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 about R gives information about K. There are many choices for R but the best 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 field k(X) = K. Such an X is called a model for K. X gives information about K. There are many choice for X so we usually impose extra conditions for example: normal/smooth, projective, etc. The theory of orders studies “integral models” for central simples algebras. In analogy 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. An R-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) ⊗R A is a central simple K(R)- algebra.

Note that torsion free implies A ֒→ K(R) ⊗R A so A is a subring of a central simple algebra. In fact, with the above notation and D := 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]. Rhx, yi A = (x2 − u, y2 − v, xy + yx) khx, yi = . (xy + yx)

1 Proposition 0.3. Let R ⊂ S be an extension of commutative noetherian normal domains and A an R-order in D. Then A ⊗R S is an S-order in D ⊗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 say S/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] ⊂ k(x)

Galois ⊂ ⊂

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 ﬁnite such 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 the integral 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 that K(S)G = K(R). Now, K(R) = K(SG) ⊆ K(S)G. For the reverse α G inclusion, simply note that that for β ∈ K(S) ,

α δ := αg (β)g (β) ··· g (β) = 1 2 n−1 β γ := βg1(β) ··· gn−1(β)

G G where G = {id,g1, ··· ,gn−1}. Since γ ∈ S , we must have δ ∈ S and α G so β ∈ K(S ). 2. Consider the skew group ring:

S∗G = Sg g∈G M with multiplication given by multiplication in S and G and skew com- 1 mutation relation g · s := g(s)g. Have an idempotent e := |G| g∈G g. Indeed it is easy to see that for all h ∈ G, he = eh = e. Note also that R ≃ eS∗Ge = eSe. P To show SG is noetherian, we consider the inclusion preserving map

left ideals / S∗G-submodules of eS∗Ge 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-submodules of 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 / left ideals of eS of S∗G M / (S∗G) M = SM

has a left inverse N 7→ eN. The same reasoning as before shows S is 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 integrally closed. Need to show S/R is finitely generated, so suffices to show S/R is 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 over R the same same is true of g(α) for all Pg ∈ G. Let V < F be a ﬁnitely generated R-submodule such that KV = F . The pairing gives an isomorphism ∗ between 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 =≃ F where the last isomorphism is given by the trace map. Now pick a

K-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. L

1.2 Cyclic Algebras Let R be a commutative, noetherian, normal domain and S/R a cyclic ring extension with Galois group G = hσ | σn = 1i. Let L be a rank 1, locally free S-module (for example L = S) and Lσ be an S-bimodule, which as a left S-module is L, but the right module action is m · s := σ(s)m. Furthermore, ⊗n suppose we are given an injection of S-bimodules φ: Lσ ֒→ S satisfying the

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

⊗(n−1) φ⊗1 / Lσ ⊗S Lσ ⊗S Lσ S ⊗S Lσ

1⊗φ / Lσ ⊗S S / Lσ To understand this condition, consider the easy case where L is free so can ⊗n n write Lσ = Su with us = σ(s)u. Note that the morphism φ: Lσ = Su → S is 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 now deﬁne the cyclic algebra

⊗(n−1) A (Lσ; φ)= S ⊕ Lσ ⊕···⊕ Lσ with multiplication given by:

⊗(i+j) ⊗i ⊗j Lσ i + j < n Lσ ⊗ Lσ −→ ⊗(i+j) 1⊗φ⊗1 ⊗(i+j−n) ( Lσ −→ Lσ i + j ≥ n It is left as an exercise to check that the overlap condition ensures multiplication is well-deﬁned and associative. Note that A is an R-algebra since R acts 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. Hence A is a finitely generated R-module. Suffice to show that K(R) ⊗R A is the classical cyclic algebra, which are known to be central simple. Note K(R) ⊗R L = K(S) ⊗S L = Su for some u. Also K(R) ⊗R A is a cyclic algebra in the above sense with

n−1 K(R) ⊗R A = K(S) ⊕ K(S)u ⊕···⊕ K(S)u . The overlap condition implies un ∈ K(S)G = K(R) so we get the classical cyclic algebra.

1.3 Crossed Product Algebras Recall the following approach to group cohomology. Let G be a ﬁnite group and M a G-module which also makes it a ZG-module. Then Hi(G, M) is the cohomology of the complex C•(G, M) where

Ci(G, M)= functions f : Gi → M 5 and di : Ci → Ci+1 is deﬁned by

i i j (d f)(g1, ··· ,gi+1)= g1.f(g2, ··· ,gi+1)+ (−1) f(g1, ··· ,gjgj+1, ··· gi+1) j=1 i+1 X +(−1) f(g1, ··· ,gi) Case of interest: let S/R be a Galois extension of commutative, noetherian domains so K(S)/K(R) Galois as well, say with Galois group G. Hence K(S)∗ is a G-module. The 2-cocycles f : G×G → K(S)∗ are those satisfying

−1 g1.f(g2,g3)= f(g1g2,g3)f(g1,g2g3) f(g1,g2) (∗∗) (note that we switched to multiplicative notation). This is called a factor set. If f has values in S then get crossed product

A(S, f) := Sug g∈G M 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 simple K(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 an order in D. The answer is yes, due to: Proposition 1.5. 1. There is a finitely generated R-submodule M < D such that K(R) ⊗R M ≃ D. 2. Given any M as in (1) the following is an R-order in D:

Ol(M) := {α ∈ D | αM ⊆ M} .

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

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

6 2 Relationship With Azumaya Algebras

An Azumaya algebra can be thought of as a central simple algebra varying over 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 equivalent conditions hold:

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

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

op ψ : A ⊗R A −→ 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.

op 4 Proof. (1) ⇐⇒ (2). Note A ⊗R A and EndRA are locally free of rank n so ψ is an isomorphism if and only if ψ ⊗ R/m is an isomorphism for all maximal m ⊳ R. Now (1) ⇐⇒ (2) follows from: Lemma 2.2. Let A be a ﬁnite dimensional k-algebra, where k is a ﬁeld. Then op A is central simple if and only if ψ : A ⊗k A → EndkA is an isomorphism.

op Proof. (⇒) A ⊗k A is central simple and so ker ψ = 0. (⇐) Note Z(A)= k. Also, given b ∈ A − {0}, we see that the ideal in A generated by b contains (im ψ)(b)= A. (3)⇒(2) by descent theory. 2 (1)⇒(3) Going sufficiently local we can assume A ≃ Rn . Then n n3 (e1, ··· ,en) ⊂ A ≃ R forms a complete set of orthogonal idempotents if 2 and only if ei = ei and eiej = 0 if i =6 j and ei = 1. These define a 3 subscheme X ⊂ An such that for any extension S/R the complete sets of R P idempotents for A ⊗R S correspond to sections of X ⊗R S → Spec S. Note is π : X → Spec R surjective by (1). Now π is smooth by Grothendieck’s criterion and the fact that you can lift idempotents. Hence it has an étale section say over the étale extension S/R. A complete set of idempotents for A ⊗R S can be used to give a Peirce decomposition of A ⊗R S which gives Mn(S)

7 Example 2.3. Consider the cyclic algebra A = A(Lσ; φ) formed from the ⊗n ∼ cyclic ´etale ring extension S/R of degree n and φ: Lσ −→ S. Then A is an Azumaya R-algebra.

Proof. It suﬃces to show A ⊗R S ≃ Mn(S). Since S/R is cyclic ´etale we n get S ⊗R S ≃ i=1 S where G = Gal(S/R) cyclically permutes the components. We can work locally and assume Lσ = Su i.e. L is free. We have an Q isomorphism A ⊗R S → Mn(S) given by

0 1 S ......   S ⊗R S −→  .  , u 7−→ . .. 1 S      α 0        where α = un ∈ (S∗)G = R∗. It is left as an exercise to show that this morphism is surjective and

1 s1 s1 n ...... u = α  .  , u  .  = σ  .  u 1 sn sn            

Proposition 2.4. Let R → S be a morphism of commutative noetherian rings and A be an Azumaya R-algebra. Then A ⊗R S is Azumaya over S. In particular, 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 and A an R-order. Then there is some r ∈ R − {0} such that A[r−1] is an Azumaya R[r−1]-algebra. More precisely, the Azumaya locus of A is dense in Spec R.

Proof. The Azumaya locus is the locus where (i) A is locally free and (ii) A⊗R op A ≃ 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 from ⊗n S/R and relation φ: Lσ → S is Azumaya on the intersection of ´etale locus of S/R and locus where φ is an isomorphism.

8 The upshot of this is that we can view an R-order as a family of central simple algebras and their degenerations. The non-Azumaya locus (or ramification locus) is an important invariant of the order, which can be used to study it. Example 2.7. [Terminal, maximal with nodal ramification] Let k be an algebraically 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 and u 7→ xn. This extension has Galois group G = hσ | σn = 1i with action σ(x)= ζx, ζ being a primitive nth root of 1 and σ(v)= v. Let L = Sy and consider the relation

⊗n n φ: Lσ = Sy −→ S yn 7−→ v ∈ SG = R

Thus we get the cyclic algebra A = A(Lσ,φ). By Example 2.6 it is Azumaya on uv =6 0. Note ﬁrst A = S ⊕ Sy ⊕···⊕ Syn−1 = R[x] ⊕ R[x]y ⊕···⊕ R[x]yn−1 Rhx, yi = (xn − u, yn − v,yx − σ(x)y) khx, yi = . (yx − ζxy) What about on uv = 0? At (u, v)=(1, 0): if m =(u − 1,v) then khx, yi A ⊗ R/m = . R (xn − 1, yn, yx − ζxy)

R Now x and y are normal since they skew-commute (normal means x A ⊗ m = A ⊗ R x) and so y A ⊗ R is a two sided ideal (in fact a nilpotent one). m R m Therefore not Azumaya at (1, 0). Similarly, we can see that the ramiﬁcation locus is all of uv = 0.

3 Reduced Trace

We need the following tool, known as the reduced trace of a central simple algebra. It generalises the trace map for field extensions. Our approach is via Galois descent. Let K be a field and D a central simple K-algebra. There is some Galois field extension F/K which splits D in the sense that D ⊗K F ≃ Mn(F ) for some n. Those split by F/K are classified using non-abelian cohomology, which we describe now.

9 3.1 Non-Abelian Cohomology Let F/K be a Galois ﬁeld extension with Galois group G and note that G acts on the non-commutative group P GL(n, F )= GL(n, F )/F ∗

≃ AutF −algMn(F ) where the last isomorphism follows from the Skolem-Noether theorem. Note that conjugation by an element of P GL(n, F ) is well-defined. We define {1-cocycles f : G → P GL(n, F )} H1(G, P GL(n, F )) := ∼ where f : G → P GL(n, F ) is a 1-cocycle if fgh = fg(g.fh) for all g,h ∈ G. ′ ′ −1 Also, f ∼ f if there is some a ∈ P GL(n, F ) with fg = a fg(g.a) for all g ∈ G. H1 (G, P GL(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 → P GL(n, F ) be a 1-cocycle. Then there is an induced G-action on Mn(F ) defined by −1 g.(aij) := fg(g.aij)fg . G Also, Mn(F ) is a central simple K-algebra. 2. The map {1-cocycles f : G → P GL(n, F )} −→ {central simple K-alegbras} G f 7−→ Mn(F ) induces a bijection of pointed sets

isomorphism classes of H1(G, P GL(n, F )) −→∼ central simple K-algebras  split 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 Galois field extension that splits D i.e. D ⊗K F ≃ Mn(F ). For a ∈ D we define the reduced trace (respectively determinant, characteristic polynomial) of a to be the trace (respectively determinant, characteristic polynomial) of a ∈ Mn(F ). This is well defined, as we shall see below. It is denoted by tr a (det a, char. pol.a).

10 Proposition 3.2. 1. Let a ∈ D. Then char. pol. a ∈ K[x] (not just F [x]). In particular, tr a, det a ∈ K. 2. char. pol. a is independent of the choice of F/K and the isomorphism D ⊗K F ≃ Mn(F ). Proof. 1. It suﬃces to show that for any g ∈ G := Gal (F/K), we have g.(char. pol. a) = char. pol. a for then char. pol. a ∈ F G[x] = K[x]. By Galois descent, there is a 1-cocycle f : G → P GL(n, F ) such that G D ≃ Mn(F ) with G-action induced from f as described in Theorem 3.1 (1). Then

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

= char. pol. g.(aij) −1 = char. pol. fg(g.aij)fg = char. pol. (g.aij) G = char. pol. (aij) since a ∈ Mn(F ) = 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 generated R-module (since R and A are noetherian) so a is a root of a monic polynomial m(x) ∈ R[x]. But p(x) has the same roots as the minimal polynomial of a so p(x) | m(x)r ∈ R[x] for r ≫ 0. Gauss’ Lemma implies the coeﬃcients of p(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 a reduced trace map tr : A → R.

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

tr : Mn(F ) × Mn(F ) −→ F (α, β) 7−→ tr (αβ)

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

∗ tr M := HomR (M,R) ֒→ HomR (M,K) = HomK (KM,K) = HomK (D,K) ≃ 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)-algebras D. 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 is based on the observation that for a commutative, noetherian domain S with K(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 domain, 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 to inclusion 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 extension of commutative, noetherian, normal domains. If A ⊗R S is maximal, so is A. Proof. Suppose A is not maximal. Then if B is an R-order strictly containing A then B ⊗R S is an S-order and B/A ⊗R S =6 0 so it strictly contains A ⊗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, note that by Definition/Proposition 2.1, étale locally, A is Mn(R) so A is maximal by Proposition 4.2 and the above example.

