Lectures on Orders

Dr Daniel Chan

May 16, 2011

Question: How does one study a field? Herre are two answers:

1. Suppose K/Q is a finite field extension. Pick a subring R ⊂ K con-sisting of algebraic integers such that K(R) = K. Information aboutR gives information about K. There are many choices for R but thebest choice is the one which is normal (i.e. integrally closed).

2. Let k be an algebraically closed field and K a field of finite transcen-dence degree over k. Pick an algebraic variety X with function fieldk(X) = K. Such an X is called a model for K. X gives informationabout K. There are many choice for X so we usually impose extraconditions for example: normal/smooth, projective, etc.

The theory of orders studies “integral models” for central simples algebras. Inanalogy with (1) this gives rise to non-commutative arithmetic (see [Rei03])and in analogy with (2) to non-commutative algebraic geometry.

Definition 0.1. Let R be a commutative, noetherian, normal domain. AnR-order (or an order over R) is an R-algebra A such that:

1. A is finitely generated as an R-module (analogue of integral over R),

2. A is a torsion free R-module and K(R)⊗RA is a central simple K(R)-algebra.

Note that torsion free implies A → K(R) ⊗R A so A is a subring of acentral simple algebra. In fact, with the above notation andD := K(R)⊗RA,we will say that A is an order in D.

Example 0.2. Let k be a field of characteristic 6= 2 and R = k[u, v].

A =R〈x, y〉

(x2 − u, y2 − v, xy + yx)

=k〈x, y〉

(xy + yx).

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Proposition 0.3. Let R ⊂ S be an extension of commutative noetheriannormal domains and A an R-order in D. Then A ⊗R S is an S-order inD ⊗K(R) K(S).

Proof. Suppose {a1, · · · , an} generates A as an R-module and let a ⊗ s ∈A⊗R S. Then for some r1, · · · , rn ∈ R, a = a1r1 + · · ·+ anrn and so

a⊗ s = (a1r1 + · · ·+ anrn)⊗ s

= a1 ⊗ r1s+ · · ·+ an ⊗ rns.

Thus, {a1 ⊗ 1, · · · , an ⊗ 1} generates A⊗R S as an S-module.Also,

A⊗R S ⊗S K(S) ≃ A⊗R K(S)

≃ A⊗R K(R)⊗K(R) K(S)

= D ⊗K(R) K(S).

The fact that D ⊗K(R) ⊗K(S) is a central simple K(S)-algebra is clear.

1 Examples

The classic cross product construction for central simple algebras “globalises”with some work. To describe it we need a notion of Galois extension pairing.

1.1 (Ramified) Galois Ring Extensions

Definition 1.1. Let R ⊂ S be commutative, noetherian domains. We sayS/R is (ramified, finite) Galois if the following holds:

1. S is a finitely generated R-module,

2. there is a finite group G < Aut S such that R = SG.

Example 1.2.k[x]

Galois

⊂ k(x)

k[x2]

⊂

⊂ k(x2)

⊂

In this case G = {1, σ} where σ(x) = −x.

Call G the Galois group of S/R. We say S/R is cyclic if G is.

2

Facts 1.3. 1. If S/R is Galois with Galois group G, then K(S)/K(R)is Galois with Galois group G.

2. Let S be a commutative, noetherian domain and G < Aut S finitesuch that |G|−1 ∈ S. Then R := SG is noetherian and S/R is Galois.Further, if S is normal then SG is also normal.

3. Let R be a commutative, normal, noetherian domain and K := K(R).Let F/K be a Galois field extension with Galois group G and S be theintegral closure of R in F . Then S/R is Galois with Galois group G.

Proof. 1. G acts on S and so G acts on K(S). Suffice now to show thatK(S)G = K(R). Now, K(R) = K(SG) ⊆ K(S)G. For the reverseinclusion, simply note that that for α

β∈ K(S)G,

α

β=δ := αg1(β)g2(β) · · · gn−1(β)

γ := βg1(β) · · · gn−1(β)

where G = {id, g1, · · · , gn−1}. Since γ ∈ SG, we must have δ ∈ SG andso α

β∈ K(SG).

2. Consider the skew group ring:

S∗G =⊕

g∈G

Sg

with multiplication given by multiplication in S and G and skew com-mutation relation g · s := g(s)g. Have an idempotent e := 1

|G|

∑g∈G g.

Indeed it is easy to see that for all h ∈ G, he = eh = e. Note also thatR ≃ eS∗Ge = eSe.

To show SG is noetherian, we consider the inclusion preserving map

{left idealsof eS∗Ge

}//

{S∗G-submodules

of S∗Ge

}

I // SI

whichwhich has a left inverse M 7→ eM . To see this, note that I = eI(since e is the identity in eS∗Ge) and so eSI = eSeI = eS∗GeI = I.S is noetherian so S ∗G is noetherian and ACC on S ∗G-submodulesof S ∗Ge gives ACC on ideals of eS ∗Ge ≃ R = SG and thus SG is

3

noetherian. To show S is noetherian over SG = eS∗Ge = eRe = Re,consider the inclusion preserving map

{eS∗Ge-submodules

of eS

}//

{left idealsof S∗G

}

M // (S∗G)M = SM

has a left inverse N 7→ eN . The same reasoning as before shows Sis noetherian as an SG module and hence finitely generated as as SG

module.

3. We have

S ⊂ F

R

⊂

⊂ K

⊂

Now, SG = S ∩K(S)G = S ∩K(SG) = S ∩K = R since R is integrallyclosed. Need to show S/R is finitely generated, so suffices to show S/Ris finite. Recall that we have a trace pairing F ×F → K with (α, β) 7→tr (αβ) where tr (αβ) =

∑g∈G g(αβ) which is non-degenerate and

symmetric. Also tr (S) ⊂ R because if α ∈ S is integral overR the samesame is true of g(α) for all g ∈ G. Let V < F be a finitely generatedR-submodule such that KV = F . The pairing gives an isomorphismbetween V ∗ := HomR (V,R) and R-submodule {α ∈ F | tr (αV ) ⊂ R}of F via α ∈ F 7→ tr (α ·). To see that this is indeed an isomorphism,note that HomR (V,R) ⊂ HomR (V,K) = HomK (KV,K) = F ∗ =≃ Fwhere the last isomorphism is given by the trace map. Now pick aK-basis B for F such that B ⊂ S. Let V =

⊕b∈B Rb. Then V ⊂ S ⊂

S∗ ⊂ V ∗ which is noetherian. Thus S is a noetherian R-module.

1.2 Cyclic Algebras

Let R be a commutative, noetherian, normal domain and S/R a cyclic ringextension with Galois group G = 〈σ | σn = 1〉. Let L be a rank 1, locallyfree S-module (for example L = S) and Lσ be an S-bimodule, which as a leftS-module is L, but the right module action is m · s := σ(s)m. Furthermore,suppose we are given an injection of S-bimodules φ : L⊗nσ → S satisfying the

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overlap condition i.e. the following diagram is commutative:

Lσ ⊗S L⊗(n−1)σ ⊗S Lσ

φ⊗1//

1⊗φ

��

S ⊗S Lσ

��

Lσ ⊗S S // Lσ

To understand this condition, consider the easy case where L is free so canwrite Lσ = Su with us = σ(s)u. Note that the morphism φ : L⊗nσ = Sun → Sis defined by φ(un) =: α. The overlap condition is equivalent to αu = uα =σ(α)u which is equivalent to to α = σ(α) i.e. α ∈ SG = R. We can nowdefine the cyclic algebra

A (Lσ;φ) = S ⊕ Lσ ⊕ · · · ⊕ L⊗(n−1)σ

with multiplication given by:

L⊗iσ ⊗ L⊗jσ −→

{L⊗(i+j)σ i+ j < n

L⊗(i+j)σ

1⊗φ⊗1−→ L

⊗(i+j−n)σ i+ j ≥ n

It is left as an exercise to check that the overlap condition ensures multipli-cation is well-defined and associative. Note that A is an R-algebra since Racts centrally on the bimodule Lσ.

Proposition 1.4. A(Lσ;φ) is an R-order.

Proof. S/R is finite and A is a finitely generated module over S. HenceA is a finitely generated R-module. Suffice to show that K(R) ⊗R A isthe classical cyclic algebra, which are known to be central simple. NoteK(R) ⊗R L = K(S) ⊗S L = Su for some u. Also K(R) ⊗R A is a cyclicalgebra in the above sense with

K(R)⊗R A = K(S)⊕K(S)u⊕ · · · ⊕K(S)un−1.

The overlap condition implies un ∈ K(S)G = K(R) so we get the classicalcyclic algebra.

1.3 Crossed Product Algebras

Recall the following approach to group cohomology. Let G be a finite groupand M a G-module which also makes it a ZG-module. Then H i(G,M) isthe cohomology of the complex C•(G,M) where

Ci(G,M) ={functions f : Gi →M

}

5

and di : Ci → Ci+1 is defined by

(dif)(g1, · · · , gi+1) = g1.f(g2, · · · , gi+1) +i∑

j=1

(−1)jf(g1, · · · , gjgj+1, · · · gi+1)

+ (−1)i+1f(g1, · · · , gi)

Case of interest: let S/R be a Galois extension of commutative, noetheriandomains so K(S)/K(R) Galois as well, say with Galois group G. HenceK(S)∗ is a G-module. The 2-cocycles f : G×G→ K(S)∗ are those satisfying

g1.f(g2, g3) = f(g1g2, g3)f(g1, g2g3)−1f(g1, g2) (∗∗)

(note that we switched to multiplicative notation). This is called a factorset. If f has values in S then get crossed product

A(S, f) :=⊕

g∈G

Sug

and multiplication ugs = g(s)ug for g ∈ G, s ∈ S and uguh = f(g, h)ugh.This is an R-order in the classic crossed product algebra over the field K(R).

1.4 Finding Orders in Central Simple Algebras

Given a commutative, noethrian, normal domain R, and a central simpleK(R)-algebra D, it is natural to ask if we can find a model for it. That is,can we find an R-order A with K(R) ⊗R A ≃ D. We call such an order anorder in D. The answer is yes, due to:

Proposition 1.5. 1. There is a finitely generated R-submodule M < Dsuch that K(R)⊗RM ≃ D.

2. Given any M as in (1) the following is an R-order in D:

Ol(M) := {α ∈ D | αM ⊆M} .

Proof. 1. Pick a K(R)-basis {m1, · · · ,mn} for D and let M =⊕n

i=1Rmi.

2. Note that Ol(M) is an R-algebra. Also, Ol(M) is finitely generated asan R-module because it is an R-submodule of the noetherian R-moduleHomR (M,M). It suffices now to show that K(R)⊗R Ol(L) ≃ D. Letα ∈ D and note that αM is a finitely generated R-submodule of D soK(R)M = D and hence αM ⊂ r−1M for some r ∈ R−{0}. Therefore,α ∈ r−1Ol(M).

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2 Relationship With Azumaya Algebras

An Azumaya algebra can be thought of as a central simple algebra varyingover a scheme.

Proposition/Definition 2.1. Let R be a commutative, noetherian ring.An R-algebra is Azumaya of degree n if any of the following equivalentconditions hold:

1. A is locally free of rank n2 and for all maximal m⊳R, we have A⊗RR/mis a central simple R/m-algebra.

2. A is locally free of rank n2 and the natural map

ψ : A⊗R Aop −→ EndRA

a1 ⊗ a2 7−→ (a 7→ a1aa2)

is an isomorphism.

3. There is a faithful etale extension R → S such that A ⊗R S ≃ Mn(S)- the full matrix algebra with entries in S.

Proof. (1) ⇐⇒ (2). Note A ⊗R Aop and EndRA are locally free of rank n4

so ψ is an isomorphism if and only if ψ ⊗ R/m is an isomorphism for allmaximal m ⊳ R. Now (1)⇐⇒ (2) follows from:

Lemma 2.2. Let A be a finite dimensional k-algebra, where k is a field. ThenA is central simple if and only if ψ : A⊗k A

op → EndkA is an isomorphism.

Proof. (⇒) A⊗k Aop is central simple and so kerψ = 0.

(⇐) Note Z(A) = k. Also, given b ∈ A− {0}, we see that the ideal in Agenerated by b contains (im ψ)(b) = A.

(3)⇒(2) by descent theory.(1)⇒(3) Going sufficiently local we can assume A ≃ Rn2

. Then(e1, · · · , en) ⊂ An ≃ Rn3

forms a complete set of orthogonal idempotents ifand only if e2i = ei and eiej = 0 if i 6= j and

∑ei = 1. These define a

subscheme X ⊂ An3

R such that for any extension S/R the complete sets ofidempotents for A ⊗R S correspond to sections of X ⊗R S → Spec S. Noteis π : X → Spec R surjective by (1). Now π is smooth by Grothendieck’scriterion and the fact that you can lift idempotents. Hence it has an etalesection say over the etale extension S/R. A complete set of idempotents forA ⊗R S can be used to give a Peirce decomposition of A ⊗R S which givesMn(S)

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Example 2.3. Consider the cyclic algebra A = A(Lσ;φ) formed from thecyclic etale ring extension S/R of degree n and φ : L⊗nσ

∼−→ S. Then A is an

Azumaya R-algebra.

Proof. It suffices to show A ⊗R S ≃ Mn(S). Since S/R is cyclic etale weget S ⊗R S ≃

∏ni=1 S where G = Gal (S/R) cyclically permutes the compo-

nents. We can work locally and assume Lσ = Su i.e. L is free. We have anisomorphism A⊗R S →Mn(S) given by

S ⊗R S −→

S

. . .

S

, u 7−→

0 1. . . . . .

. . . 1α 0

where α = un ∈ (S∗)G = R∗. It is left as an exercise to show that thismorphism is surjective and

un = α

1

. . .

1

, u

s1

. . .

sn

= σ

s1

. . .

sn

u

Proposition 2.4. Let R → S be a morphism of commutative noetherianrings and A be an Azumaya R-algebra. Then A⊗R S is Azumaya over S. Inparticular, if R is a normal domain, then A is an R-order.

Proof. Look at S = K(R).

Proposition 2.5. Let R be a commutative, noetherian, normal domain andA an R-order. Then there is some r ∈ R − {0} such that A[r−1] is anAzumaya R[r−1]-algebra. More precisely, the Azumaya locus of A is densein Spec R.

Proof. The Azumaya locus is the locus where (i) A is locally free and (ii) A⊗RAop ≃ EndRA. Both these are open conditions. Also (i) holds generically.So does (ii) since generically, A is central simple.

Example 2.6. By Example 2.3 a cyclic algebra A = A(Lσ;φ) formed fromS/R and relation φ : L⊗nσ → S is Azumaya on the intersection of etale locusof S/R and locus where φ is an isomorphism.

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The upshot of this is that we can view an R-order as a family of centralsimple algebras and their degenerations. The non-Azumaya locus (or ram-ification locus) is an important invariant of the order, which can be usedto study it.

Example 2.7. [Terminal, maximal with nodal ramification] Let k be analgebraically closed field of characteristic not diving n ∈ Z. Let R = k[u, v].We have a cyclic extension S = k[x, v] where R → S is given by v 7→ v andu 7→ xn. This extension has Galois group G = 〈σ | σn = 1〉 with actionσ(x) = ζx, ζ being a primitive nth root of 1 and σ(v) = v. Let L = Sy andconsider the relation

φ : L⊗nσ = Syn −→ S

yn 7−→ v ∈ SG = R

Thus we get the cyclic algebra A = A(Lσ, φ). By Example 2.6 it is Azumayaon uv 6= 0. Note first

A = S ⊕ Sy ⊕ · · · ⊕ Syn−1 = R[x]⊕R[x]y ⊕ · · · ⊕R[x]yn−1

=R〈x, y〉

(xn − u, yn − v, yx− σ(x)y)

=k〈x, y〉

(yx− ζxy).

What about on uv = 0? At (u, v) = (1, 0): if m = (u− 1, v) then

A⊗R R/m =k〈x, y〉

(xn − 1, yn, yx− ζxy).

Now x and y are normal since they skew-commute (normal means x(A⊗ R

m

)=(

A⊗ Rm

)x) and so y

(A⊗R

Rm

)is a two sided ideal (in fact a nilpotent one).

Therefore not Azumaya at (1, 0). Similarly, we can see that the ramificationlocus is all of uv = 0.

3 Reduced Trace

We need the following tool, known as the reduced trace of a central simplealgebra. It generalises the trace map for field extensions. Our approach isvia Galois descent.

Let K be a field and D a central simple K-algebra. There is some Galoisfield extension F/K which splits D in the sense that D ⊗K F ≃ Mn(F ) forsome n. Those split by F/K are classified using non-abelian cohomology,which we describe now.

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3.1 Non-Abelian Cohomology

Let F/K be a Galois field extension with Galois group G and note that Gacts on the non-commutative group

PGL(n, F ) = GL(n, F )/F ∗

≃ AutF−algMn(F )

where the last isomorphism follows from the Skolem-Noether theorem. Notethat conjugation by an element of PGL(n, F ) is well-defined. We define

H1(G,PGL(n, F )) :={1-cocycles f : G→ PGL(n, F )}

∼

where f : G → PGL(n, F ) is a 1-cocycle if fgh = fg(g.fh) for all g, h ∈ G.Also, f ∼ f ′ if there is some a ∈ PGL(n, F ) with f ′g = a−1fg(g.a) for allg ∈ G. H1 (G,PGL(n, F )) is a pointed set with distinguished element f ≡ 1.

Theorem 3.1 (Galois descent for central simple K-algebra). 1. Let f : G→PGL(n, F ) be a 1-cocycle. Then there is an induced G-action on Mn(F )defined by

g.(aij) := fg(g.aij)f−1g .

Also, Mn(F )G is a central simple K-algebra.

2. The map

{1-cocycles f : G→ PGL(n, F )} −→ {central simple K-alegbras}

f 7−→Mn(F )G

induces a bijection of pointed sets

H1(G,PGL(n, F ))∼−→

isomorphism classes of

central simple K-algebrassplit by F/K.

Proof. See [GS06].

3.2 Reduced Trace

Let K be a field and D a central simple K-algebra. Let F/K be a Galoisfield extension that splits D i.e. D⊗K F ≃Mn(F ). For a ∈ D we define thereduced trace (respectively determinant, characteristic polynomial)of a to be the trace (respectively determinant, characteristic polynomial) ofa ∈Mn(F ). This is well defined, as we shall see below. It is denoted by tr a(det a, char. pol. a).

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Proposition 3.2. 1. Let a ∈ D. Then char. pol. a ∈ K[x] (not justF [x]). In particular, tr a, det a ∈ K.

2. char. pol. a is independent of the choice of F/K and the isomorphismD ⊗K F ≃Mn(F ).

Proof. 1. It suffices to show that for any g ∈ G := Gal (F/K), we haveg.(char. pol. a) = char. pol. a for then char. pol. a ∈ FG[x] = K[x].By Galois descent, there is a 1-cocycle f : G → PGL(n, F ) such thatD ≃ Mn(F )G with G-action induced from f as described in Theorem3.1 (1). Then

g.(char. pol. a) = g.char. pol. (aij)

= char. pol. g.(aij)

= char. pol.(fg(g.aij)f

−1g

)

= char. pol. (g.aij)

= char. pol. (aij) since a ∈Mn(F )G

= char. pol. a

2. This is similar to the previous part and is left as an exercise.

Proposition 3.3. Let R be a commutative, noetherian domain and K :=K(R). Let A be an R-order in D. Then char. pol. a ∈ R[x], for a ∈ A.

