Etale CohomologyNotes on Chapter 1 of Milne’s book

March 22, 2013

1 Theorem on Formal Functions

1.1 Inverse Systems

We begin by a short discussion on inverse systems and the Mittag-Leffler condition. Let I be a partiallyordered set.

Definition 1. An inverse system of abelian groups indexed by I is a collection of abelian groups Aii∈I and homo-morphisms φi,j : Ai → Aj , for each pair i > j satisfying

1. φi,i = 1Ai

2. For i > j > k, φi,k = φj,k φi,j

The inverse limit of an inverse system is denoted by lim←−Ai and defined to be the subgroup of Πi∈IAi ofelements of the type (ai), satisfying φi,j(ai) = aj . The inverse limit has a universal property which is easilyexpressed using the following commutative diagram

Ai

φi,j

B

ψi

66llllllllllllllllll //___

ψj

((RRRRRRRRRRRRRRRRRR lim←−Ai

<<yyyyyyyy

""EEEE

EEEE

Aj

Note that there are natural projections lim←−Ai ⊂ Πi∈IAi → Ai. For every collection B,ψii∈I as above,such that the large triangle commutes, there is a unique dotted arrow making the whole diagram commute.

Remark 2. There is an obvious notion of morphism of inverse systems indexed by I , i.e., we need to givefi : Ai → Bi so that the obvious diagrams commute. Given a morphism, there is an induced map lim←−Ai →lim←−Bi.

For the remainder of this section we restrict our attention to the case when the set I = N (naturalnumbers).

Given three inverse systems An, Bn, Cn such that for each n ∈ N there is a short exact sequence

0→ Anfn−→ Bn

gn−→ Cn → 0

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There is an induced complex of inverse limits

0→ lim←−Aif−→ lim←−Bi

g−→ lim←−Ci (1.1)

whose exactness properties we want to study.

Lemma 3. The complex (1.1) is exact.

Proof. Define A := Πn∈NAn, B := Πn∈NBn, C := Πn∈NCn. Denote by αn the map An → An−1, andsimilarly βn and γn. Thus, there is a commutative diagram

0 // An+1fn+1 //

αn+1

Bn+1gn+1 //

βn+1

Cn+1//

γn+1

0

0 // Anfn // Bn

gn // Cn // 0

Now consider the following commutative diagram,

0 // Af //

α

Bg //

β

C //

γ

0

0 // Af // B

g // C // 0

(1.2)

where the vertical arrows are given by α(ai) = (αi+1(ai+1) − ai). Similarly, define β and γ. Observe thatthe kernel of α is exactly lim←−Ai, similarly, for β and γ. Applying snake lemma, get

0→ lim←−Ai → lim←−Bi → lim←−Ci

is exact.

Definition 4. An inverse system satisfies condition ML’ if for every n and every i, j > n, we have φi,n(Ai) =φj,n(Aj).

Lemma 5. Assume that we are in the setup of lemma 3 and further assume that the inverse system Ai satisfies ML’.Then the map g in (1.1) is surjective.

Proof. Denote by in : Dn ⊂ An the image φn+1,n(An+1) and consider the following commutative diagram

0 // Dnin //

Anjn //

An/Dn//

0

0

0 // Dn−1 // An−1 // An−1/Dn−1 // 0

Because of condition ML’ the left vertical arrow is surjective and the right vertical arrow is 0. This is againa short exact sequence of inverse systems. Denote by D := Πn∈NDn and E := Πn∈NAn/Dn, and proceed asin the proof of lemma 3, then we get a commutative diagram

0 // D //

δ

A //

α

E //

ε

0

0 // D // A // E // 0

Observe from the definition of ε that ε(ei) = (εi+1(ei+1) − ei) = −(ei) since εi = 0. Since Dn → Dn−1 issurjective, the cokernel of the map δ is 0. Combining these and applying the snake lemma, get that thecokernel of α is 0. Using this and the snake lemma for diagram (1.2), get the required surjectivity.

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Lemma 6. Let An be an inverse system and let nii∈N be a strictly increasing subsequence of N. Define Bi := Ani,

then there are natural maps Bi → Ai which makes this into a morphism of inverse systems. The induced map on theinverse limit is an isomorphism.

Proof. The surjectivity lim←−Bi → lim←−Ai is clear, for any (ai), take the sequence (bi), where bi := ani . To seeinjectivity, assume that (bi) 7→ 0, which means that φni,i(bi) = 0. In particular, get that bi = φnni

,ni(bni) =0.

Definition 7. An inverse system satisfies ML if for every n, the following decreasing sequence of subgroups stabilizes

An ⊃ φn+1,n(An+1) ⊃ φn+2,n(An+2) ⊃ . . .

Theorem 8. [Mittag-Leffler condition] Suppose we are given a short exact sequence of inverse systems and assumethat the first system Ai satisfies ML, then the map g in (1.1) is surjective.

Proof. Since Ai satisfies ML, find a strictly increasing subsequence such that the system Anisatisfies ML’.

Now replace the inverse systems Ai, Bi and Ci by Ani , Bni and Cni , respectively. Using lemma 6 it sufficesto prove the assertion for the new inverse systems. Since the inverse system Ai satisfies ML’, apply lemma5.

1.2 Theorem on Formal Functions

Theorem 9. Let f : X → Y be a projective morphism of Noetherian schemes and let F be a coherent sheaf on X . Ify ∈ Y is a point, and Xn denotes the fiber over the point SpecOy/m

ny . Let Fn denote the restriction of F to this

fiber, then there is a natural isomorphism

Rif∗(F )y∼−→ lim←−H

i(Xn,Fn)

Proof. We easily reduce to the case Y = SpecA (A a local ring) and X = PnA. The LHS now becomeslim←−H

i(X,F ) ⊗A A/mn. The assertion is true for F = O(m) from the computations of cohomology forprojective spaces.

For a general cohorent sheaf write the following short exact sequence

0→ R→O(m)⊕r → F → 0 (1.3)

Since A is faithfully flat, there is a long exact sequence

· · · → Hi(R)→ Hi(O(m))⊕r→ Hi(F )→ Hi+1(R)→ Hi+1(O(m))

⊕r→ · · ·

Pulling back the exact sequence (1.3) to Xn would yield the following two short exact sequences

0→ S n → O(m)⊕rn → Fn → 0 (1.4)

0→ T n → Rn → S n → 0 (1.5)

The resulting long exact sequences fit into the following commutative diagram

Hi(R) //

α1

Hi(O(m))⊕r //

α3

Hi(F ) //

α4

Hi+1(R) //

α5

Hi+1(O(m))⊕r

α7

lim←−nHi(Xn,Rn)

α2

lim←−nHi+1(Xn,Rn)

α6

lim←−nHi(Xn,S n) // lim←−nHi(Xn,O(m)⊕rn ) // lim←−nHi(Xn,Fn) // lim←−nHi+1(Xn,S n) // lim←−nHi+1(Xn,O(m)⊕rn )

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The bottom row is exact for the following reason. For each n there is a long exact sequence in cohomologycorresponding to (1.4). These long exact sequences can be split into various short exact sequences to getshort exact sequences of inverse systems. Each term in these short exact sequences is a finite module overan Artinian ring A/mn and so these satsify the ML condition. Thus, after taking inverse limit and patchingthe short exact sequences together, we get that the bottom row is exact. For the same reason, there are shortexact sequences

We claim that the natural map

lim←−n

Hj(Xn,Rn)→ lim←−n

Hj(Xn,S n)

is an isomorphism for every j. It suffices to show that lim←−nHj(Xn,T n) = 0. For this it suffices to show that

given n there is an n′ > n such that the natural map of sheaves T n′ → T n is 0. This is a local question. Wemay cover X by finitely many affine opens and choose the maximum n′ among those. Equation (1.3) lookslike

0→ R→ E → F → 0

Equations (1.4) and (1.5) look like

0→ Sn → E/mnE → F/mnF → 0

0→ Tn → R/mnR→ Sn → 0

Applying the inverse limit functor to the above two short exact sequences preserves the exactness of both.Observing that M ∼= lim←−nM/mnM and since completion is an exact functor, we get that lim←−n Tn = 0.Consider the decreasing sequence

Tn ⊃ φn+1,n(Tn+1) ⊃ φn+2,n+1(Tn+1) ⊃ · · ·

This stabilizes and if the stable image is nonzero, then lim←−n Tn 6= 0. Thus, there is a n′ such that Tn′ → Tn is0.

Let us assume that for all integers j > i the theorem is true. Thus, α2, α5 and α6 are isomorphisms. Alsoα3, α7 are isomorphisms. A simple diagram chase shows that α4 is surjective, which also means that α1 issurjective, by replacing F by R. Now a simple diagram chase shows that α4 is injective.

Corollary 10. Let f : X → Y be a projective morphism of Noetherian schemes such that f∗(OX) = OY , then thefibers f−1(y) are connected.

Proof. Assume the fiber f−1(y) is disconnected, then we may assume that Y = SpecA with A a local ringwith maximal ideal corresponding to y. We may write the fiber as a disjoint union of two closed subsetsX ′ tX ′′. Applying the above theorem for i = 0, we get

f∗(OX)y = (OY )y = A = lim←−n

(H0(X ′n,OX′n)⊕H0(X ′′n ,OX′′

n))

which is not possible since a local ring cannot be the direct sum of two rings as the spectrum of the latter isdisconnected.

