aa r X i v : . [ m a t h . R A ] O c t On a class of hereditary crossed-product orders ∗ John S. Kauta
Abstract
In this brief note, we revisit a class of crossed-product orders overdiscrete valuation rings introduced by D. E. Haile. We give simple butuseful criteria, which involve only the two-cocycle associated with a givencrossed-product order, for determining whether such an order is a hered-itary order or a maximal order.2010
Mathematics Subject Classification. Primary: If R is a ring, then J ( R ) will denote its Jacobson radical, U ( R ) its groupof multiplicative units, and R the subset of all the non-zero elements. Theterminology used in this paper, if not in [1], can be found in [3]. The bookby Reiner [3] is also an excellent source of literature on maximal orders andhereditary orders.Let V be a discrete valuation ring (DVR), with quotient field F , and let K/F be a finite Galois extension, with group G , and let S be the integral closure of V in K . Let f ∈ Z ( G, U ( K )) be a normalized two-cocycle. If f ( G × G ) ⊆ S ,then one can construct a “crossed-product” V -algebra A f = X σ ∈ G Sx σ , with the usual rules of multiplication ( x σ s = σ ( s ) x σ for all s ∈ S, σ ∈ G and x σ x τ = f ( σ, τ ) x στ ). Then A f is associative, with identity 1 = x , andcenter V = V x . Further, A f is a V -order in the crossed-product F -algebraΣ f = P σ ∈ G Kx σ = ( K/F, G, f ).Two such cocycles f and g are said to be cohomologous over S (respectivelycohomologous over K ), denoted by f ∼ S g (respectively f ∼ K g ), if there areelements { c σ | σ ∈ G } ⊆ U ( S ) (respectively { c σ | σ ∈ G } ⊆ K ) such that g ( σ, τ ) = c σ σ ( c τ ) c − στ f ( σ, τ ) for all σ, τ ∈ G . Following [1], let H = { σ ∈ G | f ( σ, σ − ) ∈ U ( S ) } . Then H is a subgroup of G . On G/H , the left coset space of G by H , one can define a partial ordering by the rule σH ≤ τ H if f ( σ, σ − τ ) ∈ U ( S ) . Then “ ≤ ” is well-defined and depends only on the cohomology class of f over S . Further, H is the unique least element. We call this partial ordering on G/H the graph of f. ∗ Accepted for publication by the Proceedings of the American MathematicalSociety. S is unramified over V , wherein, among other things, conditions equivalent tosuch orders being maximal orders were considered. This is the class of crossed-product orders we shall study in this paper, always assuming that S is unramifiedover V . We emphasize the fact that, since we do not require that f ( G × G ) ⊆ U ( S ), this theory constitutes a drastic departure from the classical theory ofcrossed-product orders over DVRs, such as can be found in [2].Let us now fix additional notation to be used in the rest of the paper, mostof it borrowed from [1] as before. If M is a maximal ideal of S , let D M be thedecomposition group of M , let K M be the decomposition field, and let S M bethe localization of S at M . The two-cocycle f : G × G S yields a two-cocycle f M : D M × D M S M , determined by the restriction of f to D M × D M and theinclusion of S in S M . Then A f M = P σ ∈ D M S M x σ is a crossed-product order inΣ f M = P σ ∈ D M Kx σ = ( K/K M , D M , f M ). In addition, we can obtain a twist of f , described in [1, pp. 137-138] and denoted by ˜ f , which depends on the choiceof a maximal ideal M of S , and the choice of a set of coset representatives of D M in G . We also define F : G × G S by F ( σ, τ ) = f ( σ, σ − τ ) for σ, τ ∈ G .While ˜ f is a two-cocycle, F is not.If B is a V -order of Σ f containing A f , then by [1, Proposition 1.3], B = A g = P σ ∈ G Sy σ for some two-cocycle g : G × G S , with g ∼ K f . Moreover,the proof of [1, Proposition 1.3] shows that y σ = k σ x σ for some k σ ∈ K , with k = 1, whence g is also a normalized two-cocycle.We begin with a technical result. Sublemma.
Let τ ∈ G . If I τ = Y f ( τ,τ − ) M M , where M denotes a maximal ideal of S , then I τ − τ = I τ − .Proof. We have I τ − τ = Y f ( τ,τ − ) M M τ − = Y f τ − ( τ,τ − ) M τ − M τ − = Y f ( τ − ,τ ) M τ − M τ − = I τ − . Theorem.