Proof. Let p(x) be the char. pol. a ∈ K[x]. R[a] ⊂ A is a finitely generatedR-module (since R and A are noetherian) so a is a root of a monic polynomialm(x) ∈ R[x]. But p(x) has the same roots as the minimal polynomial of aso p(x) | m(x)r ∈ R[x] for r ≫ 0. Gauss’ Lemma implies the coefficients ofp(x) are integral over R. Since R is normal, p(x) ∈ R[x].

Corollary 3.4. With the same hypotheses as in Proposition 3.3 there is areduced trace map tr : A→ R.

3.3 Trace Pairing

Let R be a commutative, noetherian, normal domain with K = K(R). Let Dbe a central simple K-algebra split by a Galois field extension F/K. Recallthat there is a non-degenerate symmetric trace pairing

tr : Mn(F )×Mn(F ) −→ F

(α, β) 7−→ tr (αβ)

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This restricts to a non-degenerate pairing tr : D × D → K as we saw inProposition 3.2. Consider a finitely generated R-submodule M < D suchthat KM = D. Then

M∗ := HomR (M,R) → HomR (M,K) = HomK (KM,K) = HomK (D,K)tr≃ D

The image of M∗ in D via this trace pairing is:

M∗ = {α ∈ D | tr (αM) ⊆ R} .

Corollary 3.5. Let A ⊆ B be R-orders in a central simple K(R)-algebrasD. Then B ⊆ A∗ ⊆ D.

Proof. Proposition 3.3 implies that tr (B) ⊆ R so B ⊆ B∗ < D. Also,A ⊆ B and so B∗ ⊆ A∗. Hence B ⊆ A∗

4 Maximal Orders. Basic Properties

We give the first generalisation of the concept of normality to orders. It isbased on the observation that for a commutative, noetherian domain S withK(S) = K, its integral closure R is the largest integral extension of S in K.

We fix some notation: let R be a commutative, noetherian normal do-main, A an R-order in a central simple K(R)-algebra D.

Definition 4.1. We say A is maximal if it is maximal with respect toinclusion amongst orders in A⊗R K(R).

Before giving examples, note:

Proposition 4.2. Let A be an R-order and S/R be a faithful finite extensionof commutative, noetherian, normal domains. If A ⊗R S is maximal, so isA.

Proof. Suppose A is not maximal. Then if B is an R-order strictly containingA then B ⊗R S is an S-order and B/A ⊗R S 6= 0 so it strictly containsA⊗R S

Example 4.3. Mn(R) is a maximal R-order in Mn(K(R)) since by Corol-lary 3.5 and R-order B containing Mn(R) satisfies B ⊆ Mn(R)∗ = Mn(R).Therefore Mn(R) is maximal.

Example 4.4. Any Azumaya R-algebra A is maximal. To see this, notethat by Definition/Proposition 2.1, etale locally, A is Mn(R) so A is maximalby Proposition 4.2 and the above example.

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If the concept of maximality indeed generalises normality to the non-commutative setting, then the procedure of normalisation should extend toto embedding in a maximal order. The following proposition shows that thisis indeed the case:

Proposition 4.5. Any R-order A embeds in a maximal order.

Proof. If B ⊆ D is an R-order containing A then by 3.5 B ⊆ A∗. But A∗

is a noetherian R-module so ACC shows that there exists a maximal ordercontaining A.

Unfortunately, unlike normalisation, the maximal order in the aboveproposition is not unique. It is easy to come up with such examples:

Example 4.6. Let R = k[x] and A =

(R RxR R

). It is contained in a

maximal orders

(R RR R

)and

(R x−1RxR R

)=

(1 00 x

)(R RR R

)(1 00 x

)−1

.

Despite this, maximality is compatible with localisation and completionin dimension 1, which we now show.

Proposition 4.7. Let A be maximal R-order and U ⊂ R be a multiplicativelyclosed set. Then A[U−1] is a maximal R[U−1]-order in D.

Proof. Suppose B ⊃ A[U−1] be an R[U−1]-order in D. Pick a finitely gen-erated A-submodule M < B such that M [U−1] = B. Note K(R) ⊗R M =K(R)⊗R[U−1] B = D and M is finitely generated as an R-module too. ThusOl(M) := {α ∈ D | αM ⊆M} is an R-order which contains A. It strictlycontains A since

Ol(M)[U−1] ={β = αu−1 | α ∈ D,αM ⊆M,u ∈ U

}

={β ∈ D | βM [U−1] ⊆M [U−1]

}

⊇ B ⊃ A[U−1] ⊆ A

which contradicts the maximality of A.

Definition 4.8. Let R be a Dedekind domain (i.e. a normal, noetheriandomain of dimension 1) with K(R) = K and V be a finite dimensional K-vector space. A full R-lattice in V is a finitely generated R-module M < Vsuch that KM = V .

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Lemma 4.9. Let R be a dvr (i.e. a local Dedekind domain or equivalently,a local PID) and R its completion. Let K, K be their field of fractions andν be the valuation on K. Let V be an n-dimensional K-vector space. LetV = K ⊗K V .

1. Let M be a full R-lattice in V . Then it is free over R with an R-basis{m1, · · · ,mn} ⊆ V .

2. There are inverse bijections

{full R-latticeM < V

}Φ

//

{full R-lattice

M < V

}

Ψoo

M // RM

M ∩ V Moo

Proof. 1. Note M is torsion free over a PID R so is free with basism1, · · · ,mn ∈ V . Pick K-basis v1, · · · , vn of V . Change of base matrix(αij) ∈ GLn(K),mj =

∑i αijvi. We are done if (αij) ∈ GLn(R) since

we can change basis m1, · · · ,mn. Thus it suffice to show that afterusing elementary column operations (ECO’s) over R and elementaryrow operations (ERO’s) over K we can get (αij) into lower triangu-lar form with 1’s on the diagonal and entries below the diagonal inR. Uniformising parameter for R is also a uniformising parameterfor R so can scale rows using ERO’s to get (αij) ∈ Mn(R) and some

αij ∈ R∗. Switch columns so that α1,1 ∈ R

∗. Using ECO’s can make

0 = α1,2 = · · · = α1,n. Now R is dense in R so can use ERO’s over Kto make ν(α2,1), · · · , ν(αn,1) >> 0. Now we can repeat.

2. We first show Ψ is well-defined and ΦΨ = id. By (1) we know that afull R-lattice M =

⊕ni=1 Rmi with mi ∈ V . Then Ψ(M) = M ∩ V =⊕

Rmi which is a full R-lattice in V . Also, ΦΨ(M) = R(⊕

Rmi) =⊕Rmi = M . Similarly ΨΦ = id

Corollary 4.10. Let R be a dvr and A a maximal R-order. Then A ⊗R Ris a maximal R-order.

Proof. By Lemma 4.9 (2) if B ⊃ A ⊗R R is an R-order in A ⊗R K(R) thenB ∩ (A⊗R K(R)) is an R-order strictly contatining A.

14

Example 4.11 (Counter example in higher dimesnion). Let R = k[u, v].We saw the following cyclic algebra is an R-order:

A =R〈x, y〉

(x2 − u, y2 − v, xy + yx).

We will show later that this is maximal. Let’s complete at m = (u − 1, v).Rm = kJw, vK. Let p(w) ∈ kJwK be such that p(w)2 = 1 + w. Then considerthe injective algebra homomorphism

A⊗R Rm =Rm〈x, y〉

(x2 − (1 + w), y2 − v, yx+ xy)−→

(Rm Rm

vRm Rm

)⊂M2(Rm)

x 7−→

(p(w) 0

0 −p(w)

)

y 7−→

(0 1v 0

)

and so A⊗R Rm is not maximal.

5 Normal Orders in Dimension 1

In Section 4 we saw that maximality was a generalisation of integrally closedfor orders. We now introduce a second such generalisation to orders overdvr’s. This is based on the fact that a dvr is a (commutative) noetherian,local domain whose unique maximal ideal is principal (nonzero).

Definition 5.1. Let R be a dvr. An R-order A is normal if its Jacobsonradical J has the form Aπ for some π ∈ A.

The definition appears to be not left-right symmetric. We will shortlyprove that it in fact it is. we first need to prove some basic properties aboutfinite extensions of local rings.

Lemma 5.2. Let (R,m) be a local ring and R ⊂ S a finite (non necessarilycommutative) ring extension with J = rad S. Then:

1. mS ⊆ J .

2. If I E S is nilpotent modulo m, i.e. there exists n ∈ N such thatIn ⊆ mS then I ⊆ rad S.

3. rad S is nilpotennt modulo m.

15

Proof. 1. Let r ∈ m, s ∈ S. Then S(1 − sr) + mS = S and so byNakayama’s lemma (Lemma 4.22 (3) of [Lam91]) we have S(1−sr) = S.Thus 1 − sr is left invertible for all s ∈ S and so by Lemma 4.1 (1)of [Lam91], r ∈ J . Thus m ⊆ J and since J is an closed under leftmultiplication mS ⊆ J .

2. By part (1), In ⊆ J and so (I + J)/J is a nilpotent ideal in S/J . ButS/J is left artinian since S/mS is finite dimensional vector space overR/m (this is where we need S/R to be finite) and S/J is a quotient ofit. Thus by Theorem 4.12 of [Lam91] (I + J)/J ⊆ rad (S/J) = 0 andso I ⊆ J .

3. Using part (1) we know J/mS E S/mS which is artinian. By Nakayama’slemma, J2/mS ⊳ J/mS or J/mS = 0. Continuing on we get a strictlydecreasing sequence

· · · ⊳J3

mS⊳J2

mS⊳S

mS

Since S/mS is artinian this must eventually reach zero. That is, forsome n ∈ N Jn ∈ mS.

We shall now prove that our definition of a so normal order is left-rightsymmetric and then show some interesting examples.

Proposition 5.3. Let A be a normal order over (R,m) with J := rad A =Aπ. Then π is invertible in D := (R)⊗RA and hence J is free. Furthermore,J = πA.

Proof. Using Lemma 5.2(1) we know that mA ⊆ J . Therefore, K(R)J =K(R)A = D, and Dπ = K(R)Aπ = K(R)J = D, so π ∈ D∗. ThusA ≃ Aπ = J as left A-modules and so J is free.

Since J is a two sided ideal, AπA = Aπ and so Aπ = J ⊇ πA whichimplies A ⊆ π−1Aπ ⊆ π−2Aπ2 ⊆ · · · . The chain terminates by the existenceof maximal orders and so Aπ = πA.

Example 5.4 (Cyclic algebras). Let (R,m) be a dvr such that the char-acteristic of R/m is coprime to n, and S/R be a cyclic extension of normalrings of degree n with Galois group G = 〈σ | σn = 1〉. Consider the cyclicalgebra A := A(S/R,Lσ, φ) built from the relation φ : L⊗nσ → S. If either

1. S/R is etale and im φ = mS, or

2. φ is an isomorphism

16

then A is normal.

Proof. We know that A is an R-order. Also S is semilocal so by [Eis95]Exercise 4.13 we can assume L ≃ S hence Lσ ≃ Su and A = S ⊕ Su⊕ · · · ⊕Sun−1.

1. Note that u is normal and un ∈ SG = R generates mS, so un is a uni-formising parameter for R. Therefore uA is an ideal which is nilpotentmodulo m. So by Lemma 5.2(2) Au = uA ⊆ rad A. To show equality,it suffices to show that A/uA is semisimple (Theorem 4.6 and Theorem4.14 of [Lam91]). But A/uA ≃ S/mS which is a product of fields sinceS/R is etale.

2. If S is semilocal, then rad S = St for some t ∈ S. Also σ(t) generatesrad S, so σ(t) = αt for some α ∈ S∗. Hence t is normal in A, we wishto show that At = rad A. Note that At ⊆ rad A, so it is enough toshow that A := A/At is semisimple.

Let I < G be the inertia group and e = |G/I|, so G/I < Aut S/St.Write A = S ⊕ Su ⊕ · · · ⊕ Sun−1. Note that ue is central in A, so Ais a κ[ue]-algebra (where κ = R/m). To check that A is semisimple,we can do base change to a separable field extension of κ. So assumeun ∈ R∗ is an n-th power modulo m. Scaling u, we can assume un = 1in A. Therefore A ⊇ κ[ue] ≃ κ

(Z/n

eZ)≃ κε1 × · · · × κεn/e where εi

are central idempotents. Now A =∏n/e

i=1 Aεi and each of these factorsAεi is a cyclic algebra, so A is central simple.

Definition 5.5. A ring A is hereditary if the global dimension gl. dim.Ais equal to 1, that is, Ext2

A (−,−) = 0 but Ext1A (−,−) 6= 0.

Normal orders, like dvrs, enjoy the following regularity property:

Proposition 5.6. Let A be an order over a dvr R. Suppose that J = rad Ais projective (e.g. when A is normal). Then any fintely generated torsion-freeA-module is projective and gl. dim. A = 1.

Proof. We have the following exact sequence

0→ J → A→ A/J → 0.

Since J is projective and A is free, Ext1A (J,−) = Ext1

A (A,−) = 0 and soExt2

A (A/J,−) = 0 and so pd A/J = 1. Since A/J is the direct sum of all thesimples (possibly repeated) we have that for any simple module S pd S = 1;

17

this is because if for any module M = N1⊕N2, pd M = max {pd N1, pd N2}(simply take the direct sum of any projective resolution of N1 and N2 to getone of M). Any torsion A-module T is an extension of simples, so pd T = 1.

Let M be a finitely generated A-module. Its torsion submodule T is anA-module too, so pd T = 1 and it suffices to show that if M is torsion-free,then it is projective.

Let p ∈ R be a uniformising element and consider the exact sequence

0 −→Mp−→M −→M/pM −→ 0

Let N be a finitely generated A-module. Since pd M/pM ≤ 1, we getsurjective maps ExtiA (M,N)→ ExtiA (M,N) for i ≥ 1, i.e. pExt1

A (M,N) =Ext1

A (M,N) where p generates the maximal ideal. Nakayama’s lemma thenimplies ExtiA (M,N) = 0, so we are done.

One nice property about normal orders is that it respects etale basechange.

Proposition 5.7. Let ϕ : (R,m) → (S, n) be a morphism of dvrs withϕ−1(n) = m such that κS := S/mS is a separable field extension of κ := R/m.If an R-order A is normal then A ⊗R S is a normal S-order. Furthermore,if κS/κ is also finite, then the converse holds.

Proof. We claim that

rad (A⊗R S) = (rad A)⊗R S. (1)

Since rad A is nilpotent modulo m, we have rad (A ⊗R S) ⊇ (radA) ⊗R S.So it suffices to show that

A⊗R S

(rad A)⊗R S≃

(A

rad A

)⊗R S

is semisimple. Since m ∈ rad A, the latter is isomorphic to (A/rad A)⊗κ κS,which is semisimple since A/rad A is semisimple and κS/κ is separable.

If rad A = Aπ then (1) implies rad (A ⊗R S) = (A ⊗R S)(π ⊗ 1) isprincipal.

Conversely, assume rad (A⊗R S) is free of rank 1. Let J := rad A Notefirst that J is projective since

Ext1A (J,−)⊗R S = Ext1

A⊗RS(J ⊗R S,−) = 0.

But S/R is faithfully flat so Ext1A (J,−) = 0 and J is projective.

18

We know that

(J/J2)⊗R S ≃ (J ⊗R S)/(J ⊗R S)2

≃ (A⊗R S)/(J ⊗R S)

≃ (A/J)⊗R S.

The proposition then follows from the next lemma which shows that J/J2 ≃A/J , so A and J are projective covers of isomorphic modules and so areisomorphic by Proposition 21.3.

Lemma 5.8. Let A be a semisimple κ-algebra and κ′ a finite dimensionκ-algebra. Then two finitely generated A-modules M,N are isomorphic ifM ⊗κ κ

′ ≃ N ⊗κ κ′.

Proof. Let A =∏n

i=1 Aεi for central idempotents εi so that Aεi is simple.Let Si be a simple A-module corresponding to εi. Need only show the directsum decomposition

M ≃n⊕

i=1

S⊕mi

i

depends only on M ⊗κ κ′. But

(M ⊗κ κ′)εi = S⊕mi

i ⊗κ κ′

so

mi =dimκ(M ⊗κ κ

′)

dimκ(Si ⊗κ κ′)

6 Criterion for maximality

In this section, we shall give the Auslander-Goldman’s criteria for an orderto be maximal. One easy way to obtain a bigger order from a given one isto take its reflexive hull as follows:

Let R be a commutative noetherian normal domain, K be the fractionfield of R, M be a finitely generated torsion-free R-module, and V = K⊗RM .Since M is torsion-free, we have a natural embedding M → V . We denoteduals

M∗ = HomR (M,R)

V ∨ = HomK (V,K) .

Recall:

19

Definition 6.1. An R-module M is reflexive if the natural map

M −→M∗∗

m 7−→ (ξ 7→ ξ(m))

is an isomorphism.

We get a commutative diagram

M //

��

M∗∗

��

V∼

// V ∨∨

so we can consider M∗∗ as an R-submodule of V .

Remark 6.2. If M is a finitely generated torsion-free R-module, thenM∗∗ =

⋂P MP where p runs over height 1 primes.

Proof. See [Bou89] chapter 7 section 4.

Lemma 6.3. Let A be an R-order in D. Then A∗∗ is also an R-order in D.

Proof. Note that from the commutative diagram A ⊆ A∗∗ ⊆ D and is finitelygenerated, so it suffices to show that it is closed under multiplication. Thisis clear from M∗∗ =

⋂P MP , but a better proof is by functoriality of ∗∗. Let

a ∈ A which induces the following commutative diagram

Aa

//

��

A

��

A∗∗a∗∗

//

��

A∗∗

��

Da

// D

so left multiplication by a preserves A∗∗, so multiplication extends toA⊗ A∗∗ −→ A∗∗. Repeat with right multiplication by a′ ∈ A∗∗.

We now give Auslander-Goldman’s criterion for maximallity. It reducesverification of maximality in general to checking it over a dvr.

Proposition 6.4. (Auslander-Goldman) Let A be an R-order, then A ismaximal if and only if A is reflexive and AP is maximal for all height 1primes P ∈ Spec R.

20

Proof. If A is maximal, then by Lemma 6.3 it is reflexive. Maximality islocal by Proposition 4.7.

Conversely, suppose B ⊇ A is a maximal order. We need to show A = B.Since A is reflexive and AP is maximal, we have A =

⋂P AP ⊇

⋂P BP =

B.

Definition 6.5. A ring A is local if A/rad A is simple artinian.

Proposition 6.6. Let (R,m) be a dvr and A be a local normal R-order.Then A is maximal.

Proof. Let B ⊃ A be a R-order. Let m = pR and rad A = Aπ.

B ⊂ K(R)⊗R A = A[p−1]

= A[π−1] by Proposition 5.3

=⋃

i∈Z

πiA

Hence π−ia ∈ B for some a ∈ A−πA. By multiplying by πi−1 we can assumethat i = 1 and so (π−1A ∩B)/A is a nonzero A-bimodule. Multiplying by πgives a nonzero ideal

(A ∩ πB)

πA≤

A

πA

which is simple by assumptions, and so A ∩ πB = A and π−1 ∈ B. Thisshows that B is not finitely generated.

Remark 6.7. The converse of Proposition 6.6 also holds as we shall provein Proposition 10.2. Unfortunately, we have not developed the theory enoughyet to give a direct proof at this stage.

Remark 6.8. Proposition 6.6 is true if we replace normal by hereditary.

