotient rings of A. Having done so, the quotient rings of A are partially ordered.

Deﬁnition 5.11. Let (X, A) be a ringed space. An ideal of A is an A-submodule I ⊆ A.

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Lemma 5.12. Let (X, A) be a ringed space. The assignment π : A B 7→ I := Ker(π) ⊆ A is an order preserving bijection from the set of (isoclasses of) quotient rings of A to the set if ideals of A. Exercise 5.13. Prove this lemma. Deﬁnition 5.14. Let (X, A) be a ringed space. (1) A closed subspace of (X, A) is a ringed space (Y, B), such that Y ⊆ X is a closed subset, and B a quotient ring of A. (2) Suppose (Z, C) is another closed subspace of (X, A). A map of closed subspaces (Z, C) → (Y, B) is an inclusion Z ⊆ Y of closed subsets of X, and a homomorphism B → C of quotient rings of A. Lemma 5.15. Let (X, A) be a ringed space, and let (Y, B) and (Z, C) be closed subspaces of (X, A). There is at most one map of closed subspaces (Y, B) → (Z, C). Exercise 5.16. Prove this lemma. In view of this lemma, we can identify isomorphic closed subspaces of (X, A). comment: [(190506) Next lem was a prop. And it was wrong. Here’s the corrected version.]

Lemma 5.17. Let (X, A) be a ringed space. The assignment I ⊆ A 7→ (Y, B) ⊆ (X, A), sending an ideal I ⊆ A to the quotient ring B := A/I and the closed set Y := Supp(B) ⊆ X, is an order reversing bijection from the set of ideals I ⊆ A to the set of (isoclasses of) closed subspaces (Y, B) ⊆ (X, A) such that Supp(B) = Y. Exercise 5.18. Prove this lemma. (Hint: Use Lemma 5.12 and Proposition 5.4.) Example 5.19. In the lemma above, suppose (Y, B) ⊆ (X, A) is such that Supp(B) = Y and Y X, a closed strict subset. Then the pair (X, B) is also a closed subspace of (X, A), and it does not match an ideal I ⊆ A. This is why the earlier version of the result above was wrong.

Lecture 8, 6 May 2019

Last week we learned about closed subspaces of ringed spaces. Lemma 5.20. Let (X, A) be a locally ringed space, and let (Y, B) be a closed subspace of it, such that Supp(B) = Y. Then (Y, B) is a locally ringed space, and the inclusion ( f , φ) : (Y, B) → (X, A) is a map of LR spaces. Exercise 5.21. Prove this lemma. (Hint: a nonzero quotient of a local ring is a local ring.) Proposition 5.22. Let (X, A) be a locally ringed ringed space. The assignment I ⊆ A 7→ (Y, B) ⊆ (X, A), sending an ideal I ⊆ A to the quotient ring B := A/I and the closed set Y := Supp(B) ⊆ X, is an order reversing bijection from the set of ideals I ⊆ A to the set of (isoclasses of) closed locally ringed subspaces (Y, B) ⊆ (X, A).

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Proof. Combine Lemma 5.17 and 5.20. Deﬁnition 5.23. A map of locally ringed spaces

( f , φ) : (Y, OY ) → (X, OX ) ' 0 is called a closed embedding if it factors through an isomorphism (Y, OY ) −→(Y , OY0 ), 0 where (Y , OY0 ) ⊆ (X, OX ) is a closed subspace.

Definition 5.24. Let (X, OX ) be a scheme. A closed subscheme of (X, OX ) a closed subspace (Y, OY ), as in Definition 5.14, such that (Y, OY ) is a scheme. The next proposition gives us the first example of a closed subscheme. Proposition 5.25. Take a ring A and an ideal a ⊆ A, and define the quotient ring B := A/a, with surjection π : A B. Consider the of affine schemes (X, OX ) := Spec(A) and (Y, OY ) := Spec(B). Then the map of schemes

( f , π˜) := Spec(π) : (Y, OY ) → (X, OX ) is a closed embedding, and f (Y) = ZerX (a). Exercise 5.26. Prove this prop. (Hint: this is easy!) Remark 5.27. Closed subschemes are delicate (much more complicated than open subschemes). Suppose (X, OX ) is a scheme. We will prove later that:

• A closed LR subspace (Y, OY ) is a closed subscheme iff the corresponding ideal sheaf IY := Ker(OX OY ) is a quasi-coherent OX -module. • Every closed subset Y of a scheme (X, OX ) has at least one closed subscheme structure: the reduced structure. See Thm 5.38 below. • If (X, OX ) is an affine scheme and (Y, OY ) is a closed subscheme, then (Y, OY ) is also affine. This is the converse of Prop 5.25. It will also require the study of quasi-coherent OX -modules. The next two exercises will show how to get closed LR subspaces of schemes that are usually not schemes.