Corollary 11. Let f : X → Y be a projective morphism of Noetherian schemes, then it may be factored as X f ′

−→Y ′

f ′′

−−→ Y where f ′ is projective with connected fibers and f ′′ is a finite morphism.

Proof. We factor f as

Xf ′

−→ Spec f∗(OX)f ′′

−−→ Y

Since f is projective, it is clear that f ′′ is finite and from the previous corollary it is clear that f ′ has connectedfibers.

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Corollary 12. Let f : X → Y be a projective morphism of Noetherian schemes with finite fibers, then f is finite.

Proof. Applying the previous corollary, we reduce to the case Y = SpecA where A = Γ(X,OX) and everyfiber contains exactly one point, which means that f is a homeomorphism at the level of topological spaces.Choose a point x ∈ Y and let SpecB ⊂ X be an affine open containing f−1(x). Denote by Z := X \ SpecB,and let f ∈ A be such that x ∈ SpecAf and f(Z)∩SpecAf = ∅. Then it is clear that f−1(SpecAf ) = SpecBf .Since f is projective, this means that Bf is a finite Af module. Covering SpecA by such open sets, we getthat f is finite.

Corollary 13. Let f : X → Y be a quasi-projective morphism of Noetherian schemes with finite fibers, then f can be

factored as X f ′

−→ Y ′f ′′

−−→ Y where f ′ is an open immersion and f ′′ is a finite morphism.

Proof. We may assume that f is dominant. Find a projective closure such that f factors as X i−→ X → Y .

Now applying corollary 11 to X → Y , we get it factors as Xf ′

−→ Y ′ → Y , where X → Y ′ has connectedfibers. Let x ∈ X and denote its images in Y ′ and Y by y′ and y. Note that the fiber over y′ is connected andits intersection with the open subset X is a finite set. Let us call the fiber F , then F is a connected projectivevariety over the field k(y′) such that it has an open subset which is a finite set. The only way this is possibleis that dim(F ) = 0, in which case F is forced to be a single point. This shows that f ′−1(f ′(X)) = X . AsX → Y ′ is a dominant projective morphism, the image of an open set is open since it is constructible andclosed under generization. It follows that f ′(X) is an open subset of Y ′. The restriction

f ′ : X = f ′−1(f ′(X))→ f ′(X)

is a projective morphism with finite fibers and so it is finite. From the definition of Y ′we have that f ′∗(OX) =OY ′ and so X → Y ′ is an open immersion.

2 Regular sequences, dimension theory and regular local rings

Throughout this sectionAwill be a Noetherian local ring with maximal ideal m and residue field k := A/m.

2.1 Projective dimension

Definition 14 (Projective Dimension). Let M be an A-module. The projective dimension of M is the length of thesmallest projective resolution (possibly∞). By the length we mean the following, if 0→ Fn → Fn−1 → · · · → F0 →M → 0 is an exact complex, then its length is defined to be n.

Definition 15 (Projective Dimension of Noetherian local rings). The projective dimension of A is defined to bethe projective dimension of k.

Lemma 16. The following are equivalent

1. k has finite projective dimension equal to n ≥ 1.

2. TorAi (k, k) 6= 0 for 0 ≤ i ≤ n and TorAi (k, k) = 0 for all i > n.

Proof. 1⇒2 Let us write the projective resolution as

0→ Fn → Fn−1 → · · · → F0 → k → 0

It is clear that TorAi (k, k) = 0 for i > n. Splitting the above complex into short exact sequences we get

0→ K0 → F0 → k → 0 0→ Ki+1 → Fi+1 → Ki → 0 i ≥ 0

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From the long exact Tor sequence, one gets isomorphisms

TorAi (k, k)→ TorAi−1(K0, k)→ · · · → TorA1 (Ki−2, k)

Since TorA0 (k, k) ∼= k, it is nonzero. If TorA1 (k, k) = 0, then A = k and projective dimension would be 0,contradicting n ≥ 1. If for 2 ≤ i ≤ n, TorAi (k, k) = 0, then TorA1 (Ki−2, k) = 0 which would mean that Ki−2is free, giving a projective resolution of length i− 1 < n, again a contradiction.

2⇒1 We write a resolution

0→ Kn−1 → Fn−1 → · · · → F0 → k → 0

Arguing as above we would get that 0 = TorAn+1(k, k) ∼= TorA1 (Kn−1, k), which gives that Kn−1 is free andhence k has finite projective dimension. If there was a resolution of length smaller than n, then that wouldcontradict the hypothesis that TorAi (k, k) 6= 0 for 0 ≤ i ≤ n.

Lemma 17. Assume that k has finite projective dimension, say, n. Then every finite A-module N has projectivedimension ≤ n.

Proof. For an exact sequence 0→ K ′ → F → K → 0, with F free, from the long exact Tor sequence, we seethat for j ≥ 1, the map TorAj+1(K,M)→ TorAj (K ′,M) is an isomorphism for any A-module M . Let N be afinite A-module and consider the following resolution where the Fi are free

0→ Kn−1 → Fn−2 → · · · → F0 → N → 0

We need to show that Kn−1 is free. Splitting the above complex into short exact sequences we get

0→ K0 → F0 → N → 0 0→ Ki+1 → Fi+1 → Ki → 0 i ≥ 0

From the above one gets TorAj+1(Ki,M)→ TorAj (Ki+1,M) is an isomorphism for j ≥ 1. Composing theseisomorphisms and taking M = k, we get

TorAn+1(N, k)→ TorAn (K0, k)→ TorAn−1(K1, k)→ · · · → TorA1 (Kn−1, k)

Since A/m has projective dimension n and TorAn+1(N, k) = TorAn+1(k,N) we get TorAn+1(N, k) = 0 and soTorA1 (Kn−1, k) = 0. Since Kn−1 is a finite A-module, this means that it is free.

Lemma 18. Suppose every element in m is a zero divisor, then A does not have finite projective dimension.

Proof. Since m consists of zerodivisors, it is an associated prime. Thus, ∃a ∈ m such that Ann(a) = m. Inparticular, we have a short exact sequence

0→ k → A→ A/(a)→ 0

Assume A had finite projective dimension n := pd(A). Writing the long exact Tor sequence, we getTorAn+1(A/(a), k) → TorAn (k, k) is an isomorphism. Using lemma 17, we know that every A modulehas a porjective resolution of length ≤ n and so TorAn+1(A/(a), k) = 0. Using lemma 16 we know thatTorAn (k, k) 6= 0, which gives a contradiction.

Lemma 19. [Prime avoidance lemma] Let I be an ideal and let P1, P2, . . . , Pn be ideals such that Pi is prime fori ≥ 3. If I ⊂ ∪ni=1Pi then I ⊂ Pi for some i.

Proof. Assume that I * Pi and for i 6= j that Pi * Pj . For n = 2, there are elements ai ∈ I \ Pi. SinceI ⊂ P1 ∪ P2, we get a1 ∈ P2 and a2 ∈ P1. Suppose a1 + a2 ∈ P1, then since a2 ∈ P1, we get a1 ∈ P1, whichis a contradiction. Similarly we get a contradiction if a1 + a2 ∈ P2.

Now assume n ≥ 3 and that Pn is prime. Since I * Pn and Pi * Pn for i < n, and Pn is prime,IP1P2 . . . Pn−1 * Pn and so ∃x ∈ IP1P2 . . . Pn−1, x /∈ Pn. Choose y ∈ I, y /∈ ∪n−1i=1 Pi, then y ∈ Pn. Theelement y+ x /∈ Pn or else we would get x ∈ Pn. Thus, y+ x ∈ ∪n−1i=1 Pi, which means y+ x ∈ Pi for some i.But x ∈ Pi and so this gives y ∈ Pi, which is a contradiction.

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Lemma 20. If every element in m \m2 is a zerodivisor, then m is an associated prime.

Proof. If every element in m \m2 is a zerodivisor, we get that m ⊂ m2 ∪⋃ni=1 Pi, where Pi are the associated

primes of A. Lemma 19 shows that m ⊂ Pi or m ⊂ m2. In the latter case, one gets that m = 0 and in theformer case one gets that m is an associated prime, which means that m consists of zerodivisors.

Definition 21 (Regular Sequence). The sequence (a1, a2, . . . , an) with ai ∈ m is called a m-regular sequence if aiis not a zerodivisor in A/(a1, a2, . . . , ai−1).

Lemma 22. Suppose A has finite projective dimension equal to n. Then n = dimkm/m2. Further, the maximal ideal

is generated by a regular sequence.

