On Representability of a Finite Local Ring

Journal of Algebra 228, 417 427 (2000) doi:10.1006/jabr.1999.8242, available online at http://www.idealibrary.com on On Representability of a Finite Local Ring A. Z. Anan in Departamento de Matemática do IMECC, Universidade Estadual de Campinas, Campinas, SP, 13083-970, Brasil E-mail: ananin@ime.unicamp.br Communicated by Efim Zelmanov Received January 10, 1999 This article is devoted to the following problem: Let R be a finite local ring with identity. Can R be embedded into a matrix ring over a commutative ring? The question is reduced to an explicitly described class of test-cases. 2000 Academic Press Key Words: finite local ring; embedding into matrices. 1. INTRODUCTION This article is devoted to the following Problem 2 from [6, p. 401]: Let R be a finite local ring with identity. Can R be embedded into a matrix ring over a commutative ring? By a local (noncommutative) ring, we mean an associative ring with identity which has a unique maximal left ideal (it has to be two-sided). The question is deeply related to the articles of Bergman [2, 5], Bergman and Vovsi [3], and Bergman et al. [4], which led the author to the mistaken belief that it was raised by Bergman. The author sincerely thanks Bergman who kindly pointed out the reference [6] and the referee who made many improvements on the style of the article and suggested some simplification of the construction. First, an example of a finite local ring: Suppose we are given a prime integer p>0, an integer n>0, and integers i n i n 1 i k i 1 0. Let E = End B denote the ring of all automorphisms of the finite abelian p-group B= /p n /p n /p n 1 /p n 1 }{{}}{{} i n times i n 1 times 417 0021-8693/00 $35.00 Copyright 2000 by Academic Press All rights of reproduction in any form reserved.

418 a. z. anan in /p k /p k /p /p }{{}}{{} i k times i 1 times = B 1 B 2 B α B r = C n C n 1 C k C 1 where B α = /p k α (with k α k β for α β), r = i n + i n 1 + + i k + +i 1, and C k = /p k /p k. As is well known, every element c E can be written in the form of a matrix, c = c αβ, where }{{} i k times c αβ Hom B β B α, and the multiplication (and addition) in E is defined using the usual rule for multiplication (and addition) of matrices and the composition (and addition) of homomorphisms. Let R = R p i1 i n E be the subset of all matrices of the form T nn T n1 a 1 + (1) T 1n T 11 where a and every block T kk End C k is lower triangular modulo p, 0 0 T kk mod p 0 I claim that R is a local ring (in fact, it is a maximal local subring in E containing the identity of E). Let N denote the ideal of matrices occurring on the right-hand side of (1). It is not hard to see from the definition that every member of N sends each B i into B i+1 + +B r +pb. Hence any member of N r sends B into pb and hence any member of N rn sends B into p n B = 0 ; son is nilpotent. Therefore, N is contained in N R (the nilpotent radical of R). Clearly, p 1 is also contained in N R. Thus, R/N R is at most /p. In some sense, the example presented is a general one. For an arbitrary finite local ring R with the nilpotent radical N R, the field R/N R can be lifted into R in the form of a local commutative subring A which can be described quite explicitly (see the next section). I have not succeeded in answering the question. The main result is the Theorem 1. Let R be a finite local ring. Then there is an embedding R Matr m R preserving identity such that R = A R p i1 i n, where A is a commutative finite local ring satisfying p n A = 0. Thus, the question is reduced to the one concerning the concrete finite local ring R p i1 i n. It seems likely to me is that it will not in general be