Exercise 5.28. Let (X, OX ) be a locally ringed space, let Y ⊆ X be a closed subset, with −1 inclusion map f : Y → X, and let OY = OX |Y := f (OX ). By Proposition 5.7 we can view OY as a sheaf of rings on X, supported on Y. Prove that:

(1) The pair (Y, OY ) is a locally ringed space. (2) Let ψ : OX → OY be the homomorphism of X -rings that corresponds to the identity of OY under the adjunction (5.6). Then ψ is surjective, and for every point y ∈ Y the homomorphism on stalks ψy : OX,y → OY,y is an isomorphism. (3) The pair ( f , ψ) is a closed embedding of locally ringed spaces (Y, OY ) → (X, OX ). (4) The functor f∗ : Mod OY → ModY OX is an equivalence.

1 Exercise 5.29. Take (X, OX ) := A = Spec([t]) for a ﬁeld . Let y ∈ X be the origin, and let Y := {y}, i.e. Y = ZerX (t) ⊆ X. By the previous exercise, (Y, OY ) := (Y, OX |Y ) is a closed LR subspace of (X, OX ). Show that the locally ringed space (Y, OY ) is not a scheme. (Hint: If (Y, OY ) were a scheme, then it would have to be aﬃne, say Spec(C) for a ring C, and then C OY,y.) Try to identify the ideal I ⊆ OX that matches OY under the bijection in Prop 5.22. (Hint: I = g!(OU ), where U := X − Y and g is the inclusion. Read about g!, called extension by 0, in [Har, Sec III.6].)

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We end this section with a general way of producing closed subschemes. A ring A is called reduced if the only nilpotent element in it√ is 0. Any ring A has a maximal reduced quotient ring Ared. It is gotten as follows: let 0√⊆ A be radical of A, i.e. the ideal consisting of all nilpotent elements. Then Ared = A/ 0. It is the universal ⊆ quotient ring of A for√ homomorphisms into reduced rings. An ideal a A is called a radical ideal if a = a; namely if the quotient ring A/a is reduced.

Lemma 5.30. If A is a reduced ring, then any localization AS is also reduced. Exercise 5.31. Prove this lem.

Deﬁnition 5.32. A scheme (X, OX ) is called a reduced scheme if for every open set U ⊆ X, the ring Γ(U, OX ) is reduced.

Proposition 5.33. Let (X, OX ) be a scheme. The following three conditions are equivalent.

(i) (X, OX ) is a reduced scheme. (ii) X has an aﬃne open covering {Ui }i∈I such that the rings Ai := Γ(Ui, OX ) are all reduced. (iii) For every point x ∈ X the local ring OX,x is reduced. Exercise 5.34. Prove Proposition 5.33.

Lemma 5.35. Let (X, OX ) = Spec(A) be an affine scheme. Then ring A is reduced iff the scheme (X, OX ) is reduced. Exercise 5.36. Prove Lemma 5.35. Example 5.37. Suppose is an algebraically closed field and Y is an affine variety over n() ≥ . This means that Y is the zero locus inside X :=√ A , for some n 0, of an ideal a ⊆ A := [t1,..., tn]. Because ZerX (a) = ZerX ( a), we can assume that a is a radical ideal.

Let B := A/a, and let (Yrsc, OYrsc ) := Spec(B). By Lemma 5.35 and Proposition 5.25 we n know that (Yrsc, OYrsc ) is a reduced closed subscheme of A = Spec(A). (The letters “rsc” stand for “reduced scheme”.) It turns out thatY = Yrsc(), i.e. the points of the variety Y are the -points of the scheme Yrsc. The Hilbert NZS implies that the inclusion f : Y → Yrsc is a “homeomorphism”, in the sense that these topological spaces have the same open sets. And the sheaf of rings −1 OY := f (OYrsc ) is the sheaf of -valued functions on the variety . The same holds for any quasi-projective variety Y over .