12 If the concept of maximality indeed generalises normality to the noncommutative setting, then the procedure of normalisation should extend to to embedding in a maximal order. The following proposition shows that this is 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 order containing A. Unfortunately, unlike normalisation, the maximal order in the above proposition is not unique. It is easy to come up with such examples: RR Example 4.6. Let R = k[x] and A = . It is contained in a xR R maximal orders

RR R x−1R 1 0 RR 1 0 −1 and = . RR xR R 0 x RR 0 x Despite this, maximality is compatible with localisation and completion in dimension 1, which we now show.

Proposition 4.7. Let A be maximal R-order and U ⊂ R be a multiplicatively closed 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- −1 erated A-submodule M < B such that M[U ] = B. Note K(R) ⊗R M = K(R) ⊗R[U −1] B = D and M is ﬁnitely generated as an R-module too. Thus Ol(M) := {α ∈ D | αM ⊆ M} is an R-order which contains A. It strictly contains A since

−1 −1 Ol(M)[U ]= β = αu | α ∈ D,αM ⊆ M, u ∈ U −1 −1 = β ∈ D | βM[U ] ⊆ M[U ] −1 ⊇ B ⊃ A[U ] ⊆ A which contradicts the maximality of A.

Definition 4.8. Let R be a Dedekind domain (i.e. a normal, noetherian domain 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 < V such that KM = V .

13 Lemma 4.9. Let R be a dvr (i.e. a local Dedekind domain or equivalently, a local PID) and Rˆ its completion. Let K, Kˆ be their ﬁeld of fractions and ν be the valuation on Kˆ . Let V be an n-dimensional K-vector space. Let Vˆ = Kˆ ⊗K V . 1. Let Mˆ be a full Rˆ-lattice in Vˆ . Then it is free over Rˆ with an Rˆ-basis {m1, ··· ,mn} ⊆ V . 2. There are inverse bijections

Φ full R-lattice / full Rˆ-lattice o M < V Ψ ( Mˆ < Vˆ )

M / RMˆ

Mˆ ∩ V o Mˆ

Proof. 1. Note Mˆ is torsion free over a PID Rˆ so is free with basis m1, ··· ,mn ∈ Vˆ . Pick K-basis v1, ··· ,vn of V . Change of base matrix ˆ ˆ (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 after using elementary columnP operations (ECO’s) over Rˆ and elementary row 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 in Rˆ. Uniformising parameter for R is also a uniformising parameter for Rˆ so can scale rows using ERO’s to get (αij) ∈ Mn(Rˆ) and some ∗ ∗ αij ∈ Rˆ . Switch columns so that α1,1 ∈ Rˆ . Using ECO’s can make 0 = α1,2 = ··· = α1,n. Now R is dense in Rˆ so can use ERO’s over K to 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 a ˆ ˆ n ˆ ˆ ˆ full R-lattice M = i=1 Rmi with mi ∈ V . Then Ψ(M) = M ∩ V = Rm which is a full R-lattice in V . Also, ΦΨ(Mˆ ) = Rˆ( Rm ) = i L i Rmˆ i = Mˆ . Similarly ΨΦ = id L L L Corollary 4.10. Let R be a dvr and A a maximal R-order. Then A ⊗R Rˆ is a maximal Rˆ-order.

Proof. By Lemma 4.9 (2) if B ⊃ A ⊗R Rˆ is an Rˆ-order in A ⊗R K(Rˆ) then B ∩ (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:

Rhx, yi A = . (x2 − u, y2 − v, xy + yx)

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

Rˆmhx, yi Rˆ Rˆ A ⊗ Rˆ = −→ m m ⊂ M (Rˆ ) R m (x2 − (1 + w), y2 − v,yx + xy) vRˆ Rˆ 2 m m m p(w) 0 x 7−→ 0 −p(w) 0 1 y 7−→ v 0 and so A ⊗R Rˆm is not maximal.

5 Normal Orders in Dimension 1

In Section 4 we saw that maximality was a generalisation of integrally closed for orders. We now introduce a second such generalisation to orders over dvr’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 Jacobson radical J has the form Aπ for some π ∈ A.

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

Lemma 5.2. Let (R, m) be a local ring and R ⊂ S a ﬁnite (non necessarily commutative) 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 that In ⊆ 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 by Nakayama’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 left multiplication mS ⊆ J.

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

3. Using part (1) we know J/mS E S/mS which is artinian. By Nakayama’s lemma, J 2/mS ⊳J/mS or J/mS = 0. Continuing on we get a strictly decreasing sequence J 3 J 2 S ··· ⊳ ⊳ ⊳ mS mS mS Since S/mS is artinian this must eventually reach zero. That is, for some n ∈ N J n ∈ mS.

We shall now prove that our deﬁnition of a so normal order is left-right symmetric 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)⊗R A 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∗. Thus A ≃ 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 which implies A ⊆ π−1Aπ ⊆ π−2Aπ2 ⊆ · · · . The chain terminates by the existence of maximal orders and so Aπ = πA. Example 5.4 (Cyclic algebras). Let (R, m) be a dvr such that the characteristic of R/m is coprime to n, and S/R be a cyclic extension of normal rings of degree n with Galois group G = hσ | σn = 1i. Consider the cyclic ⊗n algebra A := A(S/R,Lσ,φ) built from the relation φ: Lσ → 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 uniformising parameter for R. Therefore uA is an ideal which is nilpotent modulo 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 Theorem 4.14 of [Lam91]). But A/uA ≃ S/mS which is a product of fields since S/R is étale.

2. If S is semilocal, then rad S = St for some t ∈ S. Also σ(t) generates rad S, so σ(t)= αt for some α ∈ S∗. Hence t is normal in A, we wish to show that At = rad A. Note that At ⊆ rad A, so it is enough to show 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¯ ⊕···⊕ Su¯ n−1. Note that ue is central in A¯, so A¯ is 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 assume un ∈ R∗ is an n-th power modulo m. Scaling u, we can assume un = 1 ¯ ¯ e n in A. Therefore A ⊇ κ[u ] ≃ κ Z/ e Z ≃ κε1 ×···× κεn/e where εi are central idempotents. Now A¯ = n/e Aε¯ and each of these factors i=1 i Aε¯ i is a cyclic algebra, so A¯ is central simple. Q

Deﬁnition 5.5. A ring A is hereditary if the global dimension gl. dim.A 2 1 is equal to 1, that is, ExtA (−, −) = 0 but ExtA (−, −) =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 A is projective (e.g. when A is normal). Then any ﬁntely generated torsion-free A-module is projective and gl. dim.A = 1. Proof. We have the following exact sequence

0 → J → A → A/J → 0.

1 1 Since J is projective and A is free, ExtA (J, −) = ExtA (A, −) = 0 and so 2 ExtA (A/J, −) = 0 and so pd A/J = 1. Since A/J is the direct sum of all the simples (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 get one 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 an A-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

p 0 −→ M −→ M −→ M/pM −→ 0

Let N be a ﬁnitely generated A-module. Since pd M/pM ≤ 1, we get i i 1 surjective maps ExtA (M,N) → ExtA (M,N) for i ≥ 1, i.e. pExtA (M,N)= 1 ExtA (M,N) where p generates the maximal ideal. Nakayama’s lemma then i implies ExtA (M,N) = 0, so we are done. One nice property about normal orders is that it respects ´etale base change.

Proposition 5.7. Let ϕ : (R, m) ֒→ (S, n) be a morphism of dvrs with −1 ϕ (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 ⊗ S A R ≃ ⊗ S (rad A) ⊗ S rad A R R 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) is principal. Conversely, assume rad (A ⊗R S) is free of rank 1. Let J := rad A Note ﬁrst that J is projective since

1 1 ExtA (J, −) ⊗R S = ExtA⊗RS (J ⊗R S, −) = 0.

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

18 We know that 2 2 (J/J ) ⊗R S ≃ (J ⊗R S)/(J ⊗R S)

≃ (A ⊗R S)/(J ⊗R S)

≃ (A/J) ⊗R S. The proposition then follows from the next lemma which shows that J/J 2 ≃ A/J, so A and J are projective covers of isomorphic modules and so are isomorphic by Proposition 21.3. Lemma 5.8. Let A¯ be a semisimple κ-algebra and κ′ a ﬁnite dimension κ-algebra. Then two ﬁnitely generated A¯-modules M,N are isomorphic if ′ ′ M ⊗κ κ ≃ N ⊗κ κ . ¯ n ¯ ¯ Proof. Let A = 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 direct sum decompositionQ n ⊕mi M ≃ Si i=1 M ′ depends only on M ⊗κ κ . But

′ ⊕mi ′ (M ⊗κ κ )εi = Si ⊗κ κ so ′ dimκ(M ⊗κ κ ) mi = ′ dimκ(Si ⊗κ κ )

6 Criterion for maximality

In this section, we shall give the Auslander-Goldman’s criteria for an order to be maximal. One easy way to obtain a bigger order from a given one is to take its reflexive hull as follows: Let R be a commutative noetherian normal domain, K be the fraction field 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 denote duals ∗ 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, then ∗∗ M = P MP where p runs over height 1 primes. Proof. SeeT [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 ﬁnitely generated, so it suﬃces to show that it is closed under multiplication. This ∗∗ ∗∗ is clear from M = P MP , but a better proof is by functoriality of . Let a ∈ A which induces the following commutative diagram T a A / A

a∗∗ A∗∗ / A∗∗

a D / D

so left multiplication by a preserves A∗∗, so multiplication extends to A ⊗ A∗∗ −→ A∗∗. Repeat with right multiplication by a′ ∈ A∗∗. We now give Auslander-Goldman’s criterion for maximallity. It reduces veriﬁcation of maximality in general to checking it over a dvr.

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

20 Proof. If A is maximal, then by Lemma 6.3 it is reﬂexive. Maximality is local 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. T T 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.

−1 B ⊂ K(R) ⊗R A = A[p ] = A[π−1] by Proposition 5.3 = πiA Z i[∈ Hence π−ia ∈ B for some a ∈ A−πA. By multiplying by πi−1 we can assume that i = 1 and so (π−1A ∩ B)/A is a nonzero A-bimodule. Multiplying by π gives a nonzero ideal

(A ∩ πB) A ≤ πA πA which is simple by assumptions, and so A ∩ πB = A and π−1 ∈ B. This shows that B is not finitely generated. Remark 6.7. The converse of Proposition 6.6 also holds as we shall prove in Proposition 10.2. Unfortunately, we have not developed the theory enough yet 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 an algebraically closed field with characteristic not dividing n. Let R = k[u, v] and R hx, yi A = (xn − u, yn − v, xy − ζyx) where ζ ∈ k∗ is a primitive n-th root of unity. We claim that A is maximal.

21 n−1 i j Proof. We check maximlaity using Proposition 6.4. Note that A = i,j=0 Rx y is R-free, hence reﬂexive. We check that AP is maximal for all height 1 primes P . As we saw in Example 2.6 and 2.7 A is Azumaya on uv =6 0L and so by Example 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 constructed, as we saw in Example 2.7, from the cyclic algebra construction via n n φ: Sy → S, y 7→ v which is an isomorphism since v is a unit in AP . Thus by Example 5.4 AP is normal. Note also that x ∈ A is normal so xA is nilpotent modulo (u). Lemma 5.2(1) xAP ≤ rad AP . Also

AP k(v)[y] ≃ n xAP (y − v) which is a ﬁeld extension of k(v) and is simple and artinian. Thus AP is local. Hence AP is maximal by Proposition 6.6. n n If P =(v), then AP is constructed via φ : Sy → S, y 7→ v. This time, S/R is ´etale and im φ = vk[x, x](v) = mS and so again by Example 5.4 AP is normal. 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 rings and show, in particular, they are principal ideal domains. Suppose R is a complete discrete valuation ring, and A be an R-algebra, ﬁnitely generated as an R-module. For any N < rad A, A is complete with respect to the N-adic topology.

Lemma 7.1. With the above notation, let ε¯1,..., ε¯n ∈ A/N be a complete set of idempotents. Then there exists a complete set of idempotents ε1,...,εn ∈ A such that εi + N =ε ¯i for i = 1,...,n. Proof. (Sketch) See Theorem 6.7 of [CurtisReiner]. Main idea is to lift in- 2 4 8 ductivelyε ¯1 to A/N then A/N , A/N ,.... 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 division ring. Then A¯ = A/rad A is also a division ring. Proof. First A¯ is a semisimple ring, so is a product of matrix algebras over division rings. Hence A¯ fails to be a division ring if and only if there exists a nontrivial idempotentε ¯. Given one such idempotent, we can lift this to ε ∈ A (see [Lam91] Theorem 21.31). Since ε(1 − ε) = 0 ∈ K(R) ⊗R A, this contradicts 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 that A¯ := A/rad A is a division ring (e.g. R is complete and K(R) ⊗R A is a division ring). Then

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

s 2. For any b ∈ D := K(R) ⊗R A − {0}, we can write uniquely b = uπ for some 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 for some s ∈ Z.

e 5. If mA = (rad A) , then e · dimR/m A¯ = rankRA. Proof. 1. Note that J := rad A is nilpotent modulo m so there is a minimal s ∈ N such that J s ≤ mA. Pick b ∈ J s−1 − mA and let p be a uniformising parameter of m. Recall that Ol(J)= {α ∈ D | αJ ⊆ J} is an R-order in D containing A. Since A is maximal, we have Ol(J)= A, so p−1bJ ⊆ p−1J s ⊆ A. −1 −1 −1 We wish to show that p bJ = A. If p bJ ⊆ J, then p b ∈ Ol(J)= A, which is false, so p−1bJ * J. Since A/J is a division ring, the only proper right ideals of A are those contained in J. Since p−1bJ + J is a right ideal of A (and is not contained in J), p−1bJ + J = A, hence by Nakayama’s lemma p−1bJ = A and so J = b−1pA.