The crossed-product order A f is hereditary if and only if f ( τ, τ − ) M for all τ ∈ G and every maximal ideal M of S .Proof. The theorem obviously holds if H = G , in which case A f is an Azumayaalgebra over V , so let us assume from now on that H = G .Suppose A f is hereditary. First, assume A f is a maximal order and S is aDVR. Let v be the valuation corresponding to S with value group Z . Then by[1, Theorem 2.3], H is a normal subgroup of G and G/H is cyclic. Further,there exists σ ∈ G such that v ( f ( σ, σ − )) ≤ G/H = < σH > , and the graphof f is the chain H ≤ σH ≤ σ H ≤ · · · ≤ σ m − H , where m = | G/H | . Choose j maximal such that 1 ≤ j ≤ m − v ( f ( σ i , σ − i )) ≤ ∀ ≤ i ≤ j . Wealways have σH ≤ σ − j H ; but if j < m −
1, then we also have σ j H ≤ σ j +1 H .2ence if j < m −
1, then, from the cocycle identity f σ j ( σ, σ − j σ − ) f ( σ j , σ − j ) = f ( σ j , σ ) f ( σ j +1 , σ − j σ − ) , we conclude that v ( f ( σ i , σ − i )) ≤ ∀ ≤ i ≤ j + 1,a contradiction. So we must have j = m −
1, so that v ( f ( σ i , σ − i )) ≤ ∀ ≤ i ≤ m −
1. If τ is an arbitrary element of G , then τ = σ i h for some h ∈ H andsome integer i , 0 ≤ i ≤ m −
1. Therefore, by [1, Lemma 3.6], v ( f ( τ, τ − )) = v ( F ( σ i h, v ( F ( σ i , v ( f ( σ i , σ − i )) ≤
1; that is, f ( τ, τ − ) J ( S ) .We maintain the assumption that A f is a maximal order, but we now dropthe condition that S is a DVR. By [1, Theorem 3.16], there exists a twist of f , say ˜ f , such that f ∼ S ˜ f . By [1, Corollary 3.11], for any maximal ideal M of S , A f M is a maximal order in Σ f M ; hence f M ( τ, τ − ) M ∀ τ ∈ D M bythe preceding paragraph. Therefore, from the manner in which ˜ f is constructedfrom f , we infer that ˜ f ( τ, τ − ) M ∀ τ ∈ G and any maximal ideal M of S , and thus f ( τ, τ − ) M ∀ τ ∈ G and every maximal ideal M of S , since f ∼ S ˜ f .If A f is not a maximal order, then it is the intersection of finitely manymaximal orders, say A f , A f , . . . , A f l . Note that A f i = X τ ∈ G Sy ( i ) τ = X τ ∈ G Sk ( i ) τ x τ , for some k ( i ) τ ∈ K . Fix a σ ∈ G , and a maximal ideal N of S . Let v N be thevaluation corresponding to N , with value group Z . Since S = l \ i =1 Sk ( i ) σ , there exists i such that v N ( k ( i ) σ ) = 0. Let g = f i and, for τ ∈ G , let k τ = k ( i ) τ and y τ = y ( i ) τ , so that A g = P τ ∈ G Sk τ x τ = P τ ∈ G Sy τ . By [1, Proposition3.1], J ( A f ) = P τ ∈ G I τ x τ and J ( A g ) = P τ ∈ G J τ y τ , where I τ = Y f ( τ,τ − ) M M and J τ = Y g ( τ,τ − ) M M, and M denotes a maximal ideal of S . Since A f is a hereditary V -order inΣ f and A f ⊆ A g ⊆ Σ f , we have J ( A g ) ⊆ J ( A f ), from which we concludethat J σ − y σ − ⊆ I σ − x σ − and so J σ − k σ − ⊆ I σ − . We have y σ − J σ y σ = k σ − x σ − J σ k σ x σ = J σ − k σ − x σ − k σ x σ ⊆ I σ − x σ − k σ x σ = σ − ( k σ ) I σ − f ( σ − , σ )= (cid:0) k σ I σ f ( σ, σ − ) (cid:1) σ − . On the other hand, y σ − J σ y σ = J σ − σ g ( σ − , σ ) = J σ − g ( σ − , σ ). Since A g is a maximal order and therefore g ( σ − , σ ) M for every maximal ideal M of S , we see that J σ − g ( σ − , σ ) = J ( V ) S and so y σ − J σ y σ = J ( V ) S . Therefore J ( V ) S ⊆ k σ I σ f ( σ, σ − ). Since v N ( k σ ) = 0, weconclude that f ( σ, σ − ) N , and so f ( τ, τ − ) M ∀ τ ∈ G and any maximalideal M of S .Conversely, suppose that f ( τ, τ − ) M for every maximal ideal M of S and every τ ∈ G . Let B = O l ( J ( A f )), the left order of J ( A f ); that is,3 = { x ∈ Σ f | xJ ( A f ) ⊆ J ( A f ) } . Since Σ f ⊇ B ⊇ A f , B = P τ ∈ G Sk τ x τ , forsome k τ ∈ K . For each τ ∈ G , we have S ⊆ Sk τ , and we will now show that S = Sk τ . As above, write J ( A f ) = P I τ x τ , with I τ = Q M , where the productis taken over all maximal ideals M of S for which f ( τ, τ − ) M . Observethat J ( V ) S = I ⊇ k τ x τ I τ − x τ − = k τ I ττ − f ( τ, τ − ) = k τ I τ f ( τ, τ − ) . Since f ( τ, τ − ) M for every maximal ideal M of S , we must have I τ f ( τ, τ − ) = J ( V ) S, and so J ( V ) S ⊇ k τ J ( V ) S ⊇ J ( V ) S and thus S = Sk τ , as desired. Thisshows that O l ( J ( A f )) = A f and A f is hereditary.Not only can this criterion enable one to rapidly determine whether or notthe crossed-product order A f is hereditary, the utility of the theorem aboveis now demonstrated by the ease with which the following corollaries of it areobtained. Corollary 1.