Proof. Since A has finite projective dimension, using lemma 18 and lemma 20, we get that there is an a ∈m\m2 which is not a zerodivisor. Let F• be a projective resolution of k, then we have a short exact sequenceof complexes

0→ F•a−→ F• → F • → 0

In the above, F • := F• ⊗A A/(a). Writing the long exact sequence on homology we get

Hi(F•)a−→ Hi(F•)→ Hi(F •)→ Hi−1(F•)

which shows that Hi(F •) = 0 for i > 1. For i = 1, one has

0→ H1(F •)→ H0(F•)a−→ H0(F•)→ H0(F •)→ 0

Since H0(F•) = k = H0(F •) and k a−→ k is zero, we get that H1(F •) = k. We split the sequence

F 1 → F 0 → A/m→ 0

as follows

• F 1 → m/am→ 0

• 0→ (a)/am→ m/am→ m/(a)→ 0

• 0→ m/(a)→ A/(a)(= F 0)→ A/m→ 0

We claim that the second short exact sequence splits. Since a is not a zerodivisor, (a)/am ∼= A/m. Nowconsider the composite

0→ A/m→ m/am→ m/m2 ∼= a · k ⊕m/(m2 + (a))proj−−−→ a ·A/m→ 0

which is clearly the identity. Using this we see that F 1 → F 0 → A/m→ 0 is given by

F 1 → A/m⊕m/(a)proj−−−→ m/(a) → F 0 → A/m→ 0

Lifting the generator of A/m to F′1, write F 1 = F

′1 ⊕ F

′′1 . Then the following complex is exact

0→ Fn → Fn−1 → · · · → F 2 → F′1 → A/m→ 0

which shows that projective dimension of A/(a) ≤ n − 1. To show that equality holds, using lemma 16, itsuffices to show that TorA/(a)i (k, k) 6= 0 for 0 ≤ i ≤ n − 1. Observe that for TorA/(a)0 (k, k) = k. For i > 0,Tor

A/(a)i (k, k) is the homology of the complex (F

′1 being the term in degree 0)(

0→ Fn → Fn−1 → · · · → F 2 → F′1

)⊗A/(a) A/m

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These homology groups coincide with the homologies of the complex(0→ Fn → Fn−1 → · · · → F 2 → F 1

)⊗A/(a) A/m

which is the same complex as (0→ Fn → Fn−1 → · · · → F2 → F1

)⊗A A/m

Thus, we get TorA/(a)i (k, k) ∼= TorAi+1(k, k) for i > 0, which shows that projective dimension of A/(a) isn− 1. It is also clear that if we denote by m the image of m in A/(a), then dimkm/m

2 = dimkm/m2 − 1. By

induction the proof of the lemma is complete, the base case of projective dimension 0 being trivial since inthis case A is a field and so dimkm/m

2 = 0. From the proof it is also clear that m is generated by a regularsequence.

2.2 Dimension theory

Definition 23 (Krull dimension). The Krull dimension of A is defined as supn | ∃ p0 ( p1 ( · · · ( pn wherethe pi are prime ideals. It will be denoted by dim(A).

The main theorem we want to prove in this subsection is that dim(A) ≤ dimk(m/m2). This is achievedby defining the Hilbert dimension of A and comparing it with dim(A).

Let I be an ideal such that A/I is Artinian. For n ≥ 1 define

H(I, n) := l(A/In)

If I can be generated by d elements, then there is a surjection

A/I[X1, X2, . . . , Xd]⊕i≥0

Ii/Ii+1

Lemma 24. If M is a finite graded module over A/I[X1, X2, . . . , Xd] then the function l 7→ dimk(Ml) is a polyno-mial for l 0.

Proof. This can be seen by induction on the dimension of the support of M and by looking at the exactsequence

0→ K →MXd−−→M →M/XdM → 0

It follows that H(I, n) is a polynomial for n 0.

If mν ⊂ I ⊂ m, then it is clear that

H(m, n) ≤ H(I, n) ≤ H(m, νn)

and so it follows that the degree of the polynomial H(I, n) is independent of I . We will denote this degreeby d(A).

Remark 25. From the above proof it is clear that if I can be generated by d elements, then d(A) ≤ n.

Lemma 26. d(A) ≥ dim(A).

Proof. The proof is by induction on d(A). If d(A) = 0, then it is clear that the ring is Artinian and sodim(A) = 0. Let p be a minimal prime of A. We first show that d(A/p) ≥ dim(A/p). Let us denote byB := A/p and B := B/(b). We have a short exact sequence

0→ (b)/(b) ∩mn → B/mn → B/mn → 0

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By the Artin-Rees lemma there is an integer l such that for n 0 we have

(b)mn ⊂ (b) ∩mn ⊂ (b)mn−l

which shows that

dimk(B/mn) = dimk((b)/(b)mn) ≥ dimk((b)/(b) ∩mn) ≥ dimk((b)/(b)mn−l) = dimk(B/mn−l)

The above shows that dimk(B/mn) − dimk((b)/(b) ∩ mn) is a polynomial in n of degree strictly less thand(B). But then this means that d(B) < d(B) and so applying induction, we get that d(B) ≥ dim(B). Ifdim(B) > d(B), then we would get a chain of primes 0 = p0 ( p1 ( · · · ( pd(B)+1. Let 0 6= b ∈ p1, thendim(B) ≥ d(B), which is a contradiction. Since d(A) ≥ d(A/p) ≥ dim(A/p), taking supremum over all theminimal primes, we get the desired result.

Lemma 27. dim(A) ≥ d(A).

Proof. The previous lemma says that dim(A) < ∞ and so there exists a chain p0 ( p1 ( · · · ( pn = m ofmaximum length, in particular, each pi has height i. By lemma 19 p1 6⊂

⋃q, where the q are the minimal

primes of A (primes of height 0). Choose a1 ∈ p1 \⋃

q and consider the ring A1 = A/(a1). It is clear thatthis ring has dimension dim(A) − 1 and that the images of pi, i ≥ 1, are prime ideals in this ring. Choosea2 ∈ p2 \

⋃q, where the q are the minimal primes in the ring A1. Now consider the ring A2 = A1/(a2)

and proceeding in this fashion construct elements a1, a2, . . . , an such that the ring A/(a1, a2, . . . , an) hasdimension 0. This shows that rad(a1, a2, . . . , an) = m. Taking I = (a1, a2, . . . , an) and applying remark 25,we get that d(A) ≤ n.

Corollary 28. dim(A) = d(A) ≤ dimk(m/m2).

Proof. Take I = m and apply remark 25.

Corollary 29. Grm(A) is a polynomial ring iff dim(A) = d(A) = dimk(m/m2).

Proof. If d(A) = dimk(m/m2) then there is a surjection from R := k[X1, X2, . . . , Xd(A)] Grm(A). If thismap has a kernel I , then we have dimk(Id)+dimk(Grm(A)d) = dimk(Rd). For d 0, these are polynomialsand we will denote them by P1(d) := dimk(Id), P2(d) := dimk(Grm(A)d) and P (d) := dimk(Rd). Let x ∈ Ibe a nonzero homogeneous element of degree d0. Then we have inclusions

Rx→ I → R

and so P (d) ≥ P1(d) ≥ P (d− d0) for d 0, which means that P − P1 has degree strictly smaller than d(A).But since P − P1 = P2 and P , P2 have the same degree by assumption, this forces I = 0.

Conversely, suppose Grm(A) is a polynomial ring. By definition it is generated by its degree 1 ho-

mogeneous elements. For the polynomial ring k[x1, x2, . . . , xd] one has H(m, n) =

(n+ dd

)which is a

polynomial of degree d. Thus, d(A) = dimk(Grm(A)1) = dimk(m/m2).

2.3 Regular local rings

Definition 30. A Noetherian local ring A is called regular if dim(A) = dimk(m/m2).

Proposition 31. A regular local ring is a domain.

Proof. Using lemma 29 we get that Grm(A) is a polynomial ring and so is a domain. Suppose ∃x, y ∈ Asuch that xy = 0. Let x ∈ mi \ mi+1 and y ∈ mj \ mj+1. The elements x ∈ mi/mi+1 and y ∈ mi/mj+1 arenonzero elements of Grm(A) whose product is 0, which is a contradiction since Grm(A) is a domain.

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Lemma 32. A is regular iff m is generated by a regular sequence.

Proof. Assume m is generated by a regular sequence. Then m = (a1, a2, . . . , an). Since this is a regularsequence, dim(A) ≥ n. By remark 25, d(A) ≤ n. Thus, we get n ≥ d(A) = dim(A) ≥ n. It is clear thatdimk(m/m2) ≤ n. Combining this with corollary 28, we get that n = dim(A) ≤ dimk(m/m2) ≤ n.

Conversely, if A is regular, then by lifting a basis for m/m2, get that m = (a1, a2, . . . , an). Since A isa domain a1 is not a zerodivisor. We claim that A/(a1) is regular since dim(A/(a1)) = dim(A) − 1 anddimk(m/m2) = n − 1, where m denotes the image of m under the homomorphism A → A1. Thus, A1 is adomain. Proceeding in this fashion, we get that (a1, a2, . . . , an) is a regular sequence.

Theorem 33. If A has finite projective dimension iff A is a regular local ring.

Proof. From lemmas 22 and 32 it follows that if A has finite projective dimension then it is regular.

For the converse, choose a regular sequence which generates m = (a1, a2, . . . , an) and define Ai :=A/(a1, a2, . . . , ai). Consider the short exact sequence

0→ Aiai+1−−−→ Ai → Ai+1 → 0

We get the long exact sequence

→ TorAj (Ai, k)→ TorAj (Ai, k)→ TorAj (Ai+1, k)→ TorAj−1(Ai, k)→

By induction on i assume that TorAj (Ai, k) = 0 for j ≥ i+ 1. The base case for the induction being i = 1, forwhich we know that TorAj (A/(a1), k) = 0 for j ≥ 2 using the resolution 0→ A

a1−→ A→ A/(a1)→ 0. Fromthe long exact sequence we get that TorAj (Ai+1, k) = 0 for j ≥ (i+ 1) + 1. In particular, for i = n we get thatTorAn+1(k, k) = 0. This proves that A has finite projective resolution.