−1 s1 2. Recall that D = A[π ] and so bπ ∈ A for some s1 ∈ Z. Now i i πA i∈N π A = i∈N π A and so must equal 0 by Nakayama’s lemma. s1 s2+1 Thus there is a minimal s2 such that bπ 6∈ π A. Therefore u := bπs1T−s2 ∈ A − JT, so b = uπs2−s1 and u ∈ A∗ since A/J is a division ring.

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

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

5. Consider the ﬁltration A ⊃ J ⊃ J 2 ⊃ ··· ⊃ J e = mA which has factors J i/J i+1 = πiA/πi+1A which is isomorphic, as R/m-modules, to

23 A/πA = A¯, Then

rankRA = dimR/m A/mA i i+1 = dimR/m(J /J )

= Xe · dimR/m A.¯

8 Structure theory of complete division algebras

Let (R, m) be a complete dvr in this section. Let κ = R/m be the residue ﬁeld, which we will assume to be perfect starting from Section 8.2. The main 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 ring extension S/R is unramified if S/mS is a product of separable field extensions of R/m. If S is a domain, we also say that K(S) is an unramified field extension 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 two lemmas

Lemma 8.2. If D =6 K, then given a maximal R-order A ⊂ D, we have A/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 derive the contradiction that then A = R[π] which is nonsense since A is not commutative. Certainly, R[π] ⊆ A. By Nakayama’s lemma, it suﬃces to show 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 dimκ A/mA = e · dimκ A/rad A = e by assumption. Thus suﬃce to n1 nm show the map is injective. Suppose f¯ :=a ¯1π + ··· +a ¯mπ maps to 0 witha ¯i =6 0 and n1 < ··· < nm. Let choose an ai ∈ R such that ai =a ¯i in κ and note that ai ∈/ m and so is invertible. Since f¯ maps m1 nm n1 nm−n1 to 0, f := a1π + ··· + amπ = π (a1 + ··· + amπ ) ∈ mA. But nm−n1 a1 + ··· + amπ 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 division ring and it is a non-trivial extension of κ by Lemma 8.2. Pick a ∈ A − rad A and leta ¯ ∈ A¯ be its image. We claim that κ[¯a]/κ is separable. This is clear under hypothesis 1. If, on the other hand, hypothesis 2 holds, then by Section 7 Proposition 7.3-5, deg A¯ is coprime to char κ. So the minimal polynomial f¯(x) ∈ κ[x] ofa ¯ is separable. Note that R[a] is finitely generated over R so is complete. So by Hensel’s lemma, we can change a so that it is a root of a monic polynomial with the same degree as f¯(x). Hence we get a non-trivial unramified extension. We now resume the proof of Theorem 8.1. We prove, by induction on deg D, that D has a maximal unramified subfield, which must therefore also be splitting by Corollary 3.17 of [FD93]. If deg D = 1 the result is obvious, for D is then split and is a field. If not, by Lemma 8.3, there is a subfield F ⊂ D such that F/K is a non-trivial unramified extension. We seek to extend F to a maximal subfield K′ ⊂ D with 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 a central simple F -algebra (see [FD93] Theorem 3.15 for the proof). Note that hypotheses 1 or 2 carry over to C/F . Hence, by induction, we can find an unramified maximal subfield K′ ⊂ C for C/F . Note that K′/K is unramified since 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 make the 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 R consists of equivalence classes of Azumaya R-algebras, where A ∼ A′ if there ′ ′ exists n, n such that Mn(A) ≃ Mn′ (A ). Multiplication is given by tensor product. 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 assume that 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 field extension 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 an isomorphism. Finally, by replacing S with the integral closure of S in K(S)′, where K(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 since S/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 a natural map S∗ −→ κ∗.

Theorem 8.5. Assume κ is perfect ??

2 ∗ 2 ∗ 1. The canonical map H (G, S ) −→ H (G, κS) 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 to prove 2. Let mS = qS for some uniformising parameter q. Consider the ∗ i ∗ subgroups Si := 1+ q S

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

~~26 Thus the following is exact~~

~~∗ ∗ ∗ 1 −→ S1 −→ S −→ κS −→ 1.~~

~~∗ since by Lemma 8.6 s ≡ 1 mod qS if and only if s ≡ 1 mod S1 . The theorem follows from the long exact sequence in cohomology and the following~~

~~i ∗ Theorem 8.7. H (G, S1 ) = 0 for i ≥ 1.~~

~~∗ ∗ Proof. We start with the graded pieces of the ﬁltration S1 >S2 > ···~~

~~i ∗ ∗ Lemma 8.8. H (G, Sj /Sj+1) = 0 for i, j ≥ 1. Proof. We ﬁrst claim that the following is an isomorphism of G-modules:~~

~~∗ ∗ κS −→ Sj /Sj+1 α 7−→ 1+ αqj.~~

To see this, note that by Lemma 8.6 1+(α + β)qj ≡ (1+ αqj)(1+ βqj) mod ∗ 2j j+1 Sj+1 since αβq ∈ q S. 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 morphism of G-modules. It is clearly both surjective and injective and hence is an isomorphism. i So suﬃces to show that H (G, κS) = 0. Let HomZ (ZG, κ) be the abelian group of Z-linear G-module homomorphisms ZG → κ. It is a G-module, with G-action given by τ.f(σ) := f(στ) for all σ, τ ∈ G. Now κS/κ is Galois with Galois 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−→ f(σ)σ−1(β). σ∈G X To see that this is indeed a G-module homomorphism, note that~~

~~φ(τ.f)= (τ.f)(σ)σ−1(β)= f(στ)σ−1(β) σ∈G σ∈G X X letting σ′ = στ we get~~

~~f(σ′)(τσ′−1)(β)= τ f(σ)σ−1(β) = τ(φ(f)) ′ σ∈G ! σX∈G X~~

~~27 Let P• −→ Z be a projective resolution of ZG-modules. Then~~

~~i i H (G, κS) = ExtZG(Z,κS) i = H (HomZG(P•, HomZ(ZG, κ))) i = H (HomZ(P•,κ)) i = ExtZ(Z,κ) = 0 for i> 0. We resume the proof of Theorem 8.7. Note ﬁrst that~~

~~S∗ = lim S∗/S∗. 1 ←− 1 j j~~

i ∗ Given an i-cocycle f : G −→ S1 , we need to find an (i − 1)-cochain h : i−1 ∗ G −→ S1 with dh = f. It suffices to find a family of (i − 1)-cochains i−1 ∗ ∗ {hj : G −→ S1 /Sj } such that

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

~~i−1 hj+1 ∗ ∗ ∗ ∗ G −→ S1 /Sj+1 −→ S1 /Sj~~

~~2. dhj is the composite~~

~~i f ∗ ∗ ∗ G −→ S1 −→ S1 /Sj~~

~~We prove this by induction on j so assume h1,...,hj are already deﬁned. ′ i−1 ∗ ∗ First lift hj arbitrarily to hj+1 : G −→ S1 /Sj+1. Note that by 2 for hj we have~~

~~−1 ′ i ∗ ∗ f dhj+1 : G −→ Sj /Sj+1 an (i − 1)-cocycle. By Lemma 8.8,~~

~~−1 ′ ′′ f dhj+1 = dhj+1~~

~~′ ′′ −1 therefore hj+1 := hj+1(hj+1) works.~~

28 8.3 Witt exact sequence Let (R, m) be a complete dvr, κ = R/m be the residue field which we assume to be perfect, and K = K(R). Here we examine the Brauer group of K. Recall that Br K = H2(Gal(F/K),F ∗) F/K Galois[ field extension By Theorem 8.1, we need only look at the union over finite unramified splitting fields. Let F/K be a finite unramified field extension which is Galois with Galois group G. Let S/R be the integral closure of R in F and recall S/R is Galois (Fact 1.3(3)). By Chapter 1 Proposition 8 of [Ser79] S is Dedekind domain. Furthermore, S is local with maximal ideal mS for if not, since S/R is unramified S/mS would not be a field (but a non trivial product of field extensions). But S is mS-adically complete by Lemma 21.3 of [Lam91] and so one could would be able to lift the idempotents of S/mS to 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 ν 1 −→ S∗ −→ K(S)∗ −→ Z −→ 0 where ν is the valuation and G acts trivially on Z. To see this, let p be a uniformising 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 indeed a sequence of G-modules. It is clearly exact. Finally, the map ν is split by sending 1 ∈ Z to p. Hence we also have a split exact sequence 0 −→ H2(G, S∗) −→ H2(G, K(S)∗) −→ H2(G, Z) −→ 0

2 ∗ 2 ∗ By Theorem 8.5, we have H (G, S ) ≃ H (G, κS). 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 Hi(G, Q) = 0 for all i since Hi(G, Q) is torsion (Theorem 4.14 of [FD93]) and multiplication by n ∈ Z+ is an isomorphism. This gives H2(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 zero cochain 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 exact sequence 0 −→ Br κ −→ Br K → lim Hom(G, Q/Z) −→ 0 −→ G Remark 8.11. We have

~~lim Hom(G, Q/Z) = Homcont(Γ, Q/Z) −→ G where Γ = Gal(¯κ/κ) is the absolute Galois group of κ.~~

8.4 Classiﬁcation of division rings over complete discrete 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 Galois unramiﬁed extension S/R with Galois group G. We wish to describe D. We have seen that [D] ∈ Br K corresponds to some group homomorphism f : G → Q/Z. Note that the image of f is a cyclic group G′ ≃ G/H say of order 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 Hi(G/H, M H ) −→ Hi(G, M) is just the edge map in the spectral sequence~~

p,q q q p+q E2 = H (G/H, H (H, M)) =⇒ H (G, M). (can get this spectral sequence either from Grothendieck spectral sequence or change of rings for Ext) The commutative diagram shows that [D] comes from a 2-cocycle over G′, so we can assume G is cyclic of order n and f is injective. Let’s compute D by tracing through the isomorphisms

~~δ Hom(G, Q/Z) ≃ H2(G, Z) ≃ H2(G, K(S)∗).~~

30 In particular, we need δf ∈ H2(G, Z). Note that f is completely determined by G and the generator σ = f −1(1/n + Z). Pick a 1-cochain f ′ : G → Q sending σi 7−→ i/n for 0 6 i < n, and (δf)(σi, σj) = σif(σj) − f(σi+j)+ f(σi) j − i+j + i = 0 if i + j < n = n n n j − i+j−n + i = 1 if i + j > n n n n Now consider the isomorphism H2(G, Z) ≃ H2(G, K(S)∗) given by the split- ν ting map K(S)∗ → 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 the 2-cocycle corresponding to f is 1 if i + j < n g(σi, σj) = p otherwise We get a crossed prodcut algebra which is actually cyclic.

~~D = K(S)u0 ⊕ K(S)u1 ⊕···⊕ K(S)un−1~~

~~i with uiα = σ (a)ui and u if i + j < n u u = i+j i j pu otherwise i+j Note that K(S)[u; σ] D ≃ (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 some cyclic 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 division ring~~

~~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. We have another interpretation of ρ(β) as follows: pick the unique maximal order Aβ in Dβ. Then Gal(S/R) is Gal(κS/κ) which is given by κS ≃ Aβ/rad Aβ. Suppose that

S[u; σ] A = β (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 choice of 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~~

~~n n (rad Aβ) = u Aβ = mAβ.~~

Remark 8.14. If we drop the Br κ = 0 hypothesis, we still get a splitting of the map H2(G, K(S)∗) → Hom(G, Q/Z). Given a cyclic quotient G of Gal(¯κ/κ) and generator σ representing element of Hom(G, Q/Z) we map it to 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 discrete valuation rings in division rings In this section, as the title suggests, we will classify all the normal orders over complete dvrs (with perfect residue ﬁelds) inside division rings. As we shall see there is only one such order which consequently implies it must be maximal.

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 it which 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, B are local and so by Proposition 6.6 are maximal. Suppose A =6 B. Let rad 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-module containing A and bi for all i > 0 and hence K(R) which is a contradiction.

~~32 Continuing with the notation from the above proposition, if we further assume 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 étale extension say of degree n and σ is a generator of G := Gal(S/R). We wish to find all maximal orders in D. Theorem 8.16. The unique maximal (or normal) order in the cyclic division algebra A(K(S)/K,K(S)σ,p) is A := A(S/R,Sσ,p) where p is the uniformising parameter for R. Furthermore A/rad A ≃ κS the residue field of S. Proof. We know that A is an R-order in D. It suffices to show that it is n normal with residue ring κS. Recall A = S[u; σ]/(u − p) so u is a normal element which is nilpotent modulo m. Hence uA ⊆ radA. Now uA = rad A since A/uA ≃ S/p ≃ κS which is semisimple.

