´Etale Theory

Etale´ Theory

Andrew Kobin Fall 2018 – Spring 2019 Contents Contents

Contents

0 Introduction 1

1 Some Commutative Algebra 2 1.1 Projectives ...... 2 1.2 Injectives ...... 4 1.3 Derived Functors ...... 8 1.4 Flat Modules ...... 13

2 Scheme Theory 22 2.1 Aﬃne Schemes ...... 22 2.2 Schemes ...... 24 2.3 Properties of Schemes ...... 26 2.4 Sheaves of Modules ...... 27

3 Etale´ Morphisms 30 3.1 Flat Morphisms ...... 30 3.2 KählerDifferentials ...... 33 3.3 Sheaves of Relative Differentials ...... 38 3.4 Smooth Morphisms ...... 41 3.5 Unramified Morphisms ...... 48 3.6 Etale´ Morphisms ...... 51 3.7 Henselian Rings ...... 54

4 Descent 63 4.1 Galois Descent ...... 63 4.2 Fields of Deﬁnition ...... 74 4.3 Galois Descent for Varieties and Schemes ...... 76

5 Etale´ Fundamental Group 82 5.1 Covering Spaces ...... 82 5.2 Inﬁnite Galois Theory ...... 84 5.3 Galois Theory for Schemes ...... 86 5.4 The Etale´ Fundamental Group ...... 87

i 0 Introduction

0 Introduction

These notes offer an overview of different topics in commutative algebra and algebraic geometry with a common theme of étalemorphisms, which was the program for the Galois- Grothendieck Seminar at the University of Virginia in 2018 – 2019. The topics include:

Projectives, injectives and derived functors in commutative algebra

Flat modules and ﬂat morphisms

Faithfully ﬂat and Galois descent

K¨ahlerdiﬀerentials

Smooth, unramiﬁed and ´etalemorphisms

The ´etalefundamental group of a scheme

Henselian rings.

Since this is a variety of materials, there are many references for the course. The main ones are Bosch’s Commutative Algebra and Algebraic Geometry, Altman-Kleiman’s Introduc- tion to Grothendieck Duality Theory, Bosch-Lütkebohmert-Raynaud’s Néron Models, Görtz- Wedhorn’s Algebraic Geometry I, Mézard’sarticle “Fundamental group” and Raynaud’s Anneaux Locaux Henséliens.

1 1 Some Commutative Algebra

1 Some Commutative Algebra

1.1 Projectives

Deﬁnition. A left R-module P is projective if for every diagram with exact rows and solid arrows, shown below, there is a dashed arrow making the diagram commute.

M N 0

In other words, projectives allow you to lift along surjections.

Theorem 1.1.1. For an R-module P , the following are equivalent:

(a) P is projective.

(b) HomR(P, −) is an exact functor. (c) Every short exact sequence 0 → A → B → P → 0 splits.

(d) P is a direct summand of a free R-module.

Proof. (a) ⇐⇒ (b) We know HomR(P, −) is always left exact. For a sequence

0 0 → M 0 −→α M −→α M 00 → 0 applying HomR(P, −) induces an exact sequence with solid arrows:

0 00 0 → Hom(P,M ) → Hom(P,M) → Hom(P,M ) 99K 0. To get the 0 on the right, note that by deﬁnition of projective, every f : P → M 00 factors through α : M → M 00: P f˜ f α M 00 M 0

Notice that is equivalent to showing Hom(P,M) → Hom(P,M 00) is surjective. The converse follows by reversing the entire argument. (a) =⇒ (c) Suppose we have a short exact sequence

0 −→ A −→ B −→ P −→ 0. where P is projective. The identity on P induces a surjection f:

2 1.1 Projectives 1 Some Commutative Algebra

P f id

0 A B P 0

Hence the sequence splits. (c) =⇒ (d) Take a ﬁnite presentation of P :

0 −→ K −→ F −→ P −→ 0

where F is free and the map F → R takes a generating set {fi} of F to a generating set ∼ {pi} of P . By hypothesis the sequence splits, so F = P ⊕ K. (d) =⇒ (a) We ﬁrst prove that every free module is projective. Suppose F is a free module with basis {fi}i∈I . Let α : M → N be a surjection, i.e. the row of the following diagram with solid arrows is exact. F

α˜ ϕ α M N 0

If ϕ : F → N is an R-linear map, denote ϕ(fi) = ni ∈ N. Since α is surjective, there exists an mi ∈ M such that α(mi) = ni. Then we will deﬁneα ˜(fi) = mi and extend by linearity to all of F . It is clear that αα˜ and ϕ agree on {fi} which is a basis for F . Therefore the diagram commutes, so F is projective. Now assume P is a summand of a free R-module, say F = P ⊕ Q. Consider the following diagram with bottom row exact: F

π i ˜ f P g˜ g

M N 0 f

Given g surjective, we want to construct a mapg ˜ making the smaller triangle commute. Note that F = P ⊕Q induces π : F → P the natural projection and i : P,→ F the inclusion map. Since F is projective, there exists a map f˜ : F → M making the larger triangle commute. It suﬃces to show thatg ˜ = fi˜ is what we are looking for, but this is obvious since for any p ∈ P , ffi˜ (p) = ff˜(p) = g(p). Hence P is projective.

3 1.2 Injectives 1 Some Commutative Algebra

Notice that (a), (b) or (c) are valid statements in any abelian category A, so we may take any of them as our deﬁnition of a projective object in A. We say A has enough projectives if every object A ∈ A admits a projective resolution P• → A, i.e. an exact sequence

· · · → P2 → P1 → P0 → A → 0 where each Pi is projective.

Theorem 1.1.2. For any ring R, ModR has enough projectives.

Proof. Free modules are projective, so take a free presentation F0 → A with kernel K0. Then K0 admits a free presentation F1 → K0 and by construction F1 → F0 → A → 0 is exact. Continue inductively.

1.2 Injectives

There is a dual notion to that of a projective module. Deﬁnition. A left R-module E is injective if for every diagram with exact rows and solid arrows, shown below, there is a dashed arrow making the diagram commute.

0 A B

That is, injectives allow you to extend along injections. Theorem 1.2.1. For an R-module E, the following are equivalent: (a) E is injective.

(b) HomR(−,E) is an exact functor. (c) Every short exact sequence 0 → E → B → C → 0 splits. Proof. (a) ⇐⇒ (b) Suppose we have an exact sequence

p 0 → A −→i B −→ C → 0.

This induces a sequence

p∗ i∗ 0 → HomR(C,E) −→ HomR(B,E) −→ HomR(A, E) → 0

We know HomR(−,E) is always left exact, so it remains to show exactness at HomR(A, E) is equivalent to E being injective. In other words we must prove i∗ is surjective if and only if i is injective. On one hand, if f ∈ Hom(A, E) there exists g ∈ Hom(B,E) with f = i∗(g) = gi; that is,

4 1.2 Injectives 1 Some Commutative Algebra

f g

0 A B i

commutes, so i∗ is surjective. Conversely, if E is injective then for any f : A → E there exists g : B → E making the above diagram commute. Then we see that f = gi = i∗(g) ∈ im i∗ so i∗ is surjective. (a) =⇒ (c) Consider the diagram E

id g

0 E B i

Since E is injective, the identity on E induces g : B → E such that gi = idE. Hence the sequence splits. (c) =⇒ (a) Suppose E → M → M 00 → 0 is exact, so that by hypothesis M ∼= E ⊕ M 00. We will see shortly that direct summands of injectives are injective, so it follows that E is injective. There is a useful characterization of injectives that allows us to restrict our attention to ideals of R, rather than all module injections A,→ B.

Theorem 1.2.2 (Baer’s Criterion). Let E be a left R-module. E is left injective ⇐⇒ every R-linear map f : I → E, where I is an ideal of R, can be extended to R:

f g

0 I R i

Proof. ( =⇒ ) is obvious by the deﬁnition of injectives. To prove ( ⇒ = ) suppose E

0 A B i

has an exact row. Let X = {(A0, g0) | A ⊆ A0 ⊆ B and g0 extends f}. We deﬁne a partial ordering on X by (A0, g0) (A00, g00) if A0 ⊆ A00 and g00 extends g0. By construction X is 0 0 0 0 bounded above, so this makes X into a Zornable set. Suppose (A1, g1) (A2, g2) · · · is

5 1.2 Injectives 1 Some Commutative Algebra

0 [ 0 0 0 0 0 0 0 a chain in X. Define A = Ai and g (a ) = gi(a ) where a ∈ Ai. This defines an upper i≥1 bound on the chain, so by Zorn’s Lemma there exists a maximal element in the chain. Let (A0, g0) be such a maximal element. We claim A0 = B. Suppose not; then there is some b ∈ B r A0. Define a left ideal of R by I = {r ∈ R | rb ∈ A0} – this is sometimes called a colon ideal, denoted (bR : A0). We construct a map h : I → E by defining h(r) = g0(rb). This map may be lifted by E

h h∗

0 I R i

∗ Let A1 = A0 + Rb and define g1(a0 + rb) = g0(a0) + rh (1). Note that g1 may not be well- defined since a0 + rb is not necessarily the unique way to write an element of A1. However, if g1 is well-defined, we will have shown (A0, g0) ≺ (A1, g1) contradicting maximality of 0 0 (A0, g0). Thus to finish the proof we show well-definedness. Suppose a0 + rb = a0 + r b. 0 0 0 Then (r − r )b = a0 − a0 ∈ A =⇒ r − r ∈ I. This gives us

0 0 0 g0(a0 − a0) = g0((r − r )b) = h(r − r ).

Now we can’t take the constant r − r0 out of h, since this would leave 1 behind, and 1 6∈ I. But we do have h(r − r0) = h∗(r − r0) and h∗ is deﬁned on R, so h∗(r − r0) = (r − r0)h∗(1). It follows easily that g1 is well-deﬁned. Proposition 1.2.3. Direct sums and direct summands of injectives are injective.

Proof. Similar to the proof of (d) =⇒ (a) in Theorem 1.1.1.

Deﬁnition. A module M over a domain R is divisible if for all m ∈ M and nonzero r ∈ R, there is some m0 ∈ M such that m = rm0.

Informally, this says that in a divisible R-module, you can ‘divide’ by R. An example of a divisible abelian group (Z-module) is Q. Theorem 1.2.4. Every injective left R-module is divisible.

Proof. Consider the inclusion nZ ,→ Z. Then divisibility of an R-module A is equivalent to completing the diagram A y x - ↑ 1 n 0 nZ Z Of course this is immediate when A is injective.

6 1.2 Injectives 1 Some Commutative Algebra

The converse holds when R is a PID: Theorem 1.2.5. Let R be a PID. Then (1) Every divisible R-module is injective. (2) Quotients of injectives are injective. Proof. (2) follows from (1), while the proof of (1) is similar to the ﬁrst part of the proof of Baer’s Criterion.

Since Q is an injective Z-module, (2) of Theorem 1.2.5 implies Q/Z is also an injective Z- module. Next, we use this to show that every left R-module can be realized as a submodule of an injective left R-module. We begin by proving this for Z-modules (abelian groups). Let ∼ M be a Z-module. Then M = F/S where F is some free abelian group and S is the module ∼ M L of relations. By the fundamental theorem of abelian groups, M = Z/S and Z can be i∈I L embedded in Q so we have a composition ∼ M M M = Z/S ,→ Q/S. i∈I i∈I L L Now Q is divisible so Q is also divisible. Then by Theorem 1.2.5, Q/S is injective. Before proving the property for R-modules in general, we will need Proposition 1.2.6. Let ϕ : R → S be a ring homomorphism and E an injective left R- module. Then HomR(S,E) is an injective left S-module.

Proof. First let’s convince ourselves that HomR(S,E) is a left S-module at all. The map ϕ : R → S induces a left R-action on S given by r · s = ϕ(r)s. S is also a right module over itself, so there is an available action for S to act on HomR(S,E) on the left. Now, note that HomR(S,E) is an injective left S-module ⇐⇒ HomS(−, HomR(S,E)) is an exact functor. By Hom-Tensor Adjointness, ∼ ∼ HomS(−, HomR(S,E)) = HomR(S ⊗S −,E) = HomR(−,E) and HomR(−,E) is exact precisely when E is injective. In fact we have proven that the converse of the proposition holds as well.

Corollary 1.2.7. For any divisible abelian group D, HomZ(R,D) is an injective left R- module.

∗ Deﬁnition. For an abelian group A, the group A := HomZ(A, Q/Z) is called the dual of A. Lemma 1.2.8. For any abelian group A, there is an embedding A,→ A∗∗.

∗∗ Proof. Deﬁne Φ : A → A by a 7→ ϕa, where ϕa(f) = f(a) for any f : A → Q/Z. Take any nonzero a ∈ A. We must show that there exists some f : A → Q/Z such that f(a) 6= 0. Let B = Za. Then there is a map ψ : B → Q/Z given by a 7→ ψ(a) 6= 0. Now since Q/Z is injective, ψ extends along the inclusion B,→ A to a map f : A → Q/Z which by construction has f(a) 6= 0. Hence Φ is injective.

7 1.3 Derived Functors 1 Some Commutative Algebra

Theorem 1.2.9. For any left R-module M, there is an embedding M,→ E where E is an injective left R-module. ∗∗ ∗ Proof. By Proposition 1.2.6, M = HomZ(M , Q/Z) is injective. Therefore the embedding M,→ M ∗∗ realizes M as a submodule of an injective module.

1.3 Derived Functors

For an R-module M, the functor − ⊗R M : ModR → Ab is a right exact functor, but is not left exact in general.

Definition. In the category R-Mod of left R-modules, the left derived functors of − ⊗R M are called Tor: R Torn (M,N) := Ln(− ⊗R M)(N) = Hn(P• ⊗R N), where P• → M is a projective resolution of M. Likewise for a right R-module M, the left derived functors of M ⊗R − are called tor: R torn (N,M) := Ln(M ⊗R −)(N) = Hn(M ⊗R P•). R ∼ ∼ By definition, for all right R-modules M and left R-modules N, Tor0 (M,N) = M ⊗RN = R tor0 (M,N). The Comparison Theorem says that unique chain maps exist between projective resolutions of M and N when M → N is a module homomorphism, so we need not worry when defining LnT about which projective resolution we choose. This follows from a more general statement:

Theorem 1.3.1. Let P• : · · · → P2 → P1 → P0 be a projective chain complex and suppose C• : ··· C2 → C1 → C0 is an acyclic chain complex. Then for any homomorphism ϕ : H0(P•) → H0(C•), there is a chain map f : P• → C• whose induced map on H0 is ϕ, and ϕ is unique up to chain homotopy. Proof. Consider the diagram

∂2 ∂1 ∂0 P2 P1 P0 H0(P•) 0

f2 f1 f0 ϕ 0 0 0 ∂2 ∂1 ∂0 C2 C1 C0 H0(C•) 0

Since P0 is projective, there exists an f0 lifting ϕ to P0 → C0. Inductively, given fn−1 we have a diagram

∂n Pn Pn−1

fn fn−1 0 ∂n Cn Cn−1

8 1.3 Derived Functors 1 Some Commutative Algebra

0 0 0 Note that fn−1∂ has image lying in ker ∂n ⊆ Cn−1, but C• is acyclic, so ∂n(Cn) = ker ∂n−1. Since Pn is projective, we can lift fn−1∂n to the desired map fn : Pn → Cn. By construction, ∞ f = {fn}n=0 satisﬁes the desired properties. For uniqueness, suppose g : P• → C• is another chain map restricting to ϕ on H0. Since 0 0 f0 − g0 = 0 on H0, it must be that (f0 − g0)(P0) ⊆ ker ∂0 = im ∂1, so by projectivity of P0 there exists s0 : P0 → C1 making the following diagram commute:

s0 f0 − g0

0 C1 im ∂1 0

0 Inductively, given s0, . . . , sn−1 satisfying fk − gk = ∂k+1sk + sk+1∂k for all 0 ≤ k ≤ n − 1, we have

0 0 ∂n(fn − gn − sn−1∂n) = (fn−1 − gn−1)∂n − ∂nsn−1∂n since f, g are chain maps 0 0 = (∂nsn−1 − sn−2∂n−1)∂n − ∂nsn−1∂n = 0. Hence there is a commutative diagram

sn fn − gn − sn−1∂n

C ker ∂0 n+1 0 n 0 ∂n+1

0 This establishes the chain homotopy s : P• → C• such that fn − gn = ∂n+1sn + sn+1∂n for all n ≥ 0. Hence f is unique up to chain homotopy. Corollary 1.3.2 (Comparison Theorem). Let g : M → N be R-linear and pick projective resolutions P• and Q• of M and N, respectively. Then there exists a chain map f : P• → Q• such that H0(f) = g and f is unique up to chain homotopy.

Proof. Given projective resolutions P•,Q• → M, we have M = H0(P•) = H0(Q•) so let ϕ = idM . Since projective resolutions are acyclic, the comparison theorem gives us a chain map f : P• → Q•. Reversing the roles of P• and Q• gives a chain map in the opposite direction, and uniqueness forces the composition of these maps to be the identity in either direction.

Since Tor is deﬁned as the derived functors of − ⊗R M, for every short exact sequence 0 → A → B → C → 0

R in ModR, Torn (−,M) induces an exact sequence

R R R R · · · → Torn+1(C,M) → Torn (A, M) → Torn (B,M) → Torn (C,M) → · · · R · · · → Tor1 (C,M) → A ⊗R M → B ⊗R M → C ⊗R M → 0.

9 1.3 Derived Functors 1 Some Commutative Algebra

Theorem 1.3.3. For all projective resolutions P• of A and Q• of B and for all n ≥ 0, we have isomorphisms ∼ R ∼ Hn(P• ⊗R B) = Torn (A, B) = Hn(A ⊗R Q•) which are natural in each variable. Corollary 1.3.4. Let R be a commutative ring and suppose A and B are R-modules. Then R ∼ R for all n ≥ 0, Torn (A, B) = torn (A, B). ∼ Example 1.3.5. For p, q ∈ Z with d = gcd(p, q), we have Tor1(Z/pZ, Z/qZ) = Z/dZ. To see this, take the projective resolution

p 0 → Z −→ Z −→ Z/pZ → 0,

apply the functor − ⊗ Z/qZ and delete the last term to obtain: p 0 → Z/qZ −→ Z/qZ → 0. Then by deﬁnition

p d q Tor ( /p , /q ) = ker( /q −→ /q ) = ker( /q −→ /q ) = /q ∼= /d . 1 Z Z Z Z Z Z Z Z Z Z Z Z dZ Z Z Z

Z Proposition 1.3.6. An abelian group M is torsion-free if and only if Tor1 (A, M) = 0 for all abelian groups A. Proof. We prove this for the special case of M ﬁnitely generated. If M is torsion-free, then it is free and in particular projective, so automatically Tor1(A, M) = 0 for all A. Conversely, suppose nx = 0 for some nonzero x ∈ B and n ≥ 2. Applying − ⊗ M to the short exact sequence n 0 → Z −→ Z −→ Z/nZ → 0 yields a long exact sequence

n 0 → Tor1(Z/nZ,M) → B −→ B → B/nB → 0. n Then Tor1(Z/nZ,M) = ker(B −→ B) but since nx = 0, x is a nonzero element of Tor1(Z/nZ,M). In fact, for any M this identiﬁes Tor1(Z/nZ,M) as the n-torsion subgroup of M. R The functors Torn can actually be axiomatized, as the following theorem shows. R Theorem 1.3.7. For each n ≥ 0, there exists a functor Torn : R−Mod × R−Mod → R−Mod which satisﬁes

R (1) Tor0 (M,N) = M ⊗ N. (2) For any short exact sequence 0 → M 0 → M → M 00 → 0 and any R-module N, there is a long exact sequence

0 00 0 → Torn(M ,N) → Torn(M,N) → Torn(M ,N) → Torn−1(M ,N) →

which is natural in N.

10 1.3 Derived Functors 1 Some Commutative Algebra

R (3) For any free module F , Torn (F,N) = 0 for all n > 0. R Moreover, any functor satisfying these three properties is naturally isomorphic to Torn .

Proof. We have proven that Torn satisﬁes the stated properties so it remains to show that these in fact characterize Torn. We prove this inductively on n. For n = 0, uniqueness follows from the universal property of the tensor product. For n ≥ 1, take modules M and N and a free module F such that there is an exact sequence 0 → K → F → M → 0. Then by (2), there is a long exact sequence

0 = Torn(F,N) → Torn(M,N) → Torn−1(K,N) → Torn−1(F,N) → · · · ∼ When n > 1, Torn−1(F,N) = 0 as well so Torn(M,N) = Torn−1(K,N). When n = 1, ∼ Tor1(M,N) = ker(K ⊗ N → F ⊗ N). In all cases, induction implies that Torn is determined as a functor by Torn−1 so uniqueness holds.

This gives another proof of Corollary 1.3.4: consider the assignment (M,N) 7→ torn(M,N) := Torn(N,M). Then ∼ (1) tor0(M,N) = Tor0(N,M) = N ⊗ M = M ⊗ N.

(2) torn has a long exact sequence in the ﬁrst variable because Torn has a long exact sequence in the second variable.

(3) For a free module F , torn(F,N) = Torn(N,F ) = 0 for each n > 0 since − ⊗ F preserves exactness.

Hence by Theorem 1.3.7, Torn and torn are naturally isomorphic. Next we shift our focus to the functor Ext. This is in some ways the more interesting of the two derived functors, since in any module category it naturally admits an R-algebra structure. For any R-module M, the functor HomR(M, −): ModR → Ab is left exact, but not right exact in general. Similarly, the contravariant functor HomR(−,N) is left exact but not right exact in general.

Deﬁnition. In R-Mod, the right derived functors of HomR(M, −) are called Ext:

n n ExtR(M,N) := R (HomR(M, −))(N) = Hn(HomR(M,E•)),

where N → E• is an injective resolution of N. Likewise, the right derived of HomR(−,N) are denoted n n ExtR(M,N) := R (HomR(−,N))(M) = Hn(HomR(P•,N))

where P• → M is a projective resolution of M. 0 ∼ By deﬁnition, ExtR(M,N) = HomR(M,N) for all R-modules M and N. The comparison theorem (1.3.1) implies that ∼ Hn(HomR(M,E•)) = Hn(HomR(P•,N))

for any injective resolution N → E• and projective resolution P• → M, so the notational abuse above is only temporary.

11 1.3 Derived Functors 1 Some Commutative Algebra

As with Tor, any short exact sequence

0 → A → B → C → 0

in ModR determines two long exact sequences, corresponding to the covariant and contravariant versions of HomR:

1 0 → HomR(M,A) → HomR(M,B) → HomR(M,C) → ExtR(M,A) → · · · 1 and 0 → HomR(C,N) → HomR(B,N) → HomR(A, N) → ExtR(C,N) → · · ·

1 ∼ Proposition 1.3.8. For any abelian group B and for any n ≥ 2, ExtZ(Z/nZ,B) = B/nB. Proof. Consider the short exact sequence

n 0 → Z −→ Z −→ Z/nZ → 0. Applying Hom(−,B) determines a long exact sequence

n 1 0 → Hom(Z/nZ,B) → Hom(Z,B) −→ Hom(Z,B) → Ext (Z/nZ,B) → 0.

It is well-known that Hom(Z/nZ,B) = B[n] = {x ∈ B | nx = 0}, the n-torsion subgroup of ∼ B, and Hom(Z,B) = B, so the sequence above becomes

n 1 0 → B[n] → B −→ B → Ext (Z/nZ,B) → 0.

1 n By exactness, Ext (Z/nZ,B) is the cokernel of B −→ B which is precisely B/nB. As with Tor, the Extn functors can be axiomatized:

n Theorem 1.3.9. For each n ≥ 0, there exists a functor ExtR : R−Mod × R−Mod → R−Mod which satisﬁes

0 (1) ExtR(M,N) = HomR(M,N). (2) For any short exact sequence 0 → M 0 → M → M 00 → 0 and any R-module N, there is a long exact sequence

→ Extn(M 00,N) → Extn(M,N) → Extn(M 0,N) → Extn+1(M 00,N) →

which is natural in N.

n (3) For any free module F , ExtR(F,N) = 0 for all n > 0.

n Moreover, any functor satisfying these three properties is naturally isomorphic to ExtR. Proof. Reverse the arrows in the proof of Theorem 1.3.7.

12 1.4 Flat Modules 1 Some Commutative Algebra

1.4 Flat Modules

Definition. For a ring R, a right R-module M is flat if M ⊗R − is an exact functor. Flatness of left R-modules is defined similarly for the covariant functor − ⊗R M. Equivalently, a right R-module M is flat if and only if for every injection of left R-modules i : B → C, 1M ⊗ i : M ⊗R B → M ⊗R C is also an injection. Recall the following properties of − ⊗R M:

− ⊗R M commutes with direct sums (this follows from the proposition below). R − ⊗R M is right exact, with left derived functors Tor• (−,M).

− ⊗R M is not left exact in general. Proposition 1.4.1. For a ring R,

(i) Every projective right (or left) R-module is ﬂat.

(ii) A direct sum of right (or left) R-modules is ﬂat if and only if each summand is ﬂat.

Proof. (i) follows from (ii) since a free right R-module is the direct sum of copies of R, which is flat. Moreover, P is projective if and only if it is the direct summand of a free module, which, combined with the first statement says that projectives are always flat. M To prove (ii), consider a direct sum Mk of right R-modules Mk. For any family of R- k M M linear maps {fk : Mk → Nk} there is a map f : Mk → Nk taking (mk) 7→ (fk(mk)), k k and clearly f is injective if and only if each fk is injective. So if i : A → B is a left R-module injection, there is a commutative diagram

L 1 ⊗ i L ( Mk) ⊗R A ( Mk) ⊗R B

∼= ∼= L L (Mk ⊗R A) (Mk ⊗R B) f

where ϕ is defined by (mk ⊗ a) 7→ (mk ⊗ i(a)). By the above, 1 ⊗ i is an injection ⇐⇒ each M 1Mk ⊗ i is an injection. Hence Mk is flat ⇐⇒ each Mk is flat. k Lemma 1.4.2. For a left R-module M, the following are equivalent:

(a) M is ﬂat.

0 0 (b) Every inclusion 0 → N → N of R-modules induces an inclusion 0 → N ⊗R M → N ⊗R M.

13 1.4 Flat Modules 1 Some Commutative Algebra

(c) Every inclusion 0 → N 0 → N of finitely generated R-modules induces an inclusion 0 0 → N ⊗R M → N ⊗R M. Corollary 1.4.3. If every finitely generated submodule of a right R-module M is flat, then M is flat.

However, the converse to Corollary 1.4.3 is false in general:

Example 1.4.4. For a field k, let R = k[x, y]. Then R is flat as a module over itself (it is free) and the ideal M = (x, y) is a finitely generated submodule of R, but M is not flat.

Proposition 1.4.5. If R is an integral domain and A is a ﬂat R-module, then A is torsion- free.

Proof. Let Q be the ﬁeld of fractions of R. Since A is ﬂat, the functor − ⊗R A is exact, so the exact sequence 0 → R → Q induces an exact sequence

0 → R ⊗R A → Q ⊗R A. ∼ We know R ⊗R A = A, and Q ⊗R A is a vector space over Q, so it is torsion-free and it follows that A is torsion-free. The converse holds over a PID:

Proposition 1.4.6. If R is a PID, then every torsion-free R-module B is ﬂat.

Proof. The theory of modules over a PID says that every finitely generated R-module M that is torsion-free is also free. Thus every finitely generated submodule M ⊆ B is free, hence projective, hence flat. By Corollary 1.4.3, this implies that B is also flat.

Example 1.4.7. For any multiplicative set S ⊆ R, the localization S−1R is a ﬂat module.

Proposition 1.4.8. For an R-module M, the following are equivalent:

(a) M is ﬂat.

R (b) Torn (A, M) = 0 for all n ≥ 1 and R-modules A.

R (c) Tor1 (A, M) = 0 for all R-modules A.

R (d) Tor1 (R/I, M) = 0 for all ﬁnitely generated ideals I ⊂ R.

P P 0 (e) If rixi = 0 for some elements ri ∈ R and xi ∈ M, then each xi = rijxj for 0 P rij ∈ R and xj ∈ M such that rijri = 0. Definition. A left R-module M is faithfully flat provided that for all left R-modules A and B, the sequence 0 → A → B is exact if and only if 0 → A ⊗R M → B ⊗R M is exact. One can define faithful flatness similarly for right modules.