representability of a finite local ring 419 representable by matrices over a commutative ring. However, even if it is representable, the representation should be a little bit sophisticated: The ring R p 1 0 0 1 is representable and the minimal possible size of matrices is p. Such a representation is pointed out at the very end of the article. 2. SOME ELEMENTARY REMARKS Let R be a finite local ring with identity. Denote by N R the nilpotent radical (= maximal ideal) of R. As is known F = R/N R is a finite field of p m elements, where p is a prime integer. Clearly, p = p 1 N R and p n = 0inR. We take the least possible n to satisfy the last equality. The multiplicative group of F is cyclic. It is generated by y = y + N R. Therefore, y pm 1 = 1 + a, for some a N R. Since 1 + a pn = 1 for N 0, we obtain y pm 1 p N = 1. Since p N and p m 1 are coprime, we can replace y by y pn. For this new y, we have y pm 1 = 1. Denote by A the subring (with identity) generated by y. Obviously, A + N R =R. Hence, A is a commutative local ring with identity. Moreover, A/N A F. Note that the minimal polynomial for y over the prime field F p /p has the form t m f 0 t, deg f 0 <m. We can regard f 0 t as a polynomial over the integers. The commutative F p -algebra A/pA is generated by y which satisfies the equality y pm = y; therefore, A/pA satisfies the identity: z pm = z, z A/pA. In particular, A/pA is semisimple, which means that pa N A and, hence, N A =pa. Thus, we obtain y m = f 0 y +pg 1 y where g 1 t is a suitable polynomial over the integers. By use of this equality, we can replace every term of the form y m+i in g 1 y by f 0 y y i + pg 1 y y i on the right-hand side of the equality. After few such replacements, we shall acquire the equality y m = f 0 y +pf 1 y +p 2 g 2 y, where f 1 g 2 t and deg f 1 <m. After applying the same procedure to the terms in g 2 y, then to those in g 3 y, and so on, we shall obtain the equality y m = f 0 y +pf 1 y + +p n 1 f n 1 y where f i t, deg f i <m,0 i<n. Denote f t =f 0 t +pf 1 t + + p n 1 f n 1 t. We can regard f t as a polynomial over /p n, deg f<m. We will show that A /p n t / Ideal ( t m f t ) 2a Indeed, if g t is a nonzero polynomial over /p n, deg g<m, such that g y =0, then, multiplying g t by a suitable power p i, we have p i g t =

420 a. z. anan in p n 1 h t for a polynomial h t t, deg h<m, which is nonzero modulo p. Since the image of h y under the natural map A F is nonzero, h y is invertible in A. Thus, the equalities 0 = p i g y =p n 1 h y imply p n 1 = 0, contradiction. Arguing similarly, for an arbitrary 0 a A, we obtain a = p i q y, 0 i<n, deg q<m, and q t is nonzero modulo p. This means that q y is invertible. Consequently, every ideal in A has the form p i A. From this fact, we can deduce self-injectivity of A by use of the Baer criterion: If p i A A is an A-module homomorphism from an arbitrary ideal into A, 0 i n, then the image of p i has the form p k u, k i, with u A invertible. Therefore, the homomorphism is induced by multiplication by p i k u. The same arguments lead to self-injectivity of A/p k A, k>0. The homomorphism A F A/N A induces a homomorphism α Aut A Aut F of groups. I claim that it is an isomorphism. Indeed, if ϑ belongs to the kernel of α, then ϑ y y + pa, a A. Since y is nonzero in F, y is invertible in A, so we can write y + pa = y 1 + pb for a suitable b A. It follows that ϑ y pm 1 y pm 1 1 + pb pm 1 ; therefore, 1 = 1 + pb pm 1. On the other hand, 1 + pb pn = 1 for N 0. This means that 1 + pb = 1 and ϑ y y. Hence, α is an embedding. The group Aut F is generated by the Frobenius automorphism φ. It remains to show that there exists an automorphism of A which maps y to y p. Consider the group ring of the cyclic group G of order p m 1 over the ring /p n, i.e., B = /p n G /p n t / Ideal t pm 1 1. There exists an epimorphism B A which maps t to y. The well-known decomposition of a finite commutative ring into a direct sum of local rings, B = B 1 B r, for some i, yields an epimorphism B i A induced by the epimorphism B A. (Otherwise the local ring A would contain a nontrivial idempotent.) The epimorphism B i A induces an epimorphism B i /N B i A/N A of fields, which must be an isomorphism. Since p n 1 B i 0 by using the arguments which led to (2), we have B i /p n t / Ideal ( t m g t ), deg g<m, and card B i = card A; therefore, B i A is an isomorphism. The automorphism of G mapping x to x p induces an automorphism ψ of B. Every automorphism of B is formed by automorphisms of B i s and a permutation of B i s. We can find the permutation which corresponds to ψ by looking at ψ modulo p. Since B/pB F p G is a semisimple algebra and the induced automorphism is given by the rule x x p, we see that the permutation is trivial. This yields the automorphism of B i and, therefore, the desired automorphism of A. Denote this automorphism of A by ϕ. Let T = F F. For every i = 0 1 m 1, we define the ring homomorphisms h i T F onto F by the rule h i α β φ i α β. ( These homomorphisms have the kernels pairwise different, since h j α 1 1 φ i α ) = φ j α φ i α for all α F. Being semisimple (in view of the identity z pm = z), T is the direct sum of fields T = T 0 T m 1