Theorem 5.38. Let (X, OX ) be a scheme and let Y ⊆ X be a closed subset. Then there is a unique (up to isomorphism) sheaf of OX -rings OY , whose support is Y, and such that (Y, OY ) is a reduced closed subscheme of (X, OX ). comment: [(190506) The proof was not done in class. Please read it and try to understand the details. ]

Proof. Recall that for an open set U ⊆ X, an element a ∈ Γ(U, OX ) and a point x ∈ U, we denote by a(x) the image of a in the residue field k(x). Let IY ⊆ OX be the subpresheaf defined by the formula (5.39) Γ(U, IY ) := a ∈ Γ(U, OX ) a(y) = 0 for all y ∈ U ∩ Y for an open set U ⊆ X. Then IY is actually a sheaf, so it is an ideal sheaf of OX . Moreover, since k(y) is field, it follows that Γ(U, IY ) is a radical ideal of Γ(U, OX ).

34 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Define the OX -ring OY := OX /IY . By Proposition 5.22 we get a closed locally ringed subspace (Y, OY ) ⊆ (X, OX ). Let us prove that (Y, OY ) is a scheme, and that it is reduced. Both are local questions, so we may assume that (X, OX ) = Spec(A) for a ring A. Let I := Γ(X, IY ) = a ∈ A a(y) = 0 for all y ∈ Y ⊆ A. As noticed above, I is a radical ideal of A. Define B := A/I, which is a reduced ring. 0 We know that ZerX (I) = Y. According to Proposition 5.25 the affine scheme (Y, OY ) := 0 Spec(B) is a closed subscheme of (X, OX ); and by Lemma 5.35 the scheme (Y, OY ) is 0 reduced. We will prove that (Y, OY ) = (Y, OY ). This will be accomplished by proving that 0 0 0 there is equality IY = IY of ideals of OX , where IY := Ker(OX OY ). Take a principal open set U := NZerX (s) ⊆ X for some s ∈ A, so Γ(U, OX ) = As. By formula (5.39) we have (5.40) Γ(U, IY ) = a ∈ As a(y) = 0 for all y ∈ U ∩ Y ⊆ As.

This is a radical ideal of As with zero locus U ∩ Y. According to Proposition 5.7(2), used for the closed embedding U ∩ Y U, we have

Γ(U, OY0 ) = Γ(U ∩ Y, OY0 ) = Γ(NZerY (s), OY0 ) = Bs. By the left exactness of Γ(U, −) and the ﬂatness of localization we see that 0 0 (5.41) Γ(U, IY ) = Γ(U, Ker(OX → OY )) = Ker(As → Bs) = Is ⊆ As. 0 Since I is a radical ideal, so is Is. Thus Γ(U, IY ) is a radical ideal of As with zero locus 0 U ∩ Y. By Lemma 2.20 it follows that Γ(U, IY ) = Γ(U, IY ). Finally, because the principal open sets are a basis of the topology of X, we see that 0 IY = IY .

Deﬁnition 5.42. The reduced closed subscheme (Y, OY ) from Thm 5.38 is called the reduced closed subscheme of (X, OX ) supported on Y, and the sheaf OY is called the reduced subscheme structure on Y.

Lecture 9, 15 May 2019

Definition 5.43. A topological space X is called noetherian if it has the descending chain condition on closed subsets; namely every strictly descending chain of closed subsets X0 ! X1 ! X2 ! ··· in X is finite. This sort of topological space almost never occurs in “classical topology”. However it is very typical in algebraic geometry. Example 5.44. If A is a noetherian ring, then Spec(A) is a noetherian topological space. Cf. Lemma 2.20. Definition 5.45. A topological space X is called irreducible if it is nonempty, and it is not a union X = X1 ∪ X2, where Xi X are closed strict subsets. Example 5.46. Let X be a topological space and x ∈ X a point. The closure Z := {x} in X of the set {x} is irreducible. The point x is called a generic point of Z. Lemma 5.47. If X is an irreducible topological space, then every nonempty open subset U ⊆ X is irreducible, and it is is dense in X. Exercise 5.48. Prove the lemma. Proposition 5.49. Let X be a noetherian topological space. Then:

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(1) X is quasi-compact. (2) Every subset Y of X, with the induced topology, is a noetherian topological space. Ð (3) X is a ﬁnite union of irreducible closed subsets, say X = i=1,...,r Zi, such that Zi * Zj for all i , j. The collection of closed sets {Zi }i=1,...,r is unique up to permutation. Ý (4) X is a ﬁnite disjoint union of closed connected subsets X = i=1,...,r Xi. The sets Xi are also open. The collection {Xi }i=1,...,s is unique up to permutation. Exercise 5.50. Prove this proposition. (Hint: It is elementary but tricky. See [Har, Sec II.3] for proofs. Item (4) is a consequence of item (3).)

The irreducible closed subsets {Zi }i=1,...,r in item (3) of the proposition are called the irreducible components of X. The connected closed (and open) subsets {Xi }i=1,...,s in item (4) of the proposition are called the connected components of X.

Deﬁnition 5.51. A scheme (X, OX ) is called an irreducible scheme (resp. a connected scheme, resp. a quasi-compact scheme) if the topological space X is irreducible (resp. connected, resp. quasi-compact). comment: [(190522) next def improved - see Def 5.60 and Prop 5.61 ]

Definition 5.52. A scheme (X, OX ) is called a noetherian scheme if it is finite union of affine open subschemes Ui = Spec(Ai), s.t. each Ai is a noetherian ring.

Proposition 5.53. Let (X, OX ) be a noetherian scheme. Then (1) X is a noetherian topological space. (2) Every irreducible closed subset Z ⊆ X has a unique generic point. Exercise 5.54. Prove Proposition 5.53. (Hint: This is also elementary but tricky. The book [Har, Sec II.3] does not give a proof – it is an exercise there.)

Recall that a ring homomorphism φ : A → B is called flat if B is a flat A-module. The ring homomorphism φ : A → B is called faithfully flat if for every sequence

φi φi+1 S = ··· Mi −−→ Mi+1 −−−→ Mi+1 ··· in Mod A, the sequence S is exact iﬀ the sequence

i i+1 i id ⊗ φ i+1 id ⊗ φ i+1 Indφ(S) := ··· B ⊗A M −−−−−→ B ⊗A M −−−−−−→ B ⊗A M ··· in Mod B is exact. According to [Mat, Chapter 3 Section 7] the ring homomorphism φ : A → B is faithfully flat iff it is flat, and the induced map of affine schemes Spec(φ) : Spec(B) → Spec(A) is surjective on the underlying sets. Another fact that we need from commutative algebra is this: given ring homs A → B → C, if A → C flat and B → C is faithfully flat, then A → B is flat.

Lemma 5.55. Let (X, OX ) be a scheme. (1) Let V ⊆ U be aﬃne open sets in X. Then the ring homomorphism

Γ(U, OX ) → Γ(V, OX ) is ﬂat.

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(2) Let U ⊆ X be an affine open set, and let {Vi }i∈I be a finite affine open covering of U. Then the ring homomorphism Ö Γ(U, OX ) → Γ(Vi, OX ) i∈I is faithfully flat. Proof. comment: [(190515) The proof I gave in class was wrong. The assertion “or p < V, so B ⊗A Ap = 0” is false. The new correct proof is a bit long – can it be simplified? ] The proof is done is steps.

Step 1. Let A := Γ(U, OX ) and B := Γ(V, OX ) as in item 1, and assume that V is a principal open set in U, namely V = NZerU (s) for some s ∈ A. Then B As as A-rings, so it is ﬂat.

Step 2. Here we are in the situation of item 2, A := Γ(U, OX ) and Bi := Γ(Vi, OX ). Let B := Î Ý Ð i Bi. Since Spec(B) = i Vi and U = i Vi we see that the map Spec(B) → Spec(A) is surjective.