Lemma 1.4.9. A left R-module M is faithfully ﬂat if and only if M is ﬂat and for all left R-modules A, A ⊗R M = 0 implies A = 0.

14 1.4 Flat Modules 1 Some Commutative Algebra

∼ L L ∼ L Example 1.4.10. If F is free, say F = i∈I R, then A ⊗R i∈I R = i∈I A which is nonzero precisely when A is nonzero. Hence free modules are faithfully flat. Example 1.4.11. Any localization S−1R of a ring R is flat by Example 1.4.7, but is not faithfully flat in general. For example, if R is an integral domain with field of fractions K – the localization of R at the multiplicative set R r {0} – then for any nonzero torsion module M, M ⊗R K = 0. Example 1.4.12. In general, projective modules are not faithfully flat. For instance, suppose R has a nontrivial idempotent element e ∈ R, that is, e2 = e. Then e(1 − e) = 0 and 1−e is also an idempotent, and we have R ∼= eR⊗(1−e)R. So eR is a projective R-module, 2 but it is not faithfully flat since (1 − e)R ⊗R eR = (1 − e)R ⊗R e R = e(1 − e)R ⊗R eR = 0. Theorem 1.4.13. Let R be a commutative ring and M an R-module. Then the following are equivalent: (a) M is flat.

(b) For all prime ideals p ⊂ R, the localization Mp is ﬂat as an R-module (equivalently, as an Rp-module).

(c) For all maximal ideals m ⊂ R, the localization Mm is flat as an R-module (equivalently, as an Rm-module. Recall that an R-module M is finitely presented if there exists a short exact sequence 0 → K → Rn → M → 0 with n finite. Proposition 1.4.14. Let M be a finitely generated R-module. Then (a) If M is projective, then it is finitely presented. (b) If M is finitely presented and 0 → L → Rm → M → 0 is any short exact sequence with m finite, then L is finitely generated. Proof. (1) Let 0 → K → Rn → M → 0 be any presentation of M. Then since M is projective, the sequence splits, so K is a direct summand of Rn, thus a quotient of Rn and therefore finitely generated. (2) Assuming M is finitely presented, there is some short exact sequence 0 → K → Rn → M → 0 with n finite and K finitely generated. Consider the diagram 0 K Rn M 0

f g id

0 L Rm M 0

Then since Rm is free, we get a map g : Rn → Rm making the right square commute. Call f the restriction of g to K, which gives a commutative square on the left. By the Snake Lemma, 0 = ker id → coker f −→∼ coker g → coker id = 0 is exact, so the middle arrow is an isomorphism. Now coker g is finitely generated as it is a quotient of Rm, so this implies coker f = L/f(K) is finitely generated. Since K was finitely generated, f(K) is also finitely generated and it now follows that L is finitely generated.

15 1.4 Flat Modules 1 Some Commutative Algebra

We now relate finite presentability to flatness. First, we need a lemma. Lemma 1.4.15. Suppose 0 → N → F → M → 0 is a short exact sequence of left R-modules, where M and F are flat. Then (a) For any ideal I ⊆ R, N ∩ IF = IN. P (b) If F is free with R-basis {yi}i∈I , then for any n = riyi ∈ N ⊆ F , there exist P i∈I elements ni ∈ N such that n = i∈I rini.

(c) Suppose F is free. For any ﬁnite set of elements {n1, . . . , nr} ⊆ N, there exists a morphism f : F → N such that f(ni) = ni for all 1 ≤ i ≤ n. Proof. (a) Tensoring with I gives an exact sequence

I ⊗R N → I ⊗R F → I ⊗R M → 0. ∼ ∼ Notice that since M,F are ﬂat, I ⊗R F = IF and I ⊗R M = IM, and the kernel of the resulting map IF → IM is precisely N ∩ IF . Therefore we have a commutative diagram with exact rows

I ⊗R N I ⊗R F I ⊗R M 0

ψ ∼= ∼=

0 N ∩ IF IF IM 0

This induces a map ψ : I ⊗R N → N ∩IF which by a quick diagram chase is an isomorphism. This identifies N ∩ IF with im ψ = IN ⊆ IF . (b) Consider the ideal I = (ri)i∈I ⊆ R. Then by (a), X n ∈ N ∩ IF = IN = riN i∈I and the statement follows. (c) We induct on r. When r = 1, let n = n = P r y for some finite I ⊆ I and some 1 i∈I0 i i 0 r ∈ R. Applying (b), we get n = P r n0 for some n0 ∈ N. For each y in a basis of F , i i∈I0 i i i i set ( 0 ni, if i ∈ I0 f(yi) = 0, if i 6∈ I0. Then this extends uniquely to an R-linear map f : F → N such that ! X X X 0 f(n) = f riyi = rif(yi) = rini = n. i∈I0 i∈I0 i∈I0 To induct, assume the statement holds for all finite subsets of N consisting of r −1 elements. For a given {n1, . . . , nr} ⊆ N, the inductive hypothesis constructs for us two maps f1, f2 : F → N satisfying

f1(n1) = n1 and f2(ni − f1(ni)) = ni − f1(ni) for all 2 ≤ i ≤ r.

16 1.4 Flat Modules 1 Some Commutative Algebra

Set f = f1 + f2 − f2 ◦ f1, where f2 ◦ f1 is deﬁned by changing targets, i.e. considering f1 : F → N ⊆ F . Then f(n1) = f1(n1) = n1 and for each 2 ≤ i ≤ r,

f(ni) = f1(ni) + f2(ni − f1(ni)) = f1(ni) + ni − f1(ni) = ni.

Therefore the statement holds for all r by induction.

Theorem 1.4.16. For a left R-module M, the following are equivalent:

(1) M is ﬁnitely generated and projective.

(2) M is ﬁnitely presented and ﬂat.

(3) M is finitely presented and for all maximal ideals m ⊂ R, the localization Mm is free. Proof. (1) =⇒ (2) is Propositions 1.4.14 and 1.4.1. (2) =⇒ (1) If M is finitely presented, then there exists a short exact sequence 0 → N → F → M → 0 where F is finitely generated and free and N is finitely generated, say by n1, . . . , nr ∈ N. Then since M is flat, Lemma 1.4.15 says there is a morphism f : F → N with f(ni) = ni for all i, but since the ni generate N, this means f|N = idN . Hence the short exact sequence splits, so in particular M is a direct summand of F and thus projective. (1) =⇒ (3) By the above, M is finitely presented (and flat), so there exists a presentation m n R → R → M → 0 with m, n finite. Flatness implies that tensoring with Rm for any maximal ideal m yields an exact sequence of Rm-modules

m n Rm → Rm → Mm → 0.

Therefore Mm is finitely presented. Theorem 1.4.13 also shows Mm is flat. By finite presentation, Mm/m is a finite dimensional vector space over the residue field κ(m) = Rm/m. Choose a basisx ¯1,..., x¯r of Mm/m which are the images of x1, . . . , xr ∈ Mm. Then by Nakayama’s Pr Lemma, Mm = i=1 Rmxi. Thus Mm has a finite presentation of Rm-modules

N → F → Mm → 0

where F is free with basis y1, . . . , yr whose images in Mm are x1, . . . , xr. Suppose n = Pr Pr Pr i=1 riyi ∈ N. By exactness, n maps to i=1 rixi = 0 in Mm, so i=1 r¯ix¯i = 0 in Mm/m. Since thex ¯i are k(m)-linearly independent, this implies eachr ¯i = 0, that is, ri Pr lies in ker(Rm → κ(m)), and so ri ∈ m. Hence we see that n = i=1 riyi ∈ mF . By Lemma 1.4.15(a), N = N ∩ mF = mF , so by Nakayama’s Lemma, N = 0. This shows ∼ Mm = F . (3) =⇒ (1) will be proven later. We now turn our attention back to characterizing faithfully ﬂat R-modules.

Lemma 1.4.17. For an R-module M, the following are equivalent:

(a) M is faithfully ﬂat.

(b) M is ﬂat and for every maximal ideal m ⊂ R, mM 6= M.

17 1.4 Flat Modules 1 Some Commutative Algebra

Proof. (a) =⇒ (b) For any maximal ideal m ⊂ R, R/m 6= 0, so if M is faithfully flat, ∼ M/mM = R/m ⊗R M 6= 0. Hence mM 6= M. (b) =⇒ (a) Take an R-module N 6= 0. By Lemma 1.4.9, it suffices to show N ⊗R M 6= 0. Choose a nonzero element x ∈ N and consider the module homomorphism ϕ : R → N defined by setting ϕ(1) = x. Let I = ker ϕ ⊂ R. Then R/I is a submodule of N and I 6= R (since x 6= 0), so there is some maximal ideal m ⊂ R containing I. Thus IM ⊆ mM 6= M so ∼ R/I ⊗R M = M/IM 6= 0. Now by flatness, R/I ⊗R M is a submodule of N ⊗R M, so this shows N ⊗R M 6= 0 as desired. Example 1.4.18. For any ring R, the module M M = Rm m∈MaxSpec(R)

is faithfully flat. Indeed, M is flat since it is a direct sum of flat R-modules (use Proposi- tion 1.4.1(ii) and Theorem 1.4.13). On the other hand, suppose N ⊗R M = 0. Then ! M M M 0 = N ⊗R Rm = (N ⊗R Rm) = Nm. m m m

Therefore Nm = 0 for all maximal ideals m ⊂ R, so N = 0. This gives examples of modules which are faithfully ﬂat but not projective, such as the L direct sum p Z(p) over all prime integers p ∈ Z. Lemma 1.4.19. If M is a ﬁnitely presented R-module and N is any R-module, then for all multiplicative sets S ⊆ R, there is an isomorphism

−1 ∼ −1 −1 S HomR(M,N) = HomS−1R(S M,S N).

Proof. The assumption that M is ﬁnitely presented means there is an exact sequence Rm → Rn → M → 0. Localization is an exact functor, so S−1Rm → S−1Rn → S−1M → 0 is still exact, meaning S−1M is ﬁnitely presented as an S−1R-module, since S−1Rn ∼= (S−1R)n. On the other hand, since HomR(−,N) is left exact we get an exact sequence

n n m m 0 → HomR(M,N) → HomR(R ,N) = N → HomR(R ,N) = N .

Localizing this sequence gives the top row in the following diagram:

−1 −1 n −1 m 0 S HomR(M,N) S N S N

ϕ ∼= ∼=

−1 −1 −1 n −1 m 0 HomS−1R(S M,S N) (S N) (S N)

−1 −1 −1 This induces a morphism ϕ : S HomR(M,N) → HomS−1R(S M,S N) and a quick diagram chase shows that it is an isomorphism.

18 1.4 Flat Modules 1 Some Commutative Algebra

Let f : A → B be a ring homomorphism. This induces two functors on the corresponding categories of modules:

ModB −→ ModA N 7−→ N, with a · n = f(a)n,

and ModA −→ ModB

M 7−→ MB := B ⊗A M.

Lemma 1.4.20. If M is a (faithfully) flat A-module, then MB is a (faithfully) flat B-module. Definition. A ring homomorphism f : A → B is called (faithfully) flat if B is (faithfully) flat as an A-module. Corollary 1.4.21. If f : A → B is a (faithfully) flat morphism and S ⊆ A is a multiplicative set with T = f(S), then the induced morphism S−1f : S−1A → T −1B is (faithfully) flat. Proof. This follows from the fact that there is a canonical isomorphism of S−1A-modules −1 ∼ −1 T B = S A ⊗A B. On the other hand, we have: Lemma 1.4.22. Suppose f : A → B is faithfully flat and M is a faithfully flat B-module. Then M is also (faithfully) flat as an A-module.

f g Corollary 1.4.23. If A −→ B −→ C are faithfully ﬂat ring homomorphisms, then so is g ◦ f : A → C. Proposition 1.4.24. Let f : A → B be a faithfully ﬂat homomorphism. Then

(a) For every A-module M, the map M → M ⊗A B is injective. In particular, f itself is injective.

(b) If I ⊂ A is an ideal, then IB ∩ A = I.

Proof. (a) Suppose x ∈ M r {0}. Then Ax is a nonzero submodule of M, so because B is a faithfully ﬂat A-module, 0 6= Ax⊗A B,→ M ⊗A B is an inclusion of a nonzero submodule. On the other hand, Ax⊗A B = (x⊗10b, so x⊗1 6= 0 in Ax⊗A B and hence x⊗1 6= 0 in M ⊗A B. This shows that x is not an element of ker(M → M ⊗A B), so ker(M → M ⊗A B) = 0. (b) Apply (a) to the A-module M = A/I to see that ∼ A/I ,→ A/I ⊗A B = B/IB.

In general, the kernel of this map is (IB∩A)/I but because the map is injective, (IB∩A)/I = 0, that is, IB ∩ A = I. (c) Take a prime ideal p ⊂ A. We want to find a prime ideal q ⊂ B with p = q ∩ A. Let Ap be the localization at p, which is a local ring with maximal ideal pAp. Since A → B is faithfully flat, Corollary 1.4.21 shows that Ap → Bp is also faithfully flat, and hence injective

19 1.4 Flat Modules 1 Some Commutative Algebra

by (a). Applying (b) to the ideal I = pBp ⊂ Bp, we get pBp ∩ Ap = pAp 6= Ap, so pBp 6= Bp. Thus there is some maximal ideal m ⊂ Bp containing pBp. Now m ∩ Ap ⊇ pAp but 1 6∈ m, so m ∩ Ap is a proper ideal of Ap containing its unique maximal ideal pAp; hence m ∩ Ap = pAp. By the correspondence theorem for primes in a localization, there is a prime ideal q ⊂ B extending to m ⊂ Bp, and by construction q ∩ A = pAp ∩ A = p. Remark. It turns out that f : A → B is faithfully ﬂat if and only if f is ﬂat and f ∗ : Spec B → Spec A is surjective.

Theorem 1.4.25 (Generic Faithful Flatness). Let A be an integral domain with field of fractions K, B/A a finitely generated ring extension and assume B,→ K ⊗A B is injective. Then there exist nonzero elements a ∈ A and b ∈ B such that the induced morphism Aa → Bb is faithfully flat.

Proof. Suppose B = A[b1, . . . , bn] with bi ∈ B. By Noether’s normalization lemma, there exist elements x1, . . . , xm ∈ BK = K ⊗A B such that BK /K[x1, . . . , xm] is an integral extension. After multiplying through by a common denominator, we may assume x1, . . . , xm ∈ B. In

particular, b1, . . . , bn are integral over K[x1, . . . , xm], so they are integral over Aa0 [x1, . . . , xm]

for some nonzero element a0 ∈ A. This implies Ba0 /Aa0 [x1, . . . , xm] is a ﬁnite ring exten-

sion. Thus we may replace A with Aa0 and B with Bb0 to have a ﬁnite ring extension B/A[x1, . . . , xm]. Set C = A[x1, . . . , xm], E = K(x1, . . . , xm) and consider the diagram

B BK E ⊗C B

C K[x1, . . . , xm] E

After inverting an element of C, we may assume BK → E ⊗C B is injective. Note that the columns in the diagram are all ﬁnite ring extensions; also, the right column is injective 0 0 because E is a ﬂat C-module. Choose an E-basis {z1, . . . , zr} of E ⊗C B and let z1, . . . , zs be Ps 0 0 generators of B as a C-module, so that B = i=1 Czi. Then each zi is a linear combination Pr of the zj, so there is a nonzero polynomial q ∈ C with Bq = j=1 Cqzj. Observe that the map

r ϕ : Cq −→ Bq r X (c1, . . . , cr) 7−→ cjzj j=1 is surjective. Then since E ⊗ − is right exact, the base change ϕ : E ⊗ Cr → E ⊗ B = Cq b Cq q Cq q r E ⊗C B is also surjective. But both E ⊗Cq Cq and E ⊗C B are E-vector spaces of dimension r, so ϕ is also injective. Thus E ⊗ ker ϕ = 0, which implies ker ϕ = 0 since E is faithfully b Cq r ∼ flat. Hence Cq = Bq as Cq-modules. Choose a nonzero coefficient a in the polynomial q ∈ C = A[x1, . . . , xm] and set b = aq. Then Aa ,→ Bb is injective. We claim that it is faithfully flat. Consider the tower of ring

20 1.4 Flat Modules 1 Some Commutative Algebra

extensions Aa ,−→ Aa[x1, . . . , xm] ,→ Aa[x1, . . . , xm]q ,→ Bb.

The above paragraph shows that Aa[x1, . . . , xm] is a free Aa-module, so in particular it is faithfully flat by Example 1.4.10. Likewise, Bb is a free module over Aa[x1, . . . , xm]q, so the third map is also faithfully flat. Finally, by Corollary 1.4.23 we need only show the middle map is faithfully flat to complete the proof. Note that Aa[x1, . . . , xm]q is flat over Aa[x1, . . . , xm] by Example 1.4.7, so by Lemma 1.4.17 we need only check that if m ⊂ Aa is a maximal ideal, mAa[x1, . . . , xm]q 6= Aa[x1, . . . , xm]q. Since a is a coefficient of q, clearly mAa[x1, . . . , xm]q cannot contain q. Also, mAa[x1, . . . , xm] 6= Aa[x1, . . . , xm] since Aa ,→ Aa[x1, . . . , xm]q is faithfully flat (this follows from Lemma 1.4.17). Together, these imply that k mAa[x1, . . . , xm]q 6= Aa[x1, . . . , xm]q, since otherwise q ∈ mAa[x1, . . . , xm]q for some k ≥ 1, but m is radical, so we would have q ∈ mAa[x1, . . . , xm]q. Hence Aa ,→ Aa[x1, . . . , xm]q is flat.

21 2 Scheme Theory

2 Scheme Theory

2.1 Aﬃne Schemes

Hilbert’s Nullstellensatz is an important theorem in commutative algebra which is essentially the jumping oﬀ point for classical algebraic geometry (by which we mean the study of algebraic varieties in aﬃne and projective space). We recall the statement here.

Theorem 2.1.1 (Hilbert’s Nullstellensatz). If k is an algebraically closed ﬁeld, then there is a bijection

n Ak ←→ MaxSpec k[t1, . . . , tn] P = (α1, . . . , αn) 7−→ mP = (t1 − α1, . . . , tn − αn),

n n where Ak = k is aﬃne n-space over k and MaxSpec denotes the set of all maximal ideals of a ring.

Further, if f : A → B is a morphism of ﬁnitely generated k-algebras then we get a map f ∗ : MaxSpec B → MaxSpec A given by f ∗m = f −1(m) for any maximal ideal m ⊂ B. Note that if k is not algebraically closed, f −1(m) need not be a maximal ideal of A.

Lemma 2.1.2. Let f : A → B be a ring homomorphism and p ⊂ B a prime ideal. Then f −1(p) is a prime ideal of A.

Proof. Exercise. This suggests a natural replacement for MaxSpec A, called the prime spectrum:

Spec A = {p ⊂ A | p is a prime ideal}.

Deﬁnition. An aﬃne scheme is a ringed space with underlying topological space X = Spec A for some ring A.

In order to justify this deﬁnition, I will now tell you the topology on Spec A and the sheaf of rings making it into a ringed space. For any subset E ⊆ A, deﬁne

V (E) = {p ∈ Spec A | E ⊆ p}.

Lemma 2.1.3. Let A be a ring and E ⊆ A any subset. Set a = (E), the ideal generated by E. Then

(a) V (E) = V (a) = V (r(a)) where r denotes the radical of an ideal.

(b) V ({0}) = Spec A and V (A) = ∅. S T (c) For a collection of subsets {Ei} of A, V ( Ei) = V (Ei). (d) For any ideals a, b ⊂ A, V (a ∩ b) = V (ab) = V (a) ∪ V (b).

22 2.1 Aﬃne Schemes 2 Scheme Theory

As a result, the sets V (E) for E ⊆ A form the closed sets for a topology on Spec A, called the Zariski topology. Next, for any prime ideal p ⊂ A, let Ap denote the localization at p. For any open set U ⊆ Spec A, we define ( ) a f O(U) = s : U → A s(p) ∈ A , ∃ p ∈ V ⊆ U and f, g ∈ A so that s(q) = for all q ∈ V . p p g p∈U Theorem 2.1.4. (Spec A, O) is a ringed space. Moreover, ∼ (1) For any p ∈ Spec A, Op = Ap as rings. (2) Γ(Spec A, O) ∼= A as rings. (3) For any f ∈ A, define the open set D(f) = {p ∈ Spec A | f 6∈ p}. Then the D(f) ∼ form a basis for the topology on Spec A and O(D(f)) = Af as rings. Example 2.1.5. For any field k, Spec k is a single point ∗ corresponding to the zero ideal, with sheaf O(∗) ∼= k.

Example 2.1.6. Let A = k[t1, . . . , tn] be the polynomial ring in n variables over k. Then n Spec A = Ak , the aﬃne n-space over k. For example, when A = k[t] is the polynomial ring 1 in a single variable, Spec k[t] = Ak, the aﬃne line. 1 When k = C, Hilbert’s Nullstellensatz tells us that all the closed points of Ak correspond to maximal ideals of the form (t − α) for α ∈ C. But there is also a non-closed, ‘generic point’ corresponding to the zero ideal which was not detected before. closed points −2 0 1 + i generic point Spec C[t] (t + 2) (t) (t − (1 + i)) (0)

On the other hand, if k = Q or another non-algebraically closed ﬁeld, the same closed points corresponding to linear ideals (t − α) show up, as well as the generic point corresponding to (0), but there are also points corrresponding to ideals generated by higher degree irreducible polynomials like t2 + 1. Thus the structure of Spec Q[t] is much diﬀerent than the algebraically closed case. closed points −2 0 ?? generic point Spec Q[t] (t + 2) (t) (t2 + 1) (0)

Example 2.1.7. Let X be an algebraic variety over a field k, x ∈ X a point and consider the affine scheme Y = Spec(k[ε]/(ε2)). We can think of Y as a “big point” with underlying space ∗ corresponding to the zero ideal, along with a “tangent vector” extending infinitesimally in every direction around ∗. Then any map Y → X determines a unique tangent vector in TxX, the tangent space of X at x. This idea is useful in intersection theory. For example, 2 2 consider the tangency of the x-axis and the parabola y = x in Ak:

23 2.2 Schemes 2 Scheme Theory

y − x2

y (0, 0)

As a variety, this point (0, 0) corresponds to the quotient of k-algebras k[x, y]/r(y, y − x2) = k[x]/r(x2) = k[x]/(x) = k. Thus the information of tangency is lost. However, as an aﬃne scheme, (0, 0) corresponds to Spec(k[x, y]/(y, y − x2)) = Spec(k[x]/(x2)) so the intersection information is preserved.

2.2 Schemes

In this section we define a scheme and prove some basic properties resulting from this definition. Recall that a ringed space is a pair (X, F) where X is a topological space and F is a sheaf of rings on X. Definition. A locally ringed space is a ringed space (X, F) such that for all P ∈ X, ∼ there is a ring A such that FP = Ap for some prime ideal p ⊂ A. Example 2.2.1. Any affine scheme Spec A is a locally ringed space by (1) of Theorem 2.1.4. We will sometimes denote the structure sheaf O by OA. Definition. The category of locally ringed spaces is the category whose objects are locally ringed spaces (X, F) and whose morphisms are morphisms of ringed spaces (X, F) → # (Y, G) such that for each P ∈ X, the induced map fP : OY,f(P ) → OX,P is a morphism of # −1 local rings, i.e. (fP ) (mP ) = mf(P ) where mP (resp. mf(P )) is the maximal ideal of the local ring OX,P (resp. OY,f(P )). We are now able to define a scheme.

Definition. A scheme is a locally ringed space (X, OX ) that admits an open covering {Ui} ∼ such that each Ui is affine, i.e. there are rings Ai such that (Ui, OX |Ui ) = (Spec Ai, OAi ) as locally ringed spaces. The category of schemes Sch is defined to be the full subcategory of schemes in the category of locally ringed spaces. Denote the subcategory of affine schemes by AffSch. Also let CommRings denote the category of commutative rings with unity. Proposition 2.2.2. There is an isomorphism of categories AffSch −→∼ CommRingsop

(X, OX ) 7−→ OX (X) (Spec A, O) 7−→ A.

24 2.2 Schemes 2 Scheme Theory

Proof. (Sketch) First suppose we have a homomorphism of rings f : A → B. By Lemma 2.1.2 this induces a morphism f ∗ : Spec B → Spec A, p 7→ f −1(p) which is continuous since f −1(V (a)) = V (f(a)) for any ideal a ⊂ A. Now for each p ∈ Spec B, define the localization fp : Af ∗p → Bp using the universal property of localization. Then for any open set V ⊆ Spec A, we get a map # ∗ −1 f : OA(V ) −→ OB((f ) (V )). One checks that each is a homomorphism of rings and commutes with the restriction maps. # Thus f : OA → OB is defined. Moreover, the induced map on stalks is just each fp, so the ∗ # pair (f , f ) gives a morphism (Spec B, OB) → (Spec A, OA) of locally ringed spaces, hence of schemes. # Conversely, take a morphism of schemes (ϕ, ϕ ) : (Spec B, OB) → (Spec A, OA). This induces a ring homomorphism Γ(Spec A, OA) → Γ(Spec B, OB) but by (2) of Theorem 2.1.4, ∼ ∼ Γ(Spec A, OA) = A and Γ(Spec B, OB) = B so we get a homomorphism A → B. It’s easy to see that the two functors described give the required isomorphism of categories. Example 2.2.3. We saw in Example 2.1.5 that for any field k, Spec k = ∗ is a point with structure sheaf O(∗) = k. If L1,...,Lr are finite separable field extensions of k, we ` ` call A = L1 × · · · × Lr a finite étale k-algebra. Then Spec A = Spec L1 ··· Spec Lr is (schematically) a disjoint union of points.

Example 2.2.4. Let A be a DVR with residue field k. Then Spec A = {0, mA}, a closed point for the maximal ideal m and a generic point for the zero ideal. There are two open subsets here, {0} and Spec A, and we have OA({0}) = k and OA(Spec A) = A. Example 2.2.5. If k is a field and A is a finitely generated k-algebra, then the closed points of X = Spec A are in bijection with the closed points of an affine variety over k with coordinate ring A.

Example 2.2.6. Let A = Z (or any Dedekind domain). Then dim A = 1 and it turns out that dim Spec A = 1 for some appropriate notion of dimension (see Section 2.3). Explicitly, Spec Z has a closed point for every prime p ∈ Z and a generic point for (0): closed points generic point Spec Z 2 3 5 7 11 (0)

1 Example 2.2.7. Let k be a field, X1 = X2 = Ak two copies of the affine line and U1 = U2 = 1 1 Ak r {0}, where 0 is the closed point of Ak corresponding to (x) in k[x]. Then we can glue together X1 and X2 along the identity map U1 → U2 to get a scheme X which looks like the affine line with the origin “doubled”. Note that X is not affine!

1 Ak r {0}

25 2.3 Properties of Schemes 2 Scheme Theory

2.3 Properties of Schemes

Many deﬁnitions in ring theory can be rephrased for schemes. For example:

Deﬁnition. A scheme X is reduced if for all open U ⊆ X, OX (U) has no nilpotent elements.

Deﬁnition. A scheme X is integral if for all open U ⊆ X, OX (U) has no zero divisors. Lemma 2.3.1. X is integral if and only if X is reduced and irreducible as a topological space. Proof. ( =⇒ ) Clearly integral implies reduced, so we just need to prove X is irreducible. Suppose X = U ∪ V for open subsets U, V ⊆ X. Then OX (U ∪ V ) = OX (U) × OX (V ) which is not a domain unless one of OX (U), OX (V ) is 0. In that case, U or V is empty, so this shows X is irreducible. ( ⇒ = ) Suppose X is reduced and irreducible, but there exists an open set U ⊆ X and f, g ∈ OX (U) with fg = 0. Deﬁne closed sets

C = {P ∈ U | fP ∈ mP ⊂ OX,P }

D = {P ∈ U | gP ∈ mP ⊂ OX,P }.

