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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 K¨ahlerDifferentials . 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 ge- ometry with a common theme of ´etalemorphisms, 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-L¨utkebohmert-Raynaud's N´eron Models, G¨ortz- Wedhorn's Algebraic Geometry I, M´ezard'sarticle \Fundamental group" and Raynaud's Anneaux Locaux Hens´eliens. 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. P 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 ffig of F to a generating set ∼ fpig 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 ffigi2I . Let α : M ! N be a surjection, i.e. the row of the following diagram with solid arrows is exact. F α~ ' α M N 0 If ' : F ! N is an R-linear map, denote '(fi) = ni 2 N. Since α is surjective, there exists an mi 2 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 ' agree on ffig 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 2 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 2 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. E 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 2 Hom(A; E) there exists g 2 Hom(B; E) with f = i∗(g) = gi; that is, 4 1.2 Injectives 1 Some Commutative Algebra E 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) 2 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: E f g 0 I R i Proof. ( =) ) is obvious by the definition of injectives. To prove ( ) = ) suppose E f 0 A B i has an exact row. Let X = f(A0; g0) j A ⊆ A0 ⊆ B and g0 extends fg. 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 2 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 2 B r A0. Define a left ideal of R by I = fr 2 R j rb 2 A0g { 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 2 A =) r − r 2 I.
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