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 multipli- cation 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.