Then by definition of OX , we must have C ∪D = U. By irreducibility, C = U without loss of 0 0 generality. Thus for any affine open set U ⊆ U with U = Spec A, we have (OX |U 0 )(D(f)) = ∼ 0 but by (3) of Theorem 2.1.4, OU 0 (D(f)) = Af , the localization of A at powers of f. When Af = 0, f is nilpotent but by assumption this means f = 0. Hence X is integral. Definition. The dimension of a scheme X (or any topological space) is

dim X = sup{n ∈ N0 | there exists a chain of irreducible, closed sets X0 ( X1 ( ··· ( Xn ⊆ X}. Proposition 2.3.2. Let A be a noetherian ring. Then dim Spec A = dim A, the Krull dimension of A. Be warned that the converse to Proposition 2.3.2 is false in general. Deﬁnition. Let X be a scheme. Then

X is locally noetherian if each stalk OX,P is a local noetherian ring. X is noetherian if X is integral and locally noetherian.

An integral scheme X is normal if each stalk OX,P is integrally closed in its ﬁeld of fractions. 2 X is regular if each OX,P is regular as a local ring, that is, dim OX,P = dim mP /mP as OX,P /mP -vector spaces.

Deﬁnition. Let U ⊆ X be an open subset. Then (U, OX |U ) is a scheme which we call an # open subscheme of X. The natural morphism j : U,→ X, j : OX → j∗OX |U is called an open immersion.

26 2.4 Sheaves of Modules 2 Scheme Theory

Example 2.3.3. For X = Spec A, let f ∈ A and recall the open set D(f) deﬁned in Theorem 2.1.4. Then D(f) is an open subscheme of X and the open immersion D(f) ,→ X corresponds to the natural inclusion of prime ideals Spec Af ,→ Spec A (this is a property of any localization).

Deﬁnition. Let A → A/I be a quotient homomorphism of rings. Then the induced morphism Spec(A/I) → Spec A is called an aﬃne closed immersion. For a general morphism of schemes f : X → Y , f is called a closed immersion if f is injective, f(X) ⊆ Y is closed

and there exists a covering of X by aﬃne open sets {Ui} such that each f|Ui : Ui → f(Ui) is an aﬃne closed immersion. The set f(X) is called a closed subscheme of Y .

Definition. A morphism f : Y → X is locally of finite type if there exists an affine S −1 covering X = Ui, with Ui = Spec Ai, such that each f (Ui) has an open covering −1 Sni f (Ui) = j=1 Spec Bij for ni < ∞ and Bij a finitely generated Ai-algebra. Further, we say −1 f is a finite morphism if each ni = 1, i.e. f (Ui) = Spec Bi for some finitely generated Ai-algebra Bi.

2.4 Sheaves of Modules

Through Proposition 2.2.2, we are able to transfer commutative ring theory to the language of aﬃne schemes. In this section, we deﬁne a suitable setting for transferring module theory to the language of sheaves and schemes.

Deﬁnition. Let (X, OX ) be a ringed space. A sheaf of OX -modules, or an OX -module for short, is a sheaf of abelian groups F on X such that each F(U) is an OX (U)-module and for each inclusion of open sets V ⊆ U, the following diagram commutes:

OX (U) × F(U) F(U)

OX (V ) × F(V ) F(V )

If F(U) ⊆ OX (U) is an ideal for each open set U, then we call F a sheaf of ideals on X. Example 2.4.1. Let f : Y → X be a morphism of ringed spaces. Then the pushforward # sheaf f∗OY is naturally an OX -module on X via f : OX → f∗OY . Additionally, the kernel # # sheaf of f , deﬁned on open sets by (ker f )(U) = ker(OX (U) → f∗OY (U)), is a sheaf of ideals on X.

Most module terminology extends to sheaves of OX -modules. For example,

A morphism of OX -modules is a morphism of sheaves F → G such that each F(U) →

G(U) is an OX (U)-module map. We write HomX (F, G) = HomOX (F, G) for the set of morphisms F → G as OX -modules. This deﬁnes the category of OX -modules, written OX -Mod.

27 2.4 Sheaves of Modules 2 Scheme Theory

Taking kernels, cokernels and images of morphisms of OX -modules again give OX - modules.

Taking quotients of OX -modules by OX -submodules again give OX -modules.

0 00 0 An exact sequence of OX -modules is a sequence F → F → F such that each F (U) → 00 F(U) → F (U) is an exact sequence of OX (U)-modules.

Basically any functor on modules over a ring generalizes to an operation on OX -

modules, including Hom, written HomOX (F, G); direct sum F ⊕ G; tensor product Vn F ⊗OX G; and exterior powers F. The most important of these constructions for our purposes will be the direct sum operation.

∼ ⊕r Deﬁnition. An OX -module F is free (of rank r) if F = O as OX -modules. F is locally S X free if X has a covering X = Ui such that each F|Ui is free as an OX |Ui -module.

Remark. The rank of a locally free sheaf of OX -modules is constant on connected components. In particular, the rank of a locally free OX -module is well-deﬁned whenever X is connected.

Definition. A locally free OX -module of rank 1 is called an invertible sheaf. Let A be a ring, M an A-module and set X = Spec A. To extend module theory to the language of schemes, we want to define an OX -module Mf on X. To start, for each p ∈ Spec A, let Mp = M ⊗A Ap be the localization of the module M at p. Then Mp is an m Ap-module consisting of ‘formal fractions’ s where m ∈ M and s ∈ S = A r p. For each open set U ⊆ X, define ( ) a m Mf(U) = h : U → Mp s(p) ∈ Mp, ∃ p ∈ V ⊆ U, m ∈ M, s ∈ A with s(q) = for all q ∈ V . s p∈U

(Compare this to the construction of the structure sheaf OA on Spec A in Section 2.1. Also, note that necessarily the s ∈ A in the deﬁnition above must lie outside of all q ∈ V .)

Proposition 2.4.2. Let M be an A-module and X = Spec A. Then Mf is a sheaf of OX - modules on X, and moreover, ∼ (1) For any p ∈ Spec A, Mfp = Mp as rings.

(2) Γ(X, Mf) ∼= M as A-modules. ∼ (3) For any f ∈ A, Mf(D(f)) = Mf = M ⊗A Af as A-modules. The proof is similar to the proof of Theorem 2.1.4; both can be found in Hartshorne.

28 2.4 Sheaves of Modules 2 Scheme Theory

Proposition 2.4.3. Let X = Spec A. Then the association

A-Mod −→ OX -Mod M 7−→ Mf deﬁnes an exact, fully faithful functor. Proof. Similar to the proof of Proposition 2.2.2.

These Mf will be our affine model for modules over a scheme X. We next define the general notion, along with an analogue of finitely generated modules over a ring.

Deﬁnition. Let (X, OX ) be a scheme. An OX -module F is quasi-coherent if there is an S ∼ aﬃne covering X = Xi, with Xi = Spec Ai, and Ai-modules Mi such that F|Xi = Mfi as

OX |Xi -modules. Further, we say F is coherent if each Mi is a ﬁnitely generated Ai-module.

Example 2.4.4. For any scheme X, the structure sheaf OX is obviously a coherent sheaf on X.

Let QCohX (resp. CohX ) be the category of quasi-coherent (resp. coherent) sheaves of OX -modules on X.

Theorem 2.4.5. QCohX and CohX are abelian categories. Example 2.4.6. Let X = Spec A, I ⊆ A an ideal and Y = Spec(A/I). Then the natural ∼ inclusion i : Y,→ X is a closed immersion by definition, and it turns out that i∗OY = A/Ig as OX -modules, so i∗OY is a quasi-coherent, even coherent, sheaf on X. We next identify the image of the functor M 7→ Mf from Proposition 2.4.3. Theorem 2.4.7. Let X = Spec A. Then there is an equivalence of categories ∼ A-Mod −→ QCohX . Moreover, if A is noetherian, this restricts to an equivalence ∼ A-mod −→ CohX where A-mod denotes the subcategory of finitely generated A-modules. Proof. (Sketch) The association M 7→ Mf sends an A-module to a quasi-coherent sheaf on X = Spec A by definition of quasi-coherence. Further, one can prove that a sheaf F on X is ∼ a quasi-coherent OX -module if and only if F = Mf for an A-module M. The inverse functor QCohX → A-Mod is given by F 7→ Γ(X, F). When A is noetherian, the above extends to say that F is coherent if and only if F ∼= Mf for a finitely generated A-module M. The rest of the proof is identical. The following lemma generalizes Example 2.4.6. Lemma 2.4.8. Let f : Y → X be a morphism of schemes and let G be a quasi-coherent sheaf on Y . Then f∗G is a quasi-coherent sheaf on X. Further, if G is coherent and f is a finite morphism, then f∗G is also coherent. Note that the second statement is false in general.

29 3 Etale´ Morphisms

3 Etale´ Morphisms

3.1 Flat Morphisms

Definition. A morphism of schemes f : X → Y is said to be flat at x ∈ X if OX,x is a flat OY,f(x)-module. We say f is a flat morphism if it is flat at every point x ∈ X. Definition. A morphism f : X → Y is faithfully flat if it is flat and surjective.

Remark. Note that a faithfully flat morphism is necessarily faithfully flat on stalks, meaning each OX,x is faithfully flat as an OY,f(x)-module. However, faithful flatness on stalks does not guarantee surjectivity, so our definition is strictly stronger than the stalkwise definition.

Example 3.1.1. An open immersion is flat but not faithfully flat (unless it is a isomorphism). Indeed, open immersions are locally of the form Spec Af ,→ Spec A for a ring A and an element f ∈ A, and Af is a flat A-module by Example 1.4.7. Lemma 3.1.2. If f : X → Y and g : Y → Z are (faithfully) flat morphisms, then so is g ◦ f : X → Z.

Proposition 3.1.3. Suppose A and B are noetherian rings and g : A → B is a ring homomorphism making B a faithfully ﬂat A-algebra. If b ∈ B is an element such that ¯b ∈ B/mB is a nonzero divisor for all maximal ideals m ⊂ A, then B/(b) is a ﬂat A-algebra.

Proof. Let m be a maximal ideal of A. Since g : A → B is faithfully ﬂat, there exists a maximal ideal n ⊂ B retracting to m along g. Then since localization preserves ﬂatness (Corollary 1.4.21), we may replace A with Am and B with Bn so that g is a morphism of local rings. Denote the resulting maximal ideals by m ⊂ A and n ⊂ B. Suppose c ∈ B such that bc = 0. Since ¯b is not a zero divisor in B/mB, we must have c¯ = 0, i.e. c ∈ mB. We claim c ∈ mnB for all n ≥ 1. To induct, assume c ∈ mnB. Let n Pr {a1, . . . , ar} be a minimal generating set for m and write c = i=1 aibi for bi ∈ B. Then

r r X X 0 = bc = b aibi = ai(bbi). i=1 i=1

Ps 0 0 It follows from Proposition 1.4.8 that bbi = j=1 aijbj for some aij ∈ A and bj ∈ B satisfying P n aijai = 0. Now since {a1, . . . , ar} is a minimal generating set for m , we must have aij ∈ m ¯ for each i, j. This shows that bbi ∈ mB, but since b is not a zero divisor in B/mB, that n+1 T∞ n means each bi ∈ mB, so c ∈ m B. By induction, c ∈ n=1 m B but since g : A → B is a morphism of local rings, ∞ ∞ \ \ mnB = nn = 0 n=1 n=1 by Krull’s intersection theorem. Therefore c = 0, so b is not a zero divisor of B. Repeating the argument for the ring homomorphism A/I → B/IB where I ⊂ A is any ideal, we see A that b + IB is not a zero divisor in any B/IB. This implies Tor1 (A/I, B/(b)) = 0 for all ideals I ⊂ A, so by Proposition 1.4.8, B/(b) is ﬂat.

30 3.1 Flat Morphisms 3 Etale´ Morphisms

Deﬁnition. Let x, x0 ∈ X be points. We say x0 is a generalization of x, and x is a specialization of x0, if x ∈ {x0}. A point η ∈ X is called a generic point if {η} = X.

Example 3.1.4. For an aﬃne scheme X = Spec A, a prime p is a generalization of q if and only if p ⊆ q. Thus p is generic if and only if it’s the unique maximal ideal of A.

Proposition 3.1.5. Suppose f : X → Y is a ﬂat morphism of aﬃne schemes and take x ∈ X with y = f(x) ∈ Y . Then for any generalization y0 of y, there exists a point x0 ∈ f −1(y0) which is a generalization of x.

Proof. The flat condition implies that OX,x is a flat OY,y-module, and by Lemma 1.4.17 it is even faithfully flat, so by Proposition 1.4.24(c), fx : Spec OX,x → Spec OY,y is surjective. Definition. Suppose X is a noetherian space. A subset U ⊆ X is locally closed if it is an intersection of open and closed sets. A set is constructible if it is a union of finitely many locally closed subsets.

Proposition 3.1.6. Let X be a noetherian space. Then a subset U ⊆ X is constructible if and only if for every irreducible closed subset Z for which Z ∩ U is dense in Z, the set Z ∩ U contains an open subset of Z.

Theorem 3.1.7 (Chevalley). If f : X → Y is a ﬁnite morphism of noetherian schemes, then f(X) is constructible.

−1 Proof. Cover Y with aﬃne open sets V1,...,Vr and for each 1 ≤ j ≤ r, cover f (Vj) with

affine open sets U1j,...,Usj j. Then it’s enough to prove each f(Uij) is constructible in Vj, so we may reduce to the case of a finite morphism of affine schemes f : Spec B → Spec A. Now any irreducible closed subset of Spec A is of the form Z = V (p) for some prime ideal p ⊂ A. Suppose f(Spec B) ∩ V (p) is dense in V (p). Then f(Spec B) ∩ V (p) may be identified with the image of the morphism

ϕ : Spec(B/pB) −→ Spec(A/p).

Thus we may reduce to the case when A is a domain, so that B is a ﬁnitely generated A-algebra. Then B = A[x1, . . . , xn, xn+1, . . . , xm] where x1, . . . , xn are algebraically independent over A and xn+1, . . . , xm are algebraic over A[x1, . . . , xn]. Thus for each n + 1 ≤ i ≤ m, there exist polynomials pij ∈ A[x1, . . . , xn] such that

dj dj −1 p0jxj + p1jxj + ... + pdj j = 0

(and p0j 6= 0). Consider the following polynomial with coeﬃcients in A:

m Y Q = p0j. j=n+1

Then for any prime ideal p ⊂ A, the ideal P = p[x1, . . . , xn] is prime in A[x1, . . . , xn] and Q 6∈ P. Hence BP is integral over A[x1, . . . , xn]P and the result follows from the going up theorem.

31 3.1 Flat Morphisms 3 Etale´ Morphisms

Corollary 3.1.8. If f : X → Y is a ﬁnite morphism of noetherian schemes and Z is constructible in X, then f(Z) is constructible in Y . Proposition 3.1.9. Suppose X is a noetherian space such that every irreducible closed set in X has a generic point. Let U ⊆ X be a constructible set and x ∈ X. Then U contains an open neighborhood of x if and only if every generalization of x lies in U. Theorem 3.1.10. Let X be a noetherian space such that every irreducible closed set in X has a generic point. Then U ⊆ X is open if and only if the following conditions hold: (a) For all x ∈ U, every generalization of x lies in U.

(b) For all x ∈ U, U ∩ {x} contains a nonempty open subset of {x}. Proof. ( =⇒ ) is clear from Propositions 3.1.6 and 3.1.9. ( ⇒ = ) We claim U is constructible. Let Z be an irreducible closed set such that Z ∩ U is dense in Z. Since X is noetherian, Z contains a generic point η, so Z = {η}. Now by condition (b), Z ∩ U contains a nonempty open subset of Z so Proposition 3.1.6 implies U is constructible. Hence (a) and Proposition 3.1.9 imply that U is open. Proposition 3.1.11. If f : X → Y is a ﬁnite morphism of noetherian schemes, x ∈ X and y = f(x) ∈ Y , then the following are equivalent: (1) f takes every open neighborhood of x to an open neighborhood of y.

(2) For any generalization y0 of y, there exists x0 ∈ f −1(y0) which is a generalization of x. Proof. (1) =⇒ (2) Suppose y ∈ {y0}. Take Z to be the union of all irreducible components of f −1({y0}) not containing x. Then X r Z is an open neighborhood of x, so f(X r Z) is an 0 open neighborhood of y. Thus by Proposition 3.1.9, y ∈ f(X r Z). Take x1 ∈ X r Z with 0 −1 0 f(x1) = y . Then x1 lies in some irreducible component of f ({y }) which also contains x, say C. Since X is noetherian, C contains a generic point, i.e. C = {x0} for some x0 ∈ C. 0 0 0 0 Thus x is a generalization of x, so x1 ∈ {x }, which implies y = f(x1) ∈ {f(x )}. But x0 ∈ C ⊆ f −1({y0}) implies that f(x0) ∈ {y0}. Hence y0 = f(x0). (2) =⇒ (1) Take an open neighborhood U of x. Then U is constructible, so by Corol- lary 3.1.8, f(U) is constructible in Y . By Proposition 3.1.9, f(U) is an open neighborhood of y = f(x). Theorem 3.1.12. Let f : X → Y be a finite, flat morphism of noetherian schemes. Then f is an open map. Proof. This follows immediately from Propositions 3.1.11 and 3.1.5. Our main goal in this section is to prove that every finite morphism of noetherian schemes is generically flat, meaning there exists an open set of the source over which the morphism is flat. To prove this, we need the following two results in commutative algebra. Lemma 3.1.13. Let f : A → B be a homomorphism of noetherian rings such that B is a finitely generated A-algebra, let q ∈ Spec B and p = f −1(q) ∈ Spec A be prime ideals and suppose M is a finitely generated B-module such that Mq is a flat Ap-module. Then

32 3.2 K¨ahler Diﬀerentials 3 Etale´ Morphisms

(a) There exists an element g ∈ B r q such that (M/pM)g is a ﬂat A/p-module and A Tor1 (M, A/p)g = 0.

0 (b) For any q ∈ D(g) ∩ {q}, Mq0 is a ﬂat A-module. Theorem 3.1.14 (Generic Flatness). If f : X → Y is a ﬁnite morphism of noetherian schemes, then the set

U = {x ∈ X | OX,x is a flat OY,f(x)-module} is open in X. Proof. It is enough to prove the theorem when X and Y are affine, say X = Spec B and Y = Spec A. Note that Corollary 1.4.21 implies the set U contains the generalizations of all of its points. Also, it follows from Lemma 3.1.13(b) that for all x ∈ U, the set U ∩ {x} contains an open set of {x}. Hence the conditions in Theorem 3.1.10 are satisfied, so U is open.

3.2 K¨ahlerDiﬀerentials

In many areas of math, especially complex and differential geometry, smooth maps are essential tools of study. To define an appropriate notion of smoothness for schemes, we first introduce a module construction developed in commutative algebra to parallel the classical theory of differential forms. Definition. For a commutative ring A, an A-algebra B and a B-module M, an A-derivation of B into M is an A-linear homomorphism d : B → M that satisfies the Leibniz rule: for 0 0 0 0 all b, b ∈ B, d(bb ) = b(db ) + (db)b . Let DerA(B,M) denote the set of all A-derivations B → M. Lemma 3.2.1. If d : B → M is an A-derivation, then for all a ∈ A, da = 0.

Proof. Here, da denotes d(a1B) so the Leibniz rule for this expression is:

da = d(a1B) = (da)1B + a(d1B) = da + da since d is A-linear. Thus da = 2da, or da = 0. Definition. For an A-algebra B, a module of relative Kählerdifferentials (usually just 1 1 relative differentials) of B is a B-module ΩB/A together with an A-derivation d : B → ΩB/A which are universal with respect to A-derivations. That is, for any A-derivation e : B → M, 1 there exists a unique B-module homomorphism g :ΩB/A → M making the following diagram commute: e B M

d g 1 ΩB/A

33 3.2 K¨ahler Diﬀerentials 3 Etale´ Morphisms

1 As with any universal object, if it exists, ΩB/A is unique up to isomorphism. For its existence, we have:

Proposition 3.2.2. For any A-algebra B, the module of relative diﬀerentials of B may be deﬁned by 1 ΩB/A = Zhdb | b ∈ Bi/N where N is the submodule generated by elements of the form da for a ∈ A and d(b+b0)−db−db0 and d(bb0) − b(db0) − (db)b0 for b, b0 ∈ B.

1 Proof. The map d : B → ΩB/A is simply b 7→ db + N, which is a derivation by definition of 1 N. Suppose e : B → M is an A-derivation. Then there is a map g :ΩB/A → M obtained P P by setting g ( rbdb + N) = rbe(b). Note that g(N) = 0 since e is a derivation, so g is well-defined. Moreover, the diagram for the universal property of differentials commutes by construction of g, and uniqueness is proven as usual.

1 The universal property of ΩB/A implies the following nice fact. Lemma 3.2.3. For any A-algebra B and B-module M, there is an isomorphism

1 ∼ HomB(ΩB/A,M) −→ DerA(B,M) ϕ 7−→ ϕ ◦ d which is natural in M.

1 Example 3.2.4. If B = A[x1, . . . , xn], then ΩB/A is a free B-module generated by {dx1, . . . , dxn} 1 and the derivation d : B → ΩB/A is given by

n X ∂f df = dx ∂x i i=1 i

where ∂f denote the formal partial derivatives of the polynomial f. ∂xi Lemma 3.2.5. Suppose A is a commutative ring. Then

1 (a) If I ⊂ A is an ideal, then Ω(A/I)/A = 0.

1 (b) For any multiplicative subset S ⊂ A, ΩS−1A/A = 0. Proof. (a) For any a + I ∈ A/I, d(a + I) = ad(1 + I) since d is A-linear, and as we saw in the proof of Lemma 3.2.1, d(1 + I) = 0. (b) For any b ∈ S−1A, there is some s ∈ S so that sb ∈ A. Then by Lemma 3.2.1, 0 = d(sb) = s(db) + (ds)b = s(db) since s ∈ S ⊂ A. But since s 6= 0, we must have db = 0.

34 3.2 K¨ahler Diﬀerentials 3 Etale´ Morphisms

Now let ϕ : B → C be an A-algebra homomorphism. This induces a homomorphism of C-modules

1 1 dϕ :ΩB/A ⊗B C −→ ΩC/A db ⊗ c 7−→ cd(ϕ(b)).

1 On the other hand, by Lemma 3.2.3, the canonical derivation C → ΩC/A induces a C-module 1 1 map βϕ :ΩC/A → ΩC/B which is just dc 7→ dc. (One can deﬁne βϕ directly this way and then check it is well-deﬁned and preserves the C-module structure, but there is a universal property for a reason.)

Proposition 3.2.6. Let B be an A-algebra. Then

0 0 0 (a) For any other A-algebra A , set B = B ⊗A A . Then there is a natural isomorphism 0 1 ∼ 1 0 of B -modules ΩB0/A0 = ΩB/A ⊗B B . (b) For any A-algebra homomorphism ϕ : B → C, there is an exact sequence of C-modules

1 dϕ 1 βϕ 1 ΩB/A ⊗B C −→ ΩC/A −→ ΩC/B → 0.

(c) For any multiplicative subset S ⊂ B, there is an isomorphism of S−1B-modules −1 1 ∼ 1 S ΩB/A = ΩS−1B/A. (d) Suppose C = B/I for an ideal I ⊂ B. Then there is an exact sequence of C-modules

2 δ 1 dϕ 1 I/I −→ ΩB/A ⊗B C −→ ΩC/A → 0 where ϕ is the quotient map B → C and δ is deﬁned by x + I2 7→ dx ⊗ 1.

1 0 0 1 0 ∼ Proof. (a) The derivation d : B → ΩB/A induces a map d = d ⊗ idA0 : B → ΩB/A ⊗A A = 1 0 ΩB/A ⊗B B and one can check that it satisﬁes the same universal property as the canonical 0 1 1 ∼ 1 0 derivation B → ΩB0/A0 . Therefore ΩB0/A0 = ΩB/A ⊗B B . (b) By Yoneda’s Lemma, it’s equivalent to show

1 1 1 0 → HomC (ΩC/B,N) → HomC (ΩC/A,N) → HomC (ΩB/A ⊗B C,N)

1 ∼ is exact for an arbitrary C-module N. By Hom-tensor adjointness, HomC (ΩB/A ⊗B C,N) = 1 HomB(ΩB/A,N) so Lemma 3.2.3 shows that the sequence in question is really

0 → DerB(C,N) → DerA(C,N) → DerA(B,N)

which is exact by the deﬁnition of derivations. 1 (c) By Lemma 3.2.5(b), ΩS−1B/B = 0, so the sequence in (b) becomes a surjection

−1 1 1 S ΩB/A → ΩS−1B/A → 0.

db −1 1 1 b The map here sends s ∈ S ΩB/A to s ·d 1 which is clearly injective, hence an isomorphism.

35 3.2 K¨ahler Diﬀerentials 3 Etale´ Morphisms

(d) Again by Yoneda’s Lemma and Lemma 3.2.3, it’s enough to show

2 0 → DerA(B/I, N) → DerA(B,N) → HomB/I (I/I ,N)

2 ∼ is exact for any B/I-module N. Note that HomB/I (I/I ,N) = HomB(I,N) by the universal property of quotients, and the map from DerA(B,N) sends d : B → N to d|I : I → N. It is clear that the quotient of this map is DerA(B/I, N), so the sequence is exact. Corollary 3.2.7. If B is a ﬁnitely generated A-algebra or a localization of such an algebra, 1 then ΩB/A is ﬁnitely generated as a B-module. Proof. Given a presentation n M ϕ 0 → I −→ A −→ B → 0 i=1 Ln set F = i=1 A and apply Proposition 3.2.6(d) to the quotient map ϕ : F → B to get an exact sequence of B-modules

2 1 dϕ 1 I/I → ΩF/A ⊗F B −→ ΩB/A → 0.

1 1 Then by definition ΩF/A is finitely generated as an A-module, so ΩF/A ⊗F B is finitely 1 generated as a B-module and this implies ΩB/A is as well.

Example 3.2.8. Suppose B = A[x1, . . . , xn] is a polynomial ring over A, take f ∈ B and set C = B/(f). Then by Proposition 3.2.6(d) there is an exact sequence

2 1 1 (f)/(f ) → ΩB/A ⊗B C → ΩC/A → 0

so we can see that n ! 1 M ΩC/A = Chdxii /Chdfi i=1 where df denotes the formal diﬀerential

n X ∂f df = dx . ∂x i i=1 i

More generally, if C = B/(f1, . . . , fm) then Ln 1 i=1 Chdxii ΩC/A = . Chdf1, . . . , dfmi Lemma 3.2.9. Let R and S be A-algebras. Then

(a) If B = R ⊗A S then there is a canonical isomorphism of B-modules

1 1 ∼ 1 (ΩR/A ⊗R B) ⊕ (ΩS/A ⊗S B) −→ ΩB/A

(dr ⊗ b1, ds ⊗ b2) 7−→ b1d(r ⊗ 1) + b2d(1 ⊗ s).

36 3.2 K¨ahler Diﬀerentials 3 Etale´ Morphisms

(b) If S = R[x1, . . . , xn]/I for an ideal I, let ρ : R → S be the restriction of the quotient 1 1 2 map to R ⊆ R[x1, . . . , xn] and let α = dρ :ΩR/A ⊗R S → ΩS/A and δ : I/I → Ω1 ⊗ S be the maps deﬁned above. Then there is a surjective B- R[x1,...,xn]/A R[x1,...,xn] module homomorphism ker δ → ker α.

1 ∼ 1 1 ∼ 1 Proof. (a) From Proposition 3.2.6(a), ΩB/R = ΩS/A ⊗S B and ΩB/S = ΩR/A ⊗R B, so (b) of the same proposition gives us exact sequences

1 ϕ 1 1 ΩR/A ⊗R B −→ ΩB/A → ΩS/A ⊗S B (1) 1 ψ 1 1 and ΩS/A ⊗S B −→ ΩB/A → ΩR/A ⊗R B. (2)

Then the desired map is ϕ ⊕ ψ which is of the desired form by construction and is an isomorphism because the last maps in (1) and (2) are sections of ψ and ϕ, respectively. 0 0 0 0 (b) Set A = A[x1, . . . , xn] and R = R[x1, . . . , xn]. Then the fact that R = A ⊗A R implies by (a) that

1 ∼ 1 0 1 0 1 1 0 ΩR0/A = (ΩA0/A ⊗A0 R ) ⊕ (ΩR/A ⊗R R ) = ΩR0/R ⊕ (ΩR/A ⊗R R ).