3 Flatness

All rings will be Noetherian in this section.

Theorem 34 (Generic Flatness). Let A be a Noetherian ring and B be a finitely generated A algebra. Let M be afinite B module, then there is a nonzero element f ∈ A such that Mf is a free Af module.

Proof. Since M is a finite B module, there is a filtration

0 = M0 ⊂M1 ⊂ . . . ⊂Mr = M

such that Mi/Mi−1 = B/pi for some prime ideal. Since Mi is free if both Mi−1 and Mi/Mi−1 are free, wereduce to the case B is a domain. If the map A → B has a kernel, then we can take f to be any nonzeroelement in the kernel and the assertion would be trivially true. Thus, we may assume A ⊂ B which meansA is a domain.

We will prove that for a domain A, a finitely generated A algebra B and a finite B module M , thereis an f ∈ A such that Mf is a free Af module. Let K denote the function field of A, then R := K ⊗A Bis a finitely generated K algebra and Supp(M) is a closed subset of SpecR. The proof is by induction ondimK(Supp(M)). We may assume that Supp(M) = SpecB by replacing B by B/Ann(M) ir required. LetK[y1, . . . , yn] ⊂ R be a Noether normalization, where yi ∈ B. Then

R = K[y1, . . . , yn][x1, . . . , xm]

where we may choose xi ∈ B. Now each xi satisfies an integral equation with coefficients in the ringK[y1, . . . , yn] and so we may find an f ∈ A such that Bf is finite over Af [y1, . . . , yn] (since B is a finitelygenerated A algebra). Now choose the largest r such that there is an inclusion

0→ Af [y1, . . . , yn]⊕r →Mf →M ′ → 0

10

Then Supp(M ′) is a proper closed subset of SpecBf and so dimK(Supp(K ⊗A M ′)) < dimK(Supp(M)).Now applying induction, we are done.

Lemma 35. Let A be a local ring and let M be a finite A module. Then M is flat iff TorA1 (M,A/m) = 0.

Proof. If M is flat, then TorA1 (M,A/m) = 0 is clear. For the converse, choose a basis miri=1 for the A/mvector space M/mM and consider the surjection

0→ K → A⊕r →M → 0 ei 7→ mi

Tensoring with A/m would yield that K/mK = 0 and so by Nakayama’s lemma, K = 0.

The theorem below is a generalisation of the above lemma.

Theorem 36 (Local criterion of flatness). Let A→ B be a local homomorphism of local rings and let M be a finiteB module such that TorA1 (M,A/J) = 0 and M/JM is a flat A/J module. Then M is A flat.

Proof. It is given to us that the following sequence is exact

0→ J ⊗AM →M →M/JM → 0

and we want to conclude that the sequence

0→ I ⊗AM →M →M/IM → 0

is exact for any ideal I of A. First let us see that TorA1 (M,N) = 0 for any A/J module N . Let

0→ N ′ → (A/J)⊕S → N → 0

be an exact sequence of A/J modules. Since M/JM is a flat A/J module, we get

0→ N ′ ⊗AM → (A/J)⊕S ⊗AM → N ⊗AM → 0

is exact. Finally writing the long exact sequence for Tor, we get that TorA1 (M,N) = 0 since TorA1 (M,A/J) =0. Given an A/Jn module N , consider the exact sequence

0→ JN → N → N/JN → 0

Since JN is an A/Jn−1 module, writing out the long exact sequences for Tor and using induction on n weconclude that TorA1 (M,N) = 0. Using this we get an exact sequence

0→ (I ∩ Jn)⊗AM → I ⊗AM → (I/I ∩ Jn)⊗AM → 0

andI ⊗AM //

M //

M/IM //

0

0 // (I/I ∩ Jn)⊗AM // A/Jn ⊗AM // A/(I + Jn)⊗AM // 0

If α ∈ I ⊗AM maps to 0 in M , then we get that α ∈⋂n≥1(I ∩ Jn) ⊗AM , which by the Artin-Rees lemma

is⋂n≥1(JnI)⊗AM . Inside I ⊗AM we have an equality⋂

n≥1

(JnI)⊗AM =⋂n≥1

I ⊗AMJn

and by the Krull intersection theorem,⋂n≥1

I ⊗AMJn =⋂n≥1

(I ⊗AM)Jn ⊂⋂n≥1

(I ⊗AM)mnB = 0

11

Corollary 37. Let A → B be a local homomorphism of local rings and let B be flat over A. Let b ∈ B be such thatthe image b ∈ B/mB is not 0 or a zerodivisor. Then B/(b) is a flat A algebra.

Proof. We first show that the assumption above implies that b is not a zerodivisor in B. Consider thefollowing diagram in which the rows are exact because B is flat.

0 // mi+1 ⊗A B //

b

mi ⊗A B //

b

mi/mi+1 ⊗A B //

b

0

0 // mi+1 ⊗A B // mi ⊗A B // mi/mi+1 ⊗A B // 0

As mi/mi+1 ⊗A B ∼= (B/mB)⊕dimk(mi/mi+1) the kernel of the right vertical arrow is 0 for every i ≥ 0. This

means that the kernel of the middle vertical arrow is contained in the kernel of the left vertical arrow forevery i ≥ 0. Since B is flat, mi ⊗A B ∼= miB, and we conclude that the kernel of B b→ B is contained in∩i≥0miB = 0.

We need to show that TorA1 (B/(b), A/m) = 0. This is clear by tensoring the following exact sequencewith A/m

0→ Bb→ B → B/(b)→ 0

Corollary 38. Let A → B be a local homomorphism of local rings. Let M be a finite B module and let t ∈ A be anon zerodivisor which is not a unit. Then M is flat over A iff

1. t is not a zerodivisor in M

2. M/tM is flat over A/t

Proof. If M is flat over A, then the two conditions are clear. Consider the short exact sequence

0→ m/(t)→ A/(t)→ A/m→ 0 (3.1)

Tensoring this with M/t over A/t, we get

0→ m/(t)⊗AM →M/t→M/m→ 0

Tensoring (3.1) with M over A and comparing with the above sequence, we get

TorA1 (M,A/t) TorA1 (M,A/m)

Thus, it suffices to show that TorA1 (M,A/t) = 0. But this is clear since t is not a zerodivisor in M .

Corollary 39. Let f : X → Y be a morphism of finite type of schemes such that X is Cohen-Macaulay and Y isregular and every fiber has dimension dim(X)− dim(Y ). Then f is flat.

Proof. The idea is to use the previous corollary inductively. Let x ∈ X be a closed point, then its imagey = f(x) is a closed point of y. It suffices to show that OX,x is flat over OY,y . Let (a1, . . . , ar) = mybe a regular sequence. Using Hartshorne, Chapter 2, Theorem 8.21A (c), and the hypothesis on the fiberdimension, we get that (a1, . . . , ar) is a regular sequence in OX,x, and applying the previous corollaryinductively, we get that f is flat.

Corollary 40. Let Z f→ Yg→ X be morphisms of schemes which are of finite type over X . Assume g and gf are flat.

Then f is flat if for each closed point x ∈ X , Zx → Yx is flat.

12

Proof. We may work with affine schemes and so let A→ B → C be homomorphisms such that B and C areflat over A. Let η be a closed point of C, then m = η ∩ A is a closed point of A and it suffices to show thatCη is flat over Bτ , where τ = ηc. So we reduce to the case of local homomorphisms of local rings. We needto show that TorB1 (C,B/τ) = 0 and we are given that TorB/m1 (C/m, B/τ) = 0.

Suppose α ∈ τ ⊗B C is in the kernel of τ ⊗B C → C. We have a commutative diagram

τ ⊗B C //

C //

C/τ //

0

0 // τ/m⊗B C // C/m // C/τ // 0

The bottom row is exact since it may be written as 0→ τ/m⊗B/m C/m→ C/m→ C/τ → 0 and since C/τis flat over B/m. Thus, α is in the image of mB ⊗B C = m ⊗A C. Now consider the commutative diagramin which the top row is exact since C is a flat A algebra

0 // m⊗A C //

C // C/m // 0

τ ⊗B C // C

which shows that α = 0.

Lemma 41. Let X be a Noetherian topological space and let U be a subset which satisfies

1. It is closed under generization

2. If p ∈ U , then there is an open subset V of p such that V ⊂ U .

Then U is open in X .

Proof. We will show that Y := X \ U is a closed subset. Let Y = ∪ni=1Yi be the irreducible components.If Y is not closed, then there is a point u ∈ U such that u ∈ Y , and since U is closed under generization,we get that the generic point of one of the Yi is contained in U , assume y1 ∈ U . Using condition 2, thereis an open subset V1 of Y1 such that V1 ⊂ U . Further, by choosing a smaller subset if necessary, we mayassume that V1 ∩ ∪i≥2Yi = ∅. Then, V1 is an open subset of Y . Since Y is dense in Y , V1 ∩ Y 6= ∅, which is acontradiction.

Corollary 42. Let φ : X → Y be a morphism of finite type. The subset U := x ∈ X|OX,x is flat over OY,f(x) isopen in X .