Example 6.9. (Terminal, maximal, with nodal ramification) Let k be analgebraically closed field with characteristic not dividing n. Let R = k[u, v]and

A =R 〈x, y〉

(xn − u, yn − v, xy − ζyx)

where ζ ∈ k∗ is a primitive n-th root of unity. We claim that A is maximal.

21

Proof. We check maximlaity using Proposition 6.4. Note thatA =⊕n−1

i,j=0Rxiyj

is R-free, hence reflexive. We check that AP is maximal for all height 1 primesP . As we saw in Example 2.6 and 2.7 A is Azumaya on uv 6= 0 and so byExample 4.4 it is maxmial there. Thus we only need to check for P = (u)and (v) which we do using Proposition 6.6. Suppose P = (u). AP is con-structed, as we saw in Example 2.7, from the cyclic algebra construction viaφ : Syn → S, yn 7→ v which is an isomorphism since v is a unit in AP . Thusby Example 5.4 AP is normal. Note also that x ∈ A is normal so xA isnilpotent modulo (u). Lemma 5.2(1) xAP ≤ rad AP . Also

APxAP

≃k(v)[y]

(yn − v)

which is a field extension of k(v) and is simple and artinian. Thus AP islocal. Hence AP is maximal by Proposition 6.6.

If P = (v), then AP is constructed via φ : Syn → S, yn 7→ v. This time,S/R is etale and im φ = vk[x, x](v) = mS and so again by Example 5.4 AP isnormal. Same calculation as before shows AP is local and so AP is maximal.Hence A is maximal.

7 Complete normal orders in division rings

We give here some basic facts about complete normal orders in division ringsand show, in particular, they are principal ideal domains. Suppose R is acomplete discrete valuation ring, and A be an R-algebra, finitely generatedas an R-module. For any N < rad A, A is complete with respect to theN -adic topology.

Lemma 7.1. With the above notation, let ε1, . . . , εn ∈ A/N be a complete setof idempotents. Then there exists a complete set of idempotents ε1, . . . , εn ∈A such that εi +N = εi for i = 1, . . . , n.

Proof. (Sketch) See Theorem 6.7 of [CurtisReiner]. Main idea is to lift in-ductively ε1 to A/N2 then A/N4, A/N8, . . .. Then repeat with (1−ε1)A(1−ε1).

Lemma 7.2. Let (R,m) be a complete dvr and A be an R-order in a divisionring. Then A = A/rad A is also a division ring.

Proof. First A is a semisimple ring, so is a product of matrix algebras overdivision rings. Hence A fails to be a division ring if and only if there existsa nontrivial idempotent ε. Given one such idempotent, we can lift this toε ∈ A (see [Lam91] Theorem 21.31). Since ε(1 − ε) = 0 ∈ K(R) ⊗R A, thiscontradicts the fact that A is a division ring.

22

Proposition 7.3. Let (R,m) be a dvr and A be a maximal R-order such thatA := A/rad A is a division ring (e.g. R is complete and K(R) ⊗R A is adivision ring). Then

1. A is normal so rad A = Aπ for some π ∈ A.

2. For any b ∈ D := K(R)⊗RA−{0}, we can write uniquely b = uπs forsome u ∈ A∗ and s ∈ Z.

3. D is a division ring.

4. Any non-zero left (or right) ideal I < A has the form I = (rad A)s forsome s ∈ Z.

5. If mA = (rad A)e, then e · dimR/m A = rankRA.

Proof. 1. Note that J := rad A is nilpotent modulo m so there is a min-imal s ∈ N such that Js ≤ mA. Pick b ∈ Js−1 − mA and let p be auniformising parameter of m. Recall that Ol(J) = {α ∈ D | αJ ⊆ J} isan R-order in D containing A. Since A is maximal, we have Ol(J) = A,so

p−1bJ ⊆ p−1Js ⊆ A.

We wish to show that p−1bJ = A. If p−1bJ ⊆ J , then p−1b ∈ Ol(J) =A, which is false, so p−1bJ * J . Since A/J is a division ring, the onlyproper right ideals of A are those contained in J . Since p−1bJ + J is aright ideal of A (and is not contained in J), p−1bJ + J = A, hence byNakayama’s lemma p−1bJ = A and so J = b−1pA.

2. Recall that D = A[π−1] and so b πs1 ∈ A for some s1 ∈ Z. NowπA⋂i∈N

πiA =⋂i∈N

πiA and so must equal 0 by Nakayama’s lemma.Thus there is a minimal s2 such that bπs1 6∈ πs2+1A. Therefore u :=bπs1−s2 ∈ A − J , so b = uπs2−s1 and u ∈ A∗ since A/J is a divisionring.

3. Note that b = uπs is invertible in D.

4. Pick b = uπs ∈ I − {0} such that s is minimal with u ∈ A∗. Thenclearly I ⊆ πsA and also πsA ⊆ I. Therefore I = (rad A)s.

5. Consider the filtration A ⊃ J ⊃ J2 ⊃ · · · ⊃ Je = mA which hasfactors J i/J i+1 = πiA/πi+1A which is isomorphic, as R/m-modules, to

23

A/πA = A, Then

rankRA = dimR/mA/mA

=∑

dimR/m(J i/J i+1)

= e · dimR/m A.

8 Structure theory of complete division alge-

bras

Let (R,m) be a complete dvr in this section. Let κ = R/m be the residuefield, which we will assume to be perfect starting from Section 8.2. Themain goal is to classify division rings over K = K(R). We will follow Serre’s[Ser79] Galois cohomological approach.

8.1 Existence of unramified splitting fields

Consider a K-central division ring D. Recall that a finite commutative ringextension S/R is unramified if S/mS is a product of separable field exten-sions of R/m. If S is a domain, we also say that K(S) is an unramified fieldextension of K.

Theorem 8.1. There exists an unramified splitting field for D if either

1. κ is perfect, or

2. deg D is coprime to char κ.

Proof. We may as well assume D 6= K. We will need the following twolemmas

Lemma 8.2. If D 6= K, then given a maximal R-order A ⊂ D, we haveA/rad A 6= R/m.

Proof. From Section 7 Proposition 7.3, we know that A is normal so rad A =Aπ for some π. Let’s assume that A/rad A = R/m. We wish to derivethe contradiction that then A = R[π] which is nonsense since A is notcommutative. Certainly, R[π] ⊆ A. By Nakayama’s lemma, it suffices toshow that the map R[π]/mR[π] = κ[π] → A/mA is an isomorphism of κ-modules. Note that by Section 7 Proposition 7.3(5) that if e := dimκ κ[π]

24

then dimκA/mA = e · dimκA/rad A = e by assumption. Thus suffice toshow the map is injective. Suppose f := a1π

n1 + · · · + amπnm maps to

0 with ai 6= 0 and n1 < · · · < nm. Let choose an ai ∈ R such thatai = ai in κ and note that ai /∈ m and so is invertible. Since f mapsto 0, f := a1π

m1 + · · · + amπnm = πn1(a1 + · · · + amπ

nm−n1) ∈ mA. Buta1 + · · · + amπ

nm−n1 is invertible in A and so πn1A ⊆ mA. Thus n1 ≥ e i.e.f = 0

Lemma 8.3. D has a non-trivial unramified subfield.

Proof. We know from Section 7 Lemma 7.2 that A := A/radA is a divisionring and it is a non-trivial extension of κ by Lemma 8.2. Pick a ∈ A− rad Aand let a ∈ A be its image. We claim that κ[a]/κ is separable. This isclear under hypothesis 1. If, on the other hand, hypothesis 2 holds, thenby Section 7 Proposition 7.3-5, deg A is coprime to char κ. So the minimalpolynomial f(x) ∈ κ[x] of a is separable. Note that R[a] is finitely generatedover R so is complete. So by Hensel’s lemma, we can change a so that it isa root of a monic polynomial with the same degree as f(x). Hence we get anon-trivial unramified extension.

We now resume the proof of Theorem 8.1. We prove, by induction ondeg D, that D has a maximal unramified subfield, which must therefore alsobe splitting by Corollary 3.17 of [FD93].

If deg D = 1 the result is obvious, for D is then split and is a field. Ifnot, by Lemma 8.3, there is a subfield F ⊂ D such that F/K is a non-trivialunramified extension. We seek to extend F to a maximal subfield K ′ ⊂ Dwith K ′/K unramified, for as before, it will necessarily be a splitting field.Let C = CD(F ) be the centraliser of F in D. Recall that C is in fact acentral simple F -algebra (see [FD93] Theorem 3.15 for the proof). Note thathypotheses 1 or 2 carry over to C/F . Hence, by induction, we can find anunramified maximal subfield K ′ ⊂ C for C/F . Note that K ′/K is unramifiedsince it is a tower of unramified extensions K ⊂ F ⊂ K ′. Thus we have

K ⊂ F ⊆ K ′ ⊆ CD(F ) ⊆ D.

We claim that K ′ is in fact a maximal subfield of D. To see this we makethe following computation:

[K ′ : K]2 = [K ′ : F ]2[F : K]2

= [C : F ][F : K][F : K] since K ′/F is maximal

= [C : K][F : K]

= [D : K] by Theorem 3.15(3) in [FD93].

25

8.2 Brauer groups of complete discrete valuation rings

From here on we assume κ is perfect. Recall that the Brauer group Br Rconsists of equivalence classes of Azumaya R-algebras, where A ∼ A′ if thereexists n, n′ such that Mn(A) ≃ Mn′(A′). Multiplication is given by tensorproduct. There is a natural group homomorphism

Br R −→ Br κ

[A] 7−→ [A⊗R κ]

In this section, we show that this is an isomorphism. Consider an Azu-maya R-algebra A. It is split by an etale extension S/R. We can assumethat S is also a complete dvr.

Lemma 8.4. rad S = mS

Proof. Since S/R is unramified, S/mS ≃ κ1 × · · · × κn where κi are fieldextension of κ. But S is a dvr and hence has no idempotents thus m = 1.Also mS ⊆ rad S an so the result follows since S/mS → S/rad S is anisomorphism.

Finally, by replacing S with the integral closure of S in K(S)′, whereK(S)′ is the Galois closure of K(S)/K, we may assume that S/R is Galois,say with Galois group G. Let κS = S/mS be the residue field of S then sinceS/R is unramified, Gal(κS/κ) is naturally isomorphic to G. As for fields,the part of Br R split by S/R is given by H2(G,S∗). Note that we have anatural map S∗ −→ κ∗.

Theorem 8.5. Assume κ is perfect ??

1. The canonical map H2(G,S∗) −→ H2(G, κ∗S) is an isomorphism.

2. The natural map Br R −→ Br κ is an isomorphism.

Proof. Since (1) implies (2) on taking the limit over G’s, we just need toprove 2. Let mS = qS for some uniformising parameter q. Consider thesubgroups S∗i := 1 + qiS < S∗ and note that these are in fact G-submodulessince q ∈ R is fixed by G.

Lemma 8.6. Let s, s′ ∈ S∗. We have s ≡ s′ mod S∗i if and only if s ≡ s′

mod qiS.

Proof. s ≡ s′ mod S∗i precisely when s/s′ ∈ S∗ i.e. s/s′ = 1 + qis′′ for somes′′ ∈ S. Thus s − s′ = qis′′s ∈ qiS. Conversely, s ≡ s′ mod qiS meanss− s′ = qis′′ for some s′′ ∈ S and so s/s′ = 1 + qis′′/s′ ∈ S∗i .

26

Thus the following is exact

1 −→ S∗1 −→ S∗ −→ κ∗S −→ 1.

since by Lemma 8.6 s ≡ 1 mod qS if and only if s ≡ 1 mod S∗1 . The theoremfollows from the long exact sequence in cohomology and the following

Theorem 8.7. H i(G,S∗1) = 0 for i ≥ 1.

Proof. We start with the graded pieces of the filtration S∗1 > S∗2 > · · ·

Lemma 8.8. H i(G,S∗j /S∗j+1) = 0 for i, j ≥ 1.

Proof. We first claim that the following is an isomorphism of G-modules:

κS −→ S∗j /S∗j+1

α 7−→ 1 + αqj.

To see this, note that by Lemma 8.6 1 + (α+ β)qj ≡ (1 +αqj)(1 + βqj) modS∗j+1 since αβq2j ∈ qj+1S. Thus this is indeed a morphism of abelian groups.Since, σ(1 + αqj) = 1 + σ(α)qj for all σ ∈ G this is indeed a morphismof G-modules. It is clearly both surjective and injective and hence is anisomorphism.

So suffices to show that H i(G, κS) = 0. Let HomZ (ZG, κ) be the abeliangroup of Z-linear G-module homomorphisms ZG→ κ. It is a G-module, withG-action given by τ.f(σ) := f(στ) for all σ, τ ∈ G. Now κS/κ is Galois withGalois group G so that it has a normal basis, that is a κ-basis {σ(β) | σ ∈ G}of κS for some β ∈ κS.

We also have an isomorphism of G-modules

φ : HomZ(ZG, κ) −→ κS

f 7−→∑

σ∈G

f(σ)σ−1(β).

To see that this is indeed a G-module homomorphism, note that

φ(τ.f) =∑

σ∈G

(τ.f)(σ)σ−1(β) =∑

σ∈G

f(στ)σ−1(β)

letting σ′ = στ we get

∑

σ′∈G

f(σ′)(τσ′−1)(β) = τ

(∑

σ∈G

f(σ)σ−1(β)

)= τ(φ(f))

27

Let P• −→ Z be a projective resolution of ZG-modules. Then

H i(G, κS) = ExtiZG(Z, κS)

= H i(HomZG(P•,HomZ(ZG, κ)))

= H i(HomZ(P•, κ))

= ExtiZ(Z, κ)

= 0

for i > 0.

We resume the proof of Theorem 8.7. Note first that

S∗1 = lim←−j

S∗1/S∗j .

Given an i-cocycle f : Gi −→ S∗1 , we need to find an (i − 1)-cochain h :Gi−1 −→ S∗1 with dh = f . It suffices to find a family of (i − 1)-cochains{hj : Gi−1 −→ S∗1/S

∗j } such that

1. They are compatible in the sense that hj is the composite

Gi−1 hj+1

−→ S∗1/S∗j+1 −→ S∗1/S

∗j

2. dhj is the composite

Gi f−→ S∗1 −→ S∗1/S

∗j

We prove this by induction on j so assume h1, . . . , hj are already defined.First lift hj arbitrarily to h′j+1 : Gi−1 −→ S∗1/S

∗j+1. Note that by 2 for hj we

have

f−1dh′j+1 : Gi −→ S∗j /S∗j+1

an (i− 1)-cocycle. By Lemma 8.8,

f−1dh′j+1 = dh′′j+1

therefore hj+1 := h′j+1(h′′j+1)

−1 works.

28

8.3 Witt exact sequence

Let (R,m) be a complete dvr, κ = R/m be the residue field which we assumeto be perfect, and K = K(R). Here we examine the Brauer group of K.Recall that

Br K =⋃

F/K

Galois fieldextension

H2(Gal(F/K), F ∗)

By Theorem 8.1, we need only look at the union over finite unramified split-ting fields. Let F/K be a finite unramified field extension which is Galoiswith Galois group G. Let S/R be the integral closure of R in F and re-call S/R is Galois (Fact 1.3(3)). By Chapter 1 Proposition 8 of [Ser79] Sis Dedekind domain. Furthermore, S is local with maximal ideal mS for ifnot, since S/R is unramified S/mS would not be a field (but a non trivialproduct of field extensions). But S is mS-adically complete by Lemma 21.3of [Lam91] and so one could would be able to lift the idempotents of S/mSto S (Theorem 21.31 [Lam91]). This would contradict S being a domain.Also the residue field κS = S/mS is Galois over κ with Galois group G.

We claim that the following is an exact sequence of G-modules

1 −→ S∗ −→ K(S)∗ν−→ Z −→ 0

where ν is the valuation and G acts trivially on Z. To see this, let p be auniformising parameter of m. Then for all σ ∈ G and u ∈ S∗, σ(pru) = prσ(u)with σ(u) ∈ S∗. Thus ν(σ(s1/s2)) = ν(s1/s2) and so the sequence is indeeda sequence of G-modules. It is clearly exact. Finally, the map ν is split bysending 1 ∈ Z to p. Hence we also have a split exact sequence

0 −→ H2(G,S∗) −→ H2(G,K(S)∗) −→ H2(G,Z) −→ 0

By Theorem 8.5, we have H2(G,S∗) ≃ H2(G, κ∗S). We also have

Lemma 8.9. H2(G,Z) ≃ Hom(G,Q/Z).

Proof. We use the following exact sequence

0 −→ Z −→ Q −→ Q/Z −→ 0.

First note that H i(G,Q) = 0 for all i since H i(G,Q) is torsion (Theorem4.14 of [FD93]) and multiplication by n ∈ Z+ is an isomorphism. This givesH2(G,Z) ≃ H1(G,Q/Z). The latter can be identified with Hom(G,Q/Z)since a 1-cocycle f G→ Q/Z satisfies (df)(σ, τ) = σ.f(τ)− f(στ) + f(σ) =f(τ) − f(στ) + f(σ) = so f is a group homomorphism and given a zerocochain a ∈ Q/Z, (da)(σ) = σ.a− a = a− a = 0.

29

Taking the limit over G, we obtain

Theorem 8.10. (Witt) Suppose κ is perfect. Then there is a split exactsequence

0 −→ Br κ −→ Br K → lim−→G

Hom(G,Q/Z) −→ 0

Remark 8.11. We have

lim−→G

Hom(G,Q/Z) = Homcont(Γ,Q/Z)

where Γ = Gal(κ/κ) is the absolute Galois group of κ.

8.4 Classification of division rings over complete dis-crete valuation rings

We continue to assume that κ is perfect. Suppose further that Br κ = 0.Then by the Theorem 8.10 and Remark 8.11, we have Br K ≃ Homcont(Γ,Q/Z)where Γ is the absolute Galois group of κ.

Let D be a K-central division ring. We can assume it is split by a Galoisunramified extension S/R with Galois group G. We wish to describe D.We have seen that [D] ∈ Br K corresponds to some group homomorphismf : G → Q/Z. Note that the image of f is a cyclic group G′ ≃ G/H say oforder n. We have the following commutative diagram

H2(G′, K(SH)∗)∼−→ Hom(G′,Q/Z)

inf ↓ ↓inf

H2(G,K(S)∗)∼−→ Hom(G,Q/Z)

where the inflation map in Galois cohomology

H i(G/H,MH) −→ H i(G,M)

is just the edge map in the spectral sequence

Ep,q2 = Hq(G/H,Hq(H,M)) =⇒ Hp+q(G,M).

(can get this spectral sequence either from Grothendieck spectral sequenceor change of rings for Ext)

The commutative diagram shows that [D] comes from a 2-cocycle overG′, so we can assume G is cyclic of order n and f is injective. Let’s computeD by tracing through the isomorphisms

Hom(G,Q/Z)δ≃ H2(G,Z) ≃ H2(G,K(S)∗).

30

In particular, we need δf ∈ H2(G,Z). Note that f is completely determinedby G and the generator σ = f−1(1/n + Z). Pick a 1-cochain f ′ : G → Qsending σi 7−→ i/n for 0 6 i < n, and

(δf)(σi, σj) = σif(σj)− f(σi+j) + f(σi)

=

{jn− i+j

n+ i

n= 0 if i+ j < n

jn− i+j−n

n+ i

n= 1 if i+ j > n

Now consider the isomorphism H2(G,Z) ≃ H2(G,K(S)∗) given by the split-ting map K(S)∗

ν→ Z say by choosing a uniformising parameter p for R

(hence S) and using the map Z → K(S)∗ given by i 7−→ pi. Then the2-cocycle corresponding to f is

g(σi, σj) =

{1 if i+ j < np otherwise

We get a crossed prodcut algebra which is actually cyclic.