Applying − ⊗R0 S yields an isomorphism

1 ∼ 1 1 ΩR0/A ⊗R0 S = (ΩR0/R ⊗R0 S) ⊕ (ΩR/A ⊗R S) (3) which ﬁts into a commutative diagram

0 2 δ 1 1 I/I ΩR0/A ⊗R0 S ΩS/A 0

δ ∼= α

1 p 1 1 i 1 0 ΩR0/R ⊗R0 S (ΩR0/R ⊗R0 S) ⊕ (ΩR/A ⊗R S) ΩR/A ⊗R S 0

where the top row is the exact sequence from Proposition 3.2.6(d) and the bottom row is the canonical split sequence for the direct sum. A quick diagram chase reveals there is a surjection ker δ → ker α. We next study relative differentials for field extensions. Let k be a field.

Lemma 3.2.10. Let E/k be an arbitrary ﬁeld extension and K = E[x]/(p(x)) a simple algebraic extension of E. Then

1 (a) If K/E is separable, then ΩK/E = 0 and there is an isomorphism of K-vector spaces 1 ∼ 1 ΩE/k ⊗E K = ΩK/k.

1 ∼ 1 1 1 (b) If K/E is inseparable, then ΩK/E = K and dimE ΩE/k ≤ dimK ΩK/k ≤ dimE ΩE/k +1. Proof. (a) By Example 3.2.8,

1 0 ∼ 0 ΩK/E = Khdxi/(p (x) dx) = K/(p (x))

37 3.3 Sheaves of Relative Diﬀerentials 3 Etale´ Morphisms

so if the extension is separable, p0(x) is relatively prime to p(x). Thus K/(p0(x)) = 0. The 1 ∼ 1 isomorphism ΩE/k ⊗E K = ΩK/k then follows from Proposition 3.2.6(d). 0 1 0 ∼ (b) When K/E is inseparable, p (x) = 0 so ΩK/E = Khdxi/(p (x) dx) = K. From the isomorphism (3) in the proof of Lemma 3.2.9, we obtain

1 1 1 dimK (ΩE[x]/k ⊗E[x] K) = dimK Khdxi + dimK (ΩE/k ⊗E K) = 1 + dimE ΩE/k

from which the inequalities in (b) are easily obtained.

Corollary 3.2.11. Suppose K/k is a finite extension. Then K/k is separable if and only if 1 ΩK/k = 0. Proof. ( =⇒ ) is immediate from Lemma 3.2.10(a) since a separable extension is always simple. ( ⇒ = ) Suppose K/k is inseparable. Then there is a tower of field extensions K ⊇ E ⊇ k 1 ∼ where K/E is simple and inseparable so by Lemma 3.2.10(b), ΩK/E = K. However, by 1 1 1 Proposition 3.2.6 there is a surjection ΩK/k → ΩE/k so ΩK/k 6= 0. Definition. A function field K/k is said to be separable if it has a purely transcendental subextension E/k such that K/E is finite and separable.

Proposition 3.2.12. Suppose K is a function ﬁeld over k of transcendence degree n. Then 1 1 ΩK/k is a ﬁnite dimensional K-vector space of dimension dimK ΩK/k ≥ n, with equality if and only if K/k is separable.

Proof. By definition K is a finite extension of L = k(x1, . . . , xn). Moreover, since L = 1 Frac(k[x1, . . . , kn]), Example 3.2.4 and Proposition 3.2.6(c) imply dimL ΩL/k = n. Thus 1 Lemma 3.2.10 gives n ≤ dimK ΩK/k < ∞. If K/k is separable, then K/L is separable and 1 1 a simple extension, and Lemma 3.2.10(a) gives dimK ΩK/k = dimL ΩL/k = n. Conversely if 1 1 dimK ΩK/k = n, we may take α1, . . . , αn ∈ K such that ΩK/k = SpanK {dα1, . . . , dαn}. Then E = k(α1, . . . , αn) defines a subextension of K/k and the exact sequence

1 1 1 ΩE/k ⊗E K → ΩK/k → ΩK/E → 0

1 from Proposition 3.2.6(b) implies ΩK/E = 0 since the ﬁrst arrow is an isomorphism. By Lemma 3.2.10, K/E must be separable and hence so is K/k.

3.3 Sheaves of Relative Diﬀerentials

1 Sheafifying the construction of ΩB/A gives a scheme-theoretic version of differential forms that serves to define smooth morphisms in the category of schemes. We begin with the affine construction.

Definition. Let A → B be a ring homomorphism with corresponding morphism of affine schemes f : Spec B → Spec A. Then the sheaf of relative differentials for f is the sheaf 1 1 ΩSpec B/ Spec A = ΩeB/A on Spec B.

38 3.3 Sheaves of Relative Diﬀerentials 3 Etale´ Morphisms

Lemma 3.3.1. Let f : X = Spec B → Y = Spec A be a morphism of aﬃne schemes. Then

1 (a) ΩX/Y is a sheaf of OX -modules on X.

1 (b) ΩX/Y is quasi-coherent. (c) For any x ∈ B, Ω1 (D(x)) ∼ Ω1 . X/Y = Bx/A

Proof. (a) and (b) follow from the construction M 7→ Mf in Section 2.4, while (c) is an application of Proposition 3.2.6(c). To extend this construction to all schemes, we ﬁrst provide an alternate deﬁnition of 1 0 0 ΩB/A. Let m : B ⊗A B → B be the multiplication map m(b ⊗ b ) = bb . Set I = ker m. Proposition 3.3.2. The map

d : B −→ I/I2 b 7−→ 1 ⊗ b − b ⊗ 1

2 ∼ 1 is an A-derivation inducing an isomorphism of B-modules I/I = ΩB/A.

0 0 Proof. Suppose e : B → M is an A-derivation. Then the map ϕ : B⊗AB → M, b⊗b 7→ be(b ) is a homomorphism of B-modules and we have

ϕ((1 ⊗ b − b ⊗ 1)(1 ⊗ b0 − b0 ⊗ 1)) = ϕ(1 ⊗ bb0 − b0 ⊗ b − b ⊗ b0 − bb0 ⊗ 1) = ϕ(1 ⊗ bb0) − ϕ(b0 ⊗ b) − ϕ(b ⊗ b0) − ϕ(bb0 ⊗ 1) = e(bb0) − b0e(b) − be(b0) − bb0e(1) = 0 since e is a derivation. One can show that I2 is generated as a B-module by elements of the form (1 ⊗ b − b ⊗ 1)(1 ⊗ b0 − b0 ⊗ 1) for b, b0 ∈ B, so this shows ϕ induces a B-module homomorphism g : I/I2 → M making the diagram e B M

d g I/I2

commute. Uniqueness follows from a similar proof using the explicit description of I/I2. 2 1 Hence I/I is universal with respect to A-derivations, so it is isomorphic to ΩB/A. Now let f : X → Y be a morphism of schemes. Then the diagonal morphism ∆ : X → X ×Y X maps X to a closed subscheme of an open set U ⊆ X ×Y X. Let I be the ideal sheaf of ∆(X) in OU . Deﬁnition. The sheaf of relative diﬀerentials for f : X → Y is the pullback sheaf

1 ∗ 2 ΩX/Y := ∆ (I/I ).

39 3.3 Sheaves of Relative Diﬀerentials 3 Etale´ Morphisms

Example 3.3.3. Let A be a ring, B an A-algebra, X = Spec B, Y = Spec A and consider the 1 1 structure morphism f : X → Y . Then it follows from Proposition 3.3.2 that ΩX/Y = ΩeB/A, 1 in agreement with our aﬃne deﬁnition of ΩX/Y above. More generally: Proposition 3.3.4. Suppose f : X → Y is a morphism of schemes. Then for any open sets V ⊆ Y and U ⊆ f −1(V ) ⊆ X, there is an isomorphism

Ω1 | ∼ Ω1 X/Y U = eOX (U)/OY (V ) of sheaves on U. In particular, for all x ∈ X, there is an isomorphism of stalks

Ω1 ∼= Ω1 . X/Y,x OX,x/OY,f(x)

1 Corollary 3.3.5. For all schemes X → Y , ΩX/Y is a quasi-coherent OX -module.

n Example 3.3.6. Let A be a ring, put Y = Spec A and let X = AY be aﬃne n-space over 1 n A. Then by Example 3.2.4, we have ΩX/A = OX .

n Example 3.3.7. With Y = Spec A again, let X = PY be projective n-space over A. Let us 1 compute ΩX/Y carefully. Recall that X has an affine cover {U0,U1} where U0 = Spec A[t] and U = Spec A[t−1]. Then Ω1 is the sheafification of the A-module A[t]dt; likewise, Ω1 1 U0/A U1/A −1 1 −1 −1 1 ∼ is the sheafification of the A-module A[t ]d t = A[t ] t2 dt. We claim that ΩX/Y = O(−2), the twisting sheaf of degree −2 – recall that this is defined as O(−2) = Mf where M = A[t](−2) is the −2 shift of the graded ring A[t]. On the open sets U0 and U1, we have isomorphisms

1 ϕ0 :ΩU0/A −→ Mf|U0 dt 7−→ 1 1 and ϕ1 :ΩU1/A −→ Mf|U1 1 1 d 7−→ − t t2

1 So ΩX/A is a line bundle on X (a locally free sheaf of rank 1). Moreover, on the overlap −1 1 1 U0 ∩U1 = Spec A[t, t ], we have d t = − t2 dt, so we see that ϕ0 and ϕ1 determine a global 1 ∼ isomorphism ΩX/A = O(−2) as claimed. Example 3.3.8. Let X be an affine curve defined over a field k by an equation F (x, y) = 0 ∂F ∂F for F ∈ k[x, y] such that at any point on X, the formal partial derivatives ∂x and ∂y do not simultaneously vanish. Such a curve is said to be smooth in the classical sense, though we will soon define smoothness for general schemes and recover this very definition. 1 1 Since X = Spec(k[x, y]/(F )) is affine, ΩX/k is the sheafification of the module ΩB/k where B = k[x, y]/(F ), and Example 3.2.8 shows that

1 ΩB/k = (Bhdxi ⊕ Bhdyi)/hdF i.

40 3.4 Smooth Morphisms 3 Etale´ Morphisms

∂F ∂F In particular, this means that ∂x dx = − ∂y dy but since they do not vanish simultaneously, ∂F ∂F ∂x and ∂y generate the unit ideal in B. Thus

1 ∼ ΩB/k = (Bhdxi ⊕ Bhdyi)/hdF i = B

1 1 so ΩX/k is isomorphic to OX = Be. That is, ΩX/k is a free OX -module of rank 1 with basis n 1 o n 1 o ∂F/∂x dy or ∂F/∂y dx . Proposition 3.3.9. Let f : X → Y be a morphism of schemes. Then

0 0 0 1 ∼ (a) For any Y -scheme Y , set X = X ×Y Y . Then there is an isomorphism ΩX0/Y 0 = ∗ 1 0 0 p ΩX/Y of sheaves on X , where p : X → X is the canonical projection. (b) For any morphism of schemes Y → Z, there is an exact sequence of sheaves on X,

∗ 1 1 1 f ΩY/Z → ΩX/Z → ΩX/Y → 0.

1 ∼ 1 (c) For any open set U ⊆ X, there is an isomorphism ΩX/Y |U = ΩU/Y of sheaves on U. In particular, for any x ∈ X, Ω1 ∼= Ω1 . X/Y,x OX,x/OY,f(x) (d) Let Z be a closed subscheme of X with ideal sheaf I. Then there is an exact sequence of sheaves on Z, 2 1 1 I/I → ΩX/Y ⊗OX OZ → ΩZ/Y → 0. Proof. This is the global version of Proposition 3.2.6. Example 3.3.10. Let E be an elliptic curve over a ﬁeld k, i.e. a nonsingular projective 2 curve in Pk with distinguished point O ∈ E(k) and aﬃne Weierstrass equation

2 3 2 y + a1xy + a3y = x + a2x + a4x + a6 for ai ∈ k.

Then ω = dx is a diﬀerential form on E that generates Ω1 , so Ω1 is a free 2y+a1x+a3 k(E)/k E/k OE-module of rank 1 generated by ω.

3.4 Smooth Morphisms

Let k be a field. We can use ring theory to adapt the notion of smoothness from differential geometry to the setting of k-varieties, and ultimately arbitrary schemes. Although our first definition of smoothness won’t be linked to Kählerdifferentials, we will later see that the 1 property of a morphism X → Y being smooth can be phrased in terms of the sheaf ΩX/Y . This will agree with the perspective that Kählerdifferentials are an abstraction of differential forms to ring theory. Recall from commutative algebra that a local noetherian ring (A, m, k) is regular if 2 2 dim A = dimk m/m . (In general, we have dimk m/m .) One easily shows, using Nakayama’s Lemma, that A is regular of dimension n if and only if m can be generated by n elements. Also note that a regular local ring is an integrally closed integral domain.

41 3.4 Smooth Morphisms 3 Etale´ Morphisms

Deﬁnition. A locally noetherian scheme X is regular at a point x ∈ X if OX,x is a regular 2 ∗ local ring, i.e. if dim OX,x = dimk(x) TxX, where TxX = (mx/mx) is the tangent space of X at x. Otherwise X is singular at x. We say X is regular if it is regular at every x ∈ X. Note that a locally noetherian scheme is regular if and only if it is regular at all of its closed points.

n Example 3.4.1. Let X = Ak = Spec k[x1, . . . , xn] be affine n-space over a field k. Since X is an affine variety, for any closed point x ∈ X we have dim OX,x = dim X = n, and if x = (α1, . . . , αn) then the maximal ideal in OX,x is mx = (x1 − α1, . . . , xn − αn). Thus mx is generated by n elements, so OX,x is regular.

n Example 3.4.2. Likewise, Pk is regular at all of its (closed) points since each one is contained n in an affine open subscheme isomorphic to Ak . 2 Example 3.4.3. Let char k 6= 2, 3 and consider the affine curve X ⊆ Ak defined by X = Spec B = Spec(k[x, y]/(x2 − y3)).

P = (0, 0)

Then B ∼= k[t3, t2] and it’s easy to compute the dimensions of tangent spaces by localization and see that X is regular at all P 6= (0, 0) and singular at P = (0, 0), but here’s a way of seeing it using the results of the preceding sections. The local ring at any P = (x0, y0) is ∗ 2 OX,P = BmP where mP = (x − x0, y − y0). Then TX,P = mP /mP has dimension equal to the rank of Ω1 ∼= Ω1 as an O -module by Proposition 3.3.4, and as we saw in X/k,P B/k,mP X,P Example 3.2.8, Bdx ⊕ Bdy Ω1 = . B/k hd(x2 − y3)i 2 3 2 Now d(x − y ) = 2x dx − 3y dy is 0 when P = (0, 0), so dim TX,P = 2 6= 1 = dim OX,P in this case, so X is singular at P = (0, 0). On the other hand, if P = (x , y ) 6= (0, 0), Ω1 0 0 B/k,mP dx dy is a free BmP -module of rank 1 generated by either 3y2 if y0 6= 0 or by 2x if x0 6= 0. Thus X is regular on the locus {P 6= (0, 0)}. The last example is also easy to verify using the Jacobian condition for regularity, which n we recall now. Let X ⊆ Ak be an (irreducible) aﬃne variety with vanishing ideal I ⊆ k[x1, . . . , xn]. Then I = (f1, . . . , fm) for some polynomials fj ∈ k[x1, . . . , xn]. For each point x ∈ X, the Jacobian of X at x is the m × n matrix ∂fi Jx = (x) ∈ Mm×n(k). ∂xj

42 3.4 Smooth Morphisms 3 Etale´ Morphisms

n Theorem 3.4.4 (Jacobian Condition). An aﬃne variety X ⊆ Ak is regular at x ∈ X if and only if rank Jx = n − dim X.

Proof. Let m = mx be the maximal ideal of x in k[x1, . . . , xn] and let m¯ = m + I be the corresponding maximal ideal in k[X] = k[x1, . . . , xn]/I. Taking the exact sequence of k(x)- vector spaces 0 → I/(I ∩ m2) → m/m2 → m¯ /m¯ 2 → 0 and dualizing gives an exact sequence

n 2 ∗ 0 → TxX → TxA → (I/(I ∩ m )) → 0.

2 ∗ n n Thus dim(I/(I ∩ m )) = dim TxA − dim TxX = n − dim TxX (since A is regular by 2 Example 3.4.1), so it suﬃces to show rank Jx = dim I/(I ∩ m ). But there is an inclusion I/(I ∩ m2) ,→ m/m2 by the ﬁrst exact sequence above and m/m2 ∼= kn via the isomorphism

n X ∂f f 7→ d f := (x)e x ∂x j j=1 j

n 2 n where ej is the jth standard basis vector in k . Therefore the image of I/(I ∩ m ) in k is the vector subspace spanned by the columns of Jx. This completes the proof. Corollary 3.4.5. The rank of the Jacobian matrix of X at x is independent of the choice of generators f1, . . . , fm for the vanishing ideal of X. Deﬁnition. Let Y be a locally noetherian scheme. A morphism f : X → Y is smooth at a point x ∈ X if the following conditions hold:

(i) f is of ﬁnite type at x.

(ii) f is ﬂat at x.

(iii) If y = f(x), then the ﬁbre Xy := X ×Y k(y) is regular at x. Otherwise f is singular at x. We say f is a smooth morphism if it is smooth at every point x ∈ X, and call X a smooth Y -scheme. Finally, f is smooth of relative dimension n if f is smooth and for each y in the image of f, dim Xy = n.

Denote by Xsm the set of smooth points of a Y -scheme X → Y and by Xsing the set of singular points of X. Note that in general, the ﬁbres of a morphism need not have the same dimension.

Remark. To define smoothness for an arbitrary morphism of schemes, one replaces condition (i) with “f is locally of finite presentation at x”. When Y is locally noetherian, this condition is equivalent to being locally of finite type. We will assume for the rest of the section that Y is always locally noetherian.

Theorem 3.4.6. If f : X → Y is a smooth morphism and Y is regular, then so is X.

43 3.4 Smooth Morphisms 3 Etale´ Morphisms

Proof. Take x ∈ X and set y = f(x) ∈ Y , m = dim OX,x and n = dim OY,y. Since f is ﬂat, we have

dim OXy,x = dim OX,x − dim OY,y = m − n

(see Theorem 4.3.12 in Liu) so by smoothness, the maximal ideal of OXy,x is generated by r = m − n elements, say α1, . . . , αr. On the other hand, OXy,x = OX,x/myOX,x where my is # the maximal ideal of OY,y. So each αi =a ¯i for some ai ∈ OX,x. Let fx : OY,y → OX,x be the morphism of local rings induced by f at x. Then by hypothesis, my is generated by some b1, . . . , bn, and the deﬁnition of the αi shows that the set

# # {a1, . . . , ar, fx (b1), . . . , fx (bn)} generates mx. Therefore OX,x is regular as desired. Proposition 3.4.7. Suppose f : X → Y is a smooth morphism. Then

0 0 0 0 (a) If Y → Y is any morphism, then the base change X := X ×Y Y → Y is also smooth. (b) If g : Y → Z is smooth, then so is the composition g ◦ f : X → Z.

Proof. These are mostly straightforward from the definitions. For (a), note that finite gener- ation and flatness for rings are preserved under tensor product (the latter by Lemma 1.4.20). Moreover, the fibre of a base change is precisely the original fibre, so all parts of the definition of smoothness hold for X0 → Y 0. For (b), finite type is a property preserved in towers of rings. Moreover, flatness is preserved under composition by Lemma 3.1.2. To prove the regularity condition for g ◦ f, set y = f(x) and z = g(y). Then

Xz = X ×Z k(z) = (X ×Y Y ) ×Z k(z) = X ×Y (Y ×Z k(z)) = X ×Y Yz.

Since g is smooth, Yz is regular but then by Theorem 3.4.6, Xz is also regular. Example 3.4.8. For a scheme Y , one can define affine n-space over Y to be the affine scheme n AY := Spec(OY (Y )[x1, . . . , xn]).

Then the ring map OY (Y ) → OY (Y )[x1, . . . , xn] induces a natural projection morphism n π : AY → Y . We claim π is always smooth of relative dimension n. By Hilbert’s Basis n Theorem, π is of finite type. Flatness is trivial. Finally, each fibre of π is Ak(y) where k(y) is the residue field at y ∈ Y , and these are regular by Example 3.4.1 and have dimension n. Thus π is smooth of relative dimension n.

Deﬁnition. A smooth morphism of relative dimension 0 is called an ´etalemorphism.

n The next result says that a smooth Y -scheme is étale-locally a subscheme in AY . The notion of ‘étale-locally’ can be made precise using Grothendieck topologies (namely, the étale topology on Y ).

44 3.4 Smooth Morphisms 3 Etale´ Morphisms

Proposition 3.4.9. Let f : X → Y be a flat morphism of finite type. Then f is smooth at x ∈ X if and only if there exists an open neighborhood U ⊆ X of x and an étalemorphism n n g : U → AY for some n such that f|U = π ◦ g : U → AY → Y , where π is the canonical projection. Proof. ( ⇒ = ) is immediate, since g is smooth by definition, π is smooth by Example 3.4.1 and thus their composition is smooth by Proposition 3.4.7(b). ( =⇒ ) Assume f is smooth at x ∈ X and set y = f(x). By definition, Xy is a regular k(y)-scheme at x, say of dimension n, so after passing to an affine neighborhood of x in n X, we may assume there is an étalemorphism Xy → Ak(y) of k(y)-schemes. Indeed, since

Xy is regular, the maximal ideal of OXy,x = OX,x/myOX,x is generated by n elements, say n t1, . . . , tn. Write Ak(y) = Spec(k(y)[x1, . . . , xn]). Then there is a map

n k(y)[x , . . . , x ] = Γ( , O n ) −→ Γ(X , O ) 1 n Ak(y) Ak(y) y Xy

xi 7−→ ti.

n The corresponding morphism Xy → Ak(y) is smooth of relative dimension 0 by construction.

Now Γ(X, OX ) → Γ(Xy, OXy ) need not be surjective, but we can ﬁnd a small enough open neighborhood U ⊆ X of x so that t1, . . . , tn lie in the image of Γ(U, OU ) → Γ(Xy, OXy ), say 0 0 they are the images of t1, . . . , tn. Then there is a map of OY -algebras

OY [x1, . . . , xn] −→ f∗OU

0 induced by sending xi 7→ f∗(ti) which in turn determines a morphism

n g : U −→ Spec(OY (Y )[x1, . . . , xn]) = AY .

n By construction, f|U = π ◦ g : U → AY → Y . Now π is smooth of relative dimension n by Example 3.4.8 and fibre dimension adds along compositions, so it remains to show g is n smooth at x. It is clear that g is of finite type, and since each Xy → Ak(y) is flat, so is g. n For regularity, use the fact that Xy → Ak(y) is smooth together with Theorem 3.4.6. n Remark. The proof above shows that the n in g : U → AY may be chosen to be the dimension of the fibre Xy. In the language of complex geometry, Proposition 3.4.9 says that (étale-locally) every smooth morphism is a local submersion. Corollary 3.4.10. Let f : X → Y be a smooth morphism of relative dimension n. Then

1 (a) ΩX/Y is locally free of rank n. (b) For any morphism Y → Z, there is a short exact sequence of sheaves

∗ 1 1 1 0 → f ΩY/Z → ΩX/Z → ΩX/Y → 0.

n Proof. By Proposition 3.4.9 we may assume f = π ◦ g where g : X → AY is ´etaleand n π : AY → Y is the projection. From Proposition 3.3.9(b), we have an exact sequence

∗ 1 1 1 g Ω n → Ω → Ω n → 0. AY /Y X/Y X/AY

45 3.4 Smooth Morphisms 3 Etale´ Morphisms

If (b) holds for the map g, we could extend this with a zero on the left and, since rank 1 ∗ 1 is additive along short exact sequences, this would imply rank Ω = rank g Ω n + X/Y AY /Y 1 1 ∗ rank Ω n . Further, Lemma 3.4.11(a) below shows that rank Ω n = n and, since g is X/AY AY /Y ∗ 1 1 exact, rank g Ω n = n as well. Then if (a) holds for g, rank Ω n = 0 so we would get AY /Y X/AY 1 rank ΩX/Y = n. Moreover, consider the commutative diagram

∗ 1 1 1 0 f ΩY/Z ΩX/Z ΩX/Y 0

∗ ∗ 1 ∗ 1 ∗ 1 g π Ω g Ω n g Ω n 0 Y/Z AY /Y AY /Y 0

f The top row is our desired sequence for X −→ Y → Z. The bottom row comes from applying g∗ to the sequence ∗ 1 1 1 0 → π Ω → Ω n → Ω n → 0. Y/Z AY /Y AY /Y We claim that it is enough to prove that (a) and (b) hold for π and g. Indeed, if they hold for g then the short exact sequences of relative diﬀerentials associated to the compositions g n g n ∗ 1 ∼ 1 ∗ 1 ∼ X −→ → Z and X −→ → Y become isomorphisms g Ω n = Ω and g Ω n = AY AY AY /Z X/Z AY /Y 1 ΩX/Y so the middle and right columns of the diagram are isomorphisms. Then (b) for π along with exactness of g∗ imply the bottom row is exact, so we can deduce exactness of the top row. To summarize, the Corollary follows from establishing (a) and (b) for g and π. This is done in the following lemma. Lemma 3.4.11. Let Y → Z be a morphism of locally noetherian schemes. Then

n 1 (a) For every n ≥ 1, let π : → Y be the canonical projection. Then Ω n is a free AY AY /Y OY -module of rank n and there is a short exact sequence

∗ 1 1 1 0 → π Ω → Ω n → Ω n → 0. Y/Z AY /Z AY /Y

∗ 1 ∼ 1 (b) If g : X → Y is ´etalethen g ΩY/Z = ΩX/Z . 1 Proof. (a) Example 3.2.4 implies that Ω n is free of rank n. The second statement can be AY /Y deduced from the local property that the sequence

1 1 1 0 → ΩB/A ⊗B B[x1, . . . , xn] → ΩB[x1,...,xn]/A → ΩB[x1,...,xn]/B → 0

from Proposition 3.2.6(b) is in fact split exact when C = B[x1, . . . , xn]. This is obvious from the deﬁnition of the symbols df. (b) Consider the commutative diagram

∆X q X X ×Y X X ×Z X

g × g g ∆Y Y Y ×Z Y

46 3.4 Smooth Morphisms 3 Etale´ Morphisms

where ∆X and ∆Y are the diagonal maps. Let I be the OY ×Z Y -ideal corresponding to 1 ∗ 2 the image of ∆Y , so that ΩY/Z = ∆Y (I/I ). Then since g and therefore g × g is ﬂat, ∗ J := (g × g) I is the OX×Z X -ideal corresponding to the image of q. We will see that since 1 ∼ ∗ 2 g is ´etale,∆X is an open immersion. Assuming this, ΩX/Z = (q ◦ ∆X ) (J /J ) and we have

1 ∼ ∗ 2 ∗ ∗ 2 ΩX/Z = (q ◦ ∆X ) (J /J ) = ∆X ◦ q (J /J ) ∗ ∗ ∗ 2 = ∆X ◦ q ◦ (g × g) (I/I ) ∼ ∗ ∗ 2 = g ◦ ∆Y (I/I ) by commutativity ∗ 1 = g ΩY/Z by deﬁnition.

∗ 1 ∼ 1 Therefore g ΩY/Z = ΩX/Z . In fact, the converse of (a) is also true. We omit the proof. Theorem 3.4.12. Let f : X → Y be a flat morphism of finite type with fibres of dimension 1 n. Then f is smooth if and only if ΩX/Y is locally free of rank n. There are two more important properties of smoothness to discuss. The first says that every morphism is generically smooth. Theorem 3.4.13 (Generic Smoothness). For a morphism of finite type f : X → Y between locally noetherian schemes, the smooth locus Xsm is a nonempty open subset of X.

g n π Proof. By Proposition 3.4.9, f|Xsm factors locally as Xsm −→ AY −→ Y for some ´etale g. Since n the smooth locus of π is all of AY (Example 3.4.8), it’s enough to prove the theorem for ´etale morphisms. This will be done in the next section.