Step 3. Back to item 1: we can cover V by open sets {Wj }j ∈J , such that each Wj is a principal open set of U, i.e. Wj = NZerU (sj ) for some sj ∈ A. Because V is quasi-compact, we can assume that the indexing set J is finite. Note that Wj is also a principal open set in Î V, since Wj = NZerV (sj ). Let B := Γ(V, OX ), Cj := Γ(Wj , OX ) and C := j ∈J Cj . By step 1 we know that A → Cj and B → Cj are flat. So A → C and B → C are flat. By step 2 the homomorphism B → C is faithfully flat. It follows that A → B is flat. This establishes item 1. Î Step 4. Now we prove item 2. Let Bi := Γ(Vi, OX ) and B := i Bi. By item (1) the ring Ý Ð homomorphism A → B is flat. Since Spec(B) = i Vi and U = i Vi we see that the map Spec(B) → Spec(A) is surjective.

Proposition 5.56. Let (X, OX ) be a noetherian scheme. Then every aﬃne open subscheme (U, OX |U ) of (X, OX ) is a noetherian scheme.

Proof. We need to prove that the ring A := Γ(U, OX ) = Γ(U, OX |U ) is noetherian. Take a point x ∈ U. By Deﬁnition 5.52 there is an aﬃne open set V ⊆ X such that x ∈ V and B := Γ(V, OX ) is a noetherian ring. There is an element s ∈ B such that

x ∈ NZerV (s) = Spec(Bs) ⊆ U. Ð The ring Bs is noetherian too. This shows that U = i∈I Vi, where each Vi is an affine open set, and Bi := Γ(Vi, OX ) is a noetherian ring. Because U is quasi-compact (see Proposition 5.49), we can assume that the indexing set I is finite. Î Consider the ring homomorphism φ : A → B, where B := i∈I Bi. By Lemma 5.55, φ is faithfully flat. Suppose a0 a1 ··· is a strictly ascending chain of ideals in A. Then, letting bi := B ⊗A ai, we get a strictly ascending chain of ideals b0 b1 ··· in B. This is where we need faithful flatness. But the ring B is noetherian, so this chain must be finite. Remark 5.57. Following Hartshorne [Har] we will often consider only noetherian schemes, in those cases where this assumption simplifies the discussion significanly. For instance, a finite type map f : Y → X between noetherian schemes (to be defined soon) is a very useful notion; but when the schemes are not noetherian, we have to look at locally finitely presented maps. Recall that an integral domain is a nonzero ring A in which the only zero-divisor is 0. In other words, the zero ideal is a prime ideal of A.

37 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Deﬁnition 5.58. A scheme (X, OX ) is called an integral scheme if it is nonempty, and if for every nonempty open set U ⊆ X, the ring Γ(U, OX ) is an integral domain. Warning: The deﬁnitions of reduced and integral schemes in [Har, Sec II.3] are impre- cise.

Proposition 5.59. Let (X, OX ) be a scheme. The following two conditions are equivalent.

(i) (X, OX ) is an integral scheme. (ii) (X, OX ) is reduced and irreducible. Proof. (i) ⇒ (ii): Assume Clearly integral implies reduced. If X is not an irreducible topological space, then there are closed strict subsets X1, X2 X such that X = X1 ∪ X2. Let U1 := X − X2 and U2 := X − X1, which are nonempty open subsets of X, and U1 ∩ U2 = . Deﬁne U := U1 ∪ U2. We then have

Γ(U, OX ) = Γ(U1, OX ) × Γ(U2, OX ), and this is not an integral domain. comment: [(190515) The last part of the proof, below, wasn’t done in class. Please read it at home. ]

(ii) ⇒ (i): Let U ⊆ X be a nonempty open subset such that Γ(U, OX ) is not an integral domain. Then there exist nonzero elements a, b ∈ Γ(U, OX ) s.t. a·b = 0. Deﬁne Y := ZerU (a) and Z := ZerU (b). These are closed subsets of U, by Prop 4.18. We claim that Y U. Otherwise, if ZerU (a) = U, then for every nonempty aﬃne open V ⊆ U we would have ZerV (a) = V. Hence, taking B := Γ(V, OX ), we would have

Spec(Ba) = NZerV (a) = , so Ba = 0. But the ring B is reduced, so a must be zero in B, i.e. a|V = 0. This would imply a = 0, a contradiction. For the same reasons we have Z U. So we have strict closed subsets Y, Z U, and

Y ∪ Z = ZerU (a) ∪ ZerU (b) = ZerU (a·b) = ZerU (0) = U. Yet by Lem 5.47 the open set U is irreducible. We have a contradiction.