D = K(S)u0⊕K(S)u1

⊕ · · · ⊕K(S)un−1

with uiα = σi(a)ui and

uiuj =

{ui+j if i+ j < npui+j otherwise

Note that

D ≃K(S)[u;σ]

(un − p)

= A(K(S)/K,K(S)σ, p)

Note that [D] has period n and degree n so is a division algebra.

Theorem 8.12. Assume that κ is perfect and Br κ = 0. Then any K-central division ring is cyclic of the form A(K(S)/K,K(S)σ, p) for somecyclic unramified extension K(S)/K and p a uniformising parameter for R.

8.4.1 Algebraic interpretation of residue map when Br κ = 0

Recall that from the Witt exact sequence we have a surjective residue map

ρ : Br K −→ Homcont(Γ,Q/Z)

where Γ = Gal(κ/κ). Let β ∈ Br K which is represented by a cyclic divisionring

Dβ := A(K(S)/K,K(S)σ, p)

31

which, as we just saw, is possible provided Br κ = 0. We also saw that ρ(β) =group homomorphism G→ Gal(S/R) → Q/Z given by σ 7−→ 1/n + Z. Wehave another interpretation of ρ(β) as follows: pick the unique maximal orderAβ in Dβ. Then Gal(S/R) is Gal(κS/κ) which is given by κS ≃ Aβ/rad Aβ.Suppose that

Aβ =S[u;σ]

(un − p)

so an arbitrary uniformising parameter has the form vu for some v ∈ A∗.Then σ is induced by conjugation by u. This does not depend on the choiceof uniformising parameter, so σ is conjugation by any generator of rad Aβ.

Remark 8.13. If n = [κS : κ] = deg Dβ then we also have

(rad Aβ)n = unAβ = mAβ.

Remark 8.14. If we drop the Br κ = 0 hypothesis, we still get a splittingof the map H2(G,K(S)∗) → Hom(G,Q/Z). Given a cyclic quotient G ofGal(κ/κ) and generator σ representing element of Hom(G,Q/Z) we map itto A(K(S)/K,K(S)σ, p)

0 −→ Br κ −→ Br K −→ Homcont(Γ,Q/Z) −→ 0

8.5 Classification of normal orders over complete dis-crete valuation rings in division rings

In this section, as the title suggests, we will classify all the normal ordersover complete dvrs (with perfect residue fields) inside division rings. As weshall see there is only one such order which consequently implies it must bemaximal.

Proposition 8.15. Let R be a complete dvr, with a perfect residue field κ.If D is a K-central division ring then it has a unique normal R-order in itwhich is consequently the unique maximal order in it.

Proof. Let A,B be normal R-orders in D. Note that by Lemma 7.2 A,Bare local and so by Proposition 6.6 are maximal. Suppose A 6= B. Letrad A = Aπ. Pick b ∈ B −A, from Proposition 7.3(2) we know that b = uπs

for some u ∈ A∗ and s < 0. Now AB is a finitely generated R-modulecontaining A and bi for all i > 0 and henceK(R) which is a contradiction.

32

Continuing with the notation from the above proposition, if we furtherassume that Br κ = 0 then by Theorem 8.12 D is cyclic of the form

D = A(K(S)/K,K(S)σ, p)

where S/R is a cyclic etale extension say of degree n and σ is a generator ofG := Gal(S/R). We wish to find all maximal orders in D.

Theorem 8.16. The unique maximal (or normal) order in the cyclic di-vision algebra A(K(S)/K,K(S)σ, p) is A := A(S/R, Sσ, p) where p is theuniformising parameter for R. Furthermore A/rad A ≃ κS the residue fieldof S.

Proof. We know that A is an R-order in D. It suffices to show that it isnormal with residue ring κS. Recall A = S[u;σ]/(un − p) so u is a normalelement which is nilpotent modulo m. Hence uA ⊆ radA. Now uA = rad Asince A/uA ≃ S/p ≃ κS which is semisimple.

9 Classification of normal orders and maxi-

mal orders in central simple algebras.

As in the previous section, we fix a complete dvr (R,m) with perfect residuefield κ = R/m and K := K(R).

A consequence of the Artin-Wedderburn theorem is the fact that everycentral simple algebra is isomorphic to Mn(D) for some division ring D.Having shown in Proposition 8.15 that any such D has a unique normal andhence maximal order in it, it is natural to extend the classification questionto orders in Mn(D). We begin by first classifying all such maximal orders.

Let A be a maximal order in Mn(D).

Proposition 9.1. Let D be a K-central division ring and B be the (uniqueby Proposition 8.15) maximal order in D.

1. Then A is conjugate to Mn(B).

2. In particular, A is local, normal and A/rad A ≃Mn(B/rad B).

Proof. We prove 1 since 2 follows immediately from the fact that rad Mn(B) =Mn(rad B) (see Example 7 of Chapter 4 in [Lam91]).

Consider the (Mn(D), D)-bimodule

Dn =

D...D

.

33

Can also consider the (A,B)-subbimodule

M = A

B...B

⊂ Dn.

This is a finitely generated torsion-free R-module, so since B is normal,Proposition implies 5.6 M is projective over B. Now MB := M/M(rad B) ≃κmB , by the Krull-Schmidt theorem and the fact that κB := B/rad B isa simple, which we know from Lemma 7.2. Thus both MB and Bm areprojective covers of κB and so by Proposition 21.3 we must have t : MB≃B

m

(and m = n). We have an injective algebra homomorphism ι : A → End MB.Since A is maximal ι is an isomorphism and A ≃ End MB ≃ Mn(B) withisomorphism given by conjugation by t.

Having classified all the maximal orders, we turn our attention to theslightly more general case of normal orders. Before stating and proving theclassification theorem, we need some preliminary results.

Proposition 9.2. Let B be an R-algebra which is finitely generated as anR-module. Then any simple B-module S has an indecomposable projectivecover ϕ : P → S. Conversely, any indecomposable projective B-module P isthe projective cover of the simple B-module P/(rad B)P .

Proof. Let B = B/rad B. By Proposition 4.8 of [Lam91] B and B havethe same simple modules. Furthermore, since B is finitely generated overR, B/mB is finitely generated over R/m and is thus artinian. Since B isa quotient of B/mB it is also artinian and since it had trivial Jacobsonradical it must be semisimple by Theorem 4.14 in [Lam91]. Hence there isan idempotent ε ∈ B such that S ≃ Bε. We may lift the idempotent ε toε ∈ B and let P = Bε. Recall that in Example 21.2 we saw that P is aprojective cover of P/(rad B)P which is isomorphic to Bε/(rad B)ε ≃ Bε ≃S. Suppose P = P1 ⊕ P2. Then restricting ϕ : P → S to P1 and P2 we seethat at least one of them, say P1, surjects onto S (since S is simple). But ϕis a projective cover so P1 = P and P is indecomposable.

Now let P be any finitely generated projective and write P/(rad B)P ≃⊕Si where Si are simple. If Pi is the projective cover of Si, then by uniquenessof projective covers P ≃ ⊕Pi. Hence if P indecomposable then P is theprojective cover of a simple.

Let A be a normal R-order with rad A = πA.

Proposition 9.3. Let P be an indecomposable projective A-module. Thenthe Mn(D)-module K ⊗R P is isomorphic to Dn.

34

We have thus established a bijection between indecomposable projectivemodules of B and the simple B-modules: given a simple module we haveshown it has a unique indecomposable projective cover and conversely, givenan indecomposable projective module, up to isomorphism it is the projectivecover of only one simple module, namely P/(rad B)P .

Proof. Since Mn(D) is simple artinian, K⊗RP ≃ (Dn)s for some s. Consider

the composite map ψ : P → K ⊗R P∼→ (Dn)s

ρ→ Dn where ρ is projection

onto one of the factors. Then im ψ is non-zero and torsion-free so by Propo-sition 5.6 is projective. Thus P = ker ψ ⊕ im ψ. But P is indecomposableso we must have s = 1.

The following lemma describes the indecomposable projective modules ofan R-order. We will subsequently use them to prove the structure theorem.

Lemma 9.4. Suppose the number of simple A-modules (up to isomorphism)is r. If P is an indecomposable projective, then

1. up to isomorphism, the indecomposable projectives are given by P, πP, . . . , πr−1Pand P ≃ πrP , and

2. any non-zero submodule P ′ < P has the form πsP for some s.

Proof. Note that Aπ ⊗A − : A−Mod → A−Mod is a category equivalence,and that for any torsion-free module M we have Aπ ⊗A M ≃ πM . Hencemultiplication by π permutes the indecomposable projectives. From Propo-sition 9.3, we know that there are r incomposable projectives so it suffices toshow that there is only 1 orbit of them.

Let P ′ be an indecomposable projective. Then

K ⊗R HomA(P ′, P ) ≃ HomK⊗RA(K ⊗R P′, K ⊗R P ) 6= 0

so there is a non-zero homomorphism ϕ : P ′ → P . Again im ϕ is torsion-freeand therefore projective so P ′ ≃ ker ϕ⊕ im ϕ. Since P ′ is indecomposable,ker ϕ = 0 and so P ′ is isomorphic to a submodule of P . So part 1 followsfrom part 2 so we prove the latter.

Let P ′ < P . Now K⊗RP ≃ Dn, a simple K⊗RA-module, so P/P ′ mustbe torsion. Hence we can find s maximal such that P ′ 6 πsP . Consider thecomposite map ψ : P ′ → πsP → πsP/πs+1P . Note that πsP → πsP/πs+1Pis the projective cover of the simple module πs(P/πP ). Now maximality ofs ensures that im ψ 6= 0 so ψ is surjective. Then by definition of projectivecovers, P ′ = πsP .

35

Theorem 9.5. Let A be a normal R-order in Mn(D) as above with exactlyr simples. Let B be the unique maximal order in D. Then

A ≃ Mn/r(R)⊗R A0

where

A0 =

B · · · · · · B

rad B. . . . . .

......

. . . . . ....

rad B · · · rad B B

⊂Mr(B).

Furthermore

rad A = Mn/r(R)⊗R rad A0.

Proof. Let P, πP, . . . , πr−1P be the indecomposable projective and write

A ≃r−1⊕

i=0

(πiP )⊕ni .

Note that (Dn)n ≃ Mn(D) ≃ K ⊗R A ≃ ⊕r−1i=0 (K ⊗R π

iP )⊕ni ≃ ⊕(Dn)⊕ni ,so∑ni = n. We wish to show that all the ni’s are equal so in fact

A ≃

(r−1⊕

i=0

πiP

)⊕n/r.

Let Si = πiP/πi+1P be the simple corresponding to πiP . Then A/rad A ≃∏r−1i=0 Mni

(EndASi). Now Aπ ⊗A − permute the simples transitively, so con-jugation by π shows that all the factors Mni

(EndASi) are in fact isomorphic.Hence all ni = n/r.

There is a Peirce deomposition as follows

A ≃ HomA (A,A)

≃ HomA

((⊕r−1i=0π

iP)⊕n/r

,(⊕r−1i=0π

iP)⊕n/r)

≃Mn/r(R)⊗R

r−1⊕

i,j=0

HomA(πiP, πjP )

≃Mn/r(R)⊗R

r−1⊕

i,j=0

HomA(πi−jP, P ).

The theorem follows from the following lemma.

36

Lemma 9.6. With the above notation,

HomA(πiP, P ) ≃

{B if 0 6 i < rrad B if −r 6 i < 0

.

Proof. Let B′ = EndAP . Note that B′ is an R-order in D since K ⊗R B′ ≃

HomK⊗RA(K ⊗R P,K ⊗R P ) = HomMn(D)(Dn, Dn) = D. We show that it is

maximal using the criterion from Proposition 6.6 by showing it it is normaland local.

Consider the exact sequence

0 −→ πP −→ P −→ P/πP −→ 0

and we get an exact sequence

0 −→ HomA(P, πP ) −→ B′ϕ−→ HomA(P, P/πP ) −→ 0

with HomA(P, P/πP ) ≃ HomA (P/πP, P/πP ) ≃ EndAS where S := P/πP issimple by Proposition 9.2. Now ϕ is an algebra map, so J ′ := HomA(P, πP ) isan ideal. Since for any s > 0, J ′s ⊆ HomA(P, πsP ) and hence for sufficientlylarge s, J ′s ⊆ HomA (P,mP ) = HomA (P, P ) m and so we see that J ′ isnilpotent modulo m and so J ′ ⊆ rad B′. But B′/J ′ ≃ EndAS which is adivision ring by Schur’s Lemma, and is thus semisimple. Therefore B′ islocal with J ′ = rad B′.

We now show that B′ is normal. Let f ∈ HomA(P, πP ) be a non-zeromap. Then arguing as previously, we see that f gives an isomorphism of Pwith im f 6 πP . From Lemma 9.4, we know that im f = πrmP for somem ∈ Z+. Hence J ′ = HomA(P, πrP ). But P ≃ πrP so J ′ = HomA(P, πrP ) ≃HomA(P, P ) = B′ and so J ′ is local.

Thus we have shown that B′ = EndAP = B, the unique maximal orderin D. Furthermore, rad B = HomA(P, πP ).

The above argument also shows that

HomA(πiP, P ) = HomA(P, π−iP ) =

{HomA(P, P ) if 0 6 i < rHomA(P, πrP ) if −r 6 i < 0

.

10 Ramification Theory

In algebraic geometry, ramification is where a map fails to be etale. Recallthat we have introduced the concept of ramification for orders in Example

37

2.6. The ramification locus of an oder is where the order fails to be Azumaya.In this chapter we explore this concept further: first for maximal and then,more generally, for normal orders.

As before, we let (R,m) be a discrete valuation ring with perfect residuefield κ = R/m and fraction field K. Let R and K be their completions andΓ = Gal(κ/κ) be the absolute Galois group.

10.1 Ramification theory via cohomology

Consider the commutative diagram below

Br R −→ Br K↓ ↓

0 −→ Br R −→ Br Kρ−→ Homcont(Γ,Q/Z) −→ 0

where the bottom row is essentially the Witt exact sequence of Section 8.3since by Theorem 8.5 Br R ≃ Br κ and ρ is the residue map.

Proposition 10.1. Let A be a maximal R-order. Then A is Azumaya if andonly if A = R⊗R A is Azumaya if and only if ρ([K ⊗R A]) = 0.

Proof. The first part is clear since A⊗R R/m ≃ A⊗R R/m.

If ρ[K ⊗R A] = 0 then there exists an Azumaya R-algebra A′ such that[K ⊗R A

′] = [K ⊗R A]. By Corollary 4.10 A is a maximal R-order and

so by uniqueness of maximal orders (Proposition 9.1) A′ ≃ A and so A isAzumaya.

Upshot: so ρ([K ⊗R A]) = 0 measures the failure of A being Azumaya.Hence the residue map is called the ramification map.

10.2 Ramification theory via algebra

We wish to interpret the ramification map ρ in Section 10.1 algebraicallyas in Section 8.4.1 but now without the assumption Br R = 0. We will seeprecisely how ρ([K⊗RA]) measures the failure of A/mA from being a centralsimple κ-algebra.

Proposition 10.2. An R-order A is maximal if and only if it is normal andlocal.

Proof. The fact that if A is local and normal implies that it is maximal isproved in Proposition 6.6.

38

We now prove the converse. Suppose A is maximal. Let A = R ⊗RA, J = rad A = R ⊗R rad A. From Proposition 9.1, we know that A isnormal and local. Since A/rad A = A/rad A we see that A is also local.Furthermore J/J2 ≃ J/J2 which is generated by one element since J isprincipal. Nakayama’s lemma implies J is also generated by a single elementso A is normal too.

We now state the main theorem.

Theorem 10.3. Let A be a maximal R-order. Then

1. κ′ = Z(A/rad A) is a cyclic field extension of κ.

2. Let rad A = πA (possible by Proposition 10.2). Let σ ∈ Γ = Gal(κ′/κ)be the automorphism induced by conjugation by π (which induces anautomorphism of A/πA). Then ρ([K ⊗R A]) ∈ Homcont(Γ,Q/Z) is theelement corresponding to the cyclic extension κ′/κ with generator forGalois group σ ∈ Gal(κ′/κ).

We first need the following lemma.

Lemma 10.4. With the above notation,

1. κ′ is independent of the choice of maximal order A. More precisely, itdepends only on the Brauer class [K ⊗R A] ∈ Br K.

2. The automorphism σ, which is mentioned in the theorem, is indepen-dent of A and the choice of uniformising parameter π.

Proof. 1. K ⊗R A is a central simple R-algebra and so by the Artin-Wedderburn theorem there exists a unique R-division algebra D suchthat K ⊗R A = Mn(D). Let B be the unique maximal order in Dwhich is possible by Proposition 8.15. Then Proposition 9.1 impliesthat A := R ⊗R A ≃ Mn(B). Therefore Z(A/rad A) = Z(A/rad A) =Z(Mn(B/rad B)) = Z(B/rad B). But, as we just saw, B only dependson [K ⊗R A] ∈ Br K.

2. From part 1 Γ is independent of the choice of A. To show that it isindependent of the choice of uniformising parameter, suppose we changeπ to uπ ∈ A where u ∈ A×. Let u ∈ A/rad A be the image of u inA/rad A. Conjugation by uπ is the composition of σ and conjugationby u. But conjugation by u acts trivially on Z(A/rad A) so part 2 isproved.

39

Given any Brauer class β ∈ Br K, we need only prove Theorem 10.3 forsome maximal order A with [K ⊗R A] = β. This follows from the followingproposition.

Proposition 10.5. Let β ∈ Br K. Let ρ(B) ∈ Homcont(Γ,Q/Z) be givenby cyclic extension κ′/κ and generator σ ∈ Gal(κ′/κ). Let S/R be the cyclicetale extension with residue field extension κ′/κ and lift σ to S so R = S〈σ〉.Then

1. There exists an Azumaya R-algebra A0 such that

[K ⊗R A0 ⊗R A1] = β

where A1 is the cyclic algebra A(S/R, Sσ; p) and pR = m, that is, thecyclic algebra of Theorem 8.16.

2. A0 ⊗R A1 is a maximal order and rad(A0 ⊗ A1) = A0 ⊗ rad A1.

Proof. 1. We recall the Witt exact sequence 0 → Br R → Br Kρ→

Homcont(Γ,Q/Z) → 0 (where ρ([β]) = (κ′/κ, σ)) is split and in factfrom Section 8.4.1, we know ρ([K ⊗R A1]) = ρ(β). Hence ρ(β[K ⊗RA1]−1) = 0 and is represented by an Azumaya algebra A0.

The proposition and hence theorem now follows from:

Proposition 10.6. Let A0 be an Azumaya R-algebra and A1 be a maximalR-order. Then A0⊗R A1 is a maximal R-order with Jacobson radical A0⊗Rrad A1.