Let Y = Spec A be an affine scheme and Y0 ⊆ Y a closed subscheme defined by an ideal 2 I ⊂ A. Then Y0 is called an infinitesimal subscheme of Y if I = 0 (or equivalently, if I is nilpotent). Theorem 3.4.14 (Infinitesimal Lifting). Let f : X → S be a morphism of finite type over a locally noetherian scheme S. Then f is smooth if and only if for every affine S-scheme Y and every infinitesimal subscheme Y0 ⊆ Y ,

HomS(Y,X) −→ HomS(Y0,X) is surjective. The second condition in the theorem is called formal smoothness. Informally, this result says that smoothness is detected by the property of being able to lift tangent vectors. Another way to view formal smoothness is as an abstraction of Hensel’s Lemma. In the case of an arbitrary scheme, one has: Theorem 3.4.15. A morphism of schemes f : X → S is smooth if and only if f is formally smooth and locally of ﬁnite presentation. One can use the inﬁnitesimal lifting criterion to prove the following generalization of the Jacobian condition.

47 3.5 Unramiﬁed Morphisms 3 Etale´ Morphisms

Theorem 3.4.16 (Jacobian Condition for Schemes). Let f : X → Y be a morphism of locally noetherian schemes and suppose j : U,→ W is a closed immersion, where W is an n open subset of AY . Then f is smooth of relative dimension n at x ∈ U if and only if the sheaf of ideals I deﬁning U,→ W is generated on a neighborhood of j(x) by m = n − d local sections f1, . . . , fm such that ∂f rank i = n − d. ∂xj

3.5 Unramiﬁed Morphisms

In this section we defined unramified morphisms and give several different characterizations of this condition. In the next section this will be used to define étale morphisms. ∼ Qn Definition. An algebra A over a field k is called a finite étale k-algebra if A = i=1 Li as k-algebras, where each Li is a finite, separable field extension of k. Lemma 3.5.1. Let k be a field with algebraic closure k¯. Then a k-algebra A is finite étale ¯ ∼ Qm ¯ ¯ if and only if Ak¯ = A ⊗k k = i=1 k as k-algebras for some finite integer m. ∼ Qn Proof. ( =⇒ ) Suppose A = i=1 Li for finite, separable extensions Li/k. For each Li, there is a primitive element αi ∈ Li such that Li = k(αi). Fixing i, let p(t) be the minimal ¯ polynomial of αi over k, which factors over k as

r Y ¯ p(t) = (t − aj) for some aj ∈ k. j=1 ¯ Since Li/k is separable, the ideals (t − a1),..., (t − ar) are coprime in k[t], so the Chinese remainder theorem allows us to write

r r ¯ ∼ ¯ ∼ Y ¯ ∼ Y ¯ Li ⊗k k = k[t]/(p(t)) = k[t]/(t − aj) = k j=1 j=1 ¯ ¯ ¯ as k-algebras. This implies A ⊗k k is also a product of copies of k. ( ⇒ = ) is similar, using integrality and invariance of domain under integral extensions. For details, see Bosch 8.4.6.

Definition. A morphism f : X → Y of schemes which is locally of finite type is said to be unramified at a point x ∈ X if myOX,x = mx, where y = f(x), and the extension of residue fields k(x)/k(y) is finite and separable. Otherwise f is ramified at x and the set of all points at which f is ramified is called the ramification locus of f. We say f is an unramified morphism if it is unramified at all x ∈ X.

Theorem 3.5.2. Let f : X → Y which is locally of ﬁnite type and ﬁx x ∈ X. Then the following are equivalent:

(a) f is unramiﬁed at x.

48 3.5 Unramiﬁed Morphisms 3 Etale´ Morphisms

1 (b) ΩX/Y,x = 0.

(c) There is an open neighborhood U ⊆ X of x such that the diagonal morphism ∆X : X → X ×Y X restricts to an open immersion ∆X |U : U,→ X ×Y X. Proof. (a) =⇒ (b) The problem is local, so let us assume X = Spec B, Y = Spec A and x = q ⊂ B and y = p ⊂ A are prime ideals. By assumption, pBq = qBq so ∼ Bq ⊗Ap k(p) = Bq/pBq = k(q). By Proposition 3.3.4, Ω1 ∼ Ω1 and Proposition 3.2.6(a) gives X/Y,x = Bq/Ap 1 ∼ 1 1 Ω ⊗B k(p) = Ω = Ω Bq/Ap q (Bq⊗Ap k(p))/k(p) k(q)/k(p) but this vanishes by Corollary 3.2.11. Further, since f is of finite type (locally, but we may assume globally on Spec B → Spec A), Corollary 3.2.7 says Ω1 is a finitely generated Bq/Ap B -module so Nakayama’s lemma together with the above calculation imply Ω1 = 0. q Bq/Ap (b) =⇒ (c) We may again assume X = Spec B,Y = Spec A and f is of finite type. In the affine case, ∆ = ∆X is a closed immersion, say with ideal sheaf I. Then 1 ∼ 2 2 0 = ΩX/Y,x = (I/I )∆(x) = I∆(x)/I∆(x).

It is routine to check that I∆(x) is a coherent sheaf, i.e. corresponds to a ﬁnitely generated B ⊗A B-module via Theorem 2.4.7. Then by Nakayama’s lemma, I∆(x) = 0 and since it is coherent, I then vanishes in some neighborhood V ⊆ X ×Y X of ∆(x). Thus ∆ restricts to an isomorphism on U = ∆−1(V ) ⊆ X. (c) =⇒ (a) Since the question is again local, we may assume U = X = Spec B, Y = Spec A and ∆ = ∆X is an open immersion. Consider the pullback diagram

Spec(Bp/pBp) Spec B

Spec k(p) Spec A

Let ¯q be the image of qBp in Bp/pBp. The localization of Bp/pBp at ¯q is Bq/pBq and the ∼ residue field of this local ring is (Bq/pBq)/¯q = k(q). To prove f is unramified at x = q, we must show qBq/pBq = 0 and k(q)/k(p) is a finite, separable extension. Since statement (c) is preserved under base change, we may in fact assume Y = Spec k for a field k. Then Lemma 3.5.1 shows that we can assume k is algebraically closed. Let π1, π2 : X ×Y X → X be the two coordinate projections and consider the morphism

h : X −→ X ×Y X r 7−→ (r, p).

Then π1 ◦ h = idX and π2 ◦ h is the constant morphism X → Spec k ,→ X at p. Thus h−1(∆(X)) = {p} but since we are assuming ∆ is open, this actually implies the topology on X is discrete. Furthermore, since B/k is finite type, this means B is artinian and hence ∼ Spec B is finite. But OX,x = k for all k ∈ X so by the structure theory of local artinian rings, B ∼= k × · · · × k for a finite number of copies of k.

49 3.5 Unramiﬁed Morphisms 3 Etale´ Morphisms

Remark. There are good geometric interpretations of each of the conditions in Theo- rem 3.5.2:

(a) The deﬁnition of unramiﬁed is a technical way of saying that there are a ‘full’ number of preimage points above x. This is analogous to the fact that a degree n cover of Riemann surfaces is locally n-to-1.

(b) The Kählerdifferential condition in (b) says that at every point in the fibre over x, there are the same number of tangent directions as at x.

(c) Condition (c) says that ramiﬁcation points are precisely the ‘limit points’ of ∆X (X) which are not in the image of the diagonal.

Corollary 3.5.3. Let X be locally noetherian and suppose f : X → Y is a morphism locally of ﬁnite type. Then the ramiﬁcation locus of f is a proper closed subset of X.

Proof. This follows from Theorem 3.5.2(b) since the support of a coherent sheaf on a locally noetherian scheme is closed.

Corollary 3.5.4. A morphism f : X → Y locally of finite type is unramified at x ∈ X if and only if the morphism Xy = X ×Y Spec k(y) → Spec k(y) of fibres is unramified. The unramified condition is preserved under composition and base change, as we saw for flat and smooth morphisms in Lemmas 1.4.20/3.1.2 and Proposition 3.4.7, respectively.

Proposition 3.5.5. Suppose f : X → Y is unramiﬁed. Then

0 0 0 0 (a) If Y → Y is any morphism, then the base change X = X ×Y Y → Y is unramified. (b) If g : Y → Z is unramified, then the composition g ◦ f : X → Z is also unramified.

Proof. (a) follows from Theorem 3.5.2(b) and Proposition 3.3.9(a). (b) Fix x ∈ X. Set y = f(x), z = g(y) and suppose f is unramified at x and g is unramified at y. Then mzOX,x = (mzOY,y)OX,x = myOX,x = mx and a tower finite, separable extensions is again finite and separable, so k(x)/k(z) is finite and separable. Therefore g ◦ f is unramified at x.

Example 3.5.6. Let Y be the cuspidal curve from Example 3.4.3 defined by y2 − x3 = 0 1 over a field k of characteristic different from 2, 3. There is a normalization map f : Ak → Y corresponding to the ring map A → B where A = k[t2, t3] → k[t]. Then Ω1 is the A1/Y 1 2 2 sheafification of ΩB/A, but notice that B = A[s]/(s − t ) so by Example 3.2.8,

1 ∼ ∼ ΩB/A = B dx/2B dx = k[t]/(2t) = k[t]/(t).

This shows that the support of Ω1 is the point corresponding to (t), so f is unramiﬁed over A1/Y A1 r {0}. This illustrates the ‘number of tangent directions’ interpretation of ramiﬁcation in the remark above.

50 3.6 Etale´ Morphisms 3 Etale´ Morphisms

Example 3.5.7. Let us show that the deﬁnition of ramiﬁcation in this section agrees with the notion in algebraic number theory. Let B/A be an integral extension of Dedekind domains. Then for any nonzero prime p ⊂ A, the extension pB factors as a product of prime ideals

g Y eg pB = Pg i=1 where each Pi is a prime of B lying over p and ei ≥ 1. In algebraic number theory, one says that p is ramified if ei > 1 for some i and unramified if e1 = ··· = eg = 1. Fix 1 ≤ j ≤ g and consider the morphism of affine schemes f : Spec B → Spec A. Then

g Y ei ej pBPj = (pB)Pj = (PiBPj ) = (PjBPj ) . i=1

So f is unramified at Pj ∈ Spec B if and only if ej = 1. For example, the morphism Spec Z[i] → Spec Z corresponding to the extension of Dedekind domains Z[i]/Z in the quadratic extension Q(i)/Q is ramified at the point P = (1 + i) lying over p = (2) ∈ Spec Z and unramified everywhere else. More generally, f : Spec B → Spec A is ramified at P ∈ Spec B precisely when P divides the different ideal of B/A, DB/A. This ideal may be define using fractional ideals, but a more enlightening definition here is that

1 DB/A = AnnB(ΩB/A)

1 where ΩB/A is the module of Kählerdifferentials of B/A. Note that this reinforces the connection to ramification in number theory, since unramified is equivalent to Ω1 vanishing after localization (Theorem 3.5.2).

3.6 Etale´ Morphisms

We have already defined a morphism f : X → Y to be étaleif it is smooth of relative dimension 0. The following result relates this to the more common definition of étale.

Theorem 3.6.1. Let f : X → Y be a morphism of schemes. Then the following are equivalent:

(a) f is ´etale.

(b) f is smooth and unramiﬁed.

Proof. (a) =⇒ (b) follows from Theorems 3.5.2 and 3.4.12. (b) =⇒ (c) is trivial since a smooth morphism is by definition flat. (c) =⇒ (a) Suppose f is flat and unramified. We need only check that the fibres Xy are regular of dimension 0 over k(y) for each y ∈ Y . For such a y, fix x ∈ Xy. The unramified 1 condition means k(x)/k(y) is finite separable, so Ωk(x)/k(y) = 0 by Corollary 3.2.11. Therefore

51 3.6 Etale´ Morphisms 3 Etale´ Morphisms

Spec k(x) → Spec k(y) is ´etaleby Theorem 3.4.12 and as a consequence, Theorem 3.4.6 implies Xy is regular over k(y). On the other hand, myOX,x = mx which implies

OXy,x = OX,x/myOX,x = OX,x/mx = k(x).

This shows Xy has dimension 0. Corollary 3.6.2. Open immersions are ´etale.

Proof. An open immersion is locally of the form Spec Af ,→ Spec A for some ring A and nonzero element f ∈ A. This is ﬂat by Example 1.4.7 and unramiﬁed by Lemma 3.2.5(b), along with Theorem 3.5.2.

Corollary 3.6.3 (Generic Etaleness)´ . For a morphism of finite type f : X → Y between locally noetherian schemes, there is a nonempty open set Xét ⊆ X over which f is étale. Proof. This now follows from the same results for flatness (Theorem 3.1.14) and unramified- ness (Corollary 3.5.3). Note that this also proves generic smoothness (Theorem 3.4.13).

Proposition 3.6.4. Suppose f : X → Y is ´etale.Then

0 0 0 0 (a) If Y → Y is any morphism, then the base change X = X ×Y Y → Y is étale. (b) If g : Y → Z is étale,then the composition g ◦ f : X → Z is also étale.

Proof. This already follows from Proposition 3.4.7 since étalemorphisms are smooth and fibre dimension is preserved in each case, but now we obtain a second proof by applying Theorem 3.6.1, Lemmas 1.4.20 and 3.1.2 (for flat) and Proposition 3.5.5 (for unramified).

g f Proposition 3.6.5. Suppose a composition X −→ Y −→ Z is étaleand f is unramified. Then g is also étale.

Proof. Let Γg : X → X ×Z Y be the graph of g, which sends x 7→ (x, g(x)). Then g = p2 ◦Γg, where p2 : X ×Z Y → Y is the second coordinate projection. So by Proposition 3.6.4(b), it’s enough to show p2 and Γg are each ´etale.For Γg, consider the pullback diagram g X Y

Γg ∆

X × Y Y × Y Z g × 1 Z

where ∆ = ∆Y is the diagonal morphism. Since f is unramified, Theorem 3.5.2 says that ∆ is an open immersion, hence étaleby Corollary 3.6.2. Then Proposition 3.6.4(a) implies Γg is étale,too. On the other hand, p2 is also a pullback:

52 3.6 Etale´ Morphisms 3 Etale´ Morphisms

p1 X ×Z Y X

p2 f ◦ g g Y Z

So the hypothesis that f ◦ g is étaleimplies p2 is étale,again by Proposition 3.6.4(a). Proposition 3.6.6. A closed immersion of noetherian schemes which is flat is also an open immersion. Proof. Let f : X → Y be such a morphism. Since f is flat, it is open by Theorem 3.1.12 so we may replace Y by f(X) and assume f is surjective. Then since f is finite, f∗OX is a locally free OY -module by Lemma 2.4.8. Finally the closed immersion assumption means ∼ f∗OX = OY as OY -modules, so we conclude that f is an isomorphism. Example 3.6.7. Let L/k be a finite extension of fields. The morphism f : Spec L → Spec k is always flat, so f is étaleif and only if it is unramified. On the other hand, Theorem 3.5.2 and Corollary 3.2.11 say that f is unramified if and only if L/k is separable. Note that L is a trivial example of a finite étale k-algebra. For any k-algebra A, the structure morphism f : Spec A → Spec k is étaleif and only if A is a finite étale k-algebra. If A is of the form A = k[t]/(p(t)), where p(t) is a monic polynomial over k, then the étalecondition is equivalent to p(t) being separable for the same reason as the case of L/k. Example 3.6.8. Let p(t) be a monic polynomial over an arbitrary commutative ring A and consider B = A[t]/(p(t)). Then B is a free A-module of rank n = deg p and for any prime p ∈ Spec B, ∼ B ⊗A k(p) = k(p)[t]/(¯p(t)) wherep ¯(t) is the image of p(t) in k(p)[t]. Then Corollary 3.5.4 shows that B is an étale A-algebra (or equivalently Spec B → Spec A is étale)if and only if (p(t), p0(t)) = 1, where p0 is the formal derivative of p. The same statement holds when we replace B by a localization 0 Bb at a nonzero element b ∈ B; namely, Bb is étaleover A if and only if p (t) is a unit in Bb[t]. Definition. Let B = A[t]/(p(t)) for a monic polynomial p(t) ∈ A[t] and let b ∈ B be nonzero 0 such that p (t) is a unit in Bb[t]. Then Bb is called a standard étale A-algebra. In a moment we will prove that every étalemorphism is locally standard étale,meaning isomorphic to Spec Bb → Spec A for a standard étale A-algebra Bb. We will need Zariski’s Main Theorem, which we state but do not prove below. Definition. A morphism f : X → Y is called quasi-finite if f −1(y) is discrete for all y ∈ Y . Theorem 3.6.9 (Zariski’s Main Theorem). Let Y be a noetherian scheme and f : X → Y a separated, quasi-finite morphism. Then f factors as

j g f : X −→ X0 −→ Y where j is an open immersion and g is a ﬁnite morphism.

53 3.7 Henselian Rings 3 Etale´ Morphisms

Theorem 3.6.10. Let f : X → Y be a morphism over a locally noetherian scheme Y . If f is étaleover an open neighborhood of x ∈ X, then there are affine open neighborhoods V ⊆ X of x and U ⊆ Y of y = f(x) such that f|V : V → U is a standard étalemorphism. Proof. We may assume X = Spec C, Y = Spec A and f is separated. The proof of Theo- rem 3.6.1 showed that unramified morphisms are quasi-finite, so by Zariski’s main theorem, we may assume C is a finite A-algebra. Let x = q ∈ Spec C. We will construct a standard ∼ étale A-algebra Bb and show that Bb = Cc for some c ∈ C r q. Since C/A is finite, we may localize at y = p ∈ Spec A, so assume A itself is a local ring with maximal ideal mA and we have q ∩ A = mA. Choose an element γ ∈ C whose imageγ ¯ in C/pC generates the residue field extension k(q)/k(p), which is possible since this field extension is separable. Set q0 = q ∩ A[γ]. Since f is unramified, q is the unique prime lying over q0. This ∼ means C ⊗A[γ] A[γ]q0 = Cq, but since A[γ] → C is finite and injective, so is the localization 0 A[γ]q0 → Cq. On the other hand, this map is also surjective: k(q ) → k(q) is surjective, so Nakayama’s lemma implies the cokernel of A[γ]q0 → Cq is 0. ∼ ∼ 0 Next, A[γ]q0 = Cq induces an isomorphism A[γ]c0 = Cc for some c ∈ C r q and c ∈ A[γ] r q0. Since A[γ] is finite as an A-algebra, we may replace C by A[γ], so that q = q0. Set n = [k(q): k(p)]. Then {1, γ,¯ . . . , γ¯n−1} is a basis for the field extension k(q)/k(p), so Nakayama’s lemma implies {1, γ, . . . , γn−1} generates C as an A-algebra. Thus there is a monic polynomial p(t) ∈ A[t] such that

B := A[t]/(p(t)) −→ A[γ] = C is surjective. Note that the imagep ¯(t) of p(t) in k(q)[t] is exactly the characteristic polynomial ofγ ¯ over k(p), sop ¯(t) is separable. Therefore for some b ∈ B not landing in q ⊂ C, Example 3.6.8 shows that Bb is a standard ´etalealgebra over A. For the corresponding choice of c ∈ C, we get a surjection

h : Bb −→ Cc. We ﬁnish by showing h is an isomorphism. Notice that we have a composition

h∗ r Spec Cc −→ Spec Bb −→ Spec A where r and h∗ ◦ r are both étale. Therefore Proposition 3.6.5 shows that h∗ is also étale. But since h is surjective, h∗ is an étaleclosed immersion and hence an open immersion by Proposition 3.6.6. This proves h is an isomorphism.

3.7 Henselian Rings

Definition. A local ring (A, m, k) is henselian if it satisfies Hensel’s Lemma, i.e. if whenever f(x) ∈ A[x] is a monic polynomial and α ∈ A such that f(α) ∈ m but f 0(α) 6∈ m, then there exists an element a ∈ A for which a ≡ α mod m and f(a) = 0. The condition that f(α) ∈ m but f 0(α) 6∈ m is the same as sayingα ¯ is a simple root of f¯(x) over k = A/m. As with Hensel’s Lemma, there are multiple equivalent characterizations of henselian rings. We need to set up a few things first before stating the main theorem.

54 3.7 Henselian Rings 3 Etale´ Morphisms

Definition. A local homomorphism of local rings ϕ :(A, m) → (B, n) is called a pointed étaleneighborhood if B is an étale A-algebra and A/m ∼= B/n as A-modules. Note that by Theorem 3.6.10, any pointed étaleneighborhood A → B is of the form B = (A[x]/(f(x)))q where f(x) ∈ A[x] is monic and q = (x − a, mB) for some a ∈ A such that f(a) ∈ m and f 0(a) 6∈ m. ∼ Q` Definition. A ring B is decomposable if B = i=1 Bi for local rings (Bi, mi). Proposition 3.7.1. Let (A, m, k) be a local ring and B an A-module. Then Q` (1) If B = i=1 Bi is decomposable, then MaxSpec(B) = {M1,...,M`} where Mi = Q ∼ mi × j6=i Bj. Moreover, BMi = Bi for each 1 ≤ i ≤ `. ∼ (2) Let ei = (0,..., 0, 1, 0,..., 0) be the ith orthogonal idempotent in B. Then B = Q` ∼ i=1 Bei and Bei = Bei for each 1 ≤ i ≤ `.

(3) For any ﬁnite ring map A → B, B ⊗A k is decomposable.

(4) If MaxSpec(B) = {M1,...,M`}, then the following are equivalent: (i) B is decomposable. ∼ Q` (ii) B = i=1 Mi.

(iii) For any ﬁnite ring map A → B, the decomposition of B ⊗A k lifts to a decomposition of B. Proof. (1) and (2) are routine. For (3), use the Chinese remainder theorem along with the fact that B/mB is artinian. In (4), (ii) =⇒ (iii) =⇒ (i) are trivial and (i) =⇒ (ii) follows directly from (1).

As above, let (A, m, k) be a local ring and A → B a ﬁnite ring map. Set B = B ⊗A k. Then every idempotent of B determines an idempotent of B, say e 7→ e¯. Lemma 3.7.2. The assignment {idempotents in B} → {idempotents in B} is one-to-one. Moreover, it is onto if either of the following equivalent conditions hold: (a) B is decomposable.

(b) The orthogonal idempotents e¯1,..., e¯` ∈ B = B1 × · · · × B` lift to a set of idempotents of B. Proof. Let e and f be idempotents in B, set y = e − f and assumey ¯ =e ¯ − f¯ = 0 in B. Notice that y3 = (e − f)3 = e3 + 3e2f − 3ef 2 − f 3 = e + 3ef − 3ef − f = e − f = y, or in other words, y3 − y = 0, so y(y2 − 1) = 0. However, y ∈ J(B) so y2 − 1 is a unit and therefore y = 0. This shows e 7→ e¯ is one-to-one in general. Now suppose ei are idempotent lifts of thee ¯i to B. Then for each 1 ≤ i, j ≤ `, eiej is an idempotent lift ofe ¯ie¯j bute ¯ie¯j = 0 whenever i 6= j, so by the ﬁrst paragraph, eiej = 0 for all i 6= j. This then implies e1 + ... + e` is an idempotent, and it is a lift ofe ¯1 + ... +e ¯` = 1 ¯ from B to B, so again by the one-to-one property, e1 + ... + e` = 1 in B. This proves ∼ B = Be1 × · · · × Be` which is a lifting of B = B1 × · · · × B`.

55 3.7 Henselian Rings 3 Etale´ Morphisms

We can now state and prove the main theorem characterizing henselian rings.

Theorem 3.7.3. For a local ring (A, m, k), the following are equivalent:

(1) A is henselian.

(2) Every pointed ´etaleneighborhood (A, m) → (B, n) is an isomorphism of local rings.

(3) For all F1,...,Fn ∈ A[x1, . . . , xn], if there exist elements a1, . . . , an ∈ A such that Fi(a1, . . . , an) ∈ m for all 1 ≤ i ≤ n and the Jacobian determinant ∂Fi J(a1, . . . , an) := det (a1, . . . , an) ∂xj

does not lie in m, then there exist elements b1, . . . , bn ∈ A with bi ≡ ai mod m and Fi(b1, . . . , bn) = 0 for all 1 ≤ i ≤ n. (4) If F (x) ∈ A[x] is monic and its reduction mod m factors as F = gh for some relatively prime (necessarily monic) polynomials g, h ∈ k[x], then F = GH for some monic polynomials G, H ∈ A[x] with G = g and H = h.

(5) Every module-ﬁnite extension of rings A → B is decomposable.

Proof. (1) =⇒ (2) Write B = (A[x]/(f(x)))q as above, with q = (x − α, mB) for α ∈ A such that f(α) ∈ m but f 0(α) 6∈ m. By assumption, there exists a lift a ∈ A such that a ≡ α mod m and f(a) = 0. Then f(x) = (x − a)g(x) for some g ∈ A[x]; since f 0(a) 6= 0, g is not divisible by x − a. Thus g 6∈ q, so it follows that q = (x − a, mB). Localizing at q then gives ∼ B = (A[x]/(f(x)))q = (A[x]/(x − a))q = A.

(2) =⇒ (3) Set B = (A[x1, . . . , xn]/(F1,...,Fn))q where q = (x1 − a1, . . . , xn − an, mB). Then B/q ∼= k = A/m, showing B is a pointed ´etaleneighborhood of A, so by hypothesis ∼ B = A. For 1 ≤ i ≤ n, denote the image of xi under this isomorphism by bi. Then Fj(b1, . . . , bn) = 0 and by construction bi ≡ ai mod m for all j. (3) =⇒ (4) Let F, g and h be as in the statement of (4) and write

n−1 n X i F (x) = x + aix , i=0 n−1 n X i f(x) = F (x) = x + λix where λi =a ¯i, i=0 s−1 s X i g(x) = x + αix i=0 t−1 t X i and h(x) = x + βix , where s + t = n. i=0

56 3.7 Henselian Rings 3 Etale´ Morphisms

Then the condition f = gh is equivalent to the existence of a solution to the system of equations

α0β0 − λ0 = 0

α1β0 + α0β1 − λ1 = 0 . .

αs−1 + βt−1 − λn−1 = 0. Likewise, F = GH corresponds to a solution in A to the system

F1 : z0y0 − a0 = 0

F2 : z1y0 + z0y1 − a1 = 0 . .

Fn : zs−1 + yt−1 − an−1 = 0. By (3), such a solution exists as long as the Jacobian condition ∂Fi det (α0, . . . , αs−1, β0, . . . , βt−1) 6= 0 ∂zi, yi is satisﬁed. Explicitly, this matrix is the determinant of the resultant of f and g:   y0 y1 ······ yt−1 1 0 ··· 0  0 y0 ······ yt−2 yt−1 1 ··· 0   ......   ......   ......  Res(f, g) =    0 0 ······ y0 y1 y2 ··· yt−1   z0 z1 ··· zs−1 ············ 0  ......