Proof. We may reduce to the case X = SpecB and Y = SpecA and a ring homomorphism φ : A → B. Wewill show that the set U has the two properties in lemma 41. It is clear that U is closed under generization.Let x ∈ U , then it corresponds to a prime ideal q ⊂ B and B/qcB is a finitely generated A/qc algebra. Thus,there is a nonzero element f ∈ A/qc such that Bf/q

cB is flat over Af/qc. Since f /∈ qc = φ−1(q) ∩ A, we

have that φ(f) /∈ q. Thus, we may replace A by Af and B by Bf and we have that B/qcB is a flat A/qc

module. We have the following exact sequence

0→ TorA1 (B,A/qc)→ qc ⊗A B → B → B/qcB → 0

Tensoring this withBq yields TorA1 (B,A/qc)⊗BBq = 0 and so there is an s /∈ q such that TorA1 (Bs, A/qc) =

0. If p is a prime ideal in q∩SpecBs, then we get TorApc

1 (Bp, Apc/qc) = 0. Now we apply the local criterionfor flatness and we are done since Bp/q

cBp is a flat Apc/qc module. This proves the second condition inlemma 41, which completes the proof of the corollary.

13

Proposition 43. Let A→ B be a faithfully flat extension. Then A is a domain (respectively, normal) if B is.

Proof. If B is a domain, then since A → B is an inclusion (as a consequence of being faithfully flat), we getthat A is a domain.

Assume that B is normal, in particular it is a domain (and so is A). Suppose a/s ∈ K(A) is integral overA, then since it is an element of K(B) which is integral over B, we get that a/s ∈ B. We have in B ⊗A Bthat

s(a/s⊗A 1− 1⊗A a/s

)= 0

Since B ⊗A B is a faithfully flat extension of B, the element s ∈ B cannot be a 0 divisor. This shows that

a/s⊗A 1 = 1⊗A a/s

and so we have that a/s ∈ A.

Proposition 44. Let A→ B be a faithfully flat extension. Then for any ideal I ⊂ A, Iec = I .

Proof. Since A→ B is flat, it is clear that for any ideal I , the natural map I ⊗A B∼→ IB is an isomorphism.

Tensoring the following short exact sequence with B

0→ I → Iec → Iec/I → 0

and observing that IecB = IB, we get that desired result.

Proposition 45. Let A→ B be a faithfully flat extension of local rings. Then A is regular if B is regular.

Proof. From the proof of lemma 16 it suffices to show that TorAi (k, k) = 0 for some i > 0. Since B is afaithfully flat extension, TorAi (k, k) ⊗A B = TorBi (B ⊗A k,B ⊗A k) and since B is regular, by theorem 33,∃i > 0 such that this is 0.

Proposition 46. Let SpecB → SpecA be a faithfully flat morphism of finite type. Then ∃ a scheme X , a faithfullyflat quasi-finite morphism X → SpecA and a SpecA morphism X → SpecB.

Proof. If A → B is quasi-finite, then there is nothing to be done since we may take A = SpecB. So assumethat dim(B/mB) is positive for the fiber over the closed point given by the maximal ideal m, and so forall fibers (since fiber dimension does not change for a flat family). Let m be a maximal ideal of A. ThenAm → Bm is faithfully flat of finite type. Since Bm/mBm is a finitely generated algebra over k = Am/mAm

of positive dimension, there are only finitely many associated primes and so we can find b ∈ B such thatb is not a unit or a zerodivisor in Bm/mBm. Applying corollary 37 we get that Am → Bm/(b) is faithfullyflat. Repeating this we can find a sequence (b1, b2, . . . , br) such that Am → Bm/(b1, b2, . . . , br) is faithfullyflat and quasi-finite.

Consider the homomorphism A→ B/(b1, b2, . . . , br). The points of B/(b1, b2, . . . , br) which are flat overA form an open set U whose image contains m. Let g ∈ A be such that SpecAg is contained in the image ofU and contains m. Then we have

π−1(SpecAg) ∩ U //

U // SpecB/(b1, b2, . . . , br)

π

SpecAg

// SpecA

It is clear that π−1(SpecAg) ∩ U → SpecAg is faithfully flat quasi-finite whose image is open and containsm. Let us denote this morphism by Xm. Putting everything together we have a diagram

SpecB/(b1, b2, . . . , br) // SpecB

Xm

//

66nnnnnnnnnnnnnSpecA

14

The map Xm → SpecA is flat, of finite type and quasi-finite. In particular, its image is open and contains m.Covering SpecA by finitely many such Xm, we take X = tni=1Xmi

and then X → SpecA is also faithfullyflat.

4 Smooth morphisms

Definition 47 (Relative dimension). Let f : SpecB → SpecA be a morphism of finite type. The relative dimensionof f is defined as the minimum among the fiber dimensions. (We mean fiber over a geometric point.)

Definition 48 (Smooth morphism). In the above setting, f is called smooth of relative dimension l (l being therelative dimension as defined above) if it is flat and ΩB/A is a locally free B module of rank l.

Theorem 49. f is smooth of relative dimension l in a neighbourhood of p ∈ SpecB (a closed point) iff there is aneighbourhood around p of the type A[x1, x2, . . . , xn+l]g/(P1, P2, . . . , Pn) and the matrix (∂Pi/∂xj)(p) has rank n.Here g denotes an element of B and we are inverting it.

Proof. Assume that there is a neighbourhood of p of the type A[x1, x2, . . . , xn+l]g/(P1, P2, . . . , Pn) wherel is the relative dimension and the matrix (∂Pi/∂xj)(p) has rank n. Denote by C := A[x1, x2, . . . , xn],I := (P1, P2, . . . , Pn) and q := f(p). Tensoring

I/I2d→ ΩC/A ⊗C C/I → ΩB/A → 0

with ⊗Bk(p), we get an exact sequence

(I/I2)⊗B k(p)d→ ΩC/A ⊗C C/I ⊗B k(p)→ ΩB/A ⊗B k(p)→ 0

Let us denote the images of the generators of I/I2 in ΩC/A by v1, v2, . . . , vn. The condition on rank showsthat the vi are linearly independent in ΩC/A⊗C k(p). Thus, we may extend these to a basis for ΩC/A⊗C k(p)

and then lift this to a free basis for ΩC/A ⊗C Cmp. Thus, we may write ΩC/A ⊗C Cmp

/I = ⊕n+li=1Bmp· vi,

where the first n generate the image of I/I2. This clearly shows that (ΩB/A)mpis free of rank l and since

(I/I2)mp is generated by n elements and surjects onto a free module of rank n, it is free, and so both theseare free in some neighbourhood of p, which looks like B = A[x1, x2, . . . , xn+l]g/(P1, P2, . . . , Pn). To showthat Bmp

is flat over Amq, we use corollary 37. Since the local ring Bmp

/mq has dimension atleast l andΩ(Bmp/mq)/k(q) is locally free of rank l, we get that this ring is regular. This clearly shows that P i is not azerodivisor in k(q)[x1, x2, . . . , xn]mp

/(P 1, P 2, . . . , P i−1). Applying corollary 37 repeatedly, we get that Bmp

is flat over Amq.

Assume that f is smooth of relative dimension l. As f is of finite type, we may write B = C/I whereC := A[x1, x2, . . . , xr]. Let us denote by X := SpecB, Y := SpecA, then the fiber Xq := SpecB ⊗A k(q)has dimension atleast l and ΩXq/k(q) is locally free of rank l. This forces that the local ring OXq,p is regularsince the rank of ΩXq/k(q) (in case it is locally free) is always larger than the dimension of the variety. Letus denote by R the ring C/m = k(q)[x1, x2, . . . , xr] and the maximal ideal in R corresponding to the pointp by mp. Then Rmp is a regular local ring with dimk(p)(mp/m

2p) = r and Rmp/I is a regular local ring with

dimk(p)(mp/m2p+I) = l. This shows that we can find a1, a2, . . . , ar ∈ R which generate mp such that the first

r−l of these are in I . Let J be the ideal generated by a1, . . . , ar−l. ThenRmp/J is regular local of dimension l

and it admits a surjective homomorphism to Rmp/I which is also regular local of the same dimension. This

forces that J = I . By abuse of notation, we denote by mp the maximal ideal in C corresponding to the pointp. Then Cmp

/m = Rmpand we have just showed that I is generated by r − l elements. Choose lifts of these

elements in I and denote them by Pi. We will show that Pi generate I . Denote by J the ideal generated byPi. Tensoring the following short exact sequence with A/m and using the flatness of Cmp

/I = Bmp

0→ Imp/Jmp

→ Cmp/J → Cmp

/I → 0

15

we get that Imp= Jmp

+mImp. Using Nakayama’s lemma get that Imp

is generated by r− l elements and soit is generated in a neighbourhood by these elements. It only remains to show that the matrix (∂Pi/∂xj)(p)is of rank r − l. This can be seen using the exact sequence

I/I2d→ ΩC/A ⊗C C/I → ΩB/A → 0 (4.1)

Since ΩC/A ⊗C C/I is locally free of rank r and ΩB/A is locally free of rank l, the kernel is locally free ofrank r− l. Since I is generated by r− l elements, this forces that I/I2 is locally free and the above sequenceis a short exact sequence of locally free sheaves. The image of Pi under the differential is

∑j(∂Pi/∂xj)dxj .