Proof. We use the criterion for maximality proved in Proposition 6.6. Notethat from Lemma 5.2 we know that A0 ⊗R rad A1 is nilpotent modulo m, soA0 ⊗R rad A1 ⊆ rad(A0 ⊗R A1). Also

B :=A0 ⊗R A1

A0 ⊗R rad A1

≃ A0 ⊗RA1

rad A1

≃A0

mA0

⊗κA1

rad A1

which we claim is a central simple κ′-algebra where κ′ = Z(A1/rad A1). Tosee this, recall that by defintion of Azumaya, A0/mA0 is a central simpleκ-algebra and so A0/mA0 ⊗κ κ

′ is a central simple κ′-algebra. Furthermore,A1/rad A1 is also a central simple κ′-algebra and so B must be a centralsimple κ′-algebra also since it is the tensor product of two such algebras.Hence A0 ⊗R rad A1 = rad(A0 ⊗R A1) and A0 ⊗ A1 is local. Furthermore,rad A1 is a principal ideal of A1 so A0 ⊗R rad A1 is a principal ideal ofA0 ⊗R A1. Hence A0 ⊗R A1 is maximal.

40

10.3 Ramification Index

We have seen that a maximal R-order A which is not Azumaya fails to beso: A/mA is not central simple over κ but in fact A/rad A is central simpleover some nontrivial field extension κ′/κ. It also turns out that we fail tohave rad A = mA. The ramification index measures both these failuressimultaneously.

Theorem 10.7. Let A be a maximal R-order and κ′ = Z(A/rad A). Lete = [κ′ : κ]. Then mA = (rad A)e. We call e the ramification index of A. Ife > 1 then we say that A is ramified.

Proof. Note that e and the equation mA = (rad A)e remain unchanged ifwe replace A with its completion, so we will assume that R is complete.Then A ≃ Mn(B) where B is a maximal R-order in a division ring D. Weknow from Proposition ? Section 7 that mB = (rad B)e

′

for some e′. HencemA = (rad A)e

′

, so we may replace A with any maximal order A′ suchthat A′ ⊗R K has the same Brauer class as D. From Section 10.2, one suchmaximal order is A′ = A0 ⊗ A1 where A0 is Azumaya over R and A1 is thecyclic algebra A1 = A(S/R, Sσ; p) where p = mR. In Section 8.5 we saw that(rad A1)

e = mA1, hence

rad(A0 ⊗R A1)e = (A0 ⊗R rad A1)

e

= A0 ⊗R mA1

= m(A0 ⊗R A1).

10.4 Ramification Theory for Normal Orders

Most of the ramification theory for maximal orders extend to normal orders.We sketch it here

Theorem 10.8. Let A be a normal R-order with r simples.

1. Then Z(A/rad A) is a cyclic extension of κ of the form κ′ =∏r

i=1 κ′′

where κ′′/κ is a cyclic field extension. If πA = rad A then conju-gation by π induces an automorphism σ ∈ Gal(κ′/κ) which generatesGal(κ′/κ). The ramification data of A is the cyclic extension κ′/κ andthe Galois automorphism σ. The ramification index of A is the integere = [κ′ : κ] = r[κ′′ : κ].

2. mA = (rad A)e.

41

Proof. (Sketch) As usual, we may assume that R is complete in which case wehave a structure theory for A (c.f. Theorem 9.5). Suppose that A⊗RK(R) =Mn(D) and B is the unique maximal order in D. Then

A ≃ Mn/r(R)⊗R

B B · · · B

rad B. . . . . .

......

. . . . . . Brad B · · · rad B B

also the Jacobson radical is generated by (1⊗ π)A where

π =

0 1 0 · · · 0

0. . . . . . . . .

......

. . . . . . . . . 0

0. . . . . . . . . 1

πB 0 · · · 0 0

and πBB = rad B. For part 1, we see that

A/rad A ≃ Mn/r(B)⊗R

B/rad B 0 · · · · · · 0

0. . . . . . . . .

......

. . . . . . . . ....

.... . . . . . . . . 0

0 · · · · · · 0 B/rad B

so

Z(A/rad A) =r∏

i=1

Z(B/rad B)

=r∏

i=1

κ′′

where κ′′ is a cyclic field extension (c.f. Theorem 10.3). Also conjugation byπ permutes the r copies of κ′′ cyclically and πr = πBIr. Part 2 follows from

(rad A)e = (1⊗ π)eA

= (1⊗ πr)[κ′′:κ]A

= (1⊗ π[κ′′:κ]B )A

= mA.

42

Corollary 10.9. (Boris: change this to Scholium) Let A be a normal R-orderwith ramification data κ′ =

∏ri=1 κ

′′ where κ′′/κ is a cyclic field extension andσ ∈ Gal(κ′/κ). Then the ramification data of any maximal order containingA is the cyclic field extension κ′′/κ and automorphism σr ∈ Gal(κ′′/κ).

We can now give the etale local structure of normal orders.

Proposition 10.10. Let A be a normal R-order with ramification index e.

1. Let S/R be an etale extension of discrete valuation rings. Then A⊗RSis a normal S-order with ramification index e too.

2. There exists an etale extension of discrete valuation rings S/R suchthat

A⊗R S ≃ Mn/e(S)⊗S

S S · · · S

n. . . . . .

......

. . . . . . Sn · · · n S

⊂Mn/e(S)⊗S Me(S)(2)

where n is the unique maximal ideal of S.

Proof. (Sketch)

1. We know from Section 5, that A⊗R S is normal. Let κS = S/n then

Z

(A⊗R S

rad(A⊗R S)

)≃ Z

(A

rad A⊗R S

)

= Z

(A

rad A⊗κ κS

)

= Z

(A

rad A

)⊗κ κS

which is a degree e extension of κS where e is the ramification index ofA.

2. Let A/rad A ≃∏r

i=1 κ′′ where κ′′ is a cyclic field extension of κ of degree

d. The primitive element theorem implies that κ′′ = κ[x]/(f) for somemonic f ∈ κ[x]. We lift f to a monic f ∈ R[x], then S := R[x]/(f)is a discrete valuation ring such that S/R is etale with residue fieldextension κ′′/κ. Then since κ′ ⊗κ κ

′ ≃∏

g∈G κ′, we have

A⊗R S

rad(A⊗R S)=

(r∏

i=1

κ′

)⊗κ κ

′

=r d∏

i=1

κ′.

43

Hence the maximal S-order B containing A⊗RS is unramified (becausethe ramification data of B is just κ′/κ′) and hence Azumaya. Hencewe can replace S with an etale extension so that B = Mn(S) for someS.

Now we try to repeat the proof in Section 9 as follows: A⊗RS ⊆Mn(S)so Sn is an indecomposable projective. If π generates rad(A⊗RS), thenthe other indecomposable projectives are πSn, π2Sn, so we can now usethe proof in Section 9.

10.5 Canonical Bimodule

In algebraic geometry, ramification theory can be studied via the canonicalbundle. Then same applies for orders. Let R be a commutative noetheriansemi-local ring and ωR be its canonical module. The only cases we will needare where R is regular local, in which case we can take ωR = R and moregenerally if R is an R0-algebra where R0 is regular local and R0 is a finite freeR-module. Then ωR is given by the adjunction formula ωR = HomR0

(R,ωR0).

(see Eisenbud or Bruns-Herzog). This suggests the following definition

Definition 10.11. Let R be a commutative noetherian normal local domainand A be an R-order, which is free as an R-module. We define its canonicalbimodule to be ωA := HomR(A,ωR). This is an A-bimodule.

Given an R-order A such that A is free over R, then ωA = HomR(A,R)and via the trace pairing, this is an A-subbimodule of K(R) ⊗R A. Weidentify this in a special case.

Proposition 10.12. Let (R,m) be a discrete valuation ring and A be anormal R-order with ramification index e. Let π ∈ A such that πA = rad A.Then the subbimodule ωA of D = K(R)⊗R A is

ωA = π−(e−1)A.

Proof. We seek to find the largest submodule N ⊂ D such that tr(N) ⊆ R.A submodule N ⊆ D satisfies tr(N) ⊆ R if and only if tr(N ⊗R S) ⊆ S.

This follows from linearity of trace and tr(N) ⊆ tr(N ⊗R S)∩K(R) = R. Sowe may replace R by an extension S.

By Proposition 10.10, we know that there is an etale extension of discretevaluation rings S/R such that (2) holds. Recall that rad(A⊗RS) = πA⊗RS.

44

Hence

π−(e−1)A⊗R S = rad(A⊗R S)−(e−1)

= q−1rad(A⊗R S)

= Mn/e(S)⊗S

q−1S q−1S · · · q−1S

S. . . . . .

......

. . . . . . q−1SS · · · S q−1S

. (3)

Let ω′ be the right hand side of (3), to finish the proof, we show that ωA = ω′.Note that tr(ω′) ⊆ S so ω′ ⊆ ωA. Now A has n idempotents ε1, . . . , εncorresponding to “diagonal” entries. This gives a Peirce decomposition

ωA =n⊕

i,j=1

εiωAεj

Since Mn/e(S) is a tensor factor of A this must have the form

ωA = Mn/e(S)⊗S

e⊕

i,j=1

qlijS ⊆Mn/e(S)⊗S Mn(K(S))

where dij 6 0 and if i < j we have dij 6 −1. Furthermore, tr(ωA) = S solii = 0. A quick computation using the fact that ωA is an A-bimodule showsthat we must have lij = 0 for i > j and lij = −1 for i < j, so ωA = ω′.

11 Uniqueness Theorem for Normal Orders

We show here that maximal orders over discrete valuation rings in a fixedcentral simple algebra are conjugate.

Theorem 11.1. Let (R,m) be a discrete valuation ring, K = K(R) and Dbe a central simple K-algebra.

1. Any two normal orders in D with the same ranification index are con-jugate (in D)

2. Any two maximal orders in D are conjugate.

Proof. Any two maximal orders in D have the same ramification index sopart 2 follows from part 1. By the structure theory in Section 9, we knowthat any two normal orders A,B in D have the same ramification index

45

satisfy A ≃ B ⊂ D. By Skolem-Noether there is a unit t ∈ D× such thatB = t At−1. By Lemma ?, t factorises as t = u v with u ∈ D× and v ∈ A×.Also by Section 4 proposition ??,

B = B ∩D

= t At−1 ∩D

= u v Av−1u−1 ∩D

= u Au−1 ∩ uD u−1

= u(A ∩D)u−1

= uAu−1.

Lemma 11.2. Given t ∈ D× we can factorise t as t = u v with u ∈ D× andv ∈ A×.

Proof. Note that A ⊂ D is a finitely generated R-module so t−1A ⊆ m−n+1Afor n≫ 0. Since D is dense in D, we can find u ∈ D such that u− t ∈ mnA.Note that t−1u− 1 ∈ t−1mnA ⊆ mA. Hence t−1u is an invertible element ofA say with inverse v ∈ A×. Since t−1u v = 1 we have u v = t. Note that u isinvertible in D, hence in D, so done.

Part 2.Most of the results and definitions in Sections 1 to 4 extend to the scheme

setting, that is, with the commutative noetherian normal domain replacedwith a normal integral noetherian separated scheme Z.

In particular, one can talk about Azumaya algebras, (maximal) orders,cyclic algebras, and Brauer groups. For example

Definition 11.3. Given a noetherian normal integral separated scheme Z,an OZ-order or order on Z, is a sheaf of algebras A on Z such that

1. A is coherent as a sheaf on Z

2. A is a torsion-free sheaf with A⊗ZK a central simple K algebra, whereK is the function field of Z.

For such orders we have results like the existence of maximal orders,maximality preserved when restricting to open sets, existence of reducedmap tr : A→ OZ etc. The definition of normality needs further explanation.

Definition 11.4. An order A on Z (as above) is normal if

46

1. at every codimension 1 point p ∈ Z, the localisation A ⊗Z OZ,p is anormal OZ,p-order.

2. A is reflexive as a sheaf on Z.

The motivation comes from Serre’s criterion for normality, which statesthat normality corresponds to

R1 regularity in codimension one

S2 Serre’s condition 2

The latter corresponds to reflexivity in our case.Standing hypotheses: In this part we will always work over an alge-

braically closed field k of characteristic zero, so all residue fields are perfectand the results from Sections 8 to 11 are valid.

We will assume that all schemes are noetherian and separated, so we willomit these adjectives in the future.

Also, by a surface we mean a 2-dimensional excellent scheme Z such thatfor any closed point p has residue field k so if Z is also smooth at p thenOZ,p ≃ k[[x, y]]. (resolutions of singularities exist for excellent schemes ofdimension 2 - see Lipman, Desingularization of two-dimensional schemes,Ann. Math. 107 (1978) 151-207. Also see Chapter XI of Cornell, Silverman;Arithmetic Geometry. See chapter 13 of Matsumura’s Commutative Algebrafor information on excellent rings).

12 Unramified Orders

We will define ramification in the scheme setting and relate unramified ordersto Azumaya algebras. Let Z be a normal integral scheme and Z1 be the setof codimension 1 points. Let A be a normal order on Z.

Example 12.1. Any maximal order on Z is normal. Why? We know thatA is reflexive. Let p ∈ Z1. Then A ⊗Z OZ,p is a maximal OZ,p-order, hencenormal.

We define the ramification data of a normal order A on Z to be theramification data of all the normal orders A⊗ZOZ,p as p ranges over Z1.Notethat A is generically Azumaya so these data are equivalent to given a finiteset of points p1, . . . , pn ∈ Z

1 together with nontrivial extensions

κ1/κ(p1), . . . , κn/κ(pn)

47

and generators σi for Gal(κi/κ(pi)). We say that A ramifies with ramificationindex ei = [κi : κ(pi)]. We say that A is unramified if n = 0, that is,A ⊗Z OZ,p is Azumaya for all p ∈ Z1. The reason we consider ramificationin codimension 1 only is the following purity result.

Proposition 12.2. Let A be an order on Z which is locally free as a sheaf.The locus where A fails to be Azumaya is empty or has codimension 1.

Proof. Since A is locally free, the non-Azumaya locus is where the map ϕ :A ⊗Z A

op → En dZ(A) fails to be an isomorphism. Suppose rank A = n2.We compute this locus locally so assume Z is affine and A ≃ On×nZ . Thenϕ : On

4

Z → On4

Z and the non-Azumaya locus is where det ϕ = 0.

Remark 12.3. This means that if A is locally free and is not Azumaya ata point p, then it fails to be Azumaya along some divisor containing p.

Corollary 12.4. Let A be a normal order on Z which is locally free as asheaf. If A is unramified, then it is Azumaya.

Theorem 12.5. Let Z be a smooth scheme of dim 6 2. Then there is anexact sequence

0 −→ Br Z −→ Br k(Z)a−→

⊕

p∈Z1

Homcont(Γp,Q/Z)

where Γp = Gal(k(p)/k(p)) and a comes from residue maps of Section 8 withrespect to the discrete valuation rings OZ,p.

Remark 12.6. In fact, for any smooth scheme Br Z injects into Br k(Z),but need etale cohomology, and we can’t use the cheap proof here.

Proof. Note first that the sequence above is a complex, e.g. from the Wittexact sequence. We show exactness at Br k(Z). Let D be a k(Z)-centraldivision ring with a([D]) = 0. Let A ⊂ D be a maximal order. Since A isreflexive, so by Auslander-Buchsbaum it is locally free. By Corollary 12.4, Ais Azumaya, so is in the image of Br(Z)→ Br k(Z).

We next show exactness at Br Z. Let A be an Azumaya algebra on Zsuch that A ⊗Z k(Z) ≃ Mn(k(Z)). By uniqueness theorem from Section 11for the discrete valuation ring case, we know that for each p ∈ Z1, we haveA ⊗Z OZ,p ≃ Mn(OZ,p). Hence there is an open cover U1, . . . , Un of Z − Ywhere Y is a closed subset of codimension > 2 such that

A|Ui= En dUi

Vi

48

where Vi is some rank n vector bundle on Ui. The lemma below shows thatwe can assume n = 1. But Z is smooth of dimension 6 2, so V1 extendsuniquely to a rank n vector bundle on Z. Thus A ≃ En dZV since both sidesare locally free so are determined by codimension one.

Lemma 12.7. 1. Let V1 and V2 be vector bundles on a scheme Z suchthat En dZV1 ≃ En dZV2. Then there exists a line bundle L such thatV2 ≃ L⊗Z V1.

2. Let Z be a smooth scheme of dimension 6 2 with open cover Z =U1 ∪ U2. Let Vi be a vector bundle on Ui, for i = 1, 2, such that onU12 = U1 ∩ U2 we have

En dZV1|U12≃ En dZV2|U12

.

Then there is a bundle V on Z such that

En dZV |Ui≃ En dUi

Vi

for i = 1, 2.

Proof. 1. This is standard and can be seen from the exact sequence

1 −→ Gm −→ GLn −→ P GLn −→ 0

which induces an exact sequence

Pic Z −→ H1(Z,Gm) −→ H1(Z,GLn)ϕ−→ H1(Z, P GLn)

and given γ ∈ H1(Z, P GLn), the set ϕ−1(γ) is a Pic Z-torsor.

2. By part 1, we know there is a line bundle on U12 such that V2|U12≃

L⊗U12V1|U12

. Now L is a coherent subsheaf of the constant sheaf k(Z)so by the extension lemma [Hartshorne, Ch II, Section 5, exercises] wecan extend L to a coherent subsheaf L1 of k(Z) on U1. Taking reflexivehulls, we can assume that L1 is a line bundle. So we define a vectorbundle V on Z by setting

V |U2= V2 and V |U1

= L1 ⊗U1V1

and gluing using ϕ.

49

13 Normal orders on curves

Let Z be a smooth quasi-projective curve. The key result is

Theorem 13.1. (Tsen) Br(k(Z)) = 0.

Proof. We can assume that Z is projective, let K = k(Z) and D be a K-central division ring of degree n > 1. We seek a contradiction. Recall wehave a reduced determinant map det : D → K. Picking a K-basis for Dallows us to view det as a degree n polynomial map from Kn×n → K. Letf ∈ K[xi j | i, j = 1, . . . , n] be this polynomial. Note that det 0 = 0, butsince det is multiplicative, we have det a 6= 0 if a 6= 0.

We consider an effective divisor ∆ > 0 (to be determined later) andrestriction of f to H0(OZ(∆))n×n. If the coefficients of f lie in H0(OZ(∆f ))then det gives a map of affine k-spaces

ϕ : H0(OZ(∆))n×n −→ H0(OZ(n∆ + ∆f ))

‖ ‖

V −→ W

Riemann-Roch implies that for ∆≫ 0

dim V = n2(deg ∆ + 1− g(Z))

> n deg ∆ + deg ∆f + 1− g(Z)

= dim W

Now ϕ−1(0) = 0 is a non-empty fibre containing of dimension 0 which con-tradicts Chevalley’s theorem.

Corollary 13.2. Let A be a maximal order on a smooth quasi-projectivecurve Z. Then A ≃ En dZV for some vector bundle V .

Proof. By Tsen’s theorem, Br(k(Z)) = 0 so A is an unramified order, henceit is Azumaya. Also Br Z injects into Br(k(Z)) so Br(Z) = 0 and A is trivialAzumaya.

Recall Morita theorey for sheaves of algebras (ref. [Artin-Zhang, non-commutative projective schemes])

Definition 13.3. Let Z be a scheme and A be a coherent sheaf of algebras onZ. Let P be a progenerator for A by which we mean a right A-module suchthat for some affine open cover U1, . . . , Un, we have, P(Ui) is a progenerator

50

for A(Ui) for each i. Then A is Morita equivalent to B = En dAP in thesense that

BPA ⊗A − : A−Mod −→ B−Mod

is a category equivalence.

Corollary 13.4. Maximal orders on smooth curves are Morita equivalent totheir centres.

There is a similar result for normal orders

Theorem 13.5. Let Z be a smooth quasi-projective curve and A,B be twonormal orders on Z with the same ramification data. Then A and B areMorita equivalent.