~~9 Classiﬁcation of normal orders and maximal orders in central simple algebras.~~

As in the previous section, we fix a complete dvr (R, m) with perfect residue field κ = R/m and K := K(R). A consequence of the Artin-Wedderburn theorem is the fact that every central 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 and hence maximal order in it, it is natural to extend the classification question to 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 (unique by 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 D n . D =  .  . D     33 Can also consider the (A, B)-subbimodule B . n M = A  .  ⊂ D . B     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) ≃ m κB , by the Krull-Schmidt theorem and the fact that κB := B/rad B is m a simple, which we know from Lemma 7.2. Thus both MB and B are m projective covers of κB and so by Proposition 21.3 we must have t: MB≃B .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) with isomorphism given by conjugation by t. Having classified all the maximal orders, we turn our attention to the slightly more general case of normal orders. Before stating and proving the classification theorem, we need some preliminary results. Proposition 9.2. Let B be an R-algebra which is finitely generated as an R-module. Then any simple B-module S has an indecomposable projective cover ϕ : P → S. Conversely, any indecomposable projective B-module P is the 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¯ have the same simple modules. Furthermore, since B is finitely generated over R, B/mB is finitely generated over R/m and is thus artinian. Since B¯ is a quotient of B/mB it is also artinian and since it had trivial Jacobson radical it must be semisimple by Theorem 4.14 in [Lam91]. Hence there is an idempotentε ¯ ∈ B¯ such that S ≃ B¯ε¯. We may lift the idempotentε ¯ to ε ∈ B and let P = Bε. Recall that in Example 21.2 we saw that P is a projective cover of P/(rad B)P which is isomorphic to Bε/(rad B)ε ≃ B¯ε¯ ≃ S. Suppose P = P1 ⊕ P2. Then restricting ϕ : P → S to P1 and P2 we see that at least one of them, say P1, surjects onto S (since S is simple). But ϕ is 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 uniqueness of projective covers P ≃ ⊕Pi. Hence if P indecomposable then P is the projective 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. Then n the Mn(D)-module K ⊗R P is isomorphic to D .

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

n s Proof. Since Mn(D) is simple artinian, K ⊗R P ≃ (D ) for some s. Consider ∼ n s ρ n the composite map ψ : P → K ⊗R P → (D ) → D 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 indecomposable so we must have s = 1. The following lemma describes the indecomposable projective modules of an 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−1P and 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. Hence multiplication by π permutes the indecomposable projectives. From Propo- sition 9.3, we know that there are r incomposable projectives so it suﬃces to show 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-free and 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 follows from part 2 so we prove the latter. ′ n ′ Let P < P . Now K ⊗R P ≃ D , a simple K ⊗R A-module, so P/P must be torsion. Hence we can ﬁnd s maximal such that P ′ 6 πsP . Consider the composite map ψ : P ′ → πsP → πsP/πs+1P . Note that πsP → πsP/πs+1P is the projective cover of the simple module πs(P/πP ). Now maximality of s ensures that im ψ =6 0 so ψ is surjective. Then by deﬁnition of projective covers, P ′ = πsP .

~~35 Theorem 9.5. Let A be a normal R-order in Mn(D) as above with exactly r simples. Let B be the unique maximal order in D. Then~~

~~A ≃ Mn/r(R) ⊗R A0 where B ······ B .. .. .  rad B . . .  A0 = . . . . ⊂ Mr(B)......    rad B ··· rad B B      Furthermore~~

~~rad A = Mn/r(R) ⊗R rad A0. Proof. Let P,πP,...,πr−1P be the indecomposable projective and write~~

r−1 A ≃ (πiP )⊕ni . i=0 M n n r−1 i ⊕ni n ⊕ni Note that (D ) ≃ Mn(D) ≃ K ⊗R A ≃ ⊕i=0 (K ⊗R π P ) ≃ ⊕(D ) , so ni = n. We wish to show that all the ni’s are equal so in fact

P r−1 ⊕n/r A ≃ πiP . i=0 ! M i i+1 i Let Si = π P/π P be the simple corresponding to π P . Then A/rad A ≃ r−1 i=0 Mni (EndASi). Now Aπ ⊗A − permute the simples transitively, so conjugation by π shows that all the factors Mn (EndASi) are in fact isomorphic. Q i Hence all ni = n/r. There is a Peirce deomposition as follows

A ≃ HomA (A, A) r−1 i ⊕n/r r−1 i ⊕n/r ≃ HomA ⊕i=0 π P , ⊕i=0 π P r−1 i j ≃ Mn/r(R) ⊗R HomA(π P,π P ) i,j=0 M r−1 i−j ≃ Mn/r(R) ⊗R HomA(π P, P ). i,j=0 M The theorem follows from the following lemma.

~~36 Lemma 9.6. With the above notation,~~

B if 0 6 i < r Hom (πiP, P ) ≃ . A rad B if −r 6 i < 0 ′ ′ ′ Proof. Let B = EndAP . Note that B is an R-order in D since K ⊗R B ≃ n n HomK⊗RA(K ⊗R P,K ⊗R P ) = HomMn(D)(D ,D )= D. We show that it is maximal using the criterion from Proposition 6.6 by showing it it is normal and local. Consider the exact sequence

~~0 −→ πP −→ P −→ P/πP −→ 0 and we get an exact sequence~~

′ ϕ 0 −→ HomA(P,πP ) −→ B −→ HomA(P, P/πP ) −→ 0 with HomA(P, P/πP ) ≃ HomA (P/πP, P/πP ) ≃ EndAS where S := P/πP is ′ simple by Proposition 9.2. Now ϕ is an algebra map, so J := HomA(P,πP ) is ′s s an ideal. Since for any s> 0, J ⊆ HomA(P,π P ) and hence for suﬃciently ′s ′ large s, J ⊆ HomA (P, mP ) = HomA (P, P ) m and so we see that J is ′ ′ ′ ′ nilpotent modulo m and so J ⊆ rad B . But B /J ≃ EndAS which is a division ring by Schur’s Lemma, and is thus semisimple. Therefore B′ is local with J ′ = rad B′. ′ We now show that B is normal. Let f ∈ HomA(P,πP ) be a non-zero map. Then arguing as previously, we see that f gives an isomorphism of P with im f 6 πP . From Lemma 9.4, we know that im f = πrmP for some ′ r r ′ r m ∈ Z+. Hence J = HomA(P,π P ). But P ≃ π P so J = HomA(P,π P ) ≃ ′ ′ HomA(P, P )= B and so J is local. ′ Thus we have shown that B = EndAP = B, the unique maximal order in D. Furthermore, rad B = HomA(P,πP ). The above argument also shows that

~~Hom (P, P ) if 0 6 i < r Hom (πiP, P ) = Hom (P,π−iP ) = A A A Hom (P,πrP ) if −r 6 i < 0 A .~~

~~10 Ramiﬁcation Theory~~

~~In algebraic geometry, ramification is where a map fails to be étale. Recall that 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 residue field κ = R/m and fraction field K. Let Rˆ and Kˆ be their completions and Γ = Gal(¯κ/κ) be the absolute Galois group.

~~10.1 Ramiﬁcation theory via cohomology Consider the commutative diagram below~~

Br R −→ Br K ↓ ↓ ρ 0 −→ Br Rˆ −→ Br Kˆ −→ Homcont(Γ, Q/Z) −→ 0 where the bottom row is essentially the Witt exact sequence of Section 8.3 since by Theorem 8.5 Br Rˆ ≃ Br κ and ρ is the residue map.

Proposition 10.1. Let A be a maximal R-order. Then A is Azumaya if and only if Aˆ = Rˆ ⊗R A is Azumaya if and only if ρ([Kˆ ⊗R A]) = 0. ˆ ˆ Proof. The ﬁrst part is clear since A ⊗Rˆ R/m ≃ A ⊗R R/m. ′ If ρ[Kˆ ⊗R A] = 0 then there exists an Azumaya Rˆ-algebra A such that ˆ ′ ˆ ˆ ˆ ˆ [K ⊗Rˆ A ] = [K ⊗Rˆ A]. By Corollary 4.10 A is a maximal R-order and so by uniqueness of maximal orders (Proposition 9.1) A′ ≃ Aˆ and so Aˆ is Azumaya.

~~Upshot: so ρ([Kˆ ⊗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 algebraically as in Section 8.4.1 but now without the assumption Br Rˆ = 0. We will see precisely how ρ([Kˆ ⊗R A]) measures the failure of A/mA from being a central simple κ-algebra.

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

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

38 We now prove the converse. Suppose A is maximal. Let Aˆ = Rˆ ⊗R A, Jˆ = rad Aˆ = Rˆ ⊗R rad A. From Proposition 9.1, we know that Aˆ is normal and local. Since A/rad A = A/ˆ rad Aˆ we see that A is also local. Furthermore J/ˆ Jˆ2 ≃ J/J 2 which is generated by one element since Jˆ is principal. Nakayama’s lemma implies J is also generated by a single element so 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 an automorphism of A/πA). Then ρ([Kˆ ⊗R A]) ∈ Homcont(Γ, Q/Z) is the element corresponding to the cyclic extension κ′/κ with generator for Galois 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, it depends only on the Brauer class [Kˆ ⊗R A] ∈ Br Kˆ . 2. The automorphism σ, which is mentioned in the theorem, is independent of A and the choice of uniformising parameter π.

Proof. 1. Kˆ ⊗R A is a central simple Rˆ-algebra and so by the Artin- Wedderburn theorem there exists a unique Rˆ-division algebra D such that Kˆ ⊗R A = Mn(D). Let B be the unique maximal order in D which is possible by Proposition 8.15. Then Proposition 9.1 implies that Aˆ := Rˆ ⊗R A ≃ Mn(B). Therefore Z(A/rad A)= Z(A/ˆ rad Aˆ)= Z(Mn(B/rad B)) = Z(B/rad B). But, as we just saw, B only depends on [Kˆ ⊗R A] ∈ Br Kˆ . 2. From part 1 Γ is independent of the choice of A. To show that it is independent of the choice of uniformising parameter, suppose we change π to uπ ∈ A where u ∈ A×. Letu ¯ ∈ A/rad A be the image of u in A/rad A. Conjugation by uπ is the composition of σ and conjugation byu ¯. But conjugation byu ¯ acts trivially on Z(A/rad A) so part 2 is proved.

~~39 Given any Brauer class β ∈ Br Kˆ , we need only prove Theorem 10.3 for some maximal order A with [Kˆ ⊗R A] = β. This follows from the following proposition.~~

Proposition 10.5. Let β ∈ Br Kˆ . Let ρ(B) ∈ Homcont(Γ, Q/Z) be given by cyclic extension κ′/κ and generator σ ∈ Gal(κ′/κ). Let S/ˆ Rˆ be the cyclic ´etale extension with residue ﬁeld extension κ′/κ and lift σ to Sˆ so Rˆ = Sˆhσi. Then

~~1. There exists an Azumaya Rˆ-algebra A0 such that~~

~~[Kˆ ⊗R A0 ⊗R A1] = β~~

~~where A1 is the cyclic algebra A(S/ˆ R,ˆ Sˆσ; p) and p R = m, that is, the cyclic algebra of Theorem 8.16.~~

2. A0 ⊗R A1 is a maximal order and rad(A0 ⊗ A1)= A0 ⊗ rad A1. ρ Proof. 1. We recall the Witt exact sequence 0 → Br Rˆ → Br Kˆ → ′ Homcont(Γ, Q/Z) → 0 (where ρ([β]) = (κ /κ, σ)) is split and in fact from Section 8.4.1, we know ρ([Kˆ ⊗R A1]) = ρ(β). Hence ρ(β[Kˆ ⊗R −1 A1] ) = 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 maximal R-order. Then A0 ⊗R A1 is a maximal R-order with Jacobson radical A0 ⊗R rad A1. Proof. We use the criterion for maximality proved in Proposition 6.6. Note that from Lemma 5.2 we know that A0 ⊗R rad A1 is nilpotent modulo m, so A0 ⊗R rad A1 ⊆ rad(A0 ⊗R A1). Also

A0 ⊗R A1 A1 B := ≃ A0 ⊗R A0 ⊗R rad A1 rad A1 A0 A1 ≃ ⊗κ mA0 rad A1 ′ ′ which we claim is a central simple κ -algebra where κ = Z(A1/rad A1). To see this, recall that by deﬁntion of Azumaya, A0/mA0 is a central simple ′ ′ κ-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 central simple 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 of A0 ⊗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 be so: A/mA is not central simple over κ but in fact A/rad A is central simple over some nontrivial field extension κ′/κ. It also turns out that we fail to have rad A = mA. The ramification index measures both these failures simultaneously.