The crossed-product order A f is hereditary if and only if f ( τ, γ ) M for all τ, γ ∈ G and every maximal ideal M of S .Proof. This follows from the cocycle identity f τ ( τ − , τ γ ) f ( τ, γ ) = f ( τ, τ − ).In other words, the order A f is hereditary if and only if the values of thetwo-cocycle f are all square-free.Since A f is a maximal order if and only if it is hereditary and primary, bycombining our result and results in [1], we immediately have the following. Corollary 2.
Given a crossed-product order A f ,1. it is a maximal order if and only if for every maximal ideal M of S , f ( τ, τ − ) M for all τ ∈ G , and there exists a set of right coset repre-sentatives g , g , . . . , g r of D M in G (i.e., G is the disjoint union ∪ i D M g i )such that for all i , f ( g i , g − i ) M .2. if S is a DVR, then it is a maximal order if and only if f ( τ, τ − ) J ( S ) for all τ ∈ G .Proof. In either case, the primarity of A f is guaranteed by [1, Theorem 3.2] (seealso [1, Proposition 2.1(b)] when S is a DVR).The Theorem above can readily be put to effective use with the crossed-product orders in [1, § S . Using our criterion, it nowbecomes a straightforward process to determine which of those orders are max-imal orders and which are not, by simply consulting, in each case, the giventable of values for the two-cocycle; the table whose entries are all square-freerepresents a maximal order. This determination can be made with little effort!In fact, if one knows that the crossed-product order A f is a primary order, thendetermining whether or not it is a maximal order could even be easier, as thefollowing result shows. 4 orollary 3. Suppose the crossed-product order A f is primary. Then it is amaximal order if and only if there exists a maximal ideal M of S such that f ( τ, τ − ) M for all τ ∈ D M .Proof. This follows from [1, Corollary 3.11 and Proposition 2.1(b)].Let L be an intermediate field of F and K , let G L be the Galois group of K over L , let U be a valuation ring of L lying over V , and let T be the integralclosure of U in K . Then one can obtain a two-cocycle f L,U : G L × G L T from f by restricting f to G L × G L and embedding S in T . As before, A f L,U = P τ ∈ G L T x τ is a U -order in Σ f L,U = P τ ∈ G L Kx τ = ( K/L, G L , f L,U ). Corollary 4.
Suppose the crossed-product order A f is hereditary. Then A f L,U is a hereditary order in Σ f L,U for each intermediate field L of F and K and forevery valuation ring U of L lying over V . This leads to the following.
Corollary 5.
Suppose the crossed-product order A f is hereditary. Then A f M is a maximal order in Σ f M for each maximal ideal M of S .Proof. The order A f M is always primary, by [1, Proposition 2.1(b)].The following example illustrates two limitations of our theory, however. Example.
We give two crossed-product orders A f and A f with f ∼ K f andthe graphs of f and f identical, but A f is hereditary while A f is not. Also,we give an example to demonstrate that the converse of Corollary 5 does notalways hold.Let F = Q ( x ), and let K = Q ( i )( x ). Then the Galois group G = < σ > isa group of order two, where σ is induced by the complex conjugation on Q ( i ).If V = Q [ x ] ( x +1) , then S has two maximal ideals, namely M = ( x + i ) S and M = ( x − i ) S , and D M = D M = { } . Let f , f : G × G S be two-cocycles defined by f j (1 ,
1) = f j (1 , σ ) = f j ( σ,
1) = 1 and f ( σ, σ ) = ( x + 1) x , f ( σ, σ ) = ( x + 1) x .Then f ∼ K f , and the subgroup of G associated with either cocycle is H = { } , so that the graphs of f and f are identical. Clearly, A f is hereditarybut A f is not. We conclude that the property that a crossed-product order A f is hereditary is not an intrinsic property of the graph of f .Also, if we set f = f , we see that A f M = S M for each maximal ideal M of S , and therefore A f M is a maximal order in Σ f M = K for each maximalideal M of S , and yet A f is not even hereditary (cf. [1, Corollary 3.11], and[2, Theorem 1]). This is the case because A f is not primary, and also because f ( G × G ) U ( S ). (cid:3) eferences [1] D. E. Haile, Crossed-product orders over discrete valuation rings , J. Algebra (1987), 116–148. MR0871749 (88b:16013)[2] M. Harada,
Some criteria for hereditarity of crossed products , Osaka J.Math. (1964), 69–80. MR0174584 (30:4785)[3] I. Reiner, Maximal Orders , Academic Press, London, 1975. MR0393100(52:13910)Department of MathematicsFaculty of ScienceUniversiti Brunei DarussalamBandar Seri Begawan BE1410BRUNEI.