Lecture 10, 22 May 2019

Please note that there was an error in my proof in class of Lem 5.55; the typed proof above is correct. I hope everybody read and understood the 2nd half of the proof of Prop 5.59. Here is an abbreviation that we will use from now on (unless we need to be fully explicit): a scheme (X, OX ) will be be referred to as X; the structure sheaf OX will be implicit. We will start by improving upon Def 5.52. The subsequent prop will say that the two defs are equivalent. Definition 5.60. A scheme X is called noetherian if it is quasi-compact, and for every affine open set U ⊆ X the ring A := Γ(U, OX ) is noetherian. Proposition 5.61. TFAE for a scheme X : (i) X is noetherian. Ð (ii) There is a finite affine open covering X = i∈I Ui, such that each of the rings Ai := Γ(Ui, OX ) is noetherian.

38 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

Proof. Ð (i) ⇒ (ii): Since X is quasi-compact, there is a ﬁnite aﬃne open covering X = i∈I Ui. By assumption the rings Ai := Γ(Ui, OX ) are noetherian.

(ii) ⇒ (i): Each affine open set Ui is quasi-compact, and a finite union of quasi-compacts is quasi-compact. Thus X is quasi-compact. Take some affine open set U ⊆ X. We need to prove that the ring A := Γ(U, OX ) is Ð noetherian. We can find a finite affine open covering U = j ∈J Vj , where Vj := NZerUi (sj ) for some i ∈ I and some sj ∈ Γ(Ui, OX ). Now Bj := Γ(Vj , OX ) (Ai)sj is noetherian. Î Let B := j ∈J Bj , which is noetherian. The ring homomorphism A → B is faithfully flat. Hence, by faithfully flat descent (the argument used in the proof of Prop 5.56) we see that A is noetherian. The fact that a closed subscheme of an affine scheme is affine is something that I never understood well, and it always intrigued me. In Rem 5.27 I said that we need the concept of quasi-coherent sheaf to prove this result; but today I’ll give a more direct proof; see Thm ?? below. It uses the criterion for affineness Thm 5.62, that appears as [Har, Exer II.2.17]. But I think it deserves to be a theorem. comment: [(190522) Thm ?? will be next week... ]

A sequence (s1,..., sn) of elements of a ring A is called a covering sequence if A = Í i A·si. In other words, if Ø Ø Spec(A) = NZerSpec(A)(si) = Spec(Asi ). i∈[1,n] i∈[1,n]

Theorem 5.62 (Criterion for Affineness). Let X be a scheme and A := Γ(X, OX ). Assume there is a covering sequence (s1,..., sn) of A such that for every i the open set NZerX (si) is affine. Then X is an affine scheme. comment: [(190520) is there a reference for this thm? Maybe in [EGA] or in [SP]?] We first need a lemma (it is [Har, Exer II.2.16]).

Lemma 5.63. Let X be a scheme, with a ﬁnite aﬃne open covering {Ui }i∈I , such that every intersection Ui ∩ Uj is quasi-compact. Let A := Γ(X, OX ) and s ∈ A. Then Γ NZerX (s), OX As as A-rings. Note that such an A-ring isomorphism is unique (if it exists). Proof. comment: [(190520) draw picture? ]

For each pair of indices i, j ∈ I let us choose a ﬁnite aﬃne open covering Ø Ui ∩ Uj = Vk .

k ∈Ki, j Ý Let K := i, j Ki, j , Ai := Γ(Ui, OX ) and Bk := Γ(Vk , OX ). Note that Ö Γ(Ui ∩ Uj , OX ) ⊆ Γ(Vk , OX ).

k ∈Ki, j Hence, by the sheaf property, we have an exact sequence of A-modules Ö Ö (5.64) 0 → A → Ai → Bk . i∈I k ∈K

39 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

These products are ﬁnite, so they are direct sums. Applying As ⊗A (−) to (5.64) we get the exact sequence of As-modules Ö Ö (5.65) 0 → As → (As ⊗A Ai) → (As ⊗A Bk ). i∈I k ∈K