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

Proof. Note that e and the equation mA = (rad A)e remain unchanged if we 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. We ′ know from Proposition ? Section 7 that mB = (rad B)e for some e′. Hence ′ mA = (rad A)e , so we may replace A with any maximal order A′ such ′ that A ⊗R K has the same Brauer class as D. From Section 10.2, one such ′ maximal order is A = A0 ⊗ A1 where A0 is Azumaya over R and A1 is the cyclic algebra A1 = A(S/R,Sσ; p) where p = mR. In Section 8.5 we saw that e (rad A1) = mA1, hence

~~e e rad(A0 ⊗R A1) = (A0 ⊗R rad A1)~~

~~= 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.~~

′ r ′′ 1. Then Z(A/rad A) is a cyclic extension of κ of the form κ = i=1 κ where κ′′/κ is a cyclic field extension. If πA = rad A then conjugation by π induces an automorphism σ ∈ Gal(κ′/κ) which generatesQ Gal(κ′/κ). The ramification data of A is the cyclic extension κ′/κ and the Galois automorphism σ. The ramification index of A is the integer e = [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 we have a structure theory for A (c.f. Theorem 9.5). Suppose that A⊗R K(R)= Mn(D) and B is the unique maximal order in D. Then B B ··· B .. .. .  rad B . . .  A ≃ Mn/r(R) ⊗R ...... B    rad B ··· rad B B      also the Jacobson radical is generated by (1 ⊗ π)A where 0 1 0 ··· 0 ......  0 . . . .  . π = ......  . 0   . . .   0 ...... 1     π 0 ··· 0 0   B    and πBB = rad B. For part 1, we see that B/rad B 0 ······ 0 ......  0 . . . .  . . . . . A/rad A ≃ Mn/r(B) ⊗R ......  . .   . . . .   ...... 0     0 ······ 0 B/rad B    so   r Z(A/rad A) = Z(B/rad B) i=1 Yr = κ′′ i=1 Y where κ′′ is a cyclic ﬁeld extension (c.f. Theorem 10.3). Also conjugation by ′′ r π permutes the r copies of κ cyclically and π = πBIr. Part 2 follows from (rad A)e = (1 ⊗ π)eA ′′ = (1 ⊗ πr)[κ :κ]A [κ′′:κ] = (1 ⊗ πB )A = mA.

42 Corollary 10.9. (Boris: change this to Scholium) Let A be a normal R-order ′ r ′′ ′′ with ramification data κ = i=1 κ where κ /κ is a cyclic field extension and σ ∈ Gal(κ′/κ). Then the ramification data of any maximal order containing A is the cyclic field extensionQ κ′′/κ and automorphism σr ∈ Gal(κ′′/κ). We can now give the étale 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 étale extension of discrete valuation rings. Then A⊗R S is a normal S-order with ramification index e too. 2. There exists an étale extension of discrete valuation rings S/R such that S S ··· S .. .. .  n . . .  A ⊗R S ≃ Mn/e(S) ⊗S . . . ⊂ Mn/e(S) ⊗S Me(S)(2) . .. .. S    n ··· n S    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 A ⊗ S A Z R ≃ Z ⊗ S rad(A ⊗ S) rad A R R A = Z ⊗ κ rad A κ S A = Z ⊗ κ rad A κ S which is a degree e extension of κS where e is the ramiﬁcation index of A.

r ′′ ′′ 2. Let A/rad A ≃ 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 some monic f¯ ∈ κ[x].Q We lift f¯ to a monic f ∈ R[x], then S := R[x]/(f) is a discrete valuation ring such that S/R is étale with residue field ′′ ′ ′ ′ extension κ /κ. Then since κ ⊗κ κ ≃ g∈G κ , we have A ⊗ S Qr R = κ′ ⊗ κ′ rad(A ⊗ S) κ R i=1 ! r dY = κ′. i=1 Y 43 Hence the maximal S-order B containing A⊗R S is unramified (because the ramification data of B is just κ′/κ′) and hence Azumaya. Hence we can replace S with an étale extension so that B = Mn(S) for some S.

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

10.5 Canonical Bimodule In algebraic geometry, ramiﬁcation theory can be studied via the canonical bundle. Then same applies for orders. Let R be a commutative noetherian semi-local ring and ωR be its canonical module. The only cases we will need are where R is regular local, in which case we can take ωR = R and more generally if R is an R0-algebra where R0 is regular local and R0 is a ﬁnite free

~~R-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 domain and A be an R-order, which is free as an R-module. We deﬁne its canonical bimodule 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. We identify this in a special case.~~

~~Proposition 10.12. Let (R, m) be a discrete valuation ring and A be a normal 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~~

~~−(e−1) ωA = π 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. So we may replace R by an extension S. By Proposition 10.10, we know that there is an étale extension of discrete valuation rings S/R such that (2) holds. Recall that rad(A⊗R S)= πA⊗R S.

~~44 Hence~~

−(e−1) −(e−1) π A ⊗R S = rad(A ⊗R S) −1 = q rad(A ⊗R S) q−1S q−1S ··· q−1S .. .. .  S . . .  = Mn/e(S) ⊗S . . . . (3) . .. .. q−1S    S ··· S q−1S      ′ ′ 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,...,εn corresponding to “diagonal” entries. This gives a Peirce decomposition

~~ωA = εiωAεj i,j=1 M~~

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

e lij ωA = Mn/e(S) ⊗S q S ⊆ Mn/e(S) ⊗S Mn(K(S)) i,j=1 M where dij 6 0 and if i < j we have dij 6 −1. Furthermore, tr(ωA) = S so lii = 0. A quick computation using the fact that ωA is an A-bimodule shows ′ that 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 ﬁxed central simple algebra are conjugate.~~

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

~~1. Any two normal orders in D with the same raniﬁcation index are conjugate (in D)~~

~~2. Any two maximal orders in D are conjugate.~~

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

45 satisfy Aˆ ≃ Bˆ ⊂ Dˆ. By Skolem-Noether there is a unit t ∈ Dˆ × such that Bˆ = t Atˆ −1. By Lemma ?, t factorises as t = uv with u ∈ D× and v ∈ Aˆ×. Also by Section 4 proposition ??,

~~B = Bˆ ∩ D = t Atˆ −1 ∩ D = uv Avˆ −1u−1 ∩ D = u Auˆ −1 ∩ u D u−1 = u(Aˆ ∩ D)u−1 = u A u−1.~~

~~Lemma 11.2. Given t ∈ Dˆ × we can factorise t as t = uv with u ∈ D× and v ∈ Aˆ×.~~

Proof. Note that Aˆ ⊂ Dˆ is a finitely generated Rˆ-module so t−1Aˆ ⊆ m−n+1Aˆ for n ≫ 0. Since D is dense in Dˆ, we can find u ∈ D such that u − t ∈ mnAˆ. Note that t−1u − 1 ∈ t−1mnAˆ ⊆ mAˆ. Hence t−1u is an invertible element of Aˆ say with inverse v ∈ Aˆ×. Since t−1uv = 1 we have uv = t. Note that u is invertible in Dˆ, 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 replaced with 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⊗Z K a central simple K algebra, where K 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 reduced map 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 a normal OZ,p-order. 2. A is reﬂexive as a sheaf on Z.~~

~~The motivation comes from Serre’s criterion for normality, which states that 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 algebraically closed field k of characteristic zero, so all residue fields are perfect and the results from Sections 8 to 11 are valid. We will assume that all schemes are noetherian and separated, so we will omit these adjectives in the future. Also, by a surface we mean a 2-dimensional excellent scheme Z such that for any closed point p has residue field k so if Z is also smooth at p then OZ,p ≃ k[[x, y]]. (resolutions of singularities exist for excellent schemes of dimension 2 - see Lipman, Desingularization of two-dimensional schemes, Ann.b Math. 107 (1978) 151-207. Also see Chapter XI of Cornell, Silverman; Arithmetic Geometry. See chapter 13 of Matsumura’s Commutative Algebra for information on excellent rings).

~~12 Unramiﬁed Orders~~

We will define ramification in the scheme setting and relate unramified orders to Azumaya algebras. Let Z be a normal integral scheme and Z1 be the set of codimension 1 points. Let A be a normal order on Z. Example 12.1. Any maximal order on Z is normal. Why? We know that 1 A is reflexive. Let p ∈ Z . Then A ⊗Z OZ,p is a maximal OZ,p-order, hence normal. We define the ramification data of a normal order A on Z to be the 1 ramification data of all the normal orders A⊗Z OZ,p as p ranges over Z .Note that A is generically Azumaya so these data are equivalent to given a finite 1 set of points p1,...,pn ∈ Z 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 ramification index ei = [κi : κ(pi)]. We say that A is unramified if n = 0, that is, 1 A ⊗Z OZ,p is Azumaya for all p ∈ Z . The reason we consider ramification in 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 : op 2 A ⊗Z A → EndZ (A) fails to be an isomorphism. Suppose rank A = n . n×n We compute this locus locally so assume Z is aﬃne and A ≃ OZ . Then n4 n4 ϕ : OZ →OZ 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 at a 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 a sheaf. If A is unramiﬁed, then it is Azumaya.~~

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

a 0 −→ Br Z −→ Br k(Z) −→ Homcont(Γp, Q/Z) 1 pM∈Z where Γp = Gal(k(p)/k(p)) and a comes from residue maps of Section 8 with respect 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 étale cohomology, and we can’t use the cheap proof here. Proof. Note first that the sequence above is a complex, e.g. from the Witt exact sequence. We show exactness at Br k(Z). Let D be a k(Z)-central division ring with a([D]) = 0. Let A ⊂ D be a maximal order. Since A is reflexive, so by Auslander-Buchsbaum it is locally free. By Corollary 12.4, A is 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 Z such that A ⊗Z k(Z) ≃ Mn(k(Z)). By uniqueness theorem from Section 11 for the discrete valuation ring case, we know that for each p ∈ Z1, we have A ⊗Z OZ,p ≃ Mn(OZ,p). Hence there is an open cover U1,...,Un of Z − Y where Y is a closed subset of codimension > 2 such that

~~A|Ui = EndUi Vi~~

48 where Vi is some rank n vector bundle on Ui. The lemma below shows that we can assume n = 1. But Z is smooth of dimension 6 2, so V1 extends uniquely to a rank n vector bundle on Z. Thus A ≃ EndZ V since both sides are locally free so are determined by codimension one.

Lemma 12.7. 1. Let V1 and V2 be vector bundles on a scheme Z such that EndZ V1 ≃ EndZ V2. Then there exists a line bundle L such that V2 ≃ 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 on U12 = U1 ∩ U2 we have

~~EndZ V1|U12 ≃ EndZ V2|U12 .~~

~~Then there is a bundle V on Z such that~~

~~EndZ V |Ui ≃ EndUi Vi~~

~~for i = 1, 2.~~

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

~~1 −→ Gm −→ GLn −→ PGLn −→ 0~~

~~which induces an exact sequence~~

~~1 1 ϕ 1 Pic Z −→ H (Z, Gm) −→ H (Z,GLn) −→ H (Z,P GLn)~~

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

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

L ⊗U12 V1|U12 . Now L is a coherent subsheaf of the constant sheaf k(Z) so by the extension lemma [Hartshorne, Ch II, Section 5, exercises] we can extend L to a coherent subsheaf L1 of k(Z) on U1. Taking reﬂexive hulls, we can assume that L1 is a line bundle. So we deﬁne a vector bundle V on Z by setting

~~V |U2 = V2 and V |U1 = L1 ⊗U1 V1~~

~~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 we have a reduced determinant map det : D → K. Picking a K-basis for D allows us to view det as a degree n polynomial map from Kn×n → K. Let f ∈ K[xij | i, j = 1,...,n] be this polynomial. Note that det 0 = 0, but since det is multiplicative, we have det a =6 0 if a =6 0. We consider an effective divisor ∆ > 0 (to be determined later) and 0 n×n 0 restriction of f to H (OZ (∆)) . If the coefficients of f lie in H (OZ (∆f )) then det gives a map of affine k-spaces

~~0 n×n 0 ϕ : H (OZ (∆)) −→ H (OZ (n∆+∆f )) k k 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 contradicts Chevalley’s theorem.~~

Corollary 13.2. Let A be a maximal order on a smooth quasi-projective curve Z. Then A ≃ EndZ V for some vector bundle V . Proof. By Tsen’s theorem, Br(k(Z)) = 0 so A is an unramiﬁed order, hence it is Azumaya. Also Br Z injects into Br(k(Z)) so Br(Z)=0and A is trivial Azumaya. Recall Morita theorey for sheaves of algebras (ref. [Artin-Zhang, noncommutative projective schemes])

Deﬁnition 13.3. Let Z be a scheme and A be a coherent sheaf of algebras on Z. Let P be a progenerator for A by which we mean a right A-module such that 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 = EndAP in the sense that~~

~~BPA ⊗A − : A−Mod −→ B−Mod is a category equivalence.~~

~~Corollary 13.4. Maximal orders on smooth curves are Morita equivalent to their centres.~~

~~There is a similar result for normal orders~~

~~Theorem 13.5. Let Z be a smooth quasi-projective curve and A, B be two normal orders on Z with the same ramiﬁcation data. Then A and B are Morita equivalent.~~

Remark 13.6. Since the residue fields of codimension one points p are all k = k¯, there are no non-trivial cyclic field extensions of k(p). Hence the ramification dta of the normal order A in this dimension one case consists of a finite set of ramification points p1,...,pl and their ramification indices e1,...,el. Proof. We show that the normal order A is Morita equivalent to some order A0 which depends only on ramification data. We achieve this by picking a progenerator as follows. We embed A in a maximal order which by Corol- lary 13.2 has the form EndZ V for some vector bundle V . Now A is locally hereditary so V is locally projective as an A-module. Away from ramification points, we have A = EndZ V so V is a progenerator there. Pick a ramification point p with ramification index e. From the structure theory of normal orders we know that Aˆp = A ⊗Z OZ,p ≃ Mr(C) where b OZ,p ······ OZ,p .. .. .  mZ,p . . .  C ≃ ⊆ M (O ). b. .. .. b. ep Z,p  . . . .     mZ,p ··· mZ,p OZ,p  b     Let b