Remark 13.6. Since the residue fields of codimension one points p are allk = k, there are no non-trivial cyclic field extensions of k(p). Hence theramification dta of the normal order A in this dimension one case consistsof a finite set of ramification points p1, . . . , pl and their ramification indicese1, . . . , el.

Proof. We show that the normal order A is Morita equivalent to some orderA0 which depends only on ramification data. We achieve this by picking aprogenerator as follows. We embed A in a maximal order which by Corol-lary 13.2 has the form En dZV for some vector bundle V . Now A is locallyhereditary so V is locally projective as an A-module.

Away from ramification points, we have A = En dZV so V is a progener-ator there. Pick a ramification point p with ramification index e. From thestructure theory of normal orders we know that Ap = A ⊗Z OZ,p ≃ Mr(C)where

C ≃

OZ,p · · · · · · OZ,p

mZ,p. . . . . .

......

. . . . . ....

mZ,p · · · mZ,p OZ,p

⊆Mep

(OZ,p).

Let

Jp = ker

(A −→ A⊗Z OZ,p −→

OZ,p

rad(A⊗Z OZ,p)

)

51

From Section 9, we know that multiplication by Jp permutes the indecompos-

able projective Ap-modules cyclically. We may thus consider the progenerator

P = V ⊕⊕

p ram pt

ep−1⊕

j=0

J jpV.

We define A0 = En dAP and note that this depends only on the ramificationpoints and their ramification indices (but not V ) since the correspondingPeirce decomposition has components

HomA(J jpV, Jj′

p′V ) =

OO(−p)O(−p′)

as in Section 9, where the right hand side does not depend on V . Thiscompletes the proof.

Remark 13.7. Normality is not a Morita invariant in the sense that anon-normal order can be Morita equivalent to a normal order.

So this classifies normal orders on (smooth, quasi-projective) curves upto Morita equivalence.

If one wants to know more about normal orders one can try to pick anormal order with given ramification data and try to classify progeneratorsfor it. This is not unreasonable in the Fano case below

Theorem 13.8. Let A be a normal order on Z = P1 that is ramified at 6 2points. Then H0(A) has a complete set of n idempotents such that the Peircedecomposition with respect to the idempotents has the form

A ≃

n⊕

i,j=1

O(pi j)

for divisors pi j ∈ Div Z.

Proof. If A is unramified then A ≃ En dZV for some vector bundle V .Grothendieck splitting theorem implies V ≃ O(pi) for some pi ∈ Div Z.Hence A ≃

⊕HomZ(O(pi),O(pj)) =

⊕O(pj − pi). General proof follows

similar idea but a splitting theorem for locally projective modules over nor-mal orders ramified at 6 2 points.

52

13.1 Behavoiur under completions

We continue the above notation. Consider completions R and S = R ⊗R S.Note that S/R is still Galois with Galois group G. Hensel’s lemma implies

S =

g∏

i=1

Si

where Si is the completion of Sni. Further, Si are isomorphic to each other.

Let D = stab ni, the decomposition group. Then D acts on Si and in factSi/R is Galois with Galois group D. We can see this for example from thefact that

[κ(Si) : κ(R)] =1

g[κ(R)⊗R S : κ(R)]

=|G|

g

= |D|

14 Cyclic covers of curves

Given a normal order on a surface, ramification occurs at curves C, andthe ramification data involves cyclic field extensions of k(C). These lattercorrespond to some ramified cyclic covers of (the normalisation) of C. Weclassify these cohomologically.

Notation: let C be a smooth projective curve and Γ be the absoluteGalois group Gal(k(C)/k(C)). Let γ ∈ Homcont(Γ,Q/Z) corresponding tosay a cyclic degree n field extension F/k(C) and generator σ ∈ Gal(F/k(C)).

Let C be the normalisation of C in F so there is a degree n finite morphismπ : C → C. We wish to interpret the ramification of π cohomologically. Fixp ∈ C. Let R = OC,p ≃ k[[z]] and S = OC ⊗OC,p

R ≃∏g

i=1 Si where Si arecomplete local rings.

Let G = Gal(F/k(C)) and D ⊳ G be the decomposition group, whichis unique in this case since G is cyclic. Note that D is the Galois group ofSi/R.

Let Γp be the absolute Galois group of k(R) = k((z)). Then γ determinesan element r(γ) ∈ Homcont(Γp,Q/Z) by the composite map

r(γ) : Γp −→ Gal(Si/R) = D → Gγ−→ Q/Z.

Proposition 14.1. The absolute Galois group of k((z)) is lim←−µn where the

inverse system is defined by m-th power maps µmn → µn for m,n ∈ Z+.

53

Proof. The algebraic closure of k((z)) is k((z)) =⋃n∈Z+

k((z1/n)). There isalso a group isomorphism

µn −→ Gal(k((z1/n))/k((z)))

ξ 7−→ (z1/n 7−→ ξz1/n).

Hence we may consider r(γ) as an element of

µ−1 =⋃

Hom(µn,Q/Z)

Recall that the order of r(γ) is |D|, which in turn is the ramification indexof π : C → C above p.

Theorem 14.2. There is an exact sequence

Homcont(Gal(k(C)/k(C)),Q/Z)r−→

⊕

p∈C closed

µ−1P

−→ µ−1 −→ 0

where r is defined coordinate-wise as above.

Proof. Omitted. See [Artin, Grothendiec topologies Chapter IV 2.6[ Doneusing etale cohomology.

Remark 14.3. ker r classifies etale cyclic covers with generator for Galoisgroup. This can be written using etale cohomology group H1

et(C,Q/Z) orthe etale fundamental group π1C) so

ker r = Homcont(π1(C),Q/C).

14.1 (Non)-existence of cyclic covers of curves

Example 14.4. 1. There are no cyclic covers of a smooth projectivecurve ramified at exactly 1 point or at 2 points with different ramifica-tion indices.

2. Let p1, p2, p3 ∈ C. There exists a cyclic cover of C ramified exactly atp1, p2 and p3 with ramification indices 2, 4, 4. To see thi spick an iso-morphism µ−1 ∈ Q/Z and consider element of

⊕p∈C µ

−1 whose com-ponents are zero everywhere except those indexed by p1, p2 and p3. Thecomponents there are

1

2+ Z,

1

4+ Z,

1

4+ Z

so by Theorem 14.2, there exists a cyclic field extension of k(C) withthis ramification.

54

15 Artin-Mumford sequence

A key tool in studying orders on a smooth projective surface Z is the Artin-Mumford sequence. It describes Br k(Z) in terms of ramification data.

Notation: let Z1 be the set of irreducible curves on Z and Z2 be the setof closed points on Z. For C ∈ Z1, we let

ΓC = Gal(k(C)/k(C))

Theorem 15.1. Let Z be a smooth projective surface.

1. There is a complex

0 −→ Br Z −→ Br k(Z)ρ−→

⊕

C∈Z1

Homcont(ΓC ,Q/Z)r−→

⊕

p∈Z2

µ−1P

−→ µ−1 −→ 0

which is exact everywhere except possibly at the⊕

C∈Z1 terms. Theramification map ρ is that defined for orders as in Sections 10 and 12,whilst the ramification map r is that for curves as defined in Section14.

2. The cohomology is the etale cohomology H3et(Z, µ) where µ is the group

of roots of unity. In particular, H3et(Z, µ) is zero if Z is simply con-

nected.

Proof. Omitted. It involves the coniveau spectral sequence in etale cohomol-ogy.

15.1 Application to (non)-existence of maximal orders

As in Section 14, we can use the Artin-Mumford sequence to show the (non)-existence of maximal orders with given ramification data. We first observethe useful

Corollary 15.2. Let A be a maximal order on a smooth projective surface Z.Suppose it ramifies at C ∈ Z1 with ramification given by cyclic field extensionF/k(C). Suppose further that F/j(C) is ramified at p ∈ C. Then one of thefollowing holds

Boris: pictures missing A also ramifies at another curve C ′ with

(a) p ∈ C ∩ C ′

(b) if F ′/k(C ′) determines the ramification of A at C then F ′/k(C ′)also ramifies at p

55

Boris: pictures missing C is singular. Let ψ : Csm → C be the desingularisation. If F/k(Csm)gives the ramification of A at C then F/k(Csm) ramifies at at least 2points of ψ−1(p).

Proof. This follows from the Artin-Mumford sequence in particular rρ =0.

Remark 15.3. The point p in the above corollary is called a point ofsecondary ramification of the maximal order A or its corresponding Brauerpair (Z, [A ⊗Z k(Z)]). We sometimes summarise the equation rρ = 0 bysaying that secondary ramification must cancel.

Example 15.4. 1. There are no maximal orders on Z = P2 ramified ona (possibly reducible) curve of degree 1 or 2.

Why? Suppose a maximal order A ramifies on a line L ≃ P1. Since P1

is simply connected, it has no etale covers. Any cover of P1 ramifiesat > 2 points by part 1 of Example 1. Hence there are 2 points ofsecondary ramification. But these cannot be cancelled since there areno other ramification curves.

If the ramification locus has degree 2, then it is either a smooth conicso ≃ P1 or 2 lines. A simliar argument shows that this cannot happen.

2. There is a maximal order on P2x,y,z on x y z = 0.

Example 15.5. An exotic del Pezzo order. Let Z = P2 and C ⊂ P2 bea smooth quartic. We consider an etale double cover C/C. Note that thegenus of C is

g(C) = 1 +1

2C · (K + C)

= 1 +1

24(−3 + 4)

= 3

These are classified by 2-torsion points of the Jacobian J(C). Since J(C) isan abelian variety of dimension g(C) = 3 we know that J(C)2 ≃ (Z/2Z)6.

Since P2 is simply connected, the Artin-Mumford sequence implies thatthere exist maximal orders on Z = P2 ramified only on C with ramificationdata there given by the cyclic field extension k(C)/k(C).

Theorem 15.6. (deMeyer-Ford) Let Z be a smooth projective surface withKodaira dimension −∞, that is Z is birationally ruled. Then Br Z = 0.

56

Theorem 15.7. (de Jong) Let Z be a smooth projective surface and K =k(Z). Let D be a K-central division ring, then deg D = period D.

Note that by deMeyer-Ford theorem and the Artin-Mumford sequence,the division ring D with ramification k(C)/k(C) as above as period 2 so byde Jong’s theorem, you can pick the maximal order in D which is degree 2(i.e. rank 4) over P2.

We can even construct some examples explicitly using cyclic algebras asfollows. Let Z = P2 and Y be the double cover of Z ramified on the smoothquartic C. We can describe this as

Y = Sp e cZ(O ⊕O(−2))

where O⊕O(−2) is the commutative cyclic algebra with multiplication givenby any isomorphism

O(−2)⊗Z O(−2)∼−→ O(−C)

ATTENTION:Boris: picture of quartic missingIt can be shown that Y is the blowup of P2 at 7 points. Let σ ∈ Aut Y

be the covering involution. Let H1, H2 be 2 (of the 28) bitangents to C¿ Letπ : Y → Z be the covering map. Note that

π−1(Hi) = Ei ∪ σ(Ei)

for some exceptional curve Ei ⊂ Y . Note that E1 and E2 are not linearlyequivalent since E1 · σ(E1) = 2 and E2 · σ(E1) 6 1.

Let L = OY (E1 − E2). We wish to construct a cyclic algebra on

A = OY ⊕ Lσ.

One way to understand Lσ is to consider L as a sheaf on Z with an OY -module structure and then Lσ is this sheaf with the same left OY -modulestructure, but right OY -module structure is twisted through by σ. We needa relation of the form

ϕ : L⊗2σ −→ OY

Note that σ2 = 1 so L⊗2σ is an OY -central bimodule. Also

L⊗2σ ≃ L⊗Y σ

∗L

≃ OY (E1 − E2 + σ(E1)− σ(E2))

≃ OY (π∗(H1 −H2))

≃ OY

We pick ϕ to be any isomorphismClaim: ϕ satisfies the overlap condition.

57

Proof. It suffices to check this generically

Lσ∗L = OY (π∗(H1 −H2))

= αOY

where α ∈ k(Y )∗ is such that div a = π∗(H1−H2). Now α is determined upto multiplication constant in k∗ so α ∈ k(Z). Hence the overlap conditionholds.

Hence the cyclic algebra A = A(Y/Z, Lσ, φ) is a well-defined order ofrank 4 on P2. It is reflexive and is normal by Section 5 example 1b and isAzumaya away from C by Section 2 example.

Proposition 15.8. The cyclic algebra A = A(Y/Z, Lσ, φ) is a maximal orderon P2 ramified only on C with ramification given by the double cover of Cdetermined by the 2-torsion divisor p1−p2 +p′1−p2 where pi, p

′i are the points

of intersection of Ei with π−1(C).

[Fix proof max] We will compute the ramification data. Let CY =π−1(C)red so π∗C = 2CY . Note that CY is the ramification locus so σ(CY ) =CY . Let I = A⊗Y O(−CY ) whic his actually a 2-sided ideal since CY is fixedby σ. Now I2 = A⊗Y O(−2CY ) = A⊗Y O(−C). We wish to show that theramification data is given by the fraction field of A/I.

A/I = (OY ⊕OY (E1 − E2)σ)|CY

= OC ⊕OC(p1 + p′1 − p2 − p′2)

a commutative algebra since σ|CY= 1CY

. Note that p1 + p′1 − p2 − p′2 is2-torsion since the isomorphism L⊗2

σ ≃ OY restricts to isomorphism

OC(p1 + p′1 − p2 − p′2)⊗2 ≃ OC

Hence A/I defines a double cover of C which is the ramification.[have to show that A/I defines nontrivial double cover of C for maximal-

ity].First note that E ≃ P1 is simply connected, so we hvae an exact sequence

0 −→ Homcont(Gal(k(P1)/k(P1),Q/Z) −→⊕

p∈P1 closed

µ−1P

−→ µ −→ 0

as in Section 14. Hence to give ramification data on E, it suffices to give anelement of ker(

⊕p∈P1 µ−1 → µ−1).

Suppose there are exactly two ramification curves C1 and C2 passingthrough p with secondary ramification there given by ζ1, ζ2 ∈ µ

−1. Suppose

58

further that p is a smooth point of C1, C2 and these intersect transversally.So the local intersection number (C1, C2)p = 1. Let C1, C2 be the stricttransforms of C1, C2. The Artin-Mumford sequence on Z shows that thesecondary ramification cancels at p, so ζ1 + ζ2 = 0. Now we use the Artin-Mumford sequence on Z. If pi = Ci ∩ E, then the secondary ramification ofCi at pi is ζi Since the secondary ramification cancels at pi, we know thatsecondary ramification from E is pi is −ζi. Also there is no other secondaryramification on E.

Boris: pictures of blowup missing.Since (−ζ1) + (−ζ2) = 0, this determines the ramification data on E.

Suppose ζi is order e, then the ramification of (Z, α) at E is the degree ecover of E totally ramified at p1 and p2. In particular, the ramification indexat E is e.

Upshot: By blowing up, you may not be able to remove nodes in theramification locus, but you can always ensure that the ramification index ofone of (branches) of curve through node is the secondary ramification index.

Definition 15.9. Let Z be a surface and α ∈ Br k(Z). We say that (Z, α)is terminal if Z is smooth and the ramification data satisfies

1. Ramification divisor is a normal crossing divisor

2. Let p be a node of the ramification divisor and C1, C2 be the brances oframification curves passing through p. Then the secondary ramificationindex at p is equal to the ramification index of (Z, α) at either C1 orC2.

Given a maximal order A on a smooth surface Z with ramification datasatisfying 1 and 2, then we say that A is terminal.

Example 15.10. If (Z, α) is terminal and Z → Z is the blowup at a point,then (Z, α) is terminal.

Theorem 15.11. (Terminal resolutions) Let Z be a surface and α ∈ Br k(Z).There is a resolution of singularities Z → Z such that (Z, α) is terminal.

Proof. If Z is projective, then this follows form the Artin-Mumford sequenceand the usual resolution of singularities, arguing as in the example above. Ingeneral, we use a version of the Artin-Mumford sequence on the blowup ofSpec k[[u, v]] at the closed point.

Example 15.12. From Section 2, we know that the following is a maximalorder

kζ [[x, y]] =k 〈〈x, y〉〉

(y x− ζx y)

59

where ζ is a primitive e-th root of unity. Its centre is k[[u, v]] where u = xe

and v = ye. It is Azumaya on u v 6= 0 and the ramification data is given by:at u = 0, the field extension k((y))/k((v)) and chosen generator of the

Galois group is y 7−→ x y x−1 = ζ−1y; and at v = 0, the field extensionk((x))/k((u)) and chosen generator of the Galois group x 7−→ y x y−1 = ζx.Hence kζ [[u, v]] is terminal.

Cancellatio nof secondary ramification indices can be seen here by theappearance of ζ and ζ−1 in the generators of the Galois group.

16 Complete local structures of terminal or-

ders

Recall that if Z is a smooth quasi-projective surface and p ∈ Z is a closedpoint then OZ,p ≃ k[[u, v]]. We wish to give a similar description of terminalorders in the complete local case. The approach will be to use a method ofArtin’s for describing orders with given ramification data. Recall terminalorders have ramification as one of the following: Azumaya, smooth point oframification, node Boris: three pictures missing.

16.1 Completion of normal orders

Let R be a normal excellent k-domain and p be a maximal ideal such thatRp is a domain.

Proposition 16.1. Let A be a normal order, then A ⊗R Rp is a normal

Rp-order.

Proof. Note that completion commutes with duals in the following sense. ForM a finitely generated R-module, we have

HomRp(M ⊗R Rp, Rp)

nat≃ HomR(M, Rp)

≃ HomR(M,R)⊗R Rp

Hence A is reflexive implies A ⊗R Rp is reflexive. It suffices to check now

that A⊗R Rp is normal at all codimension one primes. These all come fromcodimension one primes of R. Let P be a codimension one prime of R. Notethat R/P ⊗R Rp is complete so is excellent too. Hence it is also reduced. We

may thus write PRp =⋂ri=1 Pi for primes Pi minimal over PRp.

Let Ri be the localisation of Rp at Pi. Then PRi = PiRi, the maximalideal. Then Section 5, Proposition 2 (its proof), shows that A⊗R Rp is alsonormal.

60

16.2 Method of Artin covers

Artin has a method for describing orders with given ramification data asinvariant rings.

Let Z be an integral normal noetherian scheme, π : Y → Z be a (ramified)Galois cover with abelian Galois group G, and A be a normal order on Z.Suppose that for any irreducible curve C ⊂ Z the ramification index rC ofπ : Y → Z at C divides the ramification index eC or A at C.

Theorem 16.2. (Artin) With the above set up, there is a normal orderB ⊂ A⊗Z k(Y ) on Y such that

1. B is G-stable and A = BG

2. On the etale locus U ⊂ Y of π, B = π∗A

3. At any divisor CY above C, the ramification index of B at CY is eC/rC .

We call B the Artin cover of A with respect to π : Y → C.

Remark 16.3. 1. Artin’s theorem works for arbitrary finite groups, butreduces the proof to G cyclic.

2. Artin only works in the case Z = Spec k[[u, v]] but perhaps his proofgeneralises.

Proof. For each irreducible curve C, we need only define anOY ⊗ZOZ,C-orderBC satisfying 1-3 for then we can define

B = π∗A⊗Y OU ∩⋂

C ram curve

BC

We are thus reduced to the following situation: Z = Spec R where (R,m) isa discrete valuation ring, say with residue field κ = R/m, Y = Spec S Galoiscover. It is a semilocal Dedekind domain say with Jacobson radical n.