Set J = det(Res(f, g)). Then J(α, β) := J(α0, . . . , αs−1, β0, . . . , βt−1) = 0 if and only if there is a nonzero solution to

(z0, . . . , zs−1, w0, . . . , wt−1)J(α, β) = 0

for some zi, wj. But this equation is really

s−1 t−1 (z0 + z1x + . . . zs−1x )h(x) + (w0 + ... + wt−1x )g(x) = 0 so J(α, β) = 0 precisely when g and h are not relatively prime. Hence (3) guarantees the existence of such a solution and consequently a factorization F = GH lifting the equation f = gh to A[x]. (4) =⇒ (5) Notice that condition (4) implies that every A-algebra of the form A[x]/(f(x)), for f(x) monic, is decomposable. Given a ﬁnite ring extension A → B, write B = B ⊗A k = B1 × · · · × B` using Proposition 3.7.1(3). By Lemma 3.7.2, it’s enough to lift the orthogonal idempotentse ¯1,..., e¯` ∈ B; since the argument is identical for each one, we will just construct the lift fore ¯ =e ¯1. For any lift b ∈ B ofe ¯, let f(x) ∈ A[x] be the minimal monic polynomial having b as a root. Set R = A[x]/(f(x)) and consider the diagram

57 3.7 Henselian Rings 3 Etale´ Morphisms

A A[b]

ϕ R B

As in the statement of Proposition 3.7.1, let M1,...,M` be the maximal ideals of B1,..., B`, ∗ respectively. Then b ∈ M2 ∩· · ·∩M` and bM1 = {1}. Set n = ϕ M1 ∈ Spec(R). We claim M1 is the unique maximal ideal of R lying over n – this follows from the fact that (x, M1) = R and ϕ(x) = b ∈ M2 ∩ · · · ∩ M`. Further, we know R is decomposable and Rn is one of its factors, so the idempotent corresponding to this factor liftse ¯. (5) =⇒ (1) Assume f(x) ∈ A[x] is monic and α ∈ A satisfies f(α) ∈ m and f 0(α) 6∈ m. Set B = A[x]/(f(x)) – since f is monic, B is finite over A. Then the hypothesis says that B is decomposable, say B = B1 × · · · × B`, but by Proposition 3.7.1(4), this decomposition ¯ is a lift of B = B ⊗A k = B1 × · · · × B`. We know f(x) = (x − α)g(x) for some g ∈ k[x] not ∼ ∼ divisible by x − α), so without loss of generality, assume B1 = k[x]/(x − α) = k. Then the ∼ decomposition of B lifts to a decomposition B = B1 × · · · × B` with B1 = A[x]/(x − a) for some a ≡ α mod m for which f(a) = 0. Remark. Hensel’s Lemma is often stated in the form of condition (4), from which (1) is easily derived. Example 3.7.4. The original version of Hensel’s Lemma states that the lifting criterion for T∞ n simple roots holds in any complete local ring which is separated, i.e. n=1 m = 0. Therefore separated complete local rings are henselian. Definition. For a local ring (A, m, k), a henselization of A is a local ring (Ah, mh) satisfying: (1) Ah is a henselian ring admitting a local homomorphism i : A → Ah.

(2) Ah is universal with respect to such local homomorphisms: if f : A → B is any local homomorphism to a henselian ring B, then there is a unique map g : Ah → B such that f = g ◦ i. We will show that every local ring has a henselization. To construct Ah, we need the following facts about pointed ´etaleneighborhoods. Proposition 3.7.5. Suppose B and C are pointed ´etaleneighborhoods of a local ring (A, m, k). Then

(1) If Q is the kernel of B ⊗A C → k ⊗A k = k, then Q is a prime ideal and (B ⊗A C)Q is a pointed ´etaleneighborhood of A.

(2) There is at most one local homomorphism B → C and if one exists, C is a pointed ´etaleneighborhood of B.

(3) If there exist local homomorphisms B → C and C → B, then they are both isomorphisms.

58 3.7 Henselian Rings 3 Etale´ Morphisms

Proof. (1) By Theorem 3.6.10, we may write

B = (A[x]/(f(x)))q1 and C = (A[y]/(g(y)))q2

for monic polynomials f(x) and g(y) and prime ideals q1 and q2. Then B⊗AC is a localization 0 0 of A[x, y]/(f, g) and f (x), g (y) 6∈ Q, so (B⊗A C)Q is a localization of a finite étale A-algebra. Moreover, k(Q) = k, so (B ⊗A C)Q is a pointed étaleneighborhood. (2) If there exists a map α : B → C then Proposition 3.6.5 implies α is also étale. Since the residue fields of A, B and C are all k, this implies B → C is a pointed étale neighborhood so the second statement holds. For uniqueness, suppose β : B → C is another 0 0 homomorphism. Then the map B ⊗A B → C given by b ⊗ b 7→ α(b)β(b ) takes Q from (1) b⊗1 to the maximal ideal of C and hence induces a map (B ⊗A B)Q → C, with 1 7→ α(b) and 1⊗b 1 7→ β(b). On the other hand, the natural map B⊗AB → B takes Q to the maximal ideal of B, so induces a map (B ⊗A B)Q → B. If I is the kernel of B ⊗A B → B, by Proposition 3.3.2 2 ∼ 1 we know I/I = ωB/A = 0 since A → B is étale. Therefore IQ ⊂ (B ⊗A B)Q is finitely 2 generated, lies in the maximal ideal and satisfies IQ = IQ so it is 0 by Nakayama’s lemma. That is, (B ⊗A B)Q → B is an isomorphism, but this means that under (B ⊗A B)Q → C, b⊗1 1⊗b both 1 and 1 map to the same element, so α(b) = β(b) and we conclude that α = β. (3) It follows from (2) that the only local homomorphism B → B is the identity, so any compositions B → C → B and C → B → C must be the identity.

Corollary 3.7.6. The isomorphism classes of pointed ´etaleneighborhoods of A form a directed system.

Theorem 3.7.7. Let (A, m, k) be a local ring. Then

(1) There exists a henselization of A, namely Ah = lim B where the limit is taken over −→ some choice of representatives of the isomorphism classes of pointed ´etaleneighborhoods A → B.

(2) If mh is the maximal ideal of Ah, then mAh = mh and Ah/mh ∼= A/m = k. (3) The natural map A → Ah is ´etale.

(4) If A → C is a pointed ´etaleneighborhood then Ch ∼= Ah.

(5) (Ah)h = Ah and (dAh) = Ab. (6) If A is noetherian, then so is Ah.

Proof. (1) Set A0 = lim B with maximal ideal m0, let f(x) ∈ A0[x] be monic and suppose −→ a ∈ A0 such that f(a) ∈ m0 and f 0(a) 6∈ m0. Then there exists a pointed ´etaleneighborhood B of A such that a ∈ B and f(x) ∈ B[x]. Let mB be the maximal ideal of B and deﬁne

B1 = (B[x]/(f(x)))(x−a,mB B[x]) which is a pointed ´etaleneighborhood of B by construction. Then Proposition 3.7.5(2) says 0 B1 is also a pointed ´etaleneighborhood of A. Thus there is a map ϕ : B1 → A compatible

59 3.7 Henselian Rings 3 Etale´ Morphisms

with the directed system defining A0 such that ϕ(¯x) is a lifting of a which is a root of f(x). That is, A0 is henselian. Now we show A0 is universal with respect to maps from A to henselian rings. Indeed, given such a map A → C with C henselian, take a pointed étaleneighborhood A → B. Since C is henselian, Theorem 3.7.3(2) says C is isomorphic to any pointed étaleneighborhood (C[x]/(f(x)))q, so with the right choice of q we get a local homomorphism B → C. These assemble to give a map A0 = lim B → C, so A0 = Ah. −→ (2) Similar to the analogous property for the completion (A,b mb ). (3) By Theorem 3.6.1 it’s enough to show A → Ah is flat and unramified, but this follows 1 from the fact that flatness and Ω−/A are preserved under direct limits, along with (1). (4) follows easily from the description in (1). (5) The first statement is obvious from the universal property of henselization. As with completion, it’s easy to show that for all n ≥ 1,

(A/mn)h ∼= Ah/mnAh ∼= Ah/(mh)n.

Then taking inverse limits gives Ab = (Ab)h ∼= (dAh). (6) can be found in Milne, 2.8.8. Proposition 3.7.8. Suppose (A, m) → (B, n) is a local homomorphism of local rings and B

is henselian. Let A be the integral closure of A in B and set C = An∩A, which is a local ring with maximal ideal mC = n ∩ A. Then (C, mC ) is a henselian ring.

Proof. Let f(x) be a monic polynomial in C[x] and suppose α ∈ C is a simple root mod mC , 0 that is, f(α) ∈ mC but f (α) 6∈ mC . Write a a a f(x) = xn + 1 xn−1 + ... + n−1 x + n s s s

for ai ∈ A and s ∈ A r (n ∩ A). Then since B is henselian, there exists an element a ∈ B with f(a) = 0 and a ≡ α mod n. The condition f(a) = 0 implies

n n−1 n−1 (sa) + a1(sa) + ... + an−1(sa) + s an = 0 so sa ∈ A since A is integral over A. On the other hand, s 6∈ n ∩ A forces a ∈ An∩A = C. Hence C is henselian.

Definition. A local ring (A, m) is an approximation ring if whenever F1,...,F` ∈ A[x1, . . . , xn] have a common zero (â1,..., aˆn) over the completion Ab, there exists a solution over A, i.e. there exist elements a1, . . . , an ∈ A with Fj(a1, . . . , an) = 0 for all 1 ≤ j ≤ `. It is immediate from the definitions that: complete local rings ⊆ approximation rings ⊆ henselian rings. Proposition 3.7.9. Suppose (A, m) is an approximation ring. Then for any polynomials F1,...,F` in A[x1, . . . , xn] admitting “approximate roots” aˆ1k,..., aˆnk ∈ Ab for all k ≥ 1 such k that Fj(âjk) ∈ m , there exist actual roots a1, . . . , an ∈ A with Fj(a1, . . . , an) = 0 for all k 1 ≤ j ≤ ` and aj ≡ aˆjk mod m for all k ≥ 1.

60 3.7 Henselian Rings 3 Etale´ Morphisms

Conversely: Theorem 3.7.10 (Popescu). Let (A, m) be a henselian local ring. If the formal ﬁbres of A are geometrically regular, then A is an approximation ring.

Remark. Recall that the formal fibres of A are k(p) ⊗A Ab for p ∈ Spec(A), and these are geometrically regular if k(p) ⊗A Ab are all regular, where k(p) denotes an algebraic closure. Corollary 3.7.11 (Artin Approximation). Let (A, m, k) be a henselian local ring, fix N ≥ 1 and suppose there are polynomials F1,...,Fn ∈ A[x] and elements aˆ1,..., aˆn ∈ Ab such that Fj(â1,..., aˆn) = 0 for all 1 ≤ j ≤ n. Then there exist a1, . . . , an ∈ A such that N Fj(a1, . . . , an) = 0 for all 1 ≤ j ≤ n and ai ≡ aî mod m Ab for all 1 ≤ i ≤ n. N Proof. Let m = (u1, . . . , ur) for elements u` ∈ A. For each i, choose any bi ∈ A such that N bi ≡ aî mod m Ab and write n X bi − aî = aîù` `=1 for somea î` ∈ Ab. Consider the following system of equations over A:

F1(x1, . . . , xn) = 0 . .

Fn(x1, . . . , xn) = 0 n X x1 − b1 = y1`u` `=1 . . n X xn − bn = yn`u`. `=1

Since the system has a solution over Ab, namely xi =a î and yi` =a î`, Popescu’s theorem says N there’s a solution over A, say xi = ai and yi` = ai`. Then Fj(a1, . . . , an) = 0, ai − bi ∈ m N anda î ≡ bi ≡ ai mod m Ab as required. Theorem 3.7.12. If (A, m, k) is a local noetherian ring which is excellent and normal, then Ah = A mb∩A where A is the integral closure of A in its completion Ab. Proof. By Theorem 3.7.7 and Proposition 3.7.8, Ah and B := A are henselian and Ah is mb∩A the smallest henselian extension of A, so it suffices to show B ⊆ Ah. In fact, we just need h h to show A ⊆ A since then its localization at mb ∩ A will also be contained in A . Take a ∈ Ab integral over A. Then since A is normal and excellent, there is a monic polynomial f(x) ∈ A[x] with f(a) = 0 and fintely many roots in Ab which can be distinguished mod mN Ab for any sufficiently large N. Applying Artin approximation to Ah, along with the fact (Theorem 3.7.7) that (Abh) = Ab, we get an element b ∈ Ah with f(b) = 0 and b ≡ a mod mN Ab. But this implies a = b, so a ∈ Ah.

61 3.7 Henselian Rings 3 Etale´ Morphisms

Deﬁnition. A local ring (A, m, k) is strictly henselian if it is henselian and k is separably closed, i.e. k = ksep.

Proposition 3.7.13. If A is strictly henselian, then for every finite étalemorphism A → B, ∼ Qn B is ‘strictly decomposable’: B = i=1 A. Proof. By Theorem 3.7.3(5), B is decomposable but the étale hypothesis guarantees that the residue fields of the factors are all finite separable extensions of k = A/m. Hence Theorem 3.7.3(2) implies each factor is isomorphic to A.

Deﬁnition. For a local ring (A, m, k), a strict henselization of A is a local ring (Ash, msh) satisfying:

(1) Ash is a strictly henselian ring admitting a local homomorphism j : A → Ash.

(2) Ash is universal with respect to such local homomorphisms: if f : A → B is any local homomorphism to a strictly henselian ring B, then there is a unique map g : Ash → B such that f = g ◦ j.

Similar to the properties of Ah, one can prove:

Theorem 3.7.14. Let (A, m, k) be a local ring. Then

(1) There exists a strict henselization Ash, namely Ash = lim B where the limit is over all −→ ∼ sep local rings (B, mB) admitting local ´etalemaps A → B such that B/mB = k . (2) If msh is the maximal ideal of Ash, then mAsh = msh and Ash/msh ∼= ksep.

(3) The natural maps A → Ash and Ah → Ash are ´etale.

∼ sep sh ∼ sh (4) If A → C is a local ´etalemap such that C/mC = k , then C = A . (5) (Ash)sh = Ash and (Ah)sh = Ash.

(6) If A is noetherian, then so is Ash.

62 4 Descent

4 Descent

The notion of descent can be phrased in a quite general context. Fix an object S in a category C and recall that the localized category at S (sometimes called the slice category at S) is the category C/S whose objects are arrows X → S in C and whose morphisms are commutative triangles X Y

Given any other object S0 ∈ C and a morphism S0 → S, there is a base change functor C/S → C/S0 given by (X → S) 7→ (X0 → S0) where X0 is the ﬁbre product

0 0 X = X ×S S X

S0 S

This answers the question “when does an object over S determine an object over S0?” The opposite question, namely when an object over S0 determines an object over S, is much harder to answer in general. In fact there are two interesting questions one might ask: given a morphism S0 → S and an arrow (X0 → S0) ∈ C/S0, (1) is there an arrow (X → S) ∈ C/S completing the diagram X0 X

S0 S

and if so, (2) how many ways are there to complete the diagram? The process of answering both of these questions is known as descent.

4.1 Galois Descent

Let K/k be a field extension and obj(k) some class of objects defined over k. Example 4.1.1. obj(k) could be the class of k-vector spaces, or more specifically, the class of Lie algebras over k, central simple algebras over k, etc. Notice that many examples like this admit morphisms, and so form a category over k. Example 4.1.2. A quadratic space over k is a k-vector space V together with a quadratic form q, that is, a symmetric bilinear function q : V → k. The set of quadratic spaces (V, q) may be an interesting class of objects over k. These too admit morphisms, where (V, q) → (V 0, q0) consists of a k-linear isomorphism V → V 0 which commutes with q and q0.

63 4.1 Galois Descent 4 Descent

Example 4.1.3. Important choices of obj(k) in algebraic geometry include the class of algebraic varieties or algebraic groups over k, the class of schemes over Spec k, and more generally things like algebraic spaces and algebraic stacks over k. Many of these examples admit a base change functor as in the chapter introduction, i.e. an assignment

obj(k) −→ obj(K)

X 7−→ XK . Most of the time this functor can be built using the tensor product or fibre product in the right category. Example 4.1.4. For some of the examples above, we have the following base change functors: object over k base change to K vector space V VK = V ⊗k K central simple algebra A AK = A ⊗k K quadratic space (V, q) (VK , qK ) where qK extends q variety/scheme X XK = X ×k K (defined by ⊗ on affine patches) algebraic group G GK = G ×k K In the pattern of the introduction, there are two natural questions we would like to answer about descending objects defined over K to objects over k:

(1) Given an object A ∈ obj(K), does there exist an object X ∈ obj(k) such that XK = A? ∼ (2) What are all the possible objects X ∈ obj(k) with XK = A or XK = A in obj(K)? When K/k is a Galois extension, this is known as Galois descent. ∼ Deﬁnition. An object X ∈ obj(k) with XK = A in obj(K) is called a K/k-form of A.

Example 4.1.5. In some situations, descent is trivial. For example, if VK is a K-vector space with basis {xi} then the same basis gives a k-vector space Vk for which Vk ⊗k K = VK . √ Example 4.1.6. In other situations, descent is√ impossible. Let k = Q, K = Q( 2) and 2 2 2 consider the quadratic form qK (x, y) = x − y 2 on V = K . Is there a quadratic form 2 qQ on VQ = Q which extends to qK on VQ ⊗Q K = V ? The answer in this case is no – we will see soon that there are some elementary conditions that are necessary for descent to be possible, and they are not satisfied in this situation. Example 4.1.7. (Motivation) The Inverse Galois Problem asks: for which finite groups ∼ G can one construct a field extension `/Q with Gal(`/Q) = G? By Hilbert’s Irreducibility ∼ Theorem, it is enough to construct an extension L/Q(t) with Gal(L/Q(t)) = G. Over C, one can construct Galois extensions L/C(t) by instead constructing branched covers of Riemann surfaces X → 1 . It is a general fact that every such cover X is defined over , so the PC Q question of constructing a G-extension of Q(t) comes down to being able to descend the cover X → 1 to a cover X → 1 . A general solution to this would solve the Inverse PQ Q PQ Galois Problem.

64 4.1 Galois Descent 4 Descent

Let K/k be a Galois extension with Galois group G = Gal(K/k) and suppose Vk and Wk ∼ are objects over k (e.g. k-algebras) that are isomorphic over K, that is, VK = Vk ×k K = Wk ×k K = WK . Moreover, assume that under the natural Galois action on VK and WK , G G we have Vk = VK and Wk = WK (this is true e.g. for k-algebras). When there is a notion of maps between objects over k (e.g. k-linear algebra homomorphisms), the G-action extends −1 to MapK (VK ,WK ) by (σf)(v) = σ(f(σ v)) for all v ∈ VK , σ ∈ G. f For example, let f be an equivalence VK −→ WK , e.g. a K-algebra isomorphism. Set −1 ξ(σ) = f ◦ σf ∈ GK := AutK (VK ). Notice that if ξ(σ) = 1 for all σ ∈ G, then σf = fσ G ∼ G so f descends to an isomorphism fk : Vk = VK −→ WK = Wk. That is, ξ = 1 is a necessary condition for two (K/k)-forms of VK to be isomorphic over k. In general, the automorphism ξ(σ) = f −1 · σf satisﬁes

ξ(στ) = f −1 · (στ)f = (f −1 · σf)((σf)−1στf) = ξ(σ)σ(ξ(τ)).

1 Such a map ξ : G → GK is called a 1-cocycle, and the set of all 1-cocycle is denoted Z (G,GK ). If K/k is an inﬁnite extension, we are guaranteed to have f(Vk) ⊆ Wk ⊗k ` ⊆ Wk ⊗k K = WK for some ﬁnite extension `/k. If σ ∈ H := Aut(K/`), then for all v ∈ Vk,

(σf)(v) = σ(f(σ−1v)) = σσ−1f(v) = f(v)

since f(v) ∈ Wk ⊗k `. Thus σf = f, so ξ(σ) = 1 for all σ ∈ H. Said another way, the 1-cocycle ξ is constant on cosets of H, which means ξ is in fact a continuous 1-cocycle on

G = lim Aut(`/k) −→ over all finite extensions `/k, when this direct limit is equipped with the Krull topology induced by discrete topologies on each Aut(`/k). ∼ Turning back to our K-isomorphism f : VK −→ WK , the definition of ξ may depend on 0 ∼ 0 this f, but suppose f were a different isomorphism VK −→ WK . Then f = f ◦ g for some 0 0 g ∈ GK and the cocycle ξ defined by f has the form

ξ0(σ) = (f 0)−1 · σf 0 = (f ◦ g)−1 · σ(f ◦ g) = g−1f −1 · (σf)(σg) = g−1ξ(σ)σg.

0 Deﬁnition. Two 1-cocycles ξ, ξ : G → GK are equivalent cocycles if there is some g ∈ GK such that ξ0(σ) = g−1ξ(σ)σg for all σ ∈ G. The set of equivalence classes of 1-cocycles is 1 denoted H (G,GK ).

Fix an object VK over K, set GK = AutK (VK ) and let F (K/k, VK ) be the set of k- isomorphism classes of (K/k)-forms of Vk. Then our work above shows that there is a map

1 Θ: F (K/k, VK ) −→ H (G,GK ).

Lemma 4.1.8. Θ is injective.

∼ Proof. Suppose Vk and Wk are (K/k)-forms of VK with isomorphisms f : VK −→ Vk ⊗k K and 0 ∼ 0 f : VK −→ Wk ⊗k K. Let ξ, ξ : G → GK be their corresponding 1-cocycles and suppose ξ and

65 4.1 Galois Descent 4 Descent

ξ0 are equivalent. Replacing f with f ◦g, we may assume ξ = ξ0. Then (f 0)−1 ·σf 0 = f −1 ·σf. Rearranging this, we have

0 −1 0 −1 f ◦ (f ) = σ(f ◦ (f ) ) as a map Wk ⊗k K −→ Vk ⊗k K.

0 −1 ∼ Therefore f◦(f ) descends to an isomorphism Wk −→ Vk over k, so Vk and Wk are isomorphic as (K/k)-forms of VK .

1 Theorem 4.1.9. When Vk is a k-algebra, Θ: F (K/k, VK ) → H (G,GK ) is a bijection.

1 Proof. Let ξ ∈ Z (G,GK ) be a 1-cocycle. We deﬁne a (K/k)-form Wk by deﬁning a new G∗ Galois action σ ∗ v on WK := Vk ⊗k K and taking Wk = (Vk ⊗k K) . We must do so in such a way that the identity VK → Vk ⊗k K = WK determines the cocycle ξ. Note that for this to happen, for any v ∈ VK we need to have

ξ(σ)(v) = f −1(σf)(v) = f −1(σfσ−1(v)) =⇒ ξ(σ)σ(v) = f −1(σf(v)) replacing v with σ(v) =⇒ f(ξ(σ)σ(v)) = σf(v).

Then deﬁning σ∗v = ξ(σ)σ(v), i.e. twisting by the cocycle ξ, gives a new k-semilinear action of G on Vk ⊗k K. Note that for any σ, τ ∈ G,

σ ∗ (τ ∗ v) = σ ∗ (ξ(τ)τ(v)) = ξ(σ)σ(ξ(τ)τ(v)) = ξ(σ)σ(ξ(τ))σ(τ(v)) by the cocycle condition on ξ = ξ(στ)στ(v) = (στ) ∗ v,

G∗ where f is the identity VK = WK . Therefore ∗ is a G-action on WK . Set Wk = WK , the fixed points of this Galois action. We will prove later that the identity f induces an isomorphism ∼ Wk ⊗k K = WK = VK as G-sets, but for now, it is obvious that this isomorphism (really, the identity) determines the cocycle ξ. Therefore Θ is surjective, so it is a bijection. The proof of Theorem 4.1.9 suggests a general strategy for descent in other contexts: define a new Galois action on the object VK over K and take Wk to be the fixed points of VK under this action. It will be useful to define group cohomology and Galois cohomology in the nonabelian case, so we recall the definitions here. Suppose G acts continuously on a group A (which may be nonabelian) with the discrete topology – remember that this means the stabilizers are open in G. Then a 1-cocycle of the action is a function ξ : G → A that is continuous and satisfies ξ(στ) = ξ(σ)σξ(τ) for all σ, τ ∈ G. Let Z1(G,A) denote the set of all 1-cocycles ξ : G → A. We say two cocycles ξ, ξ0 ∈ Z1(G,A) are equivalent if there exists an a ∈ A such that

ξ0(σ) = a−1ξ(σ)σ(a).

Let H1(G,A) be the set of equivalence classes of 1-cocycles. Then H1(G,A) is a pointed set with distinguished element [1], where 1 is the trivial cocycle.

66 4.1 Galois Descent 4 Descent

Suppose f : A → B is a morphism of G-groups, that is, a group homomorphism which commutes with the G-action. Then there is a map of pointed sets H1f : H1(G,A) −→ H1(G,B). Set H0(G,A) = AG. Proposition 4.1.10. For every short exact sequence of G-groups 1 → A → B → C → 1, there is an exact sequence of pointed sets 1 → H0(G,A) → H0(G,B) → H0(G,C) → H1(G,A) → H1(G,B) → H1(G,C). Furthermore, if A is central in B (and in particular is an abelian group), then the sequence may be extended by · · · → H1(G,B) → H1(G,C) → H2(G,A) and this is exact. Here, H2(G,A) is the ordinary 2nd group cohomology of G with coefficients in an abelian 1 G-group A. Suppose that α ∈ Z (G,A). Define the twist of A by α to be the G-group Aα whose underlying group is A but with G-action defined by g ∗ a = α(g)g(a)α(g)−1 = α(g) · g(a) for any g ∈ G and a ∈ A. (Here, x·y denotes the inner action of A on itself.) Let AG∗ denote the fixed points of this action.

0 0 1 ∼ 1 Lemma 4.1.11. The map ξ 7→ ξ := ξ · α yields a bijection tα : Z (G,Aα) −→ Z (G,A) which descends to a bijection of pointed sets

1 ∼ 1 τα : H (G,Aα) −→ H (G,A).

β Let 1 → A −→α B −→ C → 1 be a short exact sequence of G-groups and let

f 1 → H0(G,A) → H0(G,B) → H0(G,C) → H1(G,A) −→ H1(G,B) be the corresponding long exact sequence. For ξ ∈ H1(G,A), there is a twisted sequence

1 → Aξ → Bα(ξ) → Cξ → 1 which yields a long exact sequence

G G G 1 fξ 1 1 → Aξ → Bα(ξ) → Cξ → H (G,Aξ) −→ H (G,Bα(ξ)). Further, this ﬁts into a commutative diagram

fξ 1 1 [1] H (G,Aξ) H (G,Bα(ξ))

τξ τα(ξ) f ξ H1(G,A) H1(G,B)

67 4.1 Galois Descent 4 Descent

whose columns are bijections by Lemma 4.1.11. This shows that the fibre of f : H1(G,A) → 1 −1 H (G,B) over f(ξ), namely f (f(ξ)), may be identified with the fibre of fξ over [1], which G G is ker fξ. Explicitly, ker fξ = Bα(ξ)/Cξ which can be computed in many cases. Example 4.1.12. Let K = ksep be the separable closure of k and consider the matrix

algebra Ak = M2(k) ∈ Algk. Suppose Bk is a (K/k)-form of AK = M2(K), with k-linear ∼ isomorphism f : M2(K) −→ BK . Then Bk is a central simple k-algebra of dimension 4 over k. Assuming char k 6= 2, it is well-known that any such Bk is a quaternion algebra given by a basis {1, i, j, k} satisfying the relations i2 = a, j2 = b, k = ij, ij = −ji for some a, b ∈ k. a,b Denote the quaternion algebra with this presentation by Bk = k . For Ak = M2(k), × we have GK = AutK (M2(K)) = GL2(K)/K = P GL2(K). Given a quaternion algebra a,b Bk = k , set L = k(i) and deﬁne an embedding

Bk ,−→ M2(L) x0 + x1i b(x2 + x3i) x0 + x1i + x2j + x3k 7−→ . x2 − x3i x0 − x1i

∼ This extends to an isomorphism h : BK = Bk ⊗k K −→ M2(K) = AK ; call its inverse f. For x = x0 + x1i + x2j + x3k ∈ Bk and σ ∈ Gal(L/k), we have

−1 −1 σ(h)(x) = σ(hσ (x)) = σ(h(x)) since Bk is ﬁxed by σ x − x i b(x − x i) = 0 1 2 3 x2 + x3i x0 + x1i 0 b = gh(x)g−1 where g = . 1 0

This shows that 0 1 0 b ξ(σ) = g−1 = = = g in P GL (K). b−1 0 1 0 2

In fact, ( 1, σ| = id ξ(σ) = L g, σ|L 6= id. a,b This gives us a 1-cocycle ξ corresponding to k . Alternatively, consider the exact sequence of G-groups

× 1 → K → GL2(K) → P GL2(K) → 1.

Then the extended term in the long exact sequence from Proposition 4.1.10 is

1 2 × H (G, P GL2(K)) −→ H (G,K ) ξ 7−→ [δ]

˜ ˜ ˜ −1 ˜ where δ(σ, τ) = ξ(σ)σξ(τ)ξ(στ) for σ, τ ∈ G and ξ is any lift of ξ to GL2(K). Notice that H2(G,K×) = Br(k) is the Brauer group of k, whose elements are equivalence classes

68 4.1 Galois Descent 4 Descent

a,b of central simple k-algebras. For the cocycle ξ coming from the quaternion algebra k as above, ( b, σ| , τ| 6= id δ(σ, τ) = L L 1, otherwise.