OZ,p Jp = ker A −→ A ⊗Z OZ,p −→ rad(A ⊗Z OZ,p)! b b b 51 From Section 9, we know that multiplication by Jp permutes the indecomposable projective Aˆp-modules cyclically. We may thus consider the progenerator

~~ep−1 j P = V ⊕ Jp V. p ram pt j=0 M M~~

~~We define A0 = EndAP and note that this depends only on the ramification points and their ramification indices (but not V ) since the corresponding Peirce decomposition has components~~

O ′ j j HomA(Jp V,Jp′ V ) = O(−p)  ′  O(−p ) as in Section 9, where the right hand side does not depend on V . This completes the proof. Remark 13.7. Normality is not a Morita invariant in the sense that a non-normal order can be Morita equivalent to a normal order. So this classiﬁes normal orders on (smooth, quasi-projective) curves up to Morita equivalence. If one wants to know more about normal orders one can try to pick a normal order with given ramiﬁcation data and try to classify progenerators for 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 2 points. Then H0(A) has a complete set of n idempotents such that the Peirce decomposition with respect to the idempotents has the form

~~A ≃ O(pij) i,j=1 M for divisors pij ∈ Div Z.~~

Proof. If A is unramiﬁed then A ≃ EndZ V 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 normal orders ramiﬁedL at 6 2 points. L

~~52 13.1 Behavoiur under completions~~

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

Sˆ = Sî i=1 Y ˆ ˆ 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 Sî and in fact Sî/Rˆ is Galois with Galois group D. We can see this for example from the fact that 1 [κ(Sˆ ) : κ(Rˆ)] = [κ(Rˆ) ⊗ S : κ(Rˆ)] i g R |G| = g = |D|

~~14 Cyclic covers of curves~~

Given a normal order on a surface, ramification occurs at curves C, and the ramification data involves cyclic field extensions of k(C). These latter correspond to some ramified cyclic covers of (the normalisation) of C. We classify these cohomologically. Notation: let C be a smooth projective curve and Γ be the absolute Galois group Gal(k(C)/k(C)). Let γ ∈ Homcont(Γ, Q/Z) corresponding to say 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 morphism π : C˜ → C. We wish to interpret the ramification of π cohomologically. Fix g p ∈ C. Let R = OC,p ≃ k[[z]] and S = OC˜ ⊗OC,p R ≃ i=1 Si where Si are complete local rings. Let G = Gal(F/kb (C)) and D ⊳ G be the decompositionQ group, which is unique in this case since G is cyclic. Note that D is the Galois group of Si/R. Let Γp be the absolute Galois group of k(R)= k((z)). Then γ determines an element r(γ) ∈ Homcont(Γp, Q/Z) by the composite map

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

~~Proposition 14.1. The absolute Galois group of k((z)) is limµn where the ←− inverse system is deﬁned by m-th power maps µmn → µn for m, n ∈ Z+.~~

53 1/n Proof. The algebraic closure of k((z)) is k((z)) = n∈Z+ k((z )). There is also a group isomorphism S 1/n µn −→ Gal(k((z ))/k((z))) ξ 7−→ (z1/n 7−→ ξz1/n). Hence we may consider r(γ) as an element of

~~−1 µ = Hom(µn, Q/Z) [ Recall that the order of r(γ) is |D|, which in turn is the ramiﬁcation index of π : C˜ → C above p. Theorem 14.2. There is an exact sequence~~

r −1 P −1 Homcont(Gal(k(C)/k(C)), Q/Z) −→ µ −→ µ −→ 0 p∈CMclosed where r is defined coordinate-wise as above. Proof. Omitted. See [Artin, Grothendiec topologies Chapter IV 2.6[ Done using étale cohomology. Remark 14.3. ker r classifies étale cyclic covers with generator for Galois 1 group. This can be written using étale cohomology group Het(C, Q/Z) or the étale 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 projective curve ramified at exactly 1 point or at 2 points with different ramification indices.

2. Let p1,p2,p3 ∈ C. There exists a cyclic cover of C ramified exactly at p1,p2 and p3 with ramification indices 2, 4, 4. To see thi spick an iso- −1 −1 morphism µ ∈ Q/Z and consider element of p∈C µ whose components are zero everywhere except those indexed by p1,p2 and p3. The components there are L 1 1 1 + Z, + Z, + Z 2 4 4 so by Theorem 14.2, there exists a cyclic field extension of k(C) with this 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 set of 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~~

~~ρ r −1 P −1 0 −→ Br Z −→ Br k(Z) −→ Homcont(ΓC , Q/Z) −→ µ −→ µ −→ 0 1 Z2 CM∈Z pM∈~~

which is exact everywhere except possibly at the C∈Z1 terms. The ramification map ρ is that defined for orders as in Sections 10 and 12, whilst the ramification map r is that for curves asL defined in Section 14.

3 2. The cohomology is the ´etale cohomology Het(Z,µ) where µ is the group 3 of roots of unity. In particular, Het(Z,µ) is zero if Z is simply connected. Proof. Omitted. It involves the coniveau spectral sequence in ´etale cohomology.

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 observe the 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 extension F/k(C). Suppose further that F/j(C) is ramified at p ∈ C. Then one of the following 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 2 points 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 of secondary ramification of the maximal order A or its corresponding Brauer pair (Z, [A ⊗Z k(Z)]). We sometimes summarise the equation rρ = 0 by saying that secondary ramification must cancel. Example 15.4. 1. There are no maximal orders on Z = P2 ramified on a (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 étale covers. Any cover of P1 ramifies at > 2 points by part 1 of Example 1. Hence there are 2 points of secondary ramification. But these cannot be cancelled since there are no other ramification curves. If the ramification locus has degree 2, then it is either a smooth conic so ≃ P1 or 2 lines. A simliar argument shows that this cannot happen.

2 2. There is a maximal order on Px,y,z on x y z = 0. Example 15.5. An exotic del Pezzo order. Let Z = P2 and C ⊂ P2 be a smooth quartic. We consider an ´etale double cover C/C˜ . Note that the genus of C is 1 g(C) = 1+ C · (K + C) 2 1 = 1+ 4(−3+4) 2 = 3

These are classified by 2-torsion points of the Jacobian J(C). Since J(C) is 6 an abelian variety of dimension g(C) = 3 we know that J(C)2 ≃ (Z/2Z) . Since P2 is simply connected, the Artin-Mumford sequence implies that there exist maximal orders on Z = P2 ramified only on C with ramification data there given by the cyclic field extension k(C˜)/k(C).

~~Theorem 15.6. (deMeyer-Ford) Let Z be a smooth projective surface with Kodaira 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 by de 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 as follows. Let Z = P2 and Y be the double cover of Z ramiﬁed on the smooth quartic C. We can describe this as

Y = SpecZ (O⊕O(−2)) where O⊕O(−2) is the commutative cyclic algebra with multiplication given by any isomorphism ∼ O(−2) ⊗Z O(−2) −→ O(−C) ATTENTION:Boris: picture of quartic missing It 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¿ Let π : Y → Z be the covering map. Note that −1 π (Hi) = Ei ∪ σ(Ei) for some exceptional curve Ei ⊂ Y . Note that E1 and E2 are not linearly equivalent 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 -module structure, but right OY -module structure is twisted through by σ. We need a relation of the form ⊗2 ϕ : Lσ −→ OY 2 ⊗2 Note that σ = 1 so Lσ is an OY -central bimodule. Also ⊗2 ∗ Lσ ≃ L ⊗Y σ L

~~≃ OY (E1 − E2 + σ(E1) − σ(E2)) ∗ ≃ OY (π (H1 − H2))~~

~~≃ OY We pick ϕ to be any isomorphism Claim: 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 up to multiplication constant in k∗ so α ∈ k(Z). Hence the overlap condition holds.~~

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

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

[Fix proof max] We will compute the ramification data. Let CY = −1 ∗ π (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 fixed 2 by σ. Now I = A ⊗Y O(−2CY )= A ⊗Y O(−C). We wish to show that the ramification data is given by the fraction field of A/I.

~~A/I = (OY ⊕OY (E1 − E2)σ)|CY ′ ′ = OC ⊕OC (p1 + p1 − p2 − p2)~~

~~′ ′ a commutative algebra since σ|CY = 1CY . Note that p1 + p1 − p2 − p2 is ⊗2 2-torsion since the isomorphism Lσ ≃OY restricts to isomorphism~~

′ ′ ⊗2 OC (p1 + p1 − p2 − p2) ≃ 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 maximality]. First note that E ≃ P1 is simply connected, so we hvae an exact sequence

P 1 1 −1 0 −→ Homcont(Gal(k(P )/k(P ), Q/Z) −→ µ −→ µ −→ 0 P1 p∈ Mclosed as in Section 14. Hence to give ramification data on E, it suffices to give an −1 −1 element of ker( p∈P1 µ → µ ). Suppose there are exactly two ramification curves C1 and C2 passing L −1 through p with secondary ramification there given by ζ1, ζ2 ∈ µ . 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 C˜1, C˜2 be the strict transforms of C1,C2. The Artin-Mumford sequence on Z shows that the secondary ramification cancels at p, so ζ1 + ζ2 = 0. Now we use the Artin- Mumford sequence on Z˜. If pi = C˜i ∩ E, then the secondary ramification of C˜i at pi is ζi Since the secondary ramification cancels at pi, we know that secondary ramification from E is pi is −ζi. Also there is no other secondary ramification 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 e cover of E totally ramified at p1 and p2. In particular, the ramification index at E is e. Upshot: By blowing up, you may not be able to remove nodes in the ramification locus, but you can always ensure that the ramification index of one 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 of ramification curves passing through p. Then the secondary ramification index at p is equal to the ramification index of (Z, α) at either C1 or C2. Given a maximal order A on a smooth surface Z with ramification data satisfying 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 sequence and the usual resolution of singularities, arguing as in the example above. In general, we use a version of the Artin-Mumford sequence on the blowup of Spec k[[u, v]] at the closed point. Example 15.12. From Section 2, we know that the following is a maximal order k hhx, yii k [[x, y]] = ζ (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 uv =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 extension k((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 the appearance of ζ and ζ−1 in the generators of the Galois group.

~~16 Complete local structures of terminal orders~~

Recall that if Z is a smooth quasi-projective surface and p ∈ Z is a closed point then OZ,p ≃ k[[u, v]]. We wish to give a similar description of terminal orders in the complete local case. The approach will be to use a method of Artin’s for describingb orders with given ramification data. Recall terminal orders have ramification as one of the following: Azumaya, smooth point of ramification, 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 that Rp is a domain.~~

Proposition 16.1. Let A be a normal order, then A ⊗R Rˆp is a normal Rˆp-order. Proof. Note that completion commutes with duals in the following sense. For M a ﬁnitely generated R-module, we have ˆ ˆ nat ˆ HomRˆp (M ⊗R Rp, Rp) ≃ HomR(M, Rp)

~~≃ HomR(M,R) ⊗R Rˆp~~

Hence A is reflexive implies A ⊗R Rˆp is reflexive. It suffices to check now that A ⊗R Rˆp is normal at all codimension one primes. These all come from codimension one primes of R. Let P be a codimension one prime of R. Note that R/P ⊗R Rˆp is complete so is excellent too. Hence it is also reduced. We ˆ r ˆ ˆ ˆ may thus write P Rp = i=1 Pi for primes Pi minimal over P Rp. Let Rˆ be the localisation of Rˆ at Pˆ . Then P Rˆ = Pˆ Rˆ , the maximal i T p i i i i ideal. Then Section 5, Proposition 2 (its proof), shows that A ⊗R Rp is also normal.