Let I be the inertia group G = G/I. Consider the character groupG∨ = HomGrp(G, k

∗) and its subgroup G∨ = HomGrp(G, k∗) = {χ ∈ G∨ |

χ(I) = 1}. Consider the G-module decomposition

S =⊕

χ∈G∨

Sχ

where Sχ is a free R-module of rank 1. Note that S/SI is totally ramifiedand SI/SG is etale.

61

Also

SI =⊕

χ∈G∨ Sχ

Hence

S/n ≃ SI/mSI

≃ κG

Also n/n2 ≃ S/n as S-modules, but not as G modules. Hence there is acoset χnG

∨ ∈ G∨/G∨ such that n/n2 ≃⊕

χ∈χnG∨ Sχ ⊗R κ as κG-modules.Similarly, there is a κG-module isomorphism

ni/ni+1 =

⊕

χ∈χinG∨

Sχ ⊗R κ.

We have a filtration 0 ⊂ · · · ⊂ n2/mS ⊂ n/mS ⊂ S/mS =⊕

χ∈G∨ Sχ ⊗R κ.

What this shows is that χnG∨ generates the cyclic group G∨/G∨. We write

Si =⊕

χ∈χinG∨

Sχ

for i = 0, . . . , |I| − 1. Recall that A is a normal R-order so rad A = πAfor some π ∈ A. We can define J−i = π−iA ⊂ A ⊗R K(R). Let e be theramification index of A, r be the ramification index of S/R and s = e/r. Wedefine our desired order

B =r−1⊕

i=0

J−i s ⊗R Si ⊆ A⊗R K(S).

This is clearly G-stable with BG = A and B is finitely generated over S suchthat B ⊗S K(S) = A ⊗R K(S). The following calculation shows that B isclosed under multiplication,

(J−i s ⊗R Si)(J−(r−i)s ⊗R Sr−i) ⊆ J−r s ⊗R SiSr−i

= m−1A⊗R mS0

= A⊗R S0.

It remains to check that B is normal with ramification index s = e/r.Since the Jacobson radical of A is compatible with etale base change we canmake such a base change and assume

A ≃ Md(A0)

where A0 =

R Rm

...m · · · m R

⊂Me(R)

62

so we can further assume that A = A0.Now S0/R is etale so the Artin cover of A with respect to S/R is the

same as the Artin cover of A⊗R S0 with respect to S/S0. Therefore we canassume R = S0 and S/R is totally ramified. Thus we can also assume thatS = R[q]/(qr − p) where p ∈ m is a uniformising parameter for R. Thetheorem follows from direct computation, best illustrated by example.

Example.Let e = 6, r = 3, s = 2

A =

R R R R R Rm R R R R Rm m R R R Rm m m R R Rm m m m R Rm m m m m R

and S = R[q]/(q3 − p) = R⊕Rq ⊕Rq2 := S0 ⊕ S1 ⊕ S2. Recall that q3 = p

A⊗R S0 = A

J−2 ⊗R S1 =

R R R R (q−3) (q−3)R R R R R (q−3)R R R R R R

(q3) R R R R R(q3) (q3) R R R R(q3) (q3) (q3) R R R

⊗R R q

J−4 ⊗R S2 =

R R (q−3) (q−3) (q−3) (q−3)R R R (q−3) (q−3) (q−3)R R R R (q−3) (q−3)R R R R R (q−3)R R R R R R

(q3) R R R R R

⊗R Rq

2

63

so

B =

S S S q−1 S q−1 S q−2 S q−2

S q S S S q−1 S q−1 S q−2

S q S q S S S q−1 S q−1

S q2 S q S q S S S q−1

S q2 S q2 S q S q S SS q3 S q2 S q2 S q S q S

=

B0 B0q

−1 B0q−2

B0q B0 B0q−1

B0q2 B0q B0

≃ B3×30

where B0 =

(S SS q S

)and the last isomorphism is given by conjugation

by the diagonal matrix with entries (1, 1, q, q, q2, q2). The S-order B3×30 is

isomorphic to (S3×3 S3×3

S3×3q S3×3

)

which is obviously normal with ramification index 2.

17 Appendix: Twisting group actions by a

1-cocycle

Let S be a commutative ring and G be a finite group action on S. Let B bean S-algebra with a G-action compatible with the S-action. Let AutSB bethe group of S-algebra automorphisms of B and note that G acts on AutSBby “conjugation”. We wish to twist the given actions of G on B by some1-cocycle f : G→ AutSB.

Proposition 17.1. Consider the above setup

1. The cocycle f : G → AutSB determines a new G-action α on B com-patible with the S-action as follows. An element σ ∈ G acts on B byασ(b) := fσ(σ · b), for b ∈ B.

2. Any G-action on B, compatible with G-action on S arises from such atwist.

64

Proof. Note that ασ is a ring automorphism being composite of two such.Further, the action ασ on S is the usal on. Now check that G acts on B

ασατ (b) = fσ(σ · (fτ (τ · b))

= fσσ · (fτ (σ−1στ · b)

= (fσ(σ · fτ ))(στ · b)

by cocycle condition = fστ ((στ) · b)

= αστ (b)

For part 2, let β be the original G-action on B and α be another one. Forσ ∈ G, ασβ

−1σ is a ring automorphism which acts trivially on S. Hence

fσ = ασβ−1σ ∈ AutSB. The fact that ασ is a G-action means that we can

reverse the computation above to see that fσ is a cocycle.

Proposition 17.2. Let f, f ′ : G → AutSB be 1-cocycles and α, α′ be thetwisted G-actions on B corresponding to f, f ′. If f and f ′ are cohomologousthen the respective invariant rings A,A′ are isomorphic.

Proof. If f, f ′ are cohomologous, then there is some u ∈ AutSB such thatf ′σ = u−1fσ(σ · u). Now for all b ∈ B, we have

α′σ(b) = f ′σ(σ · b)

= (u−1fσ(σ · u))(σ · b)

= u−1fσ(σ · u(b))

= u−1ασ(u(b))

so α′σ = u−1ασu, hence A′ = u−1(A).

18 Criterion for normality for invariant rings

We give here another criterion for normality in the dimension 1 case.

Proposition 18.1. Let (R,m) be a discrete valuation ring with residue fieldκ. Let A be an R-order, then A is normal if and only if A is hereditary andA/rad A =

∏Di where the Di’s are isomorphic κ-algebras.

Proof. Recall that A is normal if and only if A := A ⊗R Rm is normal (c.f.Section 5 Proposition 2), and A is hereditary if and only if A is hereditary.Hence we can assume that R is complete. Let J = rad A.

65

Suppose that A is normal. We know that A is hereditary from Section5 Proposition 1, and if πA = J , then conjugation by π induces an automor-phism between the Di (as in Section 10.4).

Conversely, it suffices to prove that A/J ≃ J/J2 as A/J-modules. Letεi ∈ A/J be the central idempotent corresponding toDi. Let Si be the simpleDi-module and Pi ∈ A-mod be its projective cover. Now A is hereditary soJ is projective and J =

⊕P ai

i for some ai. Let A =⊕

P ai where Di = Sai

(same a because Di’s are isomorphic), so it suffices to show that ai > a forall i by rank considerations. Now J/J2 is also an A/J-module so we have aPeirce decomposition J/J2 =

⊕Bi j where Bi j = εi(J/J

2)εj is a Di⊗kDopj -

module. Note that for any i, there is some j such that Bi j 6= 0. For if not,then εi(J/J

2) = 0. But we have a natural surjection (J/J2)⊗n → Jn/Jn+1 soany simple subquotient of J/Jn must be Sj for some j 6= i. This is false sincepA/p J is a subquotient of J/Jn for n ≫ 0 and p a uniformising parameterfor m.

Lemma 18.2. Let B be a simple Di ⊗Dopj -module. Then

dimκB > dimκDi.

Proof. Let κ′ = Z(Di) and d2 = [Di : κ′].

Di ⊗κ Dopj = (Di ⊗κ′ (κ′ ⊗κ κ

′))⊗κ′⊗κκ′ ((κ′ ⊗κ κ′)⊗κ D

opj )

is Azumaya over κ′⊗κκ′ of degree d2. Now κ′/κ is separable so κ′⊗κκ

′ =∏κ′ℓ

where κ′ℓ are field extensions of κ′. Therefore any simple module B satisfies

dimκB = d2[κ′ℓ : κ]

> d2[κ′ : κ′]

= dimκDi.

To use Artin’s method to classify terminal orders, we need to considerGalois covers ramified on normal crossing lines.

Example 18.3. Let S = k[[s, t]], p, q, e be positive integers and ζ, ξ primi-tive p e-th and q e-th roots of unity. Let G = 〈σ, τ | στ = τσ, σp e = 1 = τ q e〉act on S by

σ : s 7−→ ζs

t 7−→ t

τ : s 7−→ s

t 7−→ ξt

66

Note that R := SG = k[[u, v]] where u = sp e and v = tq e. Hence S/R isa Galois cover ramified on the union of u = 0 and v = 0 with ramificationindices p e and q e. Let G be as above act on Md(S) ≃ Md(k) ⊗k S by thetensor product of the above action of G on S with some action of G on Md(k).

Notation 18.4. We use the following notation to describe the G-action onMd(k). By Skolem-Noether, σ, τ act by conjugation by matrices say bσ, bτ ∈GLd(k). Scaling these if necessary, we may assume bp εσ = 1 = bq eτ . Note thatthere is also some scalar η such that bσbτ = ηbτbσ. Now bτ = bp eσ bτ = ηp ebτ .Hence η is a p e-th root of unity and similarly a q e-th root of unity. Bychanging e we can assume that η is a primitive e-th root of unity.

Theorem 18.5 (C-Hacking-Ingalls). With the notation and hypotheses above.Let A = Md(S)G.

1. Then A is normal if and only if the dimensions of the eigenspaces ofbσ are all the same and the same is true for bτ .

2. A is ramified only only u = 0 and v = 0 and the ramification indicesthere are the number eigenspaces of bσ and bτ respectively.

3. The secondary ramification index is e.

Proof. We concentrate on 1), the other parts will follow from the proof of 1).Note that R ⊆ A and is finitely generated over R, since Md(S) is finitely

generated over R. Also K(R) ⊗R A = Md(K(S))G which is central simpleover K(R). In fact, by descent theory, it is Azumaya over u v 6= 0. Also, A isreflexive since Md(S) is free over R and by Maschke’s theorem, A is a directsummand. Hence A is a reflexive R-order. Thus part 1) follows if we can shownormality of A(u) = A ⊗R R(u) is equivalent to the eigenspace condition onbσ and normality of A(v) : A⊗RR(v) is equivalent to the eigenspace conditionon bτ . These are symmetric, so will only look at latter. Let J := rad A(v).Note that A(v) = Md(S(v))

G. Since Md(S(v)) is hereditary, so is A(v) by[McConnell-Robson, Theorem 7.8.8]. By Proposition 1, it suffices to showthat the eigenspace condition on bτ is equivalent to the fact that A(v)/J is aproduct of isomophic simple algebras.

Lemma 18.6. Let C be a k-algebra and H be a finite group acting on C.Then

rad CH ⊇ radC ∩ CH = (rad C)H .

In particular, if (C/rad C)H is semisimple, then

CH/rad CH ≃ (C/radC)H .

67

Proof. Let x ∈ (rad C)H and c ∈ CH . It suffices to check that 1 − c x isinvertible in CH . It has an inverse y ∈ C, and y is H-invariant because 1−x cis. Hence the lemma follows.

Returning to proof of theorem...Let B = Md(S(v)), we compute (B/radB)G. Note that B/rad B ≃

Md(k((s))), and G acts on B via some 1-cocycle and we may pass to some1-cohomologous cocycle if desired. We may thus conjugate simultaneouslyby bσ and bτ so that

bτ =

1ξ

ξ2

. . .

where the block sizes are d0, d1, ., dq e−1, so d = d0 + d1 + · · ·+ dq e−1. Then

(B/radB)〈τ〉 =

Md0(k((s)))

Md1(k((s))). . .

≃

q e−1∏

i=0

Mdi(k((s))).

We now compute σ-invariants of (B/radB)〈τ〉.Note that bσ permutes the eigenspaces of bτ as follows. Let v be an

eigenvector of bτ with eigenvalue ξi. Then bτbσv = η−1bσbτv = η−1ξibσv, sobσ maps the ξi-eigenspace to the η−1ξi-eigenspace. In particular σ permutescyclically the factors {Mdi

(k((s))) | i ≡ jmod q}. Therefore

(B/radB)G ≃

(q−1∏

i=0

Mdi(k((s)))

)〈σe〉

Now σe has order p so passing to a cohomologous 1-cocycle we can assumethe action of σ on Mdi

(k((s))) is such that (Boris: the matrix below is ablockmatrix)

beσ =

1ζε

ζ2e

. . .

68

We calculate Mdi(k((s)))〈σ

e〉. Note that σe : s 7−→ ζes so k((s))〈σe〉 = k((sp)).

(Boris: the matrix below is a blockmatrix)

Mdi(k((s)))〈σ

e〉 =

k((sp)) s−1k((sp))s k((sp)) k((sp))

. . .

≃ Mdi(k((sp))

as seen by conjugating by (Boris: the matrix below is a blockmatrix)

1s−1

s−2

. . .

.

Hence

BG/rad(BG) = (B/radB)G

=

q−1∏

i=0

Mdi(k((sp)))

By our criterion for normality (find reference ??), this is normal if and onlyif all nonzero di are the same. Also

Z(BG/rad(BG)) =∏

i∈[0,q−1]

di 6=0

k((sp))

hence is an extension of k((u)) = k((sp e)). This has degree q′e where q′ is{i ∈ [0, q − 1] | di 6= 0}, that is the number of eigenvalues of bτ . Also we seethat the secondary ramification index is [k((sp)) : k((u))] = e.

19 Structure Theory

In this subsection, we classify terminal orders in the complete local case. Wesaw one example in Section 16, namely, the skew power series ring

kη[[x, y]] =k 〈x, y〉

(x y − ηy x)

where η is some primitive e-th root of unity. This has centre k[[u = xe, v =ye]] and is ramified on u v = 0. Recall that the centre of a terminal order issmooth and the only complete local 2-dimensional ring which is smooth andhas residue field k is isomorphic to k[[u, v]].

69

Theorem 19.1. [C-Ingalls] (maybe Artin knew this) Let A be a terminalk[[u, v]]-order. Then one of the following occurs

1. If A is unramified then A ≃Md(k[[u, v]]).

2. If the ramification curve is smooth and the ramification index is e, thenchanging coordinates so that u = 0 is the ramification curve, then A isn× n matrices in

A0 ≃

k[[u, v]] k[[u, v]] · · · k[[u, v]]

u k[[u, v]]. . . . . .

......

. . . . . ....

u k[[u, v]] · · · u k[[u, v]] k[[u, v]]

⊂Me(k[[u, v]])

3. Suppose A is ramified on normal crossing lines. Change coordinagesso it is ramified on u = 0 with ramification index e and v = 0 withramification index q e. Then A is isomorphic to n× n matrices in

A0 =

kη[[x, y]] kη[[x, y]] · · · kη[[x, y]]

(y). . . . . .

......

. . . . . ....

(y) · · · (y) kη[[x, y]]

⊂Me(kη[[x, y]])

where η is a primitive e-th root of unity and u = xe, v = ye.

Proof. The proof of part 1 follows from the next proposition which tells usthat any Azumaya k[[u, v]]-algebra is trivial Azumaya.

Proposition 19.2. Let R be a commutative noetherian complete (also truewith Henselian) local ring with residue field κ. Then the natural map BrR→Brκ is injective.

Remark 19.3. In fact, BrR ≃ Brκ, which we saw in the case where R isa complete discrete valuation ring.

Proof. Let A be an Azumaya R-algebra such that A ⊗R κ ≃ Md(κ). Letε ∈ A⊗Rκ be a primitive idempotent so (A⊗Rκ)ε is a simple module. Lift εto ε ∈ A. Note that R[ε] is finitely generated over a complete local noetherinring R, so we can assume ε is an idempotent in R[ε] ⊂ A. The propositionwill follow if we can show that the natural map ψ : A → EndR(Aε) is anisomorphism, since Aε is a direct summand of the free R-module A.

70

Note that surjectivity of ψ follows by Nakayama’s lemma and the factthat

ψ ⊗R κ : A⊗R κ∼−→ Endκ((A⊗R κ)ε)

is an isomorphism. We check injectivity by studying I = ker ψ ⊳ A. Notethat I ∩ R = 0 since Aε is a free R-module. Suppose I 6= 0, then since Ais a free R-module, we can find an R-homomorphism ϕ : A → R such thatϕ(I) 6= 0. But A is Azumaya, so A⊗RA

op → EndRA. Hence as I is a 2-sidedideal, I ⊇ ϕ(I) 6= 0, which contradicts I ∩ P = 0. Hence ker ψ = 0 and weare done.

This finishes the proof of part 1. We will omit the proof of part 2 sinceit can be extracted from the proof of part 3. We use the method of Artincovers to prove part 3. Consider the following cover of R = k[[u, v]]. LetS = k[[s, t]] and G = 〈σ, τ | σe = τ q e = 1, στ = τσ〉 act as in example insection 17.3, that is

σ : s 7→ ζs σ : t 7→ t

τ : s 7→ s τ : t 7→ ξt

where ξ is a primitive q e-th root of unity and ζ = ξq. Note SG = k[[u =se, v = tq e]]. Recall S/R has ramification e on u = 0 and q e on v = 0. Weconsider the Artin cover B of A with respect to S/R. By choice of S, B isan unramified order on k[[s, t]] so by part 1 that in fact

B ≃ Md(k[[s, t]])

for some d. We compute A = BG. We compute A = BG. The action of Gon B = Md(S) is given by twisting the trivial coordinatewise action by some1-cocycle β : G→ PGLd(S). As in the proof of theorem ? (from section 8.2)we have an isomorphism

H1(G,PGLd(S))∼−→ H1(G,PGLd(k)).

Hence we may assume the cocycle comes from a 1-cocycle β : G→ PGLd(k),that is, σ acts on Md(S) = Md(k)⊗k S by tensor product of the usual actionon S and conjugation by an element bσ ∈ GLd(k) on Md(k). Similarly τ actsby usual action on S and conjugation by bτ ∈ GLd(k). We can scale bσ andbτ as usual so beσ = 1 = bq eτ . Then bσbτ = ηbτbσ for some root of unity η.But A is terminal with secondary ramification index e, so theorem ? (fromsection 17.3) implies η is a primitive e-th root of unity.

71

Remark 19.4. If we change bσ bσ⊗ In and bτ bτ ⊗ In, then of course,BG Mn(B

G).

Also, A is normal and the ramification index of A along v = 0 is q e, sowe may assume all the eigenspaces of bτ have the same dimension, say n,and all q e eigenvalues occur (by theorem 17.3). Hence we may pass to acohomologous 1-cocycle and assume bτ = b′τ ⊗ In where

b′τ =

1ξ

. . .

ξq e−1

.

Since bσ permutes the eigenspaces of bτ by sending the ξi-eigenspace to theη−1ξi-eigenspace. We may futher assume that bτ = b

′′

τ ⊗ b′′′τ ⊗ In where

b′′τ =

1η

. . .

ηe−1

b′′′τ =

1ξ

. . .

ξq−1

and bσ = b′σ ⊗ Iq ⊗ In where

b′σ =

0 1 0 · · · 0...

. . . . . . . . ....

.... . . . . . . . . 0

0. . . . . . . . . 1

1 0 · · · · · · 0

(note: eigenvectors: v, bσv, . . . , be−1σ v)

Now we calculate Md(S)G. We may as well assume n = 1.

