PRIMARY RINGS IN WHICH THE PRODUCT OF ANY TWO ZERO DIVISORS OF A RING IS IN ITS COEFFICIENT SUBRING

According to general terminology, a ring R is completely primary if its set of zero divisors J forms an ideal. Let R be a finite completely primary ring. It is easy to establish that J is the unique maximal ideal of R and R has a coefficient subring S (i.e. R/J isomorphic to S/pS) which is a Galois ring. In this paper we give the construction of finite completely primary rings in which the product of any two zero divisors is in S and determine their enumeration. We also show that finite rings in which the product of any two zero divisors is a power of a fixed prime p are completely primary rings with either J2=0 or their coefficient subring is Z2n with n=2 or 3. A special case of these rings is the class of finite rings, studied in [2], in which the product of any two zero divisors is zero.

positive integers m,n and with l<n<m.If n=m, then R is of the lbrm n[x]/(g) and R=Zpn[a], where Zn is the ring of integers modulo p", g is monic polynomial over Zn and irreducible modulo p and a is an element of R of multiplicative order p'-1.In this case Aut R, the automorphism group of R, is cyclic and is of order r.These rings are uniquely determined by the triplet p, n, they are called Galois rings and are denoted by GR(p",r).
Let R be a finite completely primary ring.It is already known that any two coefficient subrings of R are conjugate (cf.[4]).Also if S is a coefficient subring of R; then there exist rt, min J and o', m in Aut S such that R=S Srt (as S-modules) and rtr =r rt =1 for all in S and for all i=l m. (This result is a direct consequence of theorems 2-2 and 2-4 in [6]).
Moreover the automorphisms o, o,are uniquely determined by R and S (cf.[2]).Thus we call , , the associated automorphisms of R and the automorphism r, is called the automorphism associated with .T hroughout this paper, for a given finite completely primary nng R, we denote by T, the set of all (S, rt, ,) which come from the above description.In addition, let F=R/J, and let F" and G denote the multiplicative group of units ofF and R respectively.
CONSTRUCTION A" Let S bc a Galois r=ng of the form GR(p",r) and F be S/pS.Also assume that s, t, w. m arc non-negative integers such that m=s+t+w and suppose that is an injcctive function from {s+l, s+t} to {s+! m }. ()n the additive group R=SP', define the multiplication as follows" (r )(So, S s (roSo+pn-U s o 1' where u, arc elements of F', o, automorphisms of F such that ,2=idF lot" all i=l s and f,l=," for all =s+ s+t and r" is the image of under the canonical homomorphism from S to F.
It can hc easily verified that R is a ring and it is commutative if and only if o',=id for all i= m.
THEOREM 1-Let R be a finite completely primary ring.Then the product of any two zero divisors is an clement of its coefficient subring S if and if it is one of the rings given by construction A.
PROOF: Let R be a finite completely primary ring with .F contained in S and (S, :/ ft.,') be an clement of T,.Since SS:,'=() and the product of any two zero divisors is in S, prt,'=0 for all i=l m.
But n:,'rt,' is an clement of pS; thus 'rt,' is an element of p""S for all i,j=! m.Suppose n:,'rt,','n,' are non-zero elements of pS with j:k.Then rt,':,'S=K'rt,'S=p-'S and we get 'rt,'=rt,'' where ot is an clement of <a>.Thus rt,'-rt'c is an element of ann rt,' and subsequently it is contained in pS :m Srt' h.h=l,h l,k This implies that rt,' is an element of pS( m Srt'h, h=l,h which contradicts the assumption that (S, rt,', rt,') is an element of T,.Therefore for all i=l m, either r'rt,' is zero for all j=l m or r':,' is non-zero for only one j=l m.Similarly, we prove that for all i= m, either t,'r' is zero for all j= m or rq':,' is non-zero for exactly one j.Assume w is the number of ' such that n,'rq' is zero for all j= m and % is the number o,f other rt,'.Let us reindex n:,', ...m:m' in such a way that for each i= k there exists only one j= m with :,'r'=p"-'a,, ,where is an element of <a>, and let f be the function from } to m} determined by f(i)=j.Clearly is injective.Also, for all i= m n-o, (f (,) n-o, of (,) p ao,f(i =/l:i/l:f(i) a a 71:/t:f(i) p a if(t)' which implies that cy,,=o',-' for all i= k.Let s be the number of in k such that f(i)=i and be ,-s.We reindex rt,', rt' such that f(i)=i for all i= and suppose o,l,l=u, for all i= s.Put r,.,,=r' for all i= s and rt=rt' for all i=s+l m, where if e is in the image of f, say e=f(i), then a =Fl( e .=fh-(,) fh(i j-1 )g(h), where g(h)= (-1) J+h+l I-]c; and =1 otherwise.
It is easy to see that (S. rt, rtm) is an element of T, with rt,rt,,,=lY" for all i=s+l ..Nowi t follows that R is isomorphic to one of the rings given by construction A.
The converse is easy to check.