Let’s write W := NZerX (s). Then W ∩ Ui = NZerUi (s) and W ∩ Vk = NZerVk (s) are aﬃne open sets. We know that Γ(W ∩ Ui, OW ) = Γ NZerUi (s), OUi = As ⊗A Ai and Γ(W ∩ Vk , OW ) = Γ NZerVk (s), OVk = As ⊗A Bk . Also Ø W = (W ∩ Ui) i∈I and Ø (W ∩ Ui) ∩ (W ∩ Uj ) = (W ∩ Vk ).

k ∈Ki, j Therefore, by the sheaf property of OW , the exact sequence (5.65) gives Γ(W, OX ) = As. Proof of the theorem. We shall prove that the canonical map of schemes f : X → X 0 := Spec(A) is an isomorphism. Let Ui := NZerX (si), which by assumption is an aﬃne scheme. For i, j ∈ [1, n] we have

Ui ∩ Uj = NZerX (si ·sj ) = NZerUi (sj ), which is aﬃne and thus quasi-compact. This means that we can use Lemma 5.63, and it says Γ(Ui, OX ) = Asi . Therefore the map of aﬃne schemes 0 f |Ui : Ui → Ui := Spec(Asi ) 0 −→' | is an isomorphism. Let gi : Ui Ui be the inverse of f Ui . By the uniqueness of inverses we have g | 0 0 = g | 0 0 . Therefore, by Thm 1.22(gluing of maps), there is a map i Ui ∩Uj j Ui ∩Uj Ø 0 Ø 0 g : X = Ui → X = Ui i∈I i∈I s.t. g| 0 = g . The map g is an inverse of f , so f is an isomorphism. Ui i

40 | ﬁle: notes-190522-d1 Course Notes | Amnon Yekutieli | 22 May 2019

References

[EGA] A. Grothendieck and J. Dieudonné, “Éléments de géometrie algébrique”, collective reference for the whole series. [Gro] A. Grothendieck, Sur quelques points d’algèbre homologique, Tôhoku Math. J. 9 (1957), 119-221. [Har] R. Hartshorne, “Algebraic Geometry”, Springer-Verlag, New-York, 1977. [HltSt] P.J. Hilton and U. Stammbach, “A Course in Homological Algebra”, Springer, 1971. [Lee] John M. Lee, “Introduction to Smooth Manifolds”, LNM 218, Springer, 2013. [KaSc] M. Kashiwara and P. Schapira, “Sheaves on manifolds”, Springer-Verlag, 1990. [Mac1] S. Maclane, “Homology”, Springer, 1994 (reprint). [Mac2] S. Maclane, “Categories for the Working Mathematician”, Springer, 1978. [Mat] H. Matsumura, “Commutative Ring Theory”, Cambridge University Press, 1986. [Mil] J. Milne, “Algebraic Groups”, Ca,mbridge, 2017. [Rot] J. Rotman, “An Introduction to Homological Algebra”, Academic Press, 1979. [Row] L.R. Rowen, “Ring Theory” (Student Edition), Academic Press, 1991. [SP] The Stacks Project, an online reference, J.A. de Jong (Editor), http://stacks.math.columbia.edu. [Wei] C. Weibel, “An introduction to homological algebra”, Cambridge Studies in Advanced Math. 38, 1994. [Ye1] A. Yekutieli, “Derived Categories”, prepublication, eprint https://arxiv.org/abs/1610.09640. [Ye2] A. Yekutieli, “Commutative Algebra”, Course Notes, http://www.math.bgu.ac.il/~amyekut/ teaching/2017-18/comm-alg/course_page.html. [Ye3] A. Yekutieli, “Homological Algebra”, Course Notes, http://www.math.bgu.ac.il/~amyekut/ teaching/2017-18/hom-alg/course_page.html. [Ye4] A. Yekutieli, “Algebraic Geoemtry – Schemes 1”, Course Notes, https://www.math.bgu.ac.il/ ~amyekut/teaching/2018-19/schemes-1/course_page.html .

Department of Mathematics, Ben Gurion University, Be’er Sheva 84105, Israel. Email: [email protected], Web: http://www.math.bgu.ac.il/~amyekut.

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