From this it is clear that the rank of the matrix is r − l.

Theorem 50. Let A be a local ring with closed point q and let B be a finitely generated A-algebra which is flat andsuch that B := B/mq is smooth over A/mq . Then B is smooth over A.

Proof. Let p be a closed point in SpecB. Let us write B = C/I , where C = A[x1, x2, . . . , xn]. Then

Bmp = (k(q)[x1, x2, . . . , xn]/I)mp

We can find elements P1, P2, . . . , Pl ∈ I such that their images generate I in (k(q)[x1, x2, . . . , xn])mpas Bmp

is regular. Then the dimension of Bmp is n− l. If J denotes the ideal in C generated by the Pi, then we havean exact sequence

0→ Imp/Jmp

→ Cmp/J → Cmp

/I → 0

Tensoring the above short exact sequence with A/mq and using the flatness of Cmp/I = Bmp we get thatImp

= Jmp+ mqImp

. Using Nakayama’s lemma get that Impis generated by n − l elements and so it is

generated in a neighbourhood by these elements. Thus,

Bmp= (A[x1, x2, . . . , xn]/(P1, P2, . . . , Pl))mp

Let R be a local ring with maximal ideal m and suppose we have an exact sequence

R⊕lh−→ R⊕n →M → 0

such that M ⊗R R/m has dimension n− l, then this forces a short exact sequence

0→ R/m⊕lh−→ R/m⊕n →M ⊗R R/m→ 0

which would mean that M is free and that h is injective.

We apply the above toB⊕lmp

→ ΩCmp/A⊗C C/I → ΩBmp/A

→ 0

where the arrow on the left is the surjection B⊕lmp→ Imp

/I2mpfollowed by the natural map. Using the fact

that ΩBmp/A⊗Bmp

Bmp/mp is free of rank n − l since Bmp

is smooth over A/mq , we get that ΩBmp/Ais free

and that we have a short exact sequence of locally free sheaves

0→ Imp/I2mp

→ ΩCmp/A⊗C C/I → ΩBmp/A

→ 0

This shows that the matrix (∂Pi/∂xj)(p) has full rank and now use theorem 49.

5 Etale morphisms

Definition 51. A morphism X → Y is called unramified if it is of finite type and ΩX/Y = 0.

16

Theorem 52. [Structure theorem for unramified maps] Let A → B be of finite type and unramified. Let p be amaximal ideal in B and let q denote its image in SpecA. Then there is a neighbourhood of p of the type SpecC/I ,where C := A[T ]g/(P (T )) and P ′(T ) is a unit in C.

Proof. Let us first assume that A is a local ring with maximal ideal mq . Let B denote the ring B/mq , thensince ΩB/A = 0, we get that ΩB/k(q) = 0, which means thatB is a product of finite separable field extensionsk(q). By the Stein factorisation theorem, we get that there is a R which is a finite A-algebra such thatSpecB → SpecR is an open immersion.

Let mp denote the maximal ideal in R corresponding to the point p ∈ SpecB. Consider the surjectivemap R → R/mq . As R/mq contains R/mp as a factor and since k(p) is a finite separable extension of k(q),we can find an α ∈ R such that α generates k(p) over k(q) and α is in all the maximal ideals other thanmp. In other words, R/mq is a product of Artin local rings and the image of α corresponds to the element(0, 0, .., α, ..., 0). Consider the ring homomorphism A[T ]→ R given by T 7→ α. Let I denote its ideal. Thenthere is an inclusion A[T ]/I → R. Since R is a finite A module, this shows that A[T ]/I is a finite A moduleand that R is finite over A[T ]/I . Localizing at T−1 we get that there is an inclusion A[T, T−1]/I → R[α−1]and that R[α−1] is a finite A[T, T−1]/I module. Note that R[α−1] is a local ring with maximal ideal mp andsoA[T, T−1]/I is also a local ring. Also note thatR[α−1]

∼−→ B[α−1]. If we tensorR[α−1] with⊗AA/mq , thenwe would get k(p), which shows that the map A[T, T−1]/I → R[α−1] is surjective, since k(p) is generatedover k(q) by α. Hence, B[α−1], which is a finite type and local A algebra and contains the maximal idealmp, is isomorphic to A[T, T−1]/I .

Tensoring A[T ]/I with ⊗AA/mq , we get that I = (P (T )). If n is the degree of P (T ), then a minimal setof generators for the A module A[T ]/I is 1, T, ..., Tn−1, which shows that there is a monic polynomial P (T )of degree n in I such that it reduces to P (T ). Let C := A[T, T−1]/P (T ). We want to check that C is a localring. Thus, consider the ring

C/mq = k(q)[T, T−1]/P (T ) = A[T, T−1]/I ⊗A A/mq = B[α−1]/mq

which is local. If we denote by C := C/mq , then since ΩB/A = 0, we get that ΩC/k(q) = 0, which means that

C is a separable extension of k(q) and so P′(T ) is a unit in C, which means that P ′(T ) is a unit in C.

For the general case, we proceed as follows. From the above, there is an element α ∈ Bmqsuch that

α /∈ mp and there is a surjection Amq [T, T−1]/(P (T )) → Bmq [α−1]. Since B is finitely generated over A, bylooking at the generators, we can find an s ∈ A \ mq such that there is a surjection As[T, T

−1]/(P (T )) →Bs[α

−1] and such that P ′(T ) is a unit in As[T, T−1]/(P (T )).

Definition 53 (Etale morphisms). A morphism X → Y is called etale if it is of finite type, flat and unramified.

Theorem 54. [Structure theorem for etale maps] Let A→ B be of finite type and etale. Let p be a maximal ideal in Band let q denote its image in SpecA. Then there is a neighbourhood of p of the type C := A[T ]g/(P (T )) and P ′(T )is a unit in C.

Proof. Since an etale morphism is unramified, using the proof of theorem 52, we get there is a neighbour-hood of p of the type SpecC/I , with C = A[T ]g/(P (T )) and P ′(T ) a unit in C and such that the map

C ⊗A A/mqf−→ C/I ⊗A A/mq is an isomorphism. Consider the short exact sequence

0→ (I/(P (T )))mq→ Cmq

→ (C/I)mq→ 0

Since C/I is flat over A and f is an isomorphism, we get that (I/(P (T )))mq= 0 after tensoring the above

with ⊗AA/mq . If s /∈ mq is such that (I/(P (T )))s = 0, then SpecCs is the required neighbourhood.

Definition 55 (Normal ring). A Noetherian ringA is called normal if for every prime p, the ringAp is an integrallyclosed domain.

Definition 56 (Depth). LetA be a local ring. The depth ofA is defined as the length of the largest m regular sequence(see definition 21) in A.

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Theorem 57 (Serre’s Criterion for normality). Consider the following conditions

1. (Ri) Ap is regular for every p ∈ SpecA with ht(p) ≤ i.

2. (Si) depth(Ap) ≥ min(ht(p), i) for every p ∈ SpecA.

A Noetherian ring is normal if and only if it satisfies R1 and S2.

Proof. First assume A is normal. If p is a prime ideal of height 0, then Ap is a domain and so a field and soregular. If p is a prime ideal of height 1, then Ap being integrally closed is a dvr and so is regular. Thus, R1

holds. The condition (S2) is clear for prime ideals of height 0 and 1. Let p be a prime of height ≥ 2. Welocalize at p and abuse notation to denote the ring byA and the maximal ideal by m. SinceA is an integrallyclosed domain of dimension ≥ 2, there is a chain of prime ideals 0 ( p ( m. Assume depth(A) = 1 and let0 6= t1 ∈ p and consider the ring A/(t1). By the condition on depth, we get that m is an associated prime.Thus, ∃t2 ∈ A such that (t1 : t2) = m. Let α := t2

t1. By definition, for any a ∈ m, aα ∈ A. Then there are two

possibilities

• α : m→ m

• α : m→ A

In case of the latter, we get a proper inclusion (t1) ( (t2) while in case of the former we get that α is integralover A and so in A, which is a contradiction. Proceeding in the same fashion with t2, we get t3 and so on,contradicting the Noetherian hypothesis.

Conversely assume that A is a local ring satisfying R1 and S2, we need to show that it is an integrallyclosed domain. Let p be an associated prime, then there is an inclusion A/p → A. Localizing at p, we getthe maximal ideal is an associated prime in the ring Ap, i.e., depth(Ap) = 0. Because of S2, this shows thatht(p) = 0. Localizing at each of these minimal primes and using R1, we get that A is reduced. Let f ∈ A bea regular element (i.e. not a zerodivisor) and suppose that A/fA has an embedded prime p. Then ht(p) ≥ 2and in the ring Ap ∃f1 such that (f : f1) = pAp. If x ∈ pAp is a regular element, then

(xf : xf1) = (xf : yf) = (x : y) = pAp

which shows that depth(Ap) ≤ 1 contradicting S2. Thus, A/fA has no embedded primes and so all theassociated primes of (f) are minimal and of height 1 in A. Inverting all the regular elements, there is aninclusion

A → ΠiKi

We claim that A is integrally closed in ΠiKi, i.e., if any element a/f ∈ ΠiKi satisfies a monic polynomial,then a/f ∈ A. Let I = x ∈ A |xa/f ∈ A, then (f) ⊂ I ⊂ A and so

0→ I/(f)→ A/(f)→ A/I → 0

For a short exact sequence of A-modules

0→ N →M →M/N → 0

one has Ass(N) ⊂ Ass(M) ⊂ Ass(N) ∪ Ass(M/N). The first inclusion is clear by the definition of anassociated prime (∃ an inclusion A/p → M ). For the second inclusion, suppose A/p → M → M/N has akernel, say a 7→ 0. Then a /∈ p and if 1 7→ x ∈ M , we get ax ∈ N . It is easily checked that the annhilatorof ax is p. In particular, we get that Ass(A/I) consists of prime ideals of height 0 in A/(f) or prime idealsof height 1 in A. For any prime of height 1 we have that (A/I)p = 0 (because of R1), which shows thatA/I = 0 or I = A, i.e., a/b ∈ A. Since the elements ei ∈ ΠiKi satisfy e2i − ei = 0, we get that ei ∈ A and soA = ×iA/pi. As A is local, we get that A is a domain and integrally closed.