representability of a finite local ring 421 with the isomorphisms T i F induced by h i s. Comparing cardinalities, we conclude that T = T 0 T m 1. Since the T -module F given by h i and the T -module T i are isomorphic, we obtain α 1 t i = ( 1 φ i α ) t i for all α F and t i T i. The ring B = A A has a decomposition into the direct sum of local rings B = B 0 B m 1. Modulo the radical N B =pb this decomposition induces the decomposition of T. Therefore, for every b i B i, we have ( y 1 1 ϕ i y ) b i pb B i = pb i. Thus, u n B i = 0, where u = x z pi, x = y 1, and z = 1 y. The equalities z pi pm = z pi and x pm = x imply x u = x u pm = x + p m uv + u 2 w for suitable v w B. Hence u 1 + p m v + uw =0. Since u n B i = 0, the element p m v + uw is nilpotent modulo the annihilator of B i in B. Therefore, ub i = 0. Thus, for every b i B i, the subring a A a 1 b i = ( 1 ϕ i a ) b i contains y. In other words, a 1 b i = ( 1 ϕ i a ) b i for all aıa and b i B i. Let M be an arbitrary A A -bimodule. Regarding M as a left B-module, we obtain the bimodule decomposition M = M 0 M m 1, where M i = B i M. It is clear that am i = m i ϕ i a for all a A and m i M i Summarizing, we arrive at the Proposition 2. Let R be a finite local ring with residue field F = R/N R of cardinality p m, where p is prime. Then there exists a y R such that y pm 1 = 1. The subring A generated by y satisfies the following properties: A is a commutative finite local ring and the ring A/p k A is selfinjective for every k 0; R = A + N R and N A =A N R =pa; the group Aut A is naturally isomorphic to the cyclic group Aut F with the automorphism ϕ y y p corresponding to the Frobenius automorphism φ; every A A -bimodule M admits an A A -bimodule decomposition M = M 0 M m 1 such that am i = m i ϕ i a for all a A and m i M i. 3. THE CONSTRUCTION The idea of this and the next sections is to put A into the center of some finite local ring R and represent the initial finite local ring R by matrices over R. The construction presented here is composed with the following picture in mind: The subring A of R is living in Matr m R in a twisted form on the diagonal R A a diag ( ϕ 0 a ϕ m 1 a ) Matr m R Let R and A be as in the preceding section. Applying the proposition to the A A -bimodules N = N R and R, we obtain two decompositions

422 a. z. anan in N = N 0 N m 1 and R = R 0 R m 1. Since A + N = R and A R 0, we see that R 0 = A + N 0 and R i = N i for 0 <i<m. It is easy to see that R = R 0 R m 1 is graded by the group Aut A = Aut F. In particular, R 0 is a subring. Note that p n 1 r = 0 implies r N; otherwise, r would be invertible and p n 1 = 0. On the other hand, if p n 1 r 0, then the ideal a A ar = 0 A does not contain p n 1 and, hence, equals zero, which means that the left A-module Ar is isomorphic to A. By using the self-injectivity of A, we can split R 0 into a direct sum of left A-modules, R 0 = A B 0.Ifp n 1 B 0 0, then we can similarly split B 0 : There exists r 1 B 0 such that p n 1 r 1 0; therefore, B 0 = Ar 1 B 1 and R 0 = A Ar 1 B 1. Finally, we obtain R 0 = A Ar 1 Ar t B t with p n 1 B t = 0. The last equality implies the inclusion B t N. For every r i, there are two possibilities: either r i N or N r i A + N = R. Inthe second case r i = a + b, where a A \ N and b N. Since A is local, we can replace such an r i by a 1 r i = 1 + c, where c = a 1 b N. Note that A A 1 + c =A Ac. Thus, we can suppose that all r i s belong to N. Summarizing, we obtain the decompositions of A A -bimodules R = A R 0 R 1 R m 1 R 0 = A R 0 R i N 2b where, for 0 <i<m, we denote R i = R i. (Observe that R 0 is an A A subbimodule, since it is a left A-submodule and A is in the center of R 0.) We will view all indices modulo m. Denote by R A r r i R R i r r i the respective projections, so that, for every r R, r = r + r 0 + +r m 1. Let C be a cyclic group of order m on one generator c, let S be the skew group ring on C over A defined by the relations ca = ϕ a c, a A, and let P = S A R be the coproduct ( free product ) of S and R with amalgamation of the common subring A. By [1, Corollary 8.1], the A A -bimodule decomposition (2) of R and the A A -bimodule decomposition S = A Ac Ac 2 Ac m 1 imply the following structure of P: The ring P is the direct sum of S and the A A -bimodules c i 0 R j0 R j1 c i k R jk where k 0, 0 <i l <mfor l = 1 2 k, 0 i 0 i k+1 <m, and 0 j l <mfor l = 0 1 k. The above summand has the structure of additive group R j0 A A R jk with the lth A twisting the action of elements of A by ϕ i l (i.e., r jl 1 ar jl = r jl 1 ϕ i l a rjl for a A, r jl 1 R jl 1, and r jl R jl ), and the A-bimodule structure of the whole tensor product likewise twisted via ϕ i 0 on the left and via ϕ i k+1 on the right. The multiplication in P is given by the rules r j c i c i r j = r j c i+i r j if i + i 0 3