60 16.2 Method of Artin covers Artin has a method for describing orders with given ramification data as invariant 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 of π : 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 order B ⊂ 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, but reduces the proof to G cyclic.~~

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

~~Proof. For each irreducible curve C, we need only deﬁne an OY ⊗Z OZ,C -order BC satisfying 1-3 for then we can deﬁne~~

∗ B = π A ⊗Y OU ∩ BC C ram curve \ We are thus reduced to the following situation: Z = Spec R where (R, m) is a discrete valuation ring, say with residue ﬁeld κ = R/m, Y = Spec S Galois cover. It is a semilocal Dedekind domain say with Jacobson radical n. Let I be the inertia group G¯ = G/I. Consider the character group ∨ ∗ ∨ ∗ ∨ G = HomGrp(G, k ) and its subgroup G¯ = HomGrp(G,¯ k ) = {χ ∈ G | χ(I) = 1}. Consider the G-module decomposition

~~S = Sχ ∨ χM∈G I where Sχ is a free R-module of rank 1. Note that S/S is totally ramiﬁed and SI /SG is ´etale.~~

61 Also I S = χ∈G¯∨ Sχ Hence L S/n ≃ SI /mSI ≃ κG¯ Also n/n2 ≃ S/n as S-modules, but not as G modules. Hence there is a ∨ ∨ ∨ 2 ¯ ¯ n n ∨ coset χnG ∈ G /G such that / ≃ χ∈χnG¯ Sχ ⊗R κ as κG-modules. Similarly, there is a κG-module isomorphism L i i+1 n /n = Sχ ⊗R κ. i ¯∨ χ∈MχnG 2 We have a filtration 0 ⊂···⊂ n /mS ⊂ n/mS ⊂ S/mS = χ∈G∨ Sχ ⊗R κ. What this shows is that χ G¯∨ generates the cyclic group G∨/G¯∨. We write n L Si = Sχ i ¯∨ χ∈MχnG for i = 0,..., |I| − 1. Recall that A is a normal R-order so rad A = πA −i −i for some π ∈ A. We can define J = π A ⊂ A ⊗R K(R). Let e be the ramification index of A, r be the ramification index of S/R and s = e/r. We define our desired order r−1 −i s B = J ⊗R Si ⊆ A ⊗R K(S). i=0 M This is clearly G-stable with BG = A and B is finitely generated over S such that B ⊗S K(S) = A ⊗R K(S). The following calculation shows that B is closed under multiplication, −i s −(r−i)s −r s (J ⊗R Si)(J ⊗R Sr−i) ⊆ J ⊗R SiSr−i −1 = m A ⊗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 can make such a base change and assume~~

A ≃ Md(A0) RR m where A0 =  .  ⊂ Me(R) .    m ··· m 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 can assume R = S0 and S/R is totally ramiﬁed. Thus we can also assume that S = R[q]/(qr − p) where p ∈ m is a uniformising parameter for R. The theorem follows from direct computation, best illustrated by example. Example. Let e = 6, r = 3, s = 2

RRRRRR m RRRRR  m m RRRR  A =  m m m RRR     mmmm RR     mmmmm R      3 2 3 and S = R[q]/(q − p)= R ⊕ Rq ⊕ Rq := S0 ⊕ S1 ⊕ S2. Recall that q = p

A ⊗R S0 = A RRRR (q−3) (q−3) RRRRR (q−3)  RRRRRR  J −2 ⊗ S = ⊗ Rq R 1  (q3) RRRRR  R    (q3) (q3) RRRR     (q3) (q3) (q3) RRR     RR (q−3) (q−3) (q−3) (q−3) RRR (q−3) (q−3) (q−3)  RRRR (q−3) (q−3)  J −4 ⊗ S = ⊗ Rq2 R 2  RRRRR (q−3)  R    RRRRRR     (q3) RRRRR     

63 so S S Sq−1 Sq−1 Sq−2 Sq−2 Sq S S Sq−1 Sq−1 Sq−2  SqSq S S Sq−1 Sq−1  B =  Sq2 Sq Sq S S Sq−1     Sq2 Sq2 Sq Sq S S     Sq3 Sq2 Sq2 Sq Sq S    −1 −2  B0 B0q B0q  −1 = B0q B0 B0q  2  B0q B0q B0 3×3  ≃ B0

S S where B = and the last isomorphism is given by conjugation 0 Sq S 2 2 3×3 by the diagonal matrix with entries (1, 1,q,q,q ,q ). The S-order B0 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 be an S-algebra with a G-action compatible with the S-action. Let AutSB be the group of S-algebra automorphisms of B and note that G acts on AutSB by “conjugation”. We wish to twist the given actions of G on B by some 1-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 compatible with the S-action as follows. An element σ ∈ G acts on B by ασ(b) := fσ(σ · b), for b ∈ B. 2. Any G-action on B, compatible with G-action on S arises from such a twist.

~~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)) −1 = fσσ · (fτ (σ στ · 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. For −1 σ ∈ G, ασβσ is a ring automorphism which acts trivially on S. Hence −1 fσ = ασβσ ∈ 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 the twisted G-actions on B corresponding to f, f ′. If f and f ′ are cohomologous then the respective invariant rings A, A′ are isomorphic.

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

~~′ ′ ασ(b) = fσ(σ · b) −1 = (u fσ(σ · u))(σ · b) −1 = u fσ(σ · u(b)) −1 = u ασ(u(b))~~

~~′ −1 ′ −1 so ασ = u ασu, hence A = u (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 ﬁeld κ. Let A be an R-order, then A is normal if and only if A is hereditary and A/rad A = Di where the Di’s are isomorphic κ-algebras.

Proof. RecallQ that A is normal if and only if Aˆ := A ⊗R Rˆm is normal (c.f. Section 5 Proposition 2), and A is hereditary if and only if Aˆ 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 Section 5 Proposition 1, and if πA = J, then conjugation by π induces an automorphism between the Di (as in Section 10.4). Conversely, it suﬃces to prove that A/J ≃ J/J 2 as A/J-modules. Let εi ∈ A/J be the central idempotent corresponding to Di. Let Si be the simple Di-module and Pi ∈ A-mod be its projective cover. Now A is hereditary so ai a a J is projective and J = Pi for some ai. Let A = Pi where Di = Si (same a because Di’s are isomorphic), so it suﬃces to show that ai > a for all i by rank considerations.L Now J/J 2 is also an A/JL-module so we have a 2 2 op Peirce decomposition J/J = Bij where Bij = εi(J/J )εj is a Di ⊗k Dj - module. Note that for any i, there is some j such that Bij =6 0. For if not, 2 L 2 ⊗n n n+1 then εi(J/J ) = 0. But we have a natural surjection (J/J ) → J /J so n any simple subquotient of J/J must be Sj for some j =6 i. This is false since p A/p J is a subquotient of J/J n for n ≫ 0 and p a uniformising parameter for m.

~~op Lemma 18.2. Let B be a simple Di ⊗ Dj -module. Then~~

dimκ B > dimκ Di. ′ 2 ′ Proof. Let κ = Z(Di) and d = [Di : κ ]. op op ′ ′ ′ ′ ′ ′ ′ Di ⊗κ Dj = (Di ⊗κ (κ ⊗κ κ )) ⊗κ ⊗κκ ((κ ⊗κ κ ) ⊗κ Dj ) ′ ′ 2 ′ ′ ′ ′ is Azumaya over κ ⊗κκ of degree d . Now κ /κ is separable so κ ⊗κκ = κℓ where κ′ are ﬁeld extensions of κ′. Therefore any simple module B satisﬁes ℓ Q 2 ′ dimκ B = d [κℓ : κ] > d2[κ′ : κ′]

~~= dimκ Di.~~

To use Artin’s method to classify terminal orders, we need to consider Galois covers ramiﬁed on normal crossing lines. Example 18.3. Let S = k[[s,t]], p,q,e be positive integers and ζ, ξ primitive pe-th and qe-th roots of unity. Let G = hσ, τ | στ = τσ, σp e =1= τ q ei 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 is a Galois cover ramified on the union of u = 0 and v = 0 with ramification indices pe and qe. Let G be as above act on Md(S) ≃ Md(k) ⊗k S by the tensor 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 on Md(k). By Skolem-Noether, σ, τ act by conjugation by matrices say bσ, bτ ∈ p ε q e GLd(k). Scaling these if necessary, we may assume bσ =1= bτ . Note that p e p e there is also some scalar η such that bσbτ = ηbτ bσ. Now bτ = bσ bτ = η bτ . Hence η is a pe-th root of unity and similarly a qe-th root of unity. By changing 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. G Let A = Md(S) . 1. Then A is normal if and only if the dimensions of the eigenspaces of bσ 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 indices there 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 G generated over R. Also K(R) ⊗R A = Md(K(S)) which is central simple over K(R). In fact, by descent theory, it is Azumaya over uv =6 0. Also, A is reflexive since Md(S) is free over R and by Maschke’s theorem, A is a direct summand. Hence A is a reflexive R-order. Thus part 1) follows if we can show normality of A(u) = A ⊗R R(u) is equivalent to the eigenspace condition on bσ and normality of A(v) : A ⊗R R(v) is equivalent to the eigenspace condition on bτ . These are symmetric, so will only look at latter. Let J := rad A(v). G Note that A(v) = Md(S(v)) . Since Md(S(v)) is hereditary, so is A(v) by [McConnell-Robson, Theorem 7.8.8]. By Proposition 1, it suffices to show that the eigenspace condition on bτ is equivalent to the fact that A(v)/J is a product 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 ⊇ rad C ∩ CH = (rad C)H . In particular, if (C/rad C)H is semisimple, then CH /rad CH ≃ (C/rad C)H .

67 Proof. Let x ∈ (rad C)H and c ∈ CH . It suﬃces to check that 1 − c x is invertible in CH . It has an inverse y ∈ C, and y is H-invariant because 1−xc is. Hence the lemma follows. Returning to proof of theorem... G Let B = Md(S(v)), we compute (B/rad B) . Note that B/rad B ≃ Md(k((s))), and G acts on B via some 1-cocycle and we may pass to some 1-cohomologous cocycle if desired. We may thus conjugate simultaneously by bσ and bτ so that

~~1 ξ   bτ = ξ2  .   ..      where the block sizes are d0,d1,.,dq e−1, so d = d0 + d1 + ··· + dq e−1. Then~~

~~Md0 (k((s))) hτi M (k((s))) (B/rad B) =  d1  ...   q e−1 ~~

≃ Mdi (k((s))). i=0 Y We now compute σ-invariants of (B/rad B)hτi. Note that bσ permutes the eigenspaces of bτ as follows. Let v be an i −1 −1 i eigenvector of bτ with eigenvalue ξ . Then bτ bσv = η bσbτ v = η ξ bσv, so i −1 i bσ maps the ξ -eigenspace to the η ξ -eigenspace. In particular σ permutes cyclically the factors {Mdi (k((s))) | i ≡ j mod q}. Therefore

e q−1 hσ i G (B/rad B) ≃ Mdi (k((s))) i=0 ! Y Now σe has order p so passing to a cohomologous 1-cocycle we can assume the action of σ on Mdi (k((s))) is such that (Boris: the matrix below is a blockmatrix)

1 ζε e   bσ = ζ2e  .   ..      68 hσei e e hσei p We calculate Mdi (k((s))) . Note that σ : s 7−→ ζ s so k((s)) = k((s )). (Boris: the matrix below is a blockmatrix) k((sp)) s−1k((sp)) e M (k((s)))hσ i = s k((sp)) k((sp)) di   ...  p  ≃ Mdi (k((s ))  as seen by conjugating by (Boris: the matrix below is a blockmatrix) 1 s−1   s−2 .  .   ..    Hence   BG/rad(BG) = (B/rad B)G q−1 p = Mdi (k((s ))) i=0 Y By our criterion for normality (ﬁnd reference ??), this is normal if and only if all nonzero di are the same. Also Z(BG/rad(BG)) = k((sp)) i∈[0Y,q−1] di6=0 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 see that 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. We saw one example in Section 16, namely, the skew power series ring k hx, yi k [[x, y]] = η (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 uv = 0. Recall that the centre of a terminal order is smooth and the only complete local 2-dimensional ring which is smooth and has residue ﬁeld k is isomorphic to k[[u, v]].

~~69 Theorem 19.1. [C-Ingalls] (maybe Artin knew this) Let A be a terminal k[[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, then changing coordinates so that u = 0 is the ramification curve, then A is n × n matrices in

k[[u, v]] k[[u, v]] ··· k[[u, v]] .. .. .  u k[[u, v]] . . .  A0 ≃ . . . . ⊂ Me(k[[u, v]]) ......    u k[[u, v]] ··· u k[[u, v]] k[[u, v]]      3. Suppose A is ramified on normal crossing lines. Change coordinages so it is ramified on u = 0 with ramification index e and v = 0 with ramification index qe. Then A is isomorphic to n × n matrices in

kη[[x, y]] kη[[x, y]] ··· kη[[x, y]] .. .. .  (y) . . .  A0 = . . . . ⊂ Me(kη[[x, y]]) ......    (y) ··· (y) k [[x, y]]   η    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 us that any Azumaya k[[u, v]]-algebra is trivial Azumaya.~~

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

~~Remark 19.3. In fact, Br R ≃ Br κ, which we saw in the case where R is a complete discrete valuation ring.~~

Proof. Let A be an Azumaya R-algebra such that A ⊗R κ ≃ Md(κ). Let ε¯ ∈ A⊗R κ be a primitive idempotent so (A⊗R κ)¯ε is a simple module. Liftε ¯ to ε ∈ A. Note that R[ε] is ﬁnitely generated over a complete local noetherin ring R, so we can assume ε is an idempotent in R[ε] ⊂ A. The proposition will follow if we can show that the natural map ψ : A → EndR(Aε) is an isomorphism, 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 fact that~~

∼ ψ ⊗R κ : A ⊗R κ −→ Endκ((A ⊗R κ)¯ε) is an isomorphism. We check injectivity by studying I = ker ψ ⊳ A. Note that I ∩ R = 0 since Aε is a free R-module. Suppose I =6 0, then since A is a free R-module, we can ﬁnd an R-homomorphism ϕ : A → R such that op ϕ(I) =6 0. But A is Azumaya, so A⊗R A → EndRA. Hence as I is a 2-sided ideal, I ⊇ ϕ(I) =6 0, which contradicts I ∩ P = 0. Hence ker ψ = 0 and we are done. This ﬁnishes the proof of part 1. We will omit the proof of part 2 since it can be extracted from the proof of part 3. We use the method of Artin covers to prove part 3. Consider the following cover of R = k[[u, v]]. Let S = k[[s,t]] and G = hσ, τ | σe = τ q e = 1,στ = τσi act as in example in section 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 qe-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 qe on v = 0. We consider the Artin cover B of A with respect to S/R. By choice of S, B is an 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 G on B = Md(S) is given by twisting the trivial coordinatewise action by some 1-cocycle β : G → PGLd(S). As in the proof of theorem ? (from section 8.2) we have an isomorphism