This identifies each element of F (K/k, M2(k)) with a (Brauer class of a) central simple k- algebra of degree 2. In general, the (K/k)-forms of Mn(k) may be identified with the Brauer classes of central simple algebras of degree n. 1 Conversely, start with a quadratic extension L/k and a 1-cocycle ξ ∈ H (G, P GL2(K)). One can also construct a central simple algebra corresponding to ξ by taking BL = M2(L) as an algebra and defining a new Galois action on BL according to σ ∗ a = ξ(σ)σ(a) for all x y σ ∈ G and a ∈ M (L). In particular, if M = then 2 z w

0 b σ(x) σ(y) 0 1 σ(w) bσ(z) ξ(σ)σ(M) = = 1 0 σ(z) σ(w) b−1 0 b−1σ(y) σ(x)

G∗ for some b ∈ k. Thus we can identify Bk = BL as the matrices M ∈ M2(L) for which σ ∗ M = M, i.e. σ(w) bσ(z) x y = . b−1σ(y) σ(x) z w That is, x y Bk = x, y ∈ k . b−1σ(y) σ(x) Example 4.1.13. Assume char k 6= 2 and consider the collection of quadratic spaces (V, q) over k, with morphisms f : V1 ,→ V2 such that q2|V1 = q1. Fix a quadratic space (V, q). Then GK = Aut(VK , qK ) = O(K, 2)q, the 2 × 2 orthogonal group of K preserving the orientation induced by q. The set F (K/k, (VK , qK )) in this case classiﬁes all quadratic forms of a given rank (up to isomorphism). By Lemma 4.1.8, there is an injective map 1 Θ: F (K/k, (VK , qK )) ,→ H (G,O(K, 2)q). Descent theory shows when this is a bijection. Let q = x2 + y2 be the quadratic form on V = and q0 = ax2 + by2 the quadratic form on V 0. Deﬁne

0 f : VK −→ VK x y (x, y) 7−→ √ , √ . a b

1 The corresponding 1-cocycle ξ ∈ H (G,O(K, 2)q) is √ a ! √ 0 ξ(σ) = f −1 ◦ σ(f) = σ( a) √ . 0 √b σ( b)

0 0 0 G∗ Conversely, for a cocycle ξ one can construct a quadratic space (V , q ) with V = VK and 0 q = qK |V 0 by deﬁning the action σ ∗ v = ξ(σ)σ(v) for all σ ∈ G, v ∈ VK . Explicitly, for

69 4.1 Galois Descent 4 Descent

v = (x, y), √ a ! √ σ(x) 0 σ ∗ (x, y) = σ( a) √ . 0 √b σ(y) σ( b)

x x y y The ﬁxed points under this action are (x, y) ∈ VK such that σ √ = √ and σ √ = √ , √ √ a a b b so V 0 has k-basis {( a, 0), (0, b)}. The quadratic form in this case is precisely q0 = ax2+by2. Thus all Galois twists of the quadratic form q = x2 + y2 are of the form q0 = ax2 + by2 for some a, b ∈ k. Finally, consider the short exact sequence of G-groups

det 1 → SO(K, 2)q → O(K, 2)q −→ Z/2Z → 1.

2 2 1 For the quadratic form ax +by over k, let ξa,b be the corresponding cocycle in H (G,O(K, 2)q). By Proposition 4.1.10, there is a map

1 1 H (G,O(K, 2)q) −→ H (G, Z/2Z) √ ! ab ξa,b 7−→ det ξ : σ 7→ √ . σ( ab)

1 ∼ × × 2 Further, by Kummer theory (see below), H (G, Z/2Z) = k /(k ) , the set of nonzero elements of k modulo squares in k. Under this correspondence, ξa,b is identified with√ the coset × 2 2 2 ab(k ) . Similar√ analysis justifies the assertion in Example 4.1.6 that qK = x −y 2 defined over K = Q( 2) does not descend to a quadratic form on Q. Let K/k be a Galois extension with Galois group G = Gal(K/k) and suppose G is a linear algebraic group over k on which G acts continuously. Then

H1(G,G) = lim H1(G/U,G(KU )) −→ where the limit is over all finite index open subgroups U ⊆ G and KU is the subfield of K fixed by U.

Theorem 4.1.14 (Hilbert’s Theorem 90). For a Galois extension K/k with group G,

(1) Hi(G,K) = 0 for all i ≥ 1.

(2) H1(G,K×) = 1.

By the direct limit interpretation of H1(G,G) above, it is enough to prove Hilbert’s Theorem 90 for a ﬁnite Galois extension, where it is a classical consequence of Artin’s lemma on linear independence of characters. (For the i > 2 case in (1), one uses Shapiro’s lemma and the fact that K is induced from the representation k of the trivial subgroup {1} ⊆ G.) Note that for i ≥ 2, Hi(G,K×) need not vanish – for example, H2(G,K×) may be identiﬁed with the Brauer group of K/k which is nontrivial in general.

70 4.1 Galois Descent 4 Descent

Example 4.1.15. (Artin-Schreier Theory) Let k be a ﬁeld of characteristic p > 0, K = ksep and consider the Artin-Schreier sequence

℘ 0 → Fp −→ K −→ K → 0 where ℘(x) = xp − x. Applying H•(G, −), we get an exact sequence

℘ 1 1 0 → k −→ k → H (G, Fp) → H (G,K) = 0

1 ∼ where the last term is 0 by Hilbert’s Theorem 90. Thus H (G, Fp) = k/℘(k). On the 1 ∼ other hand, H (G, Fp) = Hom(G, Fp) which classiﬁes cyclic extensions of degree p over k (up to k-isomorphism). This demonstrates the Artin-Schreier correspondence between cyclic p-extensions of k and elements of k/℘(k); explicitly, every cyclic p-extension L/k is of the form L = k[x]/(xp − x − a) for some a ∈ k/℘(k). The general case of cyclic extensions of a characteristic p ﬁeld of order divisible by p is given by Artin-Schreier-Witt theory.

Example 4.1.16. (Kummer Theory) There is an analogous description of cyclic extensions of prime-to-p order: consider

× n × 1 → µn → K −→ K → 1

× n × n where K −→ K is the map x 7→ x and µn is the group (scheme) of nth roots of unity. The long exact sequence in Galois cohomology is

× n × 1 1 × 1 → k −→ k → H (G, µn) → H (G,K ) = 1

1 ∼ × × n by Hilbert’s Theorem 90. Thus H (G, µn) = k /(k ) . At the same time, if k contains the 1 ∼ nth roots of unity µn(k), then H (G, µn) = Hom(G, µn) classiﬁes cyclic n-extensions, and every such an extension L/k is of the form L = k[x]/(xn − b) for b ∈ k×/(k×)n.

Example 4.1.17. The additive and multiplicative groups of a ﬁeld k are often written × k = Ga and k = Gm. Consider the semidirect product a b G = = ∈ GL (k) . Ga o Gm 0 1 2

The short exact sequence 0 → Ga → G → Gm → 1 induces 1 1 1 1 = H (G, Ga) → H (G,G) → H (G, Gm) = 1 1 1 by Hilbert’s Theorem 90, so exactness implies H (G,G) = 1 as well. Note that G = Aut(Ak) 1 ∼ 1 so the injective Θ : F (K/k, Ak) −→ H (G,G) = 1 tells us that there is a unique (K/k)-form 1 1 ∼ 1 of AK , namely Ak. In simpler terms, if X is a curve over k such that X ×k K = AK , then ∼ 1 ∼ X = Ak as curves. In particular, if B is a k-algebra such that B ⊗k K = K[x], then it must ∼ 2 be that B = k[x] as k-algebras. This result also holds for Ak, or equivalently the polynomial ring k[x, y], by a result of Kambayashi.

71 4.1 Galois Descent 4 Descent

1 We next observe that H (G, GLn(K)) = 1. To do so, we need: Lemma 4.1.18. Let W be a vector space over K and assume that G acts on W by semilinear G ∼ transformations. Set W0 = W . Then W0 ⊗k K = W . Proof. By passing to the direct limit, we may assume K/k is ﬁnite. Enumerate G = {σ1, . . . , σn} and pick a basis {a1, . . . , an} for K/k. The matrix A = (σi(aj))ij ∈ Mn(K) t satisﬁes A A = (TrK/k(aiaj))ij which is the Gram matrix of the trace form TrK/k and hence invertible since the trace form is nondegenerate. This shows A ∈ GLn(k). Suppose w ∈ W and set n n X X λj(w) = σi(ajw) = σi(aj)σi(w) i=1 i=1 for each 1 ≤ j ≤ n. Then         λ1(w) σ1(w) σ1(w) λ1(w)  .  t  .   .  t −1  .   .  = A  .  =⇒  .  = (A )  .  λn(w) σn(w) σn(w) λn(w) ∼ so σi(w) is in the K-span of W0. Hence W0 spans W . By dimension counting, W = W0 ⊗k K.

1 Theorem 4.1.19. For any Galois extension K/k with group G, H (G, GLn(K)) = 1.

n 1 Proof. In this case GLn(K) = Aut(VK ) where V = k . Suppose ξ ∈ Z (G, GLn(K)). This n deﬁnes a twisted action of G on WK := K by σ ∗ w = ξ(σ)σ(w) for any σ ∈ G. Setting G∗ n ∼ Wk = WK , Lemma 4.1.18 shows that Wk ⊗k K = WK . Thus the isomorphism f : k −→ Wk n ∼ extends to an isomorphism fK : K −→ WK and additionally, f and fK each commutes with the Galois action: f(σ(v)) = σ ∗ f(v) = ξ(σ)σ(f(v)) n n −1 for all v ∈ k and σ ∈ G (likewise for fK and v ∈ K ). Replacing v with σ (v), we get f(v) = ξ(σ)σf(σ−1(v)) for all v, so f = ξ(σ)σ(f) which may be written f −1ξ(σ)σ(f) = 1. 1 This shows that ξ is equivalent to 1 in H (G, GLn(K)). Recall that by Theorem 4.1.9,

∼ 1 Θ: F (K/k, VK ) −→ H (G, GLn(K)).

Here, the proof of Theorem 4.1.19 essentially demonstrates that F (K/k, VK ) = 1 and then 1 the bijection Θ is used to conclude that H (G, GLn(K)).

1 Corollary 4.1.20. For any Galois extension K/k with group G, H (G,SLn(K)) = 1. Proof. Apply the long exact sequence in Galois cohomology to the sequence

det × 1 → SLn(K) → GLn(K) −→ K → 1 and use Hilbert’s Theorem 90 and Theorem 4.1.19.

72 4.1 Galois Descent 4 Descent

n Vn ∗ Notice that SLn(K) = AutK (V, ω) where V = K and ω is any nonzero element of V , for V ∗ the vector space dual of V . Therefore Corollary 4.1.20 says that there is only the trivial twist of (V, ω). Example 4.1.21. Let K = ksep and ﬁx a quadratic space (V, q) over K. We use de- 1 scent to show that Θ : F (K/k, (V, q)) → H (G,Oq(K, n)) is always a bijection. Given 1 ξ ∈ Z (G,Oq(K, n)), deﬁne a twisted action of G on WK := V ⊗k K by σ ∗ w = ξ(σ)σ(w). G∗ Then Wk = WK is an n-dimensional vector space over k, so it remains to produce a quadratic 0 0 form on Wk to which q restricts. Such a form q must satisfy q(w) = q (w) ∈ k for all w ∈ Wk. Indeed, if w ∈ Wk, σ ∗ w = w so

q0(w) = q0(σ ∗ w) = q0(ξ(σ)σ(w)) = q0(σ(w)) = σ(q0(w))

0 0 since ξ takes values in Oq0 (K, n) and q is defined over k. This implies q restricts to q on Wk, so descent is always possible here. Finally, we have a generalization of Hilbert’s Theorem 90 to finite dimensional k-algebras. Theorem 4.1.22. Let K/k be a Galois extension with group G, let A be a finite dimensional 1 × k-algebra and set AK = A ⊗k K. Then H (G,AK ) = 1. 1 × Proof. Since H (G,AK ) is a direct limit over finite Galois extensions, we may assume K/k 1 × is finite. Set d = [K : k]. View A and AK as right A-modules. For a cocycle ξ ∈ Z (G,AK ), G∗ define the twisted action of G on AK by σ ∗a = ξ(σ)σ(a) and put B = AK . As vector spaces, ∼ ∼ AK = B ⊗k K, so A = B as vector spaces. Note that B is also a right A-module. We claim A ∼= B as A-modules. By the Krull-Schmidt theorem, we may write

m1 mr n1 nr A = I1 ⊕ · · · ⊕ Ir and B = I1 ⊕ · · · ⊕ Ir ∼ for indecomposable submodules I1,...,Ir and integers mj, nj ≥ 1. But since A ⊗k K = B ⊗k K as AK -modules and

∼ m1d mrd ∼ n1d nrd A ⊗k K = I1 ⊕ · · · ⊕ Ir and B ⊗k K = I1 ⊕ · · · ⊕ Ir , the uniqueness part of the Krull-Schmidt theorem implies mjd = njd for all 1 ≤ j ≤ r. ∼ Therefore mj = nj for each j which shows A = B as A-modules. Fix an isomorphism of A-modules f : A −→∼ B and let c = f(1) ∈ B. Then f(x) = cx for all x ∈ A and the base change fK : AK → B ⊗k K is also an isomorphism, so c must be a G∗ unit in AK . In addition, c ∈ B = AK means that c = σ ∗ c = ξ(σ)σ(c), which can be written −1 1 × c ξ(σ)σ(c) = 1. Therefore ξ is equivalent to 1 in H (G,AK ). sep ∼ Suppose A is a central simple k-algebra and K = k . Then AK = A ⊗k K = Mn(K). The determinant map det : Mn(K) → K restricts to a map

Nrd = NrdK/k : A −→ k

called the reduced norm map. Further, there is a short exact sequence of algebraic groups

det × 1 → SL1(AK ) → GL1(AK ) −→ K → 1.

73 4.2 Fields of Deﬁnition 4 Descent

Taking Galois cohomology yields an exact sequence

× Nrd × 1 1 1 → A −−→ k → H (G,SL1(AK )) → H (G, GL1(AK )).

1 × × ∼ 1 By Theorem 4.1.22, H (G, GL1(AK )) = 1 so k / Nrd(A ) = H (G,SL1(AK )). To interpret this cohomology set, recall that for an algebraic group G, the set H1(G,G(K)) classiﬁes torsors for G deﬁned over k, up to isomorphism over K. That is, an equivalence class in H1(G,G(K)) is represented by a G-space X for which the map

G ×k X −→ X ×k X (g, x) 7−→ (x, gx) is an isomorphism. In general, a torsor X is deﬁned over k but need not have any rational points over k.

Example 4.1.23. For G = SL1(AK ), the G-torsors are given by solutions to the algebraic equation Nrd(x) = a for some a ∈ k×. Further, such a torsor has a rational point over k × × × ∼ 1 if and only if a ∈ Nrd(A ). This exhibits the bijection k / Nrd(A ) = H (G,SL1(AK )) explicitly. For an example of this correspondence, let H be the quaternion algebra over R. 2 2 2 2 Then Nrd(x0 + x1i + x2j + x3k) = x0 + x1 + x2 + x3 and the equation

2 2 2 2 x0 + x1 + x2 + x3 = a has solutions over R if and only if a > 0. Example 4.1.24. Let q be a quadratic form over k which determines a nondegenerate symmetric matrix Q ∈ Mn(k). Then the torsors for Oq(K, n), which are classiﬁed by 1 H (G,Oq(K, n)), are given by equations

XtQX = B for X = (xij) and some B ∈ Mn(k). This equation has a k-rational point if and only if Q and B are similar over k.

4.2 Fields of Deﬁnition

Let K be a field, I ⊂ K[x1, . . . , xn] an ideal and fix a subfield k ⊂ K.

Definition. We say I is defined over k, or k is a field of definition for I, if Ik := I ∩ k[x1, . . . , xn] generates I in K[x1, . . . , xn].

Theorem 4.2.1. For any ideal I ⊂ K[x1, . . . , xn], there exists a subfield k0 ⊂ K such that I is defined over k0 and k0 ⊆ L for any other field of definition L of I. Further, for an σ automorphism σ ∈ Aut(K), I = I if and only if σ|k0 = 1.

74 4.2 Fields of Deﬁnition 4 Descent

Proof. Let {Mα}α∈A be the set of all monomials in x1, . . . , xn. Pick a subset B ⊆ A so that {Mβ}β∈B is a maximal linearly independent subset mod I, i.e. a basis of K[x1, . . . , xn]/I. Put C = A r B. Then for each γ ∈ C, we can write X Mγ ≡ gβγMβ mod I β∈B

for some gβγ ∈ K. Let k0 be the subﬁeld of K generated by {gβγ}β∈B,γ∈C . Take f ∈ I and write it X f = aαMα for aα ∈ K α∈A X X = aβMβ + aγMγ. β∈B γ∈C Since f ∈ I, we have ! X 0 X X f = aβMβ + aγ Mγ − gβγMβ β∈B γ∈C β∈B

0 P P 0 with aβ ∈ K. Each Mγ − β∈B gβγMβ lies in I ∩ k[x1, . . . , xn] = Ik0 , so β∈B aβMβ ≡ 0 0 mod I. Hence by deﬁnition of B, we have aβ = 0 for each β ∈ B, so f is a linear combination

of elements in Ik0 . This shows k0 is a field of definition for I. To show k0 is minimal, let L ⊆ K be any field of definition of I and write IL = (p1, . . . , ps) for polynomials pi ∈ L[x1, . . . , xn]. Given the same notation as above, we must show that gβγ ∈ L for all β ∈ B, γ ∈ C. For a fixed γ0 ∈ C, we have X Mγ0 − gβγ0 Mβ = p1q1 + ... + psqs β∈B P for some qi ∈ K[x1, . . . , xn], say qi = α∈A yiαMα for yiα ∈ K. For 1 ≤ i ≤ s, define

∗ X qi = YiαMα ∈ K[Yiα | 1 ≤ i ≤ s, α ∈ A][x1, . . . , xn] α∈A

(here, Yiα are indeterminates). Let Y = (Yiα)1≤i≤s,α∈A. Then

∗ ∗ X p1q1 + ... + psqs = ϕα(Y )Mα α∈A

where ϕα(Y ) are linear functions of Y with coeﬃcients in L itself. Consider the following linear system over L:

ϕγ0 (Y ) = 1, ϕα(Y ) = 0 for α ∈ A r {γ0}.

Then {yiα}1≤i≤s,α∈A gives a solution to the system over K, but by linear algebra, there also exists a solution {y¯iα}1≤i≤s,α∈A over L. Evaluating the system at this solution yields X Mγ0 − g¯βγ0 Mβ = p1q¯1 + ... + psq¯s ∈ I β∈B

75 4.3 Galois Descent for Varieties and Schemes 4 Descent

∗ P ¯ whereg ¯βγ0 ∈ L andq ¯i = qi (Y ). Then Mγ0 ≡ β∈B βγ0Mβ mod I so by linear independence,

g¯βγ0 = gβγ0 for all β ∈ B. Since γ0 was arbitrary, we conclude that gβγ =g ¯βγ ∈ L for all β ∈ B, γ ∈ C. Hence k0 ⊆ L. For the ﬁnal statement, note that σ ∈ Aut(K) acts on I via its action on the coeﬃcients σ gβγ, so the equations in the above paragraph show that I = I if and only if σ|k0 = 1.

In practice, this proof is useless for explicitly ﬁnding k0. Instead, Galois descent can be used to give a more meaningful description of k0.

Definition. Such a field k0 as in Theorem 4.2.1 is called a minimal field of definition for I.

Definition. For a K-vector space V and a subfield k ⊆ K, a k-structure on V is a k- ∼ subspace Vk ⊆ V such that Vk ⊗k K = V as K-vector spaces. Similarly, a k-structure on a ∼ K-algebra A is a subalgebra Ak ⊆ A such that Ak ⊗k K = A as K-algebras. Suppose K/k is a Galois extension with group G. From Theorem 4.1.9, we know that giving a k-structure on a vector space V is the same as defining a semilinear G-action on V . Similarly, giving a k-structure on an algebra A is equivalent to defining a multiplication- preserving semilinear G-action on A.

Lemma 4.2.2. Suppose I ⊂ K[x1, . . . , xn] is an ideal deﬁned over a subﬁeld k ⊆ K. Then AK := k[x1, . . . , xn]/Ik is a k-structure on A := K[x1, . . . , xn]/I.

Definition. Suppose V is a K-vector space with k-structure Vk. Then a K-subspace W ⊆ V is defined over k if W ∩ Vk spans W over K. Lemma 4.2.3. Suppose K/k is a Galois extension with group G. Then for a vector space G V with k-structure Vk and a subspace W ⊆ V , W is defined over k if and only if W = W .

Definition. Suppose V and W are K-vector spaces with k-structures Vk and Wk, respectively, and f : V → W is a K-linear function. Then f is defined over k if f(Vk) ⊆ Wk. The same definition goes for homomorphisms of K-algebras.

Lemma 4.2.4. If K/k is Galois with group G, then a k-linear map f : V → W (or a homomorphism f : A → B of K-algebras) is deﬁned over k if and only if f σ = f for all σ ∈ G.

4.3 Galois Descent for Varieties and Schemes

n Definition. Suppose X ⊆ AK is an affine algebraic set with vanishing ideal I = I(X) ⊆ K[x1, . . . , xn]. Then X is defined over a subfield k ⊆ K, or k is a field of definition for X, if I is defined over k.

Remark. Notice that the property of being defined over a subfield depends on the explicit n embedding X,→ AK . We will see later that this definition can be made intrinsic.

76 4.3 Galois Descent for Varieties and Schemes 4 Descent

Lemma 4.3.1. Let ksep be the separable closure of a ﬁeld k, set G = Gal(ksep/k), let k¯ be ¯ the algebraic closure of k and suppose I ⊂ k[x1, . . . , xn] is an ideal. Then the following are equivalent:

(a) I is deﬁned over k.

(b) I is deﬁned over ksep and is G-invariant.

n Definition. Let k ⊆ K be a subfield and X ⊆ AK an affine algebraic set. Then X is k-closed if it can be defined by equations f1 = 0, . . . , fr = 0 for fi ∈ k[x1, . . . , xn]. Remark. The property of X being k-closed is in general not the same as X being defined over k, which can be expressed by saying that the vanishing ideal I(X) = rad((f1, . . . , fr)K[x1, . . . , xn]) is defined over k.

Example 4.3.2. When char k = 0 and K = ksep, the properties of being k-closed and deﬁned over k are equivalent.

n Lemma 4.3.3. Suppose X ⊆ AK is an aﬃne algebraic set which is k-closed for some k ⊆ K. Then X is deﬁned over k if and only if X is reduced.

Proof. Saying X is deﬁned over k is equivalent to saying the natural surjection

k[x , . . . , x ] K[x , . . . , x ] K[x , . . . , x ] 1 n = 1 n −→ 1 n (I(X)kK[x1, . . . , xn]) ⊗k K (I(X)kK[x1, . . . , xn]) I(X) is an isomorphism. On the other hand, its kernel is precisely the nil radical of the quotient ring K[x1, . . . , xn]/(I(X)kK[x1, . . . , xn]) which is 0 if and only if X is reduced. Next, consider an aﬃne scheme X = Spec A over Spec K and let k ⊂ K be a subﬁeld.

Definition. If A has a k-structure Ak, we define a k-topology on X = Spec A by declaring {V (a) | a ⊂ A is defined over k} to be the closed sets. Equivalently, this topology can be specified by declaring {D(f) | f ∈ Ak} to be a basis of open sets.

For each f ∈ Ak, set OX,k(D(f)) := (Ak)f . Since {D(f) | f ∈ Ak} is a basis for the k-topology on X, this deﬁnes a sheaf of rings OX,k on X with respect to the k-topology. This proves:

Lemma 4.3.4. Suppose A is a K-algebra with k-structure Ak and X = Spec A. Then (X, OX,k) is a k-structure on the ringed space (X, OX ). In general, we deﬁne:

Deﬁnition. A k-structure on a scheme X over K is the data of a k-topology on X,

i.e. a collection of open sets Topk(X) ⊆ Top(X), and for each k-open set U ∈ Topk(X), a k-structure on the K-algebra OX (U), which are subject to the following conditions:

(1) For every inclusion of k-open sets V,→ U, the morphism OX (U) → OX (V ) is deﬁned over k.

77 4.3 Galois Descent for Varieties and Schemes 4 Descent

S (2) For any affine open cover X = Ui, the subspace k-topology on each Ui and the specified k-structure on each OX (Ui) agree with the k-structure on the affine scheme

(Ui, OUi,k) deﬁned above. Set G = Gal(K/k). For a K-scheme X, let F (K/k, X) denote the collection of k- structures on X. Then there is a map

1 Θ: F (K/k, X) −→ H (G, AutK (X)). (To ensure the G-actions are continuous, we will always assume X is a scheme of finite type over K.) By Lemma 4.1.8, Θ is always injective. To show Θ is surjective, one might try to construct a Galois twist of X(K), but in general the fixed points of this space may be too small to recover X under base change – in many cases X(K)G is even empty! The solution for affine schemes at least is to define a new G-action on the algebra K[X] of regular functions 1 on X. Explicitly, for f ∈ K[X], σ ∈ G and ξ ∈ H (G, AutK (K[X])), set (σ ∗ f)(x) = σ(f(σ−1 ∗ x)) = σf(σ−1(ξ(σ)−1x)) = σ(f)ξ(σ)−1x.

This can be written σ ∗ f = ξg(σ)σ(f) where ∼ is the map AutK (X) → AutK (K[X]) sending −1 g 7→ g˜ defined by (˜g · f)(x) = f(g x). Now take Xk to be the affine K-scheme with algebra G∗ of regular functions K[Xk] = K[X] . Then Xk defines a k-structure on X. Example 4.3.5. Consider the linear algebraic group x 0 T = = ∈ GL (K) , Gm 0 x−1 2 which as an affine scheme corresponds to the algebra K[T ] = K[x, x−1]. Then as an algebraic −1 group, T has automorphism group AutK (T ) = {1, γ} where γ(t) = t . In this case,γ ˜ : x 7→ −1 1 x . For a Galois extension K/k with G = Gal(K/k), a 1-cocycle ξ ∈ H (G, AutK (T )) is simply a group homomorphism ξ : G → Aut√K (T ). Set H = ker ξ ⊆ G, which is a subgroup of index 2 as long as ξ 6= 1. Then KH = k( a) for some a ∈ k× and G acts on the algebra K[T ] via σ ∗ a = aσ and σ ∗ f = ξg(σ)σ(f), where ( x 7→ x, if ξ(σ) = 1 ξg(σ) = x 7→ x−1, if ξ(σ) = γ.

G∗ H∗ G/H∗ √ −1 G/H∗ So K[T ] = (K[T ] ) = k( a)[x, x ] . This quotient√ group is a 2-group G/H = {1, τ} and the nontrivial element acts on the elements of k( a)[x, x−1] by

d d X n X −n τ ∗ anx = τ(an)x . n=−d n=−d So Pd a xn is a ﬁxed point of the twisted G/H-action if and only if τ(a ) = a for all n=−d n n √ −n n ∈ Z. The algebra of such Laurent series is generated by {bxn +τ(b)x−n | b ∈ k( a), n ≥ 0}. One can in fact show that this algebra is precisely k[s, t] where x + x−1 x − x−1 s = and t = √ . 2 2 a

78 4.3 Galois Descent for Varieties and Schemes 4 Descent

So K[T ]G∗ = k[s, t]. Noting that s2 − at2 = 1, we can rewrite this algebra as

k[s, t] = k[S,T ]/(S2 − aT 2 − 1)

which is the ring of regular functions for the 1-dimensional k-scheme u av T := ∈ GL (k): u2 − av2 = 1 . a v u 2 √ √ √ ∼ ∼ × × 2 In fact, Ta(k( a)) = Gm(k( a)), but if a 6∈ k then Ta 6= Gm(k). Thus each a ∈ k r(k ) gives a distinct isomorphism class of twists of the torus Gm. One can even check that the Hopf algebra structure on K[T ] descends to k[s, t], so Ta is even an algebraic group over k. a,b Example 4.3.6. Using the notation of Example 4.1.12, let A = k be a quaternion algebra over a field k of characteristic other√ than 2. Recall that A is a K/k-form of M2(L) for the Galois extension L/k where L = k( a) and G = Gal(L/k) = {1, σ}. Explicitly, A 1 1 corresponds to an equivalence class of cocycles ξ ∈ H (G, P GL2(L)) = H (G, AutL(M2(L))) where 0 b ξ(σ) = ∈ P GL (L). 1 0 2 1 1 At the same time, P GL2 = Aut(P ) so the cocycle ξ defines a twist of P , called a Severi- Brauer curve. Here’s how to construct it. Take U = P1 r {0, ∞}, which is an affine curve invariant under ξ(σ). Then as we saw in Example 4.3.5, L[U] = L[x, x−1]; define a new action of G on L[U] by √ √ σ ∗ a = σ( a) and σ ∗ x = bx−1.