Proposition 58. Let f : Y → X be etale. Then

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1. dim(OX,x) = dim(OY,y)

2. X is normal then Y is normal

3. X is regular then Y is regular

4. X is reduced then Y is reduced

Proof. (1) Clear since locally every etale map is standard etale.

(3) Since my = mxOY,y the generators for mx will generate my and as dimensions are same, we get Y isregular.

(2) We need to show that OY,y satisfies R1 and S2 (Serre’s criterion). Let p be a prime of height 1 in OY,y andlet q = p∩OX,x. Then OY,p is etale over OX,q. Since the latter is regular, using part (3) we get OY,p is regular.Now let p be a prime on OY,y of height ≥ 2 and let q be as above. Choose a regular sequence of length 2 inOX,q. The lift of this to OY,p gives a regular sequence of length 2 as etale is flat and preserved under basechange (OY,p/a is etale over OX,q/a).

(4) OX,x is reduced if and only if it satisfies R0 and S1. We reason as in the part (2).

Theorem 59 (Etale local sections of smooth maps). Let f : Y → X be a smooth map. Then we can find asurjective etale map X ′ → X which factors through f .

Proof. The above existence theorem is local on the base and so using theorem 49 we may assume thatX = SpecA and Y = SpecB, where B = A[x1, x2, . . . , xn]g/(P1, P2, . . . , Pl), with det(∂Pi/∂xj)1≤i,j≤l a unitin B. By introducing a variable xn+1 and going modulo 1− xn+1g, it is easy to check that we may write Bas A[x1, x2, . . . , xn]/(P1, P2, . . . , Pl), with det(∂Pi/∂xj)1≤i,j≤l a unit in B. We claim that the map

A[xl+1, xl+2, . . . , xn]→ A[xl+1, xl+2, . . . , xn][x1, x2, . . . , xl]/(P1, P2, . . . , Pl)

is etale. Let us denote by R the ring A[xl+1, xl+2, . . . , xn]. By theorem 49, it suffices to check that

det(∂Pi/∂xj)1≤i,j≤l

a unit in B, where we are taking R-linear partial derivatives. But since the R-linear partial derivative(∂Pi/∂xj) in B is same as the A-linear partial derivative, we are done, as by assumption this determinantis a unit in B. Thus, we have factored the map A → B as A → A[xl+1, x2, . . . , xn] → B, the first being thepolynomial ring over A and the second an etale map.

Now we proceed by considering two cases. Let x ∈ SpecA be a closed point and denote its maximalideal by mx. First consider the case where the residue field of x, which we denote by k, is a finite field. Thering B/mx = k[x1, x2, . . . , xn]/(P 1, P 2, . . . , P l) is smooth over k and so is regular. Let y be a closed point inSpecB which maps to x and denote its maximal ideal by my . Then we can find elements P l+1, P l+2, . . . , Pnin B/mx such that these generate the maximal ideal in Bmy

/mx. Since every finite extension of a finitefield is separable, we get that the determinant det(∂P i/∂xj)1≤i,j≤n is nonzero at the point y. Consider thequotient

B B/(Pl+1, . . . , Pn) =: B1

There are neighbourhoods of x ∈ U and y ∈ V ⊂ SpecB1 such that V → U is etale since det(∂Pi/∂xj)1≤i,j≤nis nonzero at y.

Now consider the case where the residue field of x is infinite. Since k[xl+1, x2, . . . , xn]→ B/mx is etale,we get that the image of SpecB/mx → Spec k[xl+1, x2, . . . , xn] is an open subset. Since k is infinite, thisopen subset contains k-points. This means that there are elements al+1, . . . , an ∈ A such that the idealgenerated by (xl+1 − al+1, . . . , xn − an) in B/mx is a proper ideal. Define Pi = xi − ai for l + 1 ≤ i ≤ n.We have just seen that the ideal generated by (P1, P2, . . . , Pn) is not the unit ideal in B. Moreover, sinceA[xl+1, x2, . . . , xn]→ B is etale, we get that the quotient

B B/(Pl+1, . . . , Pn) =: B1

is etale over A since det(∂Pi/∂xj)1≤i,j≤n is a unit in B1.

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6 Formal properties

Theorem 60 (Cohen). Let k be a field and let R be a complete local k-algebra such that R/m is a finite separableextension of k. Then we can find a field K ⊂ R such that K → R/m is an isomorphism.

Proof. Let us denote by K := R/m. To give a map from K → R, it suffices to give compatible mapsK → R/mi. Suppose we have defined such maps for i ≤ n. Let us write R/m = k[T ]/(f(T )), for someseparable polynomial f which is irreducible over k. By assumption, we have found a lift of yn ∈ R/mn

such that f(yn) = 0, such that yn 7→ T . Here f(T ) is being treated as an element of R[T ] since k → R.Let y′n+1 ∈ R/mn+1 be a lift of yn. Then f(y′n+1) ∈ mn. Let ε ∈ mn so that ε2 = 0. Then f(y′n+1 + ε) =f(y′n+1) + εf ′(y′n+1). Since the image of f ′(y′n+1) in R/m is a unit, it is a unit in R/mn+1. Thus, if we setε := −f(y′n+1)/f ′(y′n+1), then ε ∈ mn and if we set yn+1 = y′n+1 + ε, then f(yn+1) = 0. Thus, we get a mapK → R/mn+1 which is compatible with the map K → R/mn.

Lemma 61. Let R be an Artin local ring and let x1, x2, . . . , xn be elements generating the maximal ideal m, wheren = dimk(m/m2). Suppose yi ∈ m2, then the ideal I := (x1 + y1, x2 + y2, . . . , xn + yn) is equal to m.

Proof. Consider the R module M = m/I . Then M ⊗R R/m = m/(I + m2). Since I + m2 ⊃ m, we applyNakayama to get that m = I in R.

Lemma 62. Let R be an Artin local ring with maximal ideal m and let A be a local ring with a surjective mapφ : A→ R/m2. Then any lift f : A→ R/mj of φ is surjective for every j > n.

Proof. Let f : A → R/mj be a lift. Let n = dimk(m/m2) and choose generators x1, x2, . . . , xn for m. Thensince R/mj is Artin local, it is a finite A module. Moreover, the image of mA (the maximal ideal of A) underf contains elements of the type xi + yi where yi ∈ m2. Tensor the following short exact sequence with⊗AA/mA

A→ R/mj →M → 0

and use the fact that R/(mj + f(mA)) = R/m (using lemma 61) to get that M/mAM = 0, which shows thatM = 0.

Lemma 63 (Complete Nakayama). Let A be a complete local ring with maximal ideal m and residue field k. Let Mbe a A module such that

⋂i≥0 m

iM = 0 and M/mM is a finite dimensional k vector space generated by mi. ThenM is generated by mi as an A module.

Proof. Let us denote by N the submodule generated by the mi’s. Given an element β0 ∈M , we can write itas

β0 = (

n∑i=1

ai,0mi) + α1β1 ai,0 ∈ A,α1 ∈ m, β1 ∈M

We repeat this process inductively to get

βj = (

n∑i=1

ai,jmi) + αj+1βj+1 ai,j ∈ A,αj+1 ∈ m, βj+1 ∈M

Put α0 = 1 and define ci ∈ A by ci :=∑i≥0 ai,j(Π0≤l≤jαl). The ci converge since A is complete and the

coefficient Π0≤l≤jαl ∈ mj . Then the element

β0 −n∑i=1

cimi ∈⋂i≥0

miM

Since the intersection on the right is 0, we get that β0 =∑ni=1 cimi.

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Lemma 64. Let A be a complete local ring and let A → R be a finite map. Then R is a product of complete localrings.

Proof. Denote by m the maximal ideal of A and by k the residue field A/m. We shall sketch a proof here. Letr ∈ R be an element which is not in the image of A. Then r satisfies a monic polynomial with coefficientsin A. Thus, the map A→ R factors as

A→ A[T ]/f(T )→ R

SinceA is a complete local ring, by Hensel’s lemma, the ringA[T ]/f(T ) is a product of complete local rings,depending on how f factors modulo m. Hence, replacing A by the factors in A[T ]/f(T ) and proceeding inthis fashion, yields that R is a product of complete local rings.