~~1 ∼ 1 H (G, PGLd(S)) −→ H (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 action on S and conjugation by an element bσ ∈ GLd(k) on Md(k). Similarly τ acts by usual action on S and conjugation by bτ ∈ GLd(k). We can scale bσ and e q e bτ as usual so bσ =1= bτ . Then bσbτ = ηbτ bσ for some root of unity η. But A is terminal with secondary ramiﬁcation index e, so theorem ? (from section 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, G G B Mn(B ). Also, A is normal and the ramiﬁcation index of A along v = 0 is qe, so we may assume all the eigenspaces of bτ have the same dimension, say n, and all qe eigenvalues occur (by theorem 17.3). Hence we may pass to a ′ cohomologous 1-cocycle and assume bτ = bτ ⊗ In where

1 ξ b′ =   . τ ...    ξq e−1      i Since bσ permutes the eigenspaces of bτ by sending the ξ -eigenspace to the −1 i ′′ ′′′ η ξ -eigenspace. We may futher assume that bτ = bτ ⊗ bτ ⊗ In where

~~1 η b′′ =   τ ...    ηe−1     1  ξ b′′′ =   τ ...    ξq−1      ′ and bσ = bσ ⊗ Iq ⊗ In where~~

0 1 0 ··· 0 ......  . . . . .  ′ . b = ...... σ  . 0   . . .   0 ...... 1     1 0 ······ 0      e−1 (note: eigenvectors: v, bσv, . . . , bσ v) G Now we calculate Md(S) . 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 1 η x˜ =   ...    ηe−1     0 1 0 ··· 0 ......  . . . . .  . y˜ = ......  . 0   . . .   0 ...... 1     1 0 ······ 0      Note thaty ˜x˜ = ηx˜y˜. 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 by ′′ ′′ bσ and τ by conjugation by bτ . Then the G-module decomposition of Me(k) into isotypic components is~~

~~e−1 i j Me(k) = kx˜ y˜ i,j=0 M In particular k hx,˜ y˜i M (k) = . e (˜xe − 1, y˜e−, y˜x˜ − ηx˜y˜)~~

~~Proof. Exercise. Start by showing thatx ˜iy˜j is an eigenvector of bothx ˜ and y˜. Remark 19.7. Since gcd(m,e) = 1. We also have~~

~~e−1 m i mj Me(k) = kx˜ y˜ . i,j=0 M 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~~

~~e−1 S = Rsitj i,j=0 M 73 Hence~~

~~e−1 G i −m i j mj Me(S) = Rs x˜ t y˜ i,j=0 M R hx = sx˜m, y = ty˜mi = (xe − u, ye − v,yx − ζ−mx y)~~

~~= kζ−m [[x, y]] since~~

~~y x = ty˜msx˜−m = tsy˜m−1x˜−mη−my˜ = tsx˜−mζ−my˜m = ζ−msx˜−mty˜m = ζ−mx y.~~

~~Note. To compute can use the fact for G ﬁnite abelian and G-modules V,W with G-module decomposition V = χ∈G∨ Vχ and W = χ∈G∨ Wχ, then~~

~~G L L (V ⊗ W ) = Vχ ⊗ V−χ ∨ χM∈G~~

~~Let’s now ﬁnish the proof of the theorem by considering the case of general q. Let~~

~~i −m i qj mj A0 = Rs x˜ t y˜ M R hx, yi ≃ (xe − u, ye − v,yx − ζ−mx y) where x = sx˜−m, y = tqy˜m. From the case q = 1, we see that~~

~~2 e−1 A0 t A0 t A0 t A0 q−1 m ..  t y˜ A0 A0 .  G q−2 m q−1 m Mq e(S) = t y˜ t y˜ A0  . . . .   ......     A   0   ~~

~~74 Now conjugate by the diagonal matrix with entries 1,t,t2,...,te−1 to get~~

A0 A0 q m .  t y˜ A0 A0 .  G . . . . Mq e(S) = ......  . .   . . .   .. .. .     tqy˜mA tqy˜mA A   0 0 0    19.1 Resolution of singularities We show that locally, terminal orders have global dimension two, so terminal resolutions are in a sense resolutions of singularities. Deﬁnition 19.8. Let A be a noetherian ring. We say that it is regular of dimension 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

n×n kη[[x, y]] ······ kη[[x, y]] .. .. .  (y) . . .  A = ......    (y) ··· (y) k [[x, y]]   η    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 k[[y]] ······ k[[y]] n×n .. .. . ¯  (y) . . .  A: = A/x A = ......    (y) ··· (y) k[[y]]      Also let 0 1 0 ··· 0 ......  . . . . .  . . . . y˜ = ...... ⊗ In  . 0   . . .   0 ...... 1     y 0 ······ 0      75 is a regular normal element in A¯ and

~~n×n k ¯ ¯ .. A/y˜A =  .  k     which is semisimple so has global dimension 0.~~

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

Proof. By Proposition 17.1, we know that Aˆp = A⊗Z OZ,p is a normal order. Furthermore, its ramification is induced from that of A so Ap is terminal too. Hence Aˆ is a terminal k[[u, v]]-order so is regularb of dimension 2 by the Theorem. 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 normal ∗ order 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˜ → Z be a projective birational map. Then there exist normal orders A˜ on Z˜ such that (Z,˜ A˜) → (Z,A) is a modiﬁcation. We write π#A = A˜ and call π#A a modiﬁcation of A with respect to π.

Corollary 19.13. Let A be a normal order on a surface Z. Then there is 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 a modiﬁcation a resolution of singularities.

~~20 Noncommutative Mori Minimal Model Pro- gram~~

There is a noncommutative version of Mori’s minimal model program for orders on projective surfaces. It provides an intelligent scheme for classifying orders on surfaces. The theory works on the level of Brauer pairs, so we shall work 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 noncommutative analogue. Proposition 20.1. Let Z be a projective surface and α ∈ Br k(Z). Let π : 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 A with 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 essentially the same as on Z and α is unramified on E by the Artin-Mumford sequence. Hence (Z,˜ α) is terminal. In case 2, again the Artin-Mumford sequence implies that α is unramified on E and the ramification of α on Z˜ is essentially the same as that on Z. Hence (Z,˜ α) is terminal. In case 3,Boris: picture missing, Artin-Mumford shows that α is ramified on E with ramification index e =secondary ramification index. The node of ramification has split into two and we see from Artin-Mumford that they are both terminal.

~~21 Appendix: Projective Covers~~

~~Let A be a ring. Deﬁnition 21.1. Let M be an A-module. A projective cover of M is a surjective map ϕ: P → M such that PC1 P is projective,~~

ψ ϕ PC2 given any morphism ψ : N → P such that the composite N −→ P −→ M is surjective, have ψ is surjective. That is, given any submodule N < 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/L and so N + L = P . Thus N + (rad A)P = P and so by Nakayama’s lemma N = P .

~~77 Proposition 21.3. Projective covers are unique up to isomorphism. Hence ﬁnitely generated modules P, P ′ are isomorphic if~~

~~P P ′ ≃ . (rad A)P (rad A)P ′~~

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

~~φ P / M I ||> ′ || ψ ψ | || ′ | φ P ′~~

Condition PC2 on P ′ shows that ψ is surjective. Since P ′ is projective, 1 ′ ExtA (P , −) = 0 and so since ψ is surjective, the map ψ is split, say with section ψ′. Condition PC2 on P implies ψ′ is surjective too. Therefore ψ and ψ′ are inverse isomorphisms.

~~22 Appendix: Finite Global Dimension~~

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

~~Theorem 22.1. Let A be a noetherian ring and z ∈ rad A be a regular normal 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 we have pd M 6 d and equality holds for some M. We consider the z-torsion submodule~~

~~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-torsion free whilst τ(M) has a ﬁnite ﬁltration by A/z A-modules, 0 < annM z < 2 annM z < ··· . Hence the theorem follows from the next two lemmas. We will continue to assume below that A is noetherian and z ∈ rad A is regular normal.

~~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 spectral sequence~~

~~p,q p 1 p+q E2 = ExtA/z A(M, ExtA(A/z A, N)) =⇒ ExtA (M,N) Since z is regular, we have the following A-free resolution of A/z A~~

~~0 −→ A −→z A −→ A/z A −→ 0.~~

Hence the spectral sequence has only two nonzero rows, where q = 0, 1. i+1 This tells us that if pdA/z AM = i then ExtA (M,N) = 0. Hence pdAM 6 pdA/z AM + 1. Also the spectral sequence gives a surjection

~~i+1 i 1 ExtA (M,N) −→ ExtA/z A(M, ExtA(A/z A, N))~~

i Pick an A/z A-module N suc that ExtA/z A(M,N) =6 0. Then by the free res- 1 olution of A/z A above, we have ExtA(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/z M. Proof. We first show that if M is z-torsion free and M/z M is a finitely generated free A/z A-mdoule, then M is free over A. Let m1,...,mr ∈ M be such that their images in M/z M form a free A/z A basis. It suffices to show that m1,...,mr is a free A-basis for M. They generate 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 so P that ai 6 inz A. This contradicts the fact that m1,...,mr give an A/z A-basis for M/z M. Conversely, if M is A-free then M/z M is A/z A-free. Therefore if M is A-projective then M/z M is A/z A-projective.

We proceed by induction on pdA/z AM/z M. 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/z M −→ 0.~~

~~If pdA/z AM/z M = 0, then this splits, so (M ⊕ K)/z(M ⊕ K) ≃ M/z M ⊕ 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 projective also. Suppose now pdA/z AM/zM > 0 so pdAM > 0 also. Therefore~~

~~pdA/z AM/z M = pdA/z AK/z K + 1~~

~~by induction = pdAK + 1~~

~~= pdAM~~

It remains to consider the case where pdA/z AM/z M = ∞. Suppose pdAM < ∞, then let P • → M → 0 be a finite projective A-resolution. Then a simlar argument to the above shows that P •/z P • → M/z M → 0 is a projective A/z A-resolution. Contradiction finishes proof of Lemma 22.3. Proposition 22.4. Let (R, m) be a commutative regular local ring of dimension d. Let A be an R-algebra of global dimension d which is finitely generated free as an R-module. Then any A-module M which is finitely generated free as 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~~

~~zi+1 0 −→ Mi −→ Mi −→ Mi+1 −→ 0 For any A-module N we get an exact sequence~~

j zi+1 j j+1 ExtA(Mi,N) −→ ExtA(Mi,N) −→ ExtA (Mi+1,N) j If ExtA(Mi,N) =6 0, then multiplication by zi+1 on it is not surjective by NAK. 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 assumption. We will need resolutions of singularities and to pass to completions. Ex- cellent rings provide a good setting where resolutions work and completions are well behaved. Excellent rings are special types of commutative noetherian rings. You can find the definition and basic facts about them in [EGA IV, Section 7.8] also Matsumura. Next result gives some examples. Proposition 23.1 (EGA IV, Proposition 7.8.3). 1. Fields of any characteristic are excellent as are Dedekind domains whose fields of fractions 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) open affine cover Z = ∪Spec Ri, the rings are excellent. So in particular, all quasi-projective varieties are excellent. Excellent rings are well-behaved in many ways Proposition 23.2 (EGA, Section 7.8). Let R be an excellent reduced local ring 1. Rˆ 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 Rˆ has length dim R, that is, Rˆ 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 Zˆp = Spec OZ,p. Then ˜ ˜ ˆ ˆ 1. If Z → Z is a projective birational morpihsmb then so is Z ×Z Zp → Zp. Furthermore, Z˜ is nonsingular above p if and only if Z˜ ×Z Zˆp is.

~~2. Any projective birational morphism Z˜p → Zˆp comes from base change as above.~~

~~24 Blowing up and Terminal Ramiﬁcation Data~~

We now consider the question of ﬁnding a good model for K-central division rings when K is the function ﬁeld of a surface Z. Our aim is to obtain a noncommutaive analogue of smooth models and a resolution of singularities type result. Our point of view is to note that blowing up and normalisation can be used 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 normal scheme 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 a maximal 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 exceptional 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 and C˜ 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. Recall that the integral closure S ⊂ F of R in F is such that S/R is Galois with Galois group G (c.f. Section 1). Now S is a Dedekind domain which is semilocal since the maximal ideals 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 for G i> 1. Consider b = σ∈G σ(a) ∈ S = R. Note that b is not invertible since a is not invertible. Hence b ∈ m. Thus also b ∈ ni for all i. This shows that Q σ(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 called the decomposition group and it is well-defined up to conjugacy. Also, we define κS = S/ni which is independent of i up to isomorphism. Let σ ∈ D where 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 the kernel 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 inertia group.~~

D Proof. It suffices to show that κS = κ, for then Galois theory shows that κS/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 are distinct. 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 to α˜ ∈ S such thatα ˜ ∈ ni for i> 1. Consider the polynomial

p(x) = (x − σ(˜α)) ∈ SG[x]= R[x] σ∈G Y and we denote its image modulo n1 byp ¯ ∈ κ[x]. N For σ 6∈ D we have σ(˜α) ∈ n1. Hencep ¯(x)= x 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. Q

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