representability of a finite local ring 423 r j c i c i r j = r j r j if i + i = 0 and j + j 0 r j c i c i r j = r j r j 0 + ( r j r j ) if i + i = 0 and j + j = 0 where the product ( r j r j ) can be calculated with the same rules by induction. It is easy to verify that the ring P is /m -graded: the degree of the term (3) is j 0 + +j k i 0 i k+1, so that ax i = x i ϕ i a for every a A and every x i P homogeneous of degree i. In particular, R is a /m -graded subring of P. Denote by Q the centralizer of A in P. As is easy to see c i R i j c j Q (in fact Q is generated by all the c i R i j c j s and A). There is a standard embedding of a /m -graded ring into the m m- matrices over it, ψ P Matr m P x x i j ij where x i stands for the homogeneous component of x of degree i. Let T = diag c 0 c m 1. It follows from the rule ax i = x i ϕ i a that the embedding P Matr m P, x Tψ x T 1, maps P into Matr m Q. Consequently, we obtain the embedding R Matr m Q r c i r i j c j + δ ij ϕ i r ij where δ ij is the Kronecker delta. Obviously, A is in the center of Q. 4. THE EMBEDDING Denote by I the ideal of P( generated by all R i. Here we will prove that, for M large enough, I M A + c i R i j c )= j 0, which thus yields the i j embedding R Matr m R, where R = Q/ I M Q is a finite local ring such that the subring à constructed from this ring R as A was constructed from R lies in the center of R. For some q>0, N q+1 = 0. We define R l i pa + R i for l 1by induction on l>0: For l = 1, put R 1 i = R i. For l>1, put R l i = α+β=i R l 1 α R 1 β. Clearly, R l i is an A A -subbimodule, R l 1 i 1 R l 2 i 2 R l 1+l 2 i 1 +i 2, and R l i N l. In particular, R l i = 0 for l>q. Now R l i R i if i 0. For i = 0, let r R l 0. We know that r = r + r 0 with r 0 R 0 N, r r 0 N, and r A; therefore r A N pa and thus R l 0 pa + R 0.

424 a. z. anan in It is clear that I M is the sum of c i 0 R l 1 j 0 R l 2 j 1 c i k R l k j k 4 where k 1, i 1 i k 0, q l 1 l k 1, and l 1 + +l k M (in particular, k M/q). We will reduce (4) to the canonical form (3) eliminating the factors R l t j t by using the inclusion R l t j t pa + R jt, c i 0 R l 1 j 0 R l 2 j 1 c i k R l k j k c i 0 pa + R j0 pa + R j1 c i2 c i k pa + R jk c i 0 R j0 R j1 c i k R jk + pi k 1 + p 2 I k 2 + +p n 1 I k n+1 c i 0 R j0 R j1 c i k R jk + pi k n c i 0 R j0 R j1 c i k R jk + pi M/q n since p n = 0. Note that the term c i 0 Rj0 Rj1 c i k Rjk is in the canonical form (3). Now we can apply the same procedure to I M/q n. Since p n = 0, for M large enough, we arrive finally at the canonical form (3) with k 2. This yields the fact needed. Proof of the Theorem ( the tensor division ). Now we can assume that R is an A-algebra and has a faithful finite (left) R-module M (for instance, M = R). Using induction on card M, we will show that R R p i1 i n End A M, where the ring R p i1 i n is similar to the ring R p i1 i n described in the introduction. The only difference between R p i1 i n and R p i1 i n is that, for the latter, the ring A is used in place of /p n. Let M = v M pv = 0. Since M is an R-submodule, we have the homomorphism R End A M/M. The image is a finite local A-algebra and we can apply the induction hypothesis to it. Thus, M/M has a basis b 1 b r over A such that M/M = r α=1 Ab α(as an A-module), every b α has additive order p k α,1 k α n (with k α k β if α β), and, for every c R, r cb α = a βα c b β 1 α r β=1 where a βα c A, all a αα c s have the same image in F = R/N R as c, and a βα c pa if β<αand k β =k α. Denote by b α M some preimage of b α. The sum S = r α=1 Ab α is direct: If r α=1 a α b α = 0, a α A, then a α b α = 0, hence a α = p k α d α for some d α A and every α. Now p r α=1 p k α 1 d α b α = 0, so