Definition 65 (Formally unramified, formally etale, formally smooth). Let f : Y → X be a morphism of finitetype. We say that f is formally unramified (respectively, etale or smooth) if for every affine scheme X ′ → X anda closed subscheme of X ′0 defined by a nilpotent ideal X ′0 → X ′, the set map HomX(X ′, Y ) → HomX(X ′0, Y ) isinjective (respectively, bijective or surjective).

Theorem 66. Let f : Y → X be a morphism of finite type, then f is formally unramified (respectively, etale orsmooth) iff f is unramified (respectively, etale or smooth).

Proof. • Formally unramified implies unramified. Let y ∈ Y be a closed point and let x := f(y). We will firstthat k(y) is a separable extension of k(x) and then show that mxOY,y = my . Assume that k(y) is not aseparable extension of k(x), then we can find an k(x) ⊂ E ⊂ k(y) such that k(y) = E[T ]/(f(T )) such thatf(T ) = T p − α for some α ∈ E. We define two distinct E-algebra homomorphisms k(y) → k(y)[ε]/ε2 bysending T 7→ T and T 7→ T+ε, both of which become equal on putting ε = 0. This contradicts the conditionon formally unramified in the following diagram

SpecOY,y

Spec k(y) // Spec k(y)[ε]/ε2 // SpecOX,x

Next we will show that mxOY,y = my . By Nakayama’s lemma applied to the OY,y module my/(m2y + mx), it

would suffice to show that this module is 0. Assume that my/(m2y + mx) 6= 0. Let R denote the Artin local

ring OY,y/(m2y+mx). Using Lemma 60 there is an isomorphism of rings OY,y/m2

y∼= k(y)⊕my/m

2y . Similarly,

there is an isomorphism of rings R ∼= k(y)⊕my/(m2y + mx). If my/(m2

y + mx) 6= 0, then it is clear that thereare at least 2 distinct ring homomorphisms OY,y/m2

y → R, given by (x, y) 7→ (x, 0) and (x, y) 7→ (x, xv),where 0 6= v ∈ (m2

y + mx), both of which become equal on composing with R → R/m. This contradicts thecondition on formally unramified in the following diagram

SpecOY,y

SpecR/m // SpecR // SpecOX,x

• Unramified implies formally unramified. If I is a nilpotent ideal, let n be the largest integer such thatIn 6= 0. By considering the chain A → A/In → A/In−1 → · · · , it suffices to assume that I2 = 0 and provethat any two morphisms SpecA → Y which become equal when composed with SpecA/I → SpecA, areequal to start with.

Using the structure theorem for unramified maps, we reduce to the case

Spec (B[T ]g/(P (T )))/J

SpecA/I // SpecA // SpecB

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Suppose we have two mapsB[T ]g/(P (T ))→ Awhich become equal when modulo I . This means that thereis an element α ∈ A and x ∈ I such that both α and α + x satisfy P (T ). But P (α + x) = P (α) + xP ′(α) =xP ′(α). Since P ′(T ) is a unit, this shows that xP ′(α) 6= 0, which is a contradiction.

• Smooth implies formally smooth. ??

• Formally smooth implies smooth. We first show that the fiber over a geometric point is a regular scheme.In particular, this would show that for any closed point x ∈ X , the fiber Yx → Spec k(x) is a smooth map.Consider a geometric point in X and the following Cartesian square.

Spec k ×X Y //

Y

Spec k // X

Formal smoothness of Y → X implies that Spec k×XY → Spec k is formally smooth. Thus, we may assumethat X = Spec k, where k is algebraically closed and Y is of finite type over X .

Let y ∈ Y be a closed point. Let dimk(my/m2y) = n. Let m = (x1, x2, . . . , xn) denote the maximal ideal in

the polynomial ring k[x1, x2, . . . , xn]. Then there is a surjective k-algebra homomorphism

f1 : OY,y → OY,y/m2y → k[x1, x2, . . . , xn]/m2

Assume that for l, there is a surjective k-algebra homomorphism fl

OY,yfl // //

fl+1 ((PPPPPPP k[x1, x2, . . . , xn]/ml+1

k[x1, x2, . . . , xn]/ml+2

OO

This lifts to fl+1 because of the formal smoothness assumption. We now apply lemma 62 to get that fl+1 issurjective. It is easily checked that fl factors through OY,y/ml+1

y . This proves that d(OY,y) ≥ n. Combiningthis with corollary 28 we get that d(OY,y) = dim(OY,y) = n, which says that OY,y is a regular local ring.

Next we want to show that Y → X is flat. For this, let y ∈ Y be a closed point with image x ∈ X . Denoteby A := OX,x and by B := OY,y . It is easy to check that A → B also has the lifting property. We will showthat this is a flat extension, first under the assumption that there is a section, and then in the general case.

We work in the following setup, A→ B is a local homomorphism of complete local rings which has thelifting property for Artin local rings, and which admits a section, which we denote B

g−→ A.

This implies that the residue fields of A and B are the same, which we denote by k. It is easily checkedthat k → B/mA also has the lifting property. Let us denote by R := B/mA and let n = dimk(mR/m

2R).

Let m = (x1, x2, . . . , xn) denote the maximal ideal in the polynomial ring k[x1, x2, . . . , xn]. Then there is asurjective k-algebra homomorphism

f1 : R→ R/m2R → k[x1, x2, . . . , xn]/m2

Assume that for l, there is a surjective k-algebra homomorphism fl

Rfl // //

fl+1 ''OOOOOOO k[x1, x2, . . . , xn]/ml+1

k[x1, x2, . . . , xn]/ml+2

OO

This lifts to fl+1 because of the lifting property. We now apply lemma 62 to get that fl+1 is surjective.

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We also have a k-algebra homomorphism k[[x1, x2, . . . , xn]] → R obtained by sending the xi to lifts ofa basis for mR/m

2R. This is surjective using lemma 63. Combining this with the above cnstruction gives

surjective mapsk[[x1, x2, . . . , xn]]/ml → R/mlR → k[[x1, x2, . . . , xn]]/ml

which shows that all of them are isomorphisms and so k[[x1, x2, . . . , xn]] → R is also an isomorphism. Weshall denote the images of xi ∈ R = B/mA by xi.

Let bi be lifts of xi in B. It is clear that bi are in the maximal ideal of B and that their images, ai := g(bi)are in the maximal ideal mAA. From this it is clear that bi − ai are also lifts of the xi. Replacing bi by bi − ai,we see that we can choose lifts of xi which have the property that g(bi) = 0. We now claim that the natural

map A[[X1, X2, . . . , Xn]]φ−→ B given by Xi 7→ bi is an isomorphism. This is a surjection using lemma 63.

Let I denote the kernel of this surjection. We have commutative diagram

A[[X1, X2, . . . , Xn]]

φ

Xi 7→0

''OOOOOOOOOOOO

A

77oooooooooooo // B g// A

which shows that I ⊂ (X1, X2, . . . , Xn). Further, since φ when reduced modulo mA is an isomorphism, weget that I ⊂ mAA[[X1, X2, . . . , Xn]] and so

I ⊂ mA ∩ (X1, X2, . . . , Xn) = mA(X1, X2, . . . , Xn)

Let n denote the maximal ideal of S := A[[X1, X2, . . . , Xn]], then the above shows that I ⊂ n2, and so we

have a surjection B → S/n2, such that the composition Sφ−→ B → S/n2 is the natural map. Using the

lifting property, (and lemma 62) we get surjections S B S/nj for each j, and so we get isomorphisms

S/nj∼−→ B/mjB

∼−→ S/nj . Thus, in this case Sφ−→ B is an isomorphism, which shows that B is flat over A.

Now we come to the general case where we have a local homomorphism of complete local rings A→ Bwhich has the lifting property and such that the residue field of B is a finite extension of the residue fieldof A. We want to show that B is a flat A algebra.

Denote kA := A/mA and kB := B/mB . Then we can find a series of finite extensions

kA = k0 ( k1 ( · · · ( kl = kB

such that ki+1 = ki[T ]/fi(T ) for a monic irreducible polynomial fi(T ) ∈ ki[T ]. Assume we have constructeda complete local ring Ri whose residue field is ki. Then we construct a complete local ring Ri+1 which is afinite and flat extension of Ri and whose residue field is isomorphic to ki+1. Let gi(T ) ∈ Ri[T ] be a moniclift of fi(T ) and define Ri+1 := Ri[T ]/gi(T ). It is clear that Ri+1 is finite and flat over Ri and so is semilocal.If we denote my mi the maximal ideal of Ri, then since Ri+1/mi is a field, we get that miRi+1 is the onlymaximal ideal in Ri+1. Starting with R0 = A we find R = Rl which is complete local and finite and flatover A and whose residue field is kB .

We have a commutative squareB //

%%LLLLLL B/mB = R/mR

A

OO

// R/m2R

OO

Now using the lifting property repeatedly, we get a A-algebra homomorphism B → R and so a sectionof the natural map R → B ⊗A R. Since B ⊗A R is finite over B, it is a product of complete local ringsusing lemma 64. Let us write B ⊗A R = ΠiBi, then for each i, we have maps R → Bi → R such that thecomposition is the identity. Moreover, each R→ Bi has the lifting property for Artin local rings. Using theprevious case, this means that each Bi is flat over R and so B ⊗A R is flat over R and since R is faithfullyflat over A, we get that B is flat over A. Now we are done using theorem 50.

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