One can show that L[U]G∗ = k[s, t] where x + bx−1 x − bx−1 s = and t = √ . 2 2 a

Note that s2 − at2 = b, so we may instead write L[U]G∗ = k[S,T ]/(S2 − aT 2 − b). Further, note that the aﬃne equation S2 − aT 2 = b has a solution over k if and only if A splits over k. The homogeneous equation X2 − aY 2 − bZ2 = 0 deﬁnes a smooth, projective curve SB(A) over k, which is the aforementioned Severi-Brauer 1 curve for A. Note that SB(A) is a k-structure for PL and SB(A) has a k-rational point exactly when A splits over k. More broadly: Theorem 4.3.7 (Amitsur). Suppose A and B are quaternion algebras over k with Severi- Brauer curves SB(A) and SB(B), respectively. Then A ∼= B over k if and only if SB(A) and SB(B) are birationally equivalent over k. Example 4.3.8. For a central simple algebra A of degree n > 2, one can similarly construct a Severi-Brauer variety SB(A). The most general form of Amitsur’s Theorem says that if SB(A) and SB(B) are birational as k-varieties, then [A] and [B] generate the same cyclic subgroup in Br(k). Whether the converse to this statement holds is an open question.

79 4.3 Galois Descent for Varieties and Schemes 4 Descent

Question. Suppose X is an algebraic variety over K and k ⊂ K is a subﬁeld. Is there a variety Y over K such that X ∼= Y as K-varieties and Y is deﬁned over K?

n m In the case of an aﬃne algebraic variety X ⊆ AK , if there exists such a variety Y ⊆ AK (a priori m 6= n) that has a k-structure, then we can twist K[X] ∼= K[Y ] to obtain a k-structure on X. In general, this is not possible, but there are some conditions one can place on the variety to allow for descent. The following examples illustrate these conditions for quadratic forms.

Example 4.3.9. First consider the quadratic form x2 + 2iy2 over Q(i). Notice that the v 2 linear change of variables u = x, v = (1 + i)√y yields a quadratic√ form u + v over Q. On the other hand, one can show that q = x2 − 5y2 deﬁned over Q( 5) is not equivalent to a quadratic form q0 over Q. Let 1 0 √ Q = √ ∈ GL ( ( 5)) 0 − 5 2 Q

be the symmetric matrix corresponding to q. If q were equivalent to a quadratic form t over Q, with√ matrix Q0 ∈ GL2(Q), then we would be able√ to write Q = B Q0B for some 2 B ∈ GL2(Q( √5)). Taking determinants, we see that − 5 =√rs where r = det(B) is 2 a√ square in Q(√ 5) and s = det(Q0) ∈ Q. Write r = (α + β 5) √for α, β ∈ Q. Then − 5 = (α + β 5)2s and applying the nontrivial element τ ∈ Gal(Q( 5)/Q) yields √ √ 5 = (τ(α) − τ(β) 5)2s. √ √ 2 2 Combining these√ expressions, we get that (α + β 5) = −(τ(α) − τ(β) 5) , which is impossible in Q( 5). Therefore no such Q0 exists. More generally, suppose K/k is a Galois extension with group G and Q is a symmetric t matrix in GLn(K) for which Q = B Q0B for some Q0 ∈ GLn(k) and B ∈ GLn(K). Then for any σ ∈ G, σ σ t σ t Q = (B ) Q0B = A(σ) QA(σ) for A(σ) = B−1Bσ. Notice that the A(σ) satisfy A(στ) = A(σ)σ(A(τ)) for all σ, τ ∈ G, 1 so they deﬁne a 1-cocycle in Z (G, GLn(K)). By Theorem 4.1.19, any such cocycle is a coboundary, which reﬂects the fact that A(σ) = B−1Bσ.

Example 4.3.10. For the quadratic form q = x2 + 2iy2 over Q(i) above, its symmetric matrix 1 0 Q = ∈ GL ( (i)) 0 2i 2 Q has Galois conjugate 1 0 Qσ = 0 −2i ∼ σ where Gal(Q(i)/Q) = hσi = Z/2Z, and one can see that Q ∼ Q via the cocycle 1 0 A(σ) = . 0 i

80 4.3 Galois Descent for Varieties and Schemes 4 Descent

2 Then A(1) = A(σ ) = A(σ)σ(A(σ)) = I2, the 2 × 2 identity matrix, so Q descends to a matrix over Q and therefore√ descent is possible√ for the quadratic form q. Meanwhile, for q = x2 − 5y2 over Q( 5), the symmetric matrix 1 0 Q = √ 0 − 5 has Galois conjugate 1 0 Qτ = √ 0 5 √ and one can show that Q is not similar to Qτ over Q( 5). Thus descent is not possible in this case. Example 4.3.11. The previous example shows that having a similarity Q ∼ Qσ for some Galois conjugate of the symmetric matrix is a necessary condition for descent of a quadratic√ 2 2 form, but it turns out that it is not√ suﬃcient. Consider the same quadratic form q = x − 5y ∼ but now over the ﬁeld K = Q( 5, i). Set k = Q(i) and G = Gal(K/k) = hτi = Z/2Z. Then Q and Qτ are similar via the cocycle 1 0 A(τ) = ∈ GL (K). 0 i 2

Suppose q0 is a quadratic form which is equivalent to q over Q(i). Then taking determinants t of Q = B Q0B, we once again have equations √ √ − 5 = (α + βi)2s and 5 = (τ(α) + τ(β)i)2s √ for α, β ∈ Q( 5). These imply (α + βi)2 = −(τ(α) + τ(β)i)2, or α + βi = i(τ(α) + τ(β)i) or α + βi = −i(τ(α) + τ(β)i). In the ﬁrst case, we must have τ(α) = β and τ(β) = −α, which implies τ 2 = −1, a contradiction. The second case produces the same contradiction. Therefore descent is not possible in this case. The correct characterization of descent for quadratic forms is contained in the following proposition.

Proposition 4.3.12. Let K/k be a Galois extension with Galois group G, let Q ∈ GLn(K) be a symmetric matrix representing a quadratic form q over K and assume that for every 1 σ t σ ∈ G, there exists a 1-cocycle A ∈ Z (G, GLn(K)) such that Q = A(σ) QA(σ). Then q descends to a quadratic form on k. n t Proof. Let V = K . It will be enough to deﬁne a k-structure Vk on V such that Q = B Q0B for some Q0 ∈ Aut(Vk) and B ∈ GL2(K), since then Q0 deﬁnes a quadratic form which is

equivalent to q|Vk . For σ ∈ G and v ∈ V , we have: σ(q(v)) = σ(vtQv) = σ(v)tQσσ(v) = σ(v)tA(σ)tQA(σ)σ(v) = q(A(σ)σ(v)). This suggests deﬁning a twisted G-action on V by σ ∗ v = A(σ)σ(v). Then as we have seen G∗ before, Vk = V is a k-structure on V and the rest follows.

81 5 Etale´ Fundamental Group

5 Etale´ Fundamental Group

5.1 Covering Spaces

There are two key concepts in algebraic topology that, for various reasons, one might want to consider in an algebraic setting. These are covering spaces and fundamental groups, and they are intimately connected. The more familiar concept might be that of the fundamental group, which at the beginning is usually deﬁned in terms of homotopy classes of based loops in a given topological space. To deﬁne an algebraic analogue, we will need an alternative perspective on the fundamental group. Let X be a connected, locally simply connected topological space.

Deﬁnition. A cover of X is a space Y and a map p : Y → X that is a local homeomorphism. That is, for every x ∈ X there is a neighborhood U ⊆ X of x such that p−1(U) is a disjoint union of open sets in Y and p restricts to a homeomorphism on each open set.

One consequence of this deﬁnition is that for all x, y ∈ X, p−1(x) is a discrete space and p p−1(x) ∼= p−1(y). A primary goal in topology is to study and classify all such covers Y −→ X.

p p0 Deﬁnition. A morphism of covers between Y −→ X and Y 0 −→ X is a map f : Y → Y 0 making the following diagram commute:

f Y1 Y2

p1 p2

This deﬁnes a category CovX of covers over X. In this category, we will abbreviate

HomCovX (Y,Z) by HomX (Y,Z).

Deﬁnition. A covering space π : Xe → X is a universal cover for X if for every other cover p : Y → X, there is a unique map f : Xe → Y making the diagram commute: f Xe Y

π p

It is equivalent to say that a universal cover is any simply connected cover of X, and one shows easily that universal covers are unique up to equivalence of covers. An important result is that a universal cover exists, under certain mild conditions on X.

82 5.1 Covering Spaces 5 Etale´ Fundamental Group

The topological fundamental group is deﬁned using homotopy:

homotopy classes of loops πtop(X, x) := . 1 in X based at x

The universal cover has important connections to this fundamental group. In particular, consider an automorphism α ∈ AutX (Xe) = HomX (X,e Xe). Fix x ∈ X and a liftx ˜ ∈ Xe of x. Then πα(˜x) = x. Moreover, since Xe is simply connected, any pathx ˜ → αx˜ is unique up to top homotopy. This determines a map AutX (Xe) → π1 (X, x).

top Theorem 5.1.1. For any x ∈ X, AutX (Xe) → π1 (X, x) is an isomorphism.

Unfortunately, such a space Xe does not exist in the algebraic world. Thus we describe a slightly diﬀerent interpretation of the fundamental group.

Deﬁnition. The ﬁbre functor over x ∈ X is the assignment

Fibx : CovX −→ Sets p (Y −→ X) 7−→ p−1(x).

By the universal property of a universal cover Xe ∈ CovX , to give a morphism of covers −1 f : Xe → Y is the same as to choose a point y = f(˜x) ∈ p (x). In other words, Fibx is a representable functor, i.e. for x ∈ X, there is a natural isomorphism ∼ Fibx(−) = HomX (Xex˜, −),

where the Hom set consists of morphisms based atx ˜. This ﬁbre functor is constructible in algebraic categories, though it fails to be representable. Going further, there is a natural left action of AutX (Xe) on Xe; however, it will be more op convenient to view this as a right action of AutX (Xe) on Xe. This induces a left action of top AutX (Xe) = π1 (X, x) on HomX (X,Ye ):

α · f = f ◦ α for any α ∈ AutX (Xe), f : Xe → Y.

top This action is called the monodromy action. Often, one views this as an action of π1 (X, x) on the ﬁbre Fibx(Y ) given by lifting paths. In any case, we get a map

top π1 (X, x) −→ Aut(Fibx), where Aut(Fibx) is the automorphism group of the ﬁbre functor in the following sense. For any functor F : C → D, an automorphism of F is a natural transformation of F that has a two-sided inverse. The set Aut(F ) of all automorphisms of F is then a group under composition. Moreover, Aut(F ) has a natural action on F (C) for any object C ∈ C.

top Theorem 5.1.2. For all x ∈ X, π1 (X, x) → Aut(Fibx) is an isomorphism.

83 5.2 Inﬁnite Galois Theory 5 Etale´ Fundamental Group

Theorem 5.1.3. Let X be a connected, locally simply connected space and fix x ∈ X. Then the fibre functor Fibx defines an equivalence of categories

∼ CovX −→{left π1(X, x)-sets} with connected covers corresponding to transitive π1(X, x)-sets and Galois covers to coset spaces of Xex by normal subgroups.

Proof. For a transitive π1(X, x)-set S, define YS = X/He where H = Stabπ1(X,x)(s) for any point s ∈ S. This defines a Galois cover YS → X and one can extend this to arbitrary π1(X, x)-sets orbitwise for the full correspondence. The picture gets more interesting if we restrict ourselves to finite covers. Given such a cover p : Y → X, there is an exact sequence of groups

top −1 1 → N → π1 (X, x) → AutX (p (x)) → 1 where N is some finite index kernel. This shows that the monodromy action factors through a finite quotient. As a result, this action can be defined on the level of a profinite group, top namely the profinite completion of π1 (X, x):

top\ top π1 (X, x) := lim π1 (X, x)/N, ←−

top where the inverse limit is over all ﬁnite index subgroups N ≤ π1 (X, x).

Corollary 5.1.4. The ﬁbre functor Fibx deﬁnes an equivalence of categories

∼ top\ {ﬁnite covers of X} −→{ﬁnite, continuous π1 (X, x)-sets}. Moreover, the correspondence restricts to

∼ top\ {connected covers} −→{ﬁnite π1 (X, x)-sets with transitive action} ∼ top\ and {Galois covers} −→{π1 (X, x)/N | N an open normal subgroup}.

5.2 Inﬁnite Galois Theory

Fix a field k and an algebraic closure k¯, which comes equipped with a separable closure ¯ ks ⊆ k. Set Gk = Gal(ks/s). The Galois group Gk is a topological group since it is a profinite group: Gk = lim Gal(L/k) where the inverse limit is over all finite separable extensions L/k. ←− Fix such a finite separable extension L/k.

Lemma 5.2.1. Homk(L, ks) is a ﬁnite, continuous, transitive Gk-set.

Proof. By Galois theory, # Homk(L, ks) = [L : k], so this is finite when the extension is assumed to be finite. The Gk-action is defined by σ·f = σ◦f for σ ∈ Gk and f ∈ Homk(L, ks); it is routine to verify that this is indeed a group action. Now to show the action is continuous,

84 5.2 Inﬁnite Galois Theory 5 Etale´ Fundamental Group

since Homk(L, ks) is discrete, this is equivalent to showing the stabilizer StabGk (f) is open for each f ∈ Homk(L, ks). Notice that

StabGk (f) = {σ ∈ Gk | σ ◦ f = f} = {σ ∈ Gk | σ ﬁxes f(L)}

and this is open by Galois theory / the topology of the proﬁnite group Gk. Finally, since L/k is separable, we may pick a minimal polynomial h(t) for a primitive element of L/k. Then Gk permutes the roots of h(t) transitively, so it follows that Gk acts transitively on Homk(L, ks). ∼ Corollary 5.2.2. There exists an open subgroup H ≤ Gk such that Homk(L, ks) = Gk/H as Gk-sets. Further, if L/k is Galois, H may be chosen to be an open normal subgroup.

Proof. We may pick H = StabGk (h), the stabilizer of the minimal polynomial of a primitive element of L/k. The Galois case follows from the fundamental theorem of Galois theory.

Theorem 5.2.3. The assignment L/k 7→ Homk(L, ks) is a contravariant functor

{ﬁnite separable extensions of k} −→ {ﬁnite, continuous, transitive Gk-sets}

which is an anti-equivalence of categories. Moreover, Galois extensions L/k correspond to ﬁnite quotients of Gk.

Proof. For any finite, continuous, transitive Gk-set S, define a finite separable extension

LS/k by taking the subﬁeld of ks/k ﬁxed by the stabilizer StabGk (s) of any point s ∈ S. One now checks that this is an inverse functor to the Homk(−, ks) functor.

To study all finite continuous Gk-sets, we replace separable field extensions with finite étalealgebras. Recall (Section 3.5) that a k-algebra A is a finite étalealgebra if A ∼= L1 × · · · × Lr for finite separable extensions Li/k.

Corollary 5.2.4 (Grothendieck). The assignment A/k 7→ Homk(A, ks) is an anti-equivalence of categories ∼ {finite étale k-algebras} −→{finite continuous Gk-sets} which reduces to the above case when A = L is a finite separable extension of k.

Topologically, we may view k as a point space covering the ‘smaller’ point space ks, and any intermediate extension L/k as an intermediate cover. In this setting, Gk plays the role of the deck transformations of the ‘universal cover’ k → ks and Homk(L, ks) plays the role of the fibre of a cover. Also, under this analogy, a finite separable extension L represents a connected cover while a finite étalealgebra A may be viewed as a disconnected cover. Finally, the choice of a algebraic (and separable) closure of k is analagous to the choice of a basepoint of a topological space, which also determines a universal cover. We will see that this subtlety conceals a wealth of information about the algebraic fundamental group. We have thus described the dimension 0 case of the étalefundamental group: it is the absolute Galois group of the ground field.

85 5.3 Galois Theory for Schemes 5 Etale´ Fundamental Group

5.3 Galois Theory for Schemes

Let X be a scheme. The category FétX is defined to be the full subcategory of SchX ϕ consisting of finite étalecovers Y −→ X. To compare to the topological case of a covering space, we define:

Deﬁnition. A geometric point of X is a morphism x¯ : Spec Ω → X for some algebraically closed ﬁeld Ω.

Concretely, the image ofx ¯ is some point x ∈ X for which κ(x) ⊆ Ω.

Definition. Let ϕ : Y → X be a morphism and x¯ : Spec Ω → X a geometric point. Then the geometric fibre of x¯ is the fibre product Yx¯ := Y ×X Spec Ω. Proposition 5.3.1. For a morphism ϕ : Y → X, ϕ is étaleat each P ∈ X if and only if every geometric fibre Yx¯ of ϕ is of the form Spec(Ω × · · · × Ω), where Ω ⊇ κ(P ) is an algebraically closed field.

Proof. Follows from the algebraic version, Lemma 3.5.1.

Example 5.3.2. The cover C r {0} → C r {0}, y 7→ yn is a finite étalecover. Example 5.3.3. Let X be a normal scheme of dimension 1 and let ϕ : Y → X be a finite morphism. In the affine case, with X = Spec A and Y = Spec B, this corresponds to an ∗ extension of Dedekind rings ϕ : A → B. Then points in the fibre YP of a point P ∈ X correspond to a prime factorization of ideals:

r Y ei PB = Qi i=1

where Q1,...,Qr are prime ideals of B. Then YP is a finite étalealgebra if and only if ei = 1, i.e. the prime P is unramified in the language of algebraic number theory. In the general case, a finite morphism of normal schemes of dimension 1 is étaleif and only if each affine piece corresponds to a finite, unramified extension of Dedekind rings.

Proposition 5.3.4. Let ϕ : Y → X be a ﬁnite ´etalecover, z¯ : Spec Ω → z a geometric point and Z a connected scheme over X. If f, g : Z → Y are two X-morphisms such that f ◦ z¯ = g ◦ z¯, then f = g.

Proof. By Proposition 3.6.4(a), we may assume X = Z, so that f and g are two sections of ϕ : Y → X. It follows that f and g are ﬁnite ´etalemorphisms and each induces an isomorphism of Z = X with an open and closed subscheme of Y . Since Z is connected, the images of f and g are determined by the images of any geometric point, hence f ◦ z¯ = g ◦ z¯ implies f = g.

Corollary 5.3.5. Let ϕ : Y → X be a finite étalecover. Then Aut(Y/X) is finite.

86 5.4 The Etale´ Fundamental Group 5 Etale´ Fundamental Group

Proof. Take σ ∈ Aut(Y/X), σ 6= 1, and set f = ϕ and g = ϕ ◦ σ. Then by Proposition 5.3.4, f and g send some geometric point of Y to different points. In other words, the action of Aut(Y/X) on any geometric fibre is free, or the permutation representation of Aut(Y/X) on any of the geometric fibres is faithful. Now since the map is finite étale,each geometric fibre is finite as a set, so this implies Aut(Y/X) is itself finite. Now let ϕ : Y → X be a finite étalecover and let G be a group scheme over X such that Y is a left G-torsor. Let Y/G be the quotient space with projection map π : Y → Y/G. We G define a sheaf on Y/G by OY/G := (π∗OY ) , the subsheaf of G-invariants of the pushforward of OY to Y/G along π. This makes Y/G into a ringed space.

Proposition 5.3.6. The ringed space (Y/G, OY/G) is a scheme over X. Moreover, ϕ : Y → X factors through a finite morphism ψ : Y/G → X. The following are analogues of the basic Galois theory of topological covering spaces. Proposition 5.3.7. If ϕ : Y → X is a connected, finite étalecover and G ≤ Aut(Y/X) is any finite subgroup of automorphisms, then π : Y → Y/G is a finite étalecover. Definition. A connected, finite étalecover ϕ : Y → X is a Galois cover if Aut(Y/X) acts transitively on every geometric fibre of ϕ. Theorem 5.3.8. Let ϕ : Y → X be a Galois cover and suppose ψ : Z → X is a connected, finite étalecover such that Z is a scheme over Y and the diagram Y Z

ϕ ψ

X commutes. Then (1) Y → Z is a Galois cover and Z ∼= Y/G for some subgroup G ≤ Aut(Y/X). (2) There is a bijection {subgroups G ≤ Aut(Y/X)} ←→ {intermediate covers Y → Z → X}.

(3) The correspondence is bijective on normal subgroups of Aut(Y/X) and Galois covers Z → X, and in this case Aut(Z/X) ∼= Aut(Y/X)/G as groups.

5.4 The Etale´ Fundamental Group

Let X be a scheme and FétX the category of finite étalecovers of X. Fix a geometric point x¯ : Spec Ω → X and let

Fibx¯ : F´etX −→ Sets

Y 7−→ Yx¯ = Y ×X Spec Ω be the fibre functor overx ¯. We now define the algebraic, or étale, fundamental group of a scheme.

87 5.4 The Etale´ Fundamental Group 5 Etale´ Fundamental Group

Definition. The algebraic, or étalefundamental group of a scheme of X at a geometric point x¯ : Spec Ω → X is the automorphism group of the fibre functor over x¯:

π1(X, x¯) := Aut(Fibx¯).

Theorem 5.4.1 (Grothendieck). Let X be a connected scheme and x¯ : Spec Ω → X a geometric point. Then

(1) π1(X, x¯) is a proﬁnite group and its action on Fibx¯(Y ) is continuous for all Y ∈ F´etX .

(2) Fibx¯ induces an equivalence of categories

∼ F´etX −→{ﬁnite, continuous, left π1(X, x¯)-sets}

Y 7−→ Fibx¯(Y ).

Example 5.4.2. Let k be a field and consider X = Spec k. Then finite étalecovers Y → Spec k are precisely Y = Spec A for A = L1 × · · · × Lr a finite étale k-algebra. Here, the fibre functor over anyx ¯ : Spec k¯ → Spec k (equivalent to a choice of algebraic closure k¯ of k) is exhibited by ¯ Fibx¯(Y ) = Spec(A ⊗k k) = Spec(Ω × · · · × Ω), and Spec(Ω × · · · × Ω) is a finite set of r closed points indexed by the homomorphisms ¯ ∼ ¯ A → k. Indeed, Fibx¯(Y ) = Homk(A, ks), where ks is the separable closure of k in k via the ¯ embedding k ,→ k. The action of Aut(Homk(−, ks)) on any given Homk(A, ks) is given by T · σ = σ ◦ T , or precisely the action of Gk = Gal(ks/k) on Homk(A, ks). Therefore ∼ ∼ π1(X, x¯) = Aut(Fibx¯) = Aut(Homk(−, ks)) = Gal(ks/k).

Moreover, there is an identification Homk(A, ks) = HomSpec k(Spec ks, Spec k). The above shows that although Fibx¯ is not a representable functor – Spec ks is not a finite étale k-scheme – it is pro-representable in FétSpec k. Explicitly,

0 Fibx¯(Spec A) = lim HomSpec k(Y , Spec A) −→ where the direct limit is over all finite étaleGalois covers Y 0 → Spec k ordered by the existence of a Spec k-morphism Y 0 → Y 00. In fact, there was nothing special about X being Spec k or even affine in the last paragraph. Proposition 5.4.3. For a connected scheme X and any geometric point x¯ : Spec Ω → X, the fibre functor Fibx¯ is pro-representable. Explicitly, for any finite étalecover Y → X,

0 Fibx¯(Y ) = lim HomX (Y ,Y ) −→ where the direct limit is over all ﬁnite ´etaleGalois covers Y 0 → Y .

Lemma 5.4.4. Every automorphism of Fibx¯ is determined by a unique automorphism of the direct system (Y 0 → X) of ﬁnite Galois covers of X.

88 5.4 The Etale´ Fundamental Group 5 Etale´ Fundamental Group

We can now give the proof of Theorem 5.4.1.

0 Proof. (1) By Proposition 5.4.3, Fibx¯ is pro-representable by the direct system of HomX (Y ,Y ) where Y 0 ranges over all ﬁnite Galois covers of X. Now Lemma 5.4.4 implies

0 Aut(Fibx¯) = lim Aut(Y /X) ←− where again Y 0 ranges over all finite Galois covers. Since Corollary 5.3.5 says that each 0 Aut(Y /X) is a finite group, we see that π1(X, x¯) = Aut(Fibx¯) is a profinite group. Note that π1(X, x¯) has a natural action on each Fibx¯(Y ) for any finite étalecover Y → X. It remains to show this action is continuous. Take a geometric pointy ¯ ∈ Fibx¯(Y ) lying 0 overx ¯ : Spec Ω → X. Ify ¯ comes from an element of HomX (Y ,Y ) by the direct limit 0 in Proposition 5.4.3, then the action of π1(X, x¯) factors through Aut(Y /X). This implies continuity. (2) Take a finite, continuous, left π1(X, x¯)-set S. The action of π1(X, x¯) is transitive on each orbit of S, so we may assume it is transitive on S to begin with. Let H be the stabilizer of any point in S. Then H is an open subgroup of π1(X, x¯), so it contains the open normal 0 subgroup corresponding to the kernel of π1(X, x¯) → Aut(Y /X) for some finite Galois cover Y 0 → X. Let H be the image of H in Aut(Y 0/X). Then one proves S ∼= Y 0/H. This proves essential surjectivity; fully faithfulness is routine. Theorem 5.4.5. Let X be an integral normal scheme with function field K and fix a separable closure Ks/K. Let KX be the compositum of all finite subextensions L/K in Ks such that the normalization XL of X in L is étaleover X. Then

(1) KX /K is Galois. ∼ (2) For any geometric point x¯ : Spec K → X, we have π1(X, x¯) = Gal(KX /K). Theorem 5.4.6 (Grothendieck). Let X be a connected scheme over C of ﬁnite type. Then there is an equivalence of categories

FétX −→ FCovX(C) (finite-sheeted topological covers) (Y → X) 7−→ (Y (C) → X(C)). Therefore, for any geometric point x¯ : Spec C → X with image x =x ¯(Spec C), the induced map top\an π1 (X , x) −→ π1(X, x¯) is an isomorphism. Example 5.4.7. If Y → 1 is a finite étalecover, the Riemann-Hurwitz inequality says PC that 2g(Y ) − 2 = n(0 − 2) + 0 but this is only possible if g(Y ) = 0 and n = 1. Thus Y → 1 is a birational isomorphism, PC but since Y is complete, it must be a regular isomorphism. Hence there are no nontrivial étalecovers of 1 which proves π ( 1 ) = 1. Notice that this also confirms Theorem 5.4.6 PC 1 PC when X(C) = C ∪ {∞} is the Riemann sphere, which is topologically simply connected. A similar argument shows π ( 1 ) = 1. 1 AC

89 5.4 The Etale´ Fundamental Group 5 Etale´ Fundamental Group

Example 5.4.8. Theorem 5.4.6 also implies that π ( 1 {0, ∞}) ∼ , the profinite com- 1 PC r = Zb pletion of the integers. Thus there is a finite étalecover X → 1 {0, ∞} for each n ≥ 1 n PC r having Galois group /n . Indeed, 1 {0, ∞} = × and each X is isomorphic to the Z Z PC r C n cover C× → C×, z 7→ zn. Example 5.4.9. The first really interesting case is that π ( 1 {0, 1, ∞}) is the free profinite 1 PCr group on two generators.

Theorem 5.4.6 extends to any algebraically closed field of characteristic 0. However, in ét 1 characteristic p > 0, there is no such topological description available. For example, π1 (A ) is not even finitely generated in general.