representability of a finite local ring 425 rα=1 p k α 1 d α b α M, which implies p k α 1 d α b α = 0 and d α = pf α, f α A. Since a α = p k α +1 f α and p k α +1 b α = 0, the fact follows. Moreover, we have proved that the additive order of b α is p k α +1. As an A-module, M is simply an F-vector space (recall that F = A/pA). Clearly, M = M S = r α=1 Ap k α b α and the p k α b α s form an F-basis for M. Note that M is an R-submodule: For c R, cb α = r a βα c b β + q α c β=1 where q α c M. Since k α 1, we obtain c p k α b α = r β=1 p k α a βα c b β S. Obviously, c p k α b α M. In other words, RM M. The R-module M is in fact an R/p R-module and R/p R in its turn is a local finite-dimensional F-algebra. As is known, there is an F-basis for M /M such that R/p R acts on M /M in a lower triangular way with respect to this basis. Denote by b r+1 b r+s some preimage of this basis. Now, for c R and 1 α r, we can write q α c as q α c = r g βα c p k β b β + β=1 r+s γ=r+1 where g βα c a γα c A. Forr<ϑ r + s, we have cb ϑ = r a βϑ c pk β b β + β=1 r+s γ=r+1 a γα c b γ a γϑ c b γ where a βϑ c a γϑ c A, a γϑ c =0 for r<γ<ϑ r + s, and all a ϑϑ c s have the same image in F = R/N R as c. For 1 α β r<ϑ r + s, denote a βα c =a βα c +p k β g βα c and a βϑ = a βϑ c pk β. For 1 π r + s, denote by p k π the additive order of b π. (Thus, for 1 π r, we obtain k π =k π +1.) It is straightforward that k π k δ if π δ. Summarizing, we obtain: For all c R and 1 π r + s r+s cb π = a δπ c b δ δ=1 where a δπ c A, all a ππ c s have the same image in F = R/N R as c, and a δπ c pa if δ<πand k δ =k π. The rest is clear: R R p i1 i n, M = A B (the finite abelian p-group defined in the introduction), and R p i1 i n = A R p i1 i n.

426 a. z. anan in 5. REPRESENTABILITY OF R p 1 0 0 1 Note that the ring R p 1 0 0 1, n > 1, is generated by 1, e n1 Hom /p /p n induced by,1 p n 1, and e 1n Hom /p n /p induced by, 1 1, which satisfy the relations p n = 0 pe n1 = 0 pe 1n = 0 e 2 n1 = 0 e2 1n = 0 e 1ne n1 = 0 e n1 e 1n = p n 1 Conversely, if a ring R is generated by e n1, e 1n, and 1 which satisfy the above presented relations and p n 1 0, then R is isomorphic to R p 1 0 0 1 : Every element x R can be written as a + be n1 + ce 1n with a /p n and b c /p. Ifx = 0, then 0 = px = pa and a = p n 1 a, a. Therefore, 0 = xe 1n = bp n 1 and 0 = e n1 x = cp n 1, which implies b = c = 0 and, hence, a = 0. Thus, R has p n+2 elements like R p 1 0 0 1. Suppose that R is a subring of Matr m D (with the same identity), where D is a commutative ring with identity. Then 0 = tr e 1n e n1 =tr e n1 e 1n =mp n 1 and m is divisible by p. Denote by D 0 the commutative /p -algebra (without identity) with the basis b 1 b 2 b p 1 b b z and the multiplication given by the rule b b = z b 2 i = z i = 1 2 p 1 (all the others elements in the basis have zero product). Clearly, D 3 0 = 0 and D 0 is associative. Consider D 0 as a /p n -algebra and adjoin a formal identity to it. Then identify the elements p n 1 and z in the algebra D 1 = /p n 1 + D 0 to obtain ( D = /p n 1 + /p b 1 + + /p b p 1 )/ + /p b + /p b + /p z Ideal z p n 1 Since Ideal z p n 1 = /p z p n 1, we conclude that p n 1 0 in D. In the ring D, we have b 2 1 = pn 1, b 2 2 = pn 1,, b 2 p 1 = pn 1, b b = p n 1, and b 2 1 + b2 2 + +b2 p 1 + b b = pp n 1 = 0; therefore, the matrices 0 0 b 1 0 0 b 0 0 0 0 2 e n1 = and e 1n = 0 0 b 0 0 0 0 p 1 0 0 b b 1 b 2 b p 1 b satisfy the above relations. At the moment I do not know whether the ring R 2 0 1 1 (of 256 elements) is representable.

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