Notation 19.5. Note that ζ = ξm for some m relatively prime to e. We

72

denote the matrices

x =

1η

. . .

ηe−1

y =

0 1 0 · · · 0...

. . . . . . . . ....

.... . . . . . . . . 0

0. . . . . . . . . 1

1 0 · · · · · · 0

.

Note that yx = ηxy.

Special case: we will do the case q = 1 (so ζ = ξ) first.

Lemma 19.6. Let G act on Me(k) as above, that is, σ acts by conjugation byb′′σ and τ by conjugation by b′′τ . Then the G-module decomposition of Me(k)into isotypic components is

Me(k) =e−1⊕

i,j=0

kxiyj

In particular

Me(k) =k 〈x, y〉

(xe − 1, ye−, yx− ηxy).

Proof. Exercise. Start by showing that xiyj is an eigenvector of both x andy.

Remark 19.7. Since gcd(m, e) = 1. We also have

Me(k) =e−1⊕

i,j=0

kxmiymj.

Proof of Theorem 19.1, Part 3.Case q = 1. Let R = SG = k[[u = se, v = te]] and note that the G-module

decomposition of S is

S =e−1⊕

i,j=0

R sitj

73

Hence

Me(S)G =e−1⊕

i,j=0

R six−mitj ymj

=R 〈x = sxm, y = tym〉

(xe − u, ye − v, y x− ζ−mx y)

= kζ−m [[x, y]]

since

y x = tymsx−m

= t sym−1x−mη−my

= t sx−mζ−mym

= ζ−msx−mtym

= ζ−mx y.

Note. To compute can use the fact for G finite abelian and G-modules V,Wwith G-module decomposition V =

⊕χ∈G∨ Vχ and W =

⊕χ∈G∨ Wχ, then

(V ⊗W )G =⊕

χ∈G∨

Vχ ⊗ V−χ

Let’s now finish the proof of the theorem by considering the case of generalq. Let

A0 =⊕

R six−mitq j ymj

≃R 〈x, y〉

(xe − u, ye − v, y x− ζ−mx y)

where x = sx−m, y = tqym. From the case q = 1, we see that

Mq e(S)G =

A0 t A0 t2A0 te−1A0

tq−1ymA0 A0. . .

tq−2ym tq−1ym A0...

. . . . . . . . .

A0

74

Now conjugate by the diagonal matrix with entries 1, t, t2, . . . , te−1 to get

Mq e(S)G =

A0 A0

tqymA0 A0...

.... . . . . .

.... . . . . .

...tqymA0 tqymA0 A0

19.1 Resolution of singularities

We show that locally, terminal orders have global dimension two, so terminalresolutions are in a sense resolutions of singularities.

Definition 19.8. Let A be a noetherian ring. We say that it is regular ofdimension d if gl. dim A = d.

Theorem 19.9. Any termina k[[u, v]]-order is regular of dimension d.

Proof. We say in Section 17.4 that any such order has the form

A =

kη[[x, y]] · · · · · · kη[[x, y]]

(y). . . . . .

......

. . . . . ....

(y) · · · (y) kη[[x, y]]

n×n

where η s a primitive root of unity. We apply the criterion from the Appendix-Finite global dimensions twice. Consider the regular normal element x ∈rad A and note that

A : = A/xA =

k[[y]] · · · · · · k[[y]]

(y). . . . . .

......

. . . . . ....

(y) · · · (y) k[[y]]

n×n

Also let

y =

0 1 0 · · · 0...

. . . . . . . . ....

.... . . . . . . . . 0

0. . . . . . . . . 1

y 0 · · · · · · 0

⊗ In

75

is a regular normal element in A and

A/yA =

k

. . .

k

n×n

which is semisimple so has global dimension 0.

Proposition 19.10. Let A be a terminal order on a surface Z and p ∈ Z bea closed point. Then A⊗Z OZ,p is regular of dimension 2.

Proof. By Proposition 17.1, we know that Ap = A⊗Z OZ,p is a normal order.Furthermore, its ramification is induced from that of A so Ap is terminal

too. Hence A is a terminal k[[u, v]]-order so is regular of dimension 2 by theTheorem. This shows that gl. dim A⊗Z OZ,p = 2 as well.

Definition 19.11. Let A be a normal order on a surface Z. A modification(Z, A) of (Z,A) of A is a projective birational map π : Z → Z and a normalorder A on Z with a morphism of algebras πA : π∗A→ A such that

1. πA is an isomorphism away from the exceptional locus of π and

2. A is maximal at every general point of the exceptional locus.

Remark 19.12. Let A be a normal order on a surface Z and π : Z → Zbe a projective birational map. Then there exist normal orders A on Z suchthat (Z, A) → (Z,A) is a modification. We write π#A = A and call π#A amodification of A with respect to π.

Corollary 19.13. Let A be a normal order on a surface Z. Then thereis a modification (Z, A) → (Z,A) such that A is terminal. In particular,A ⊗Z OZ,p is regular of dimension 2 for any p ∈ Z closed. We call such amodification a resolution of singularities.

20 Noncommutative Mori Minimal Model Pro-

gram

There is a noncommutative version of Mori’s minimal model program fororders on projective surfaces. It provides an intelligent scheme for classifyingorders on surfaces. The theory works on the level of Brauer pairs, so we shallwork in this setting. All results have analogues for orders.

76

20.1 Blowing Up Terminal Brauer Pairs

The blowup of a smooth surface is smooth. We have the following noncom-mutative analogue.

Proposition 20.1. Let Z be a projective surface and α ∈ Br k(Z). Letπ : Z → Z be the blowup at a closed point. If (Z, α) is terminal, so is (Z, α).Hence if A is a terminal maximal order on Z, then any modification of Awith respect to π is terminal too.

Proof. We just examine the 3 possible cases

1. α is unramified at p

2. α is ramified on a smooth curve C through p

3. α is ramified on a node C1 ∪ C2 at p.

Let E be the exceptional locus. In case 1, the ramification of α on Z is es-sentially the same as on Z and α is unramified on E by the Artin-Mumfordsequence. Hence (Z, α) is terminal. In case 2, again the Artin-Mumfordsequence implies that α is unramified on E and the ramification of α onZ is essentially the same as that on Z. Hence (Z, α) is terminal. In case3,Boris: picture missing, Artin-Mumford shows that α is ramified on E withramification index e =secondary ramification index. The node of ramifica-tion has split into two and we see from Artin-Mumford that they are bothterminal.

21 Appendix: Projective Covers

Let A be a ring.

Definition 21.1. Let M be an A-module. A projective cover of M is asurjective map ϕ : P →M such that

PC1 P is projective,

PC2 given any morphism ψ : N → P such that the composite Nψ−→ P

ϕ−→

M is surjective, have ψ is surjective. That is, given any submoduleN < P with ϕ|N surjective we have N = P .

Example 21.2. Let P be a finitely generated projective A-module and L <P be such that L ⊂ (rad A)P . Then P → P/L is a projective cover of P/L.To see this, note that if N < P with N → P/L→ 0 then (N +L)/L = P/Land so N + L = P . Thus N + (rad A)P = P and so by Nakayama’s lemmaN = P .

77

Proposition 21.3. Projective covers are unique up to isomorphism. Hencefinitely generated modules P, P ′ are isomorphic if

P

(rad A)P≃

P ′

(rad A)P ′.

Proof. Let ϕ : P → M and ϕ′ : P ′ → M be projective covers. Since P isprojective, we have a commutative diagram

Pφ

//

ψ��

M

P ′φ′

>>|

||

||

||

|

ψ′

II

Condition PC2 on P ′ shows that ψ is surjective. Since P ′ is projective,Ext1

A (P ′,−) = 0 and so since ψ is surjective, the map ψ is split, say withsection ψ′. Condition PC2 on P implies ψ′ is surjective too. Therefore ψ andψ′ are inverse isomorphisms.

22 Appendix: Finite Global Dimension

We give here a simple sufficient criterion for a ring to have finite globaldimension and some basic properties.

Theorem 22.1. Let A be a noetherian ring and z ∈ rad A be a regularnormal element. If gl. dim A/z A = d− 1 <∞, then gl. dim A = d.

Proof. It suffices to show that for any finitely generated A-module M wehave pd M 6 d and equality holds for some M . We consider the z-torsionsubmodule

τ(M) = {m ∈M | znm = 0 for n≫ 0}.

Since z is normal, this is an A-module. Note that M/τ(M) is z-torsionfree whilst τ(M) has a finite filtration by A/z A-modules, 0 < annMz <annMz

2 < · · · . Hence the theorem follows from the next two lemmas. Wewill continue to assume below that A is noetherian and z ∈ rad A is regularnormal.

Lemma 22.2. Let M be a finitely generated A/z A-module, then pdAM =pdA/z AM + 1 if right hand side is finite.

78

Proof. Let N be an A-module and consider the change of rings spectralsequence

Ep,q2 = ExtpA/z A(M,Ext1

A(A/z A,N)) =⇒ Extp+qA (M,N)

Since z is regular, we have the following A-free resolution of A/z A

0 −→ Az−→ A −→ A/z A −→ 0.

Hence the spectral sequence has only two nonzero rows, where q = 0, 1.This tells us that if pdA/z AM = i then Exti+1

A (M,N) = 0. Hence pdAM 6pdA/z AM + 1. Also the spectral sequence gives a surjection

Exti+1A (M,N) −→ ExtiA/z A(M,Ext1

A(A/z A,N))

Pick an A/z A-module N suc that ExtiA/z A(M,N) 6= 0. Then by the free res-

olution of A/z A above, we have Ext1A(A/z A,N) = N/z N = N . Therefore

pdA = pdA/z AM + 1.

Lemma 22.3. Suppose M ∈ A−Mod is finitely generated and z-torsion free.Then pdA = pdA/z AM/zM .

Proof. We first show that if M is z-torsion free and M/zM is a finitelygenerated free A/z A-mdoule, then M is free over A.

Let m1, . . . ,mr ∈ M be such that their images in M/zM form a freeA/z A basis. It suffices to show that m1, . . . ,mr is a free A-basis for M . Theygenerate M by NAK. Suppose we have a nontrivial relation

∑aimi = 0 in

M . Since M is z-torsion free, we can extract a sufficently high power of z sothat ai 6 inz A. This contradicts the fact that m1, . . . ,mr give an A/z A-basisfor M/zM .

Conversely, if M is A-free then M/zM is A/z A-free. Therefore if M isA-projective then M/zM is A/z A-projective.

We proceed by induction on pdA/z AM/zM . Consider an exact sequence

0 −→ K −→ F −→M −→ 0

of A-modules with F finitely generated free. Then since M is z-torsion free,we get an exact sequence

0 −→ K/z K −→ F/z F −→M/zM −→ 0.

If pdA/z AM/zM = 0, then this splits, so

(M ⊕K)/z(M ⊕K) ≃ M/zM ⊕K/z K

≃ F/z F

79

which is A/z A-free. The claim implies that M ⊕K is A-free, so M is pro-jective also. Suppose now pdA/z AM/zM > 0 so pdAM > 0 also. Therefore

pdA/z AM/zM = pdA/z AK/z K + 1

by induction = pdAK + 1

= pdAM

It remains to consider the case where pdA/z AM/zM =∞. Suppose pdAM <∞, then let P • →M → 0 be a finite projective A-resolution. Then a simlarargument to the above shows that P •/z P • → M/zM → 0 is a projectiveA/z A-resolution. Contradiction finishes proof of Lemma 22.3.

Proposition 22.4. Let (R,m) be a commutative regular local ring of dimen-sion d. Let A be an R-algebra of global dimension d which is finitely generatedfree as an R-module. Then any A-module M which is finitely generated freeas an R-module is also projective over A.

Proof. Let z1, . . . , zd ∈ m be a regular system of parameters. So if Mi =M/(z1, . . . , zi)M then we get an exact sequence

0 −→Mizi+1

−→Mi −→Mi+1 −→ 0

For any A-module N we get an exact sequence

ExtjA(Mi, N)zi+1

−→ ExtjA(Mi, N) −→ Extj+1A (Mi+1, N)

If ExtjA(Mi, N) 6= 0, then multiplication by zi+1 on it is not surjective byNAK. Hence pdAMi+1 > pdAMi + 1, and so pdAM + d 6 pdAMd 6 d.Therefore pdAM 6 0 and we are done.

23 Appendix: excellent rings

ATTENTION: Boris: in the appendix we drop characteristic zero assump-tion.

We will need resolutions of singularities and to pass to completions. Ex-cellent rings provide a good setting where resolutions work and completionsare well behaved. Excellent rings are special types of commutative noethe-rian rings. You can find the definition and basic facts about them in [EGAIV, Section 7.8] also Matsumura. Next result gives some examples.

Proposition 23.1 (EGA IV, Proposition 7.8.3). 1. Fields of any char-acteristic are excellent as are Dedekind domains whose fields of frac-tions are characteristic 0.

80

2. Complete local noetherian rings are excellent.

3. Any localisation of an excellent ring is excellent.

4. If R is excellent, so is any R-algebra of finite type.

A noetherian scheme Z is excellent if for some (and hence any) openaffine cover Z = ∪Spec Ri, the rings are excellent. So in particular, allquasi-projective varieties are excellent. Excellent rings are well-behaved inmany ways

Proposition 23.2 (EGA, Section 7.8). Let R be an excellent reduced localring

1. R is reduced

2. The normalisation Rnorm of R is a finite R-algebra.

3. Normalisation commutes with completion

4. If R is an integral domain then any maximal chain of of primes in Rhas length dim R, that is, R is equi-dimensional.

Theorem 23.3. (Lipman) Let Z be a 2-dimensional integral normal scheme.Then there is a proper birational morphism Z → Z with Z smooth.

Proposition 23.4. (Artin, from Chapter XI of Cornell, Silverman, Arith-metic Geometry) Let Z be a 2-dimensional excellent integral normal scheme

and p ∈ Z be a closed point. Let Zp = Spec OZ,p. Then

1. If Z → Z is a projective birational morpihsm then so is Z×Z Zp → Zp.

Furthermore, Z is nonsingular above p if and only if Z ×Z Zp is.

2. Any projective birational morphism Zp → Zp comes from base changeas above.

24 Blowing up and Terminal Ramification Data

We now consider the question of finding a good model for K-centraldivision rings when K is the function field of a surface Z. Our aim is toobtain a noncommutaive analogue of smooth models and a resolution ofsingularities type result.

Our point of view is to note that blowing up and normalisation can beused to resolove singularities, so we will see how blowing up can “improve”an order.

81

From now on a surfac will always mean a 2-dimensional integral normalscheme such that the residue fields of closed points are k.

Let Z be a smooth projective surface and α ∈ Br k(Z). Let A be amaximal order such that α = [A ⊗Z k(Z)]. Thus A and the Brauer pair(Z, α) have the same ramification data.

Let p ∈ Z be a closed point and Z → Z be the blowup at p with excep-tional curve E.

Question. How to relate the ramification data of (Z, α) to (Z, α)?Answer. Note that π∗A is an order on Z isomorphic to A away from E.

Pick a maximal order π#A containing π∗A such that π∗A|Z−E = π#A|Z−E.We call π#A a blowup of A at p ∈ Z. If C ⊂ Z is an irreducible curve andC be its strict transform, then

A⊗Z OZ,C ≃ π∗A⊗Z OZ,C

Hence the ramification of α at C ⊂ Z is the same as α at C ⊂ Z. The rami-fication of (Z, α) at E can be determined using the Artin-Mumford equence.This is best illustrated by an example.

25 Appendix: Galois extensions

Galois extensions of discrete valuation rings.Let (R,m) be a discrete valuation ring with fraction field K and perfect

residue field κ. Let F/K be a Galois extension with Galois grou pG. Recallthat the integral closure S ⊂ F of R in F is such that S/R is Galois withGalois group G (c.f. Section 1).

Now S is a Dedekind domain which is semilocal since the maximal ide-als n1, . . . , ng correspond to the maximal ideals of the finite dimensional κ-algebra S/mS. Note that G permutes the maximal ideals

Proposition 25.1. G permutes the maximal ideals of S transitively.

Proof. By the Chinese remainder theorem the natural map S → S/n1×· · ·×S/ng is surjective. Hence we can pick a ∈ S such that a ∈ n1 but a 6 inni fori > 1. Consider b =

∏σ∈G σ(a) ∈ SG = R. Note that b is not invertible since

a is not invertible. Hence b ∈ m. Thus also b ∈ ni for all i. This shows thatσ(a) ∈ ni for some i. Hence σ(n1) = ni as desired.

Let D be the stabiliser in G of one of {n1, . . . , ng}. The group D is calledthe decomposition group and it is well-defined up to conjugacy. Also, wedefine κS = S/ni which is independent of i up to isomorphism. Let σ ∈ Dwhere for definiteness D = stab n1. Then σ induces an automorphism of κS.We thus obtain a group homomorphism ψ : D → Gal(κS/κ). We define thekernel of ψ to be the inertia group.

82

Proposition 25.2. Suppose κ is perfect

1. κS/κ is Galois.

2. The map ψ is surjective, so Gal(κS/κ) ≃ D/I where I is the inertiagroup.

Proof. It suffices to show that κDS = κ, for then Galois theory shows thatκS/κ is Galois with Galois group D/ ker(ψ).

Let α ∈ κS−κ. We need to find σ ∈ D such that σ(α) 6= α. Let m ∈ κ[x]be the minimal polynomial of α. Since κS/κ is separable, the roots of m aredistinct. Thus, it suffices to show that D acts transitively on these roots.

Fix D = stab n1. The Chinese remainder theorem allows us to lift α toα ∈ S such that α ∈ ni for i > 1. Consider the polynomial

p(x) =∏

σ∈G

(x− σ(α)) ∈ SG[x] = R[x]

and we denote its image modulo n1 by p ∈ κ[x].For σ 6∈ D we have σ(α) ∈ n1. Hence p(x) = xN f(x) where N = |G|−|D|

and f(x) =∏

σ∈D(x − σ(α)). Now p(a) = 0, so f(α) = 0. Thus m(x)|f(x)and we are done.

References

[Bou89] Nicolas Bourbaki. Commutative algebra. Chapters 1–7. Elementsof Mathematics (Berlin). Springer-Verlag, Berlin, 1989. Translatedfrom the French, Reprint of the 1972 edition.

[Eis95] David Eisenbud. Commutative algebra, volume 150 of GraduateTexts in Mathematics. Springer-Verlag, New York, 1995. With aview toward algebraic geometry.

[FD93] Benson Farb and R. Keith Dennis. Noncommutative algebra, volume144 of Graduate Texts in Mathematics. Springer-Verlag, New York,1993.

[GS06] Philippe Gille and Tamas Szamuely. Central simple algebras andGalois cohomology, volume 101 of Cambridge Studies in AdvancedMathematics. Cambridge University Press, Cambridge, 2006.

[Lam91] T. Y. Lam. A first course in noncommutative rings, volume 131 ofGraduate Texts in Mathematics. Springer-Verlag, New York, 1991.

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[Rei03] I. Reiner. Maximal orders, volume 28 of London Mathematical So-ciety Monographs. New Series. The Clarendon Press Oxford Uni-versity Press, Oxford, 2003. Corrected reprint of the 1975 original,With a foreword by M. J. Taylor.

[Ser79] Jean-Pierre Serre. Local fields, volume 67 of Graduate Texts inMathematics. Springer-Verlag, New York, 1979. Translated fromthe French by Marvin Jay Greenberg.

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