FINITE RINGS IN WHICH THE PRODUCT OF ANY TWO ZERO DIVISORS IS A POWER
OF A FIXED PRIME.
LEMMA 1" Let R be a finite ring of characteristic p" in which the product of any two zero divisors is a power of p. Then R is completely primary.
PROOF: Let x and y be zero divisors in R. To show that x+y is a zero divisor, we can use the distributive properties to write (x+y) as a sum of products, each containing 2n factors (which are x's or y's).Since each xy or yx is of the form p,, each of the summands of (x+y) is product of the form p'p...pn=0.Therefore x+y is zero divisor and hence R is completely primary.
PROPOSITION 1' Let R be a finite ring of characteristic p" in which the product of any two zero divisors is a power of p. Then R is completely primary with either J-'=0 or the coefficient subring of R is Z.n, where n=2,3.
PROOF: Suppose JZ-4:0; then there exist x,y in with xy=p0.Since for any unit o in R, otx is a zero divisor, we have 03tx)y=p.".On the other hand, xy=p implies that ct, xy=otp and so otp-p".Without loss of generality, we can assume la>__X and deduce that p(ot-p"-)--0.Since p-0, we have ot-p "-x is an element of J.If ta this would imply that ot is an element of J which is not possible; hence la=k and c is an element of l+J.However ot is an arbitrary unit and therefore G,= l+J.Since R=G,J (disjoint union), we have IRI IG,I+IJI I1 +Jl+lJI 21JI Thus 2 divides IRI and consequently char R is 2".If n>4, then 2,6 are zero divisors of R with (2)(6)=12 which is not a power of 2. Also n= implies that J2--0.Thus n=2,3.Let S=7_an[a] be a coefficient subring of R, where a is an element of R of multiplicative order 2'-1 and let x,y be elements of J with xy=2X0.
4. THE ENUMERATION.NOTATIONS: Retaining the above notations, assume k is the number of elements in {s+t+l m} which are not in the image of f.Let all the t, in which is not in the image of f be renamed as 0, 0k and assume x, % are the respective automorphisms associated with them.Thus we suppose that (S, n,, n, 0, 00 is an element of TR and , .,-k,x, "q, are the automophisms associated with n, r.,., 0. 0k respectively.We call (p, n, r, s, t, k, m, f) the invariants of R. In what follows we shall use these notations.
PROPOSITION 2: Let R be a finite completely primary ring in which the product of any two zero divisors is an element of its coefficient subring.Then (S, n,', r_', 0,', 0k' is an element of TR if and only if 0 =,la.O +p o. = , (after possi bl e rei ndexi ng), (after possi bl e rei ndexi ng ).
where ., are elements of F" and , , ,, , la, are elements of F such that , is zero if , is not the trivial automorphism and is zero if x, is not the trivial automorphism.
PROOF: Using the fact that K'a=a , K', we deduce that for all i= m-k, we have where )%,,, and , are elements ofF such that , is zero if o', is not the trivial automorphism.For all i= s+t, lann ,, I=IJI/tY and so rt,'r,=0 for all but one j, say j=h.Thus 'rt,, is a non-zero element of p"-'S, 'r=0 for all j#f(i) and o,=(of,)"=o,.Thus ,=0 for all except j=h.Let us put ),,,--.k, and redenotc ft,' by r'.Therefore We can prove the rest of the proposition by using a similar argument.
THEOREM 2: Let R,R' be finite completely primary rings constructed over the same coefficient subring S and having the same associated automorphisms.Suppose that (J(R)) and (J(R')) are contained in S and R,R' have the same invariants p, n, r, s, t, k, m, f.Also suppose that (S, :,', rr.', 0,', 0) is an element of TR. with rt'-" =p"-'v, for all i=l s.Then R is isomorphic to R' if and only if there exist isomorphisms , from SSr, to SS' (after possible reindexing) for all i=l m-k such that where k, are elements of F" such that h and 'hf(h) =I for all i= s and h=s+l s+t, o<j<r.
PROOF: Let be an isomorphism from R to R'.Then (S) is a coefficient subring of R' and hence there exists a unit x in R' such that (S)=xSx".Let be the composition of the conjugation by x and .
In view of the last theorem the required number is =1 =s+l =1 COROLLARY: The finite ring of characteristic pn in which the product of any two zero divisors ts a power of p is completely determined by its associated automorphisms and its invariants.
REMARK: Let R be a finite ring which has a p-ring as its coefficient subring and the product of any two zero divisors of R is in its coefficient subring.By using similar argument as in the proof of lemma 1, ne can prove that R is completely primary.Thus the construction and the enumeration of such rings is determined.
ACKNOWLEDEMENT: The author would like to thank B. Corbas for his suggestions which enabled the author to make some improvements in the contents of the paper.