A note on some properties of finite rings

Abstract

PROOF. Let R have *(k). We assert that R has a prime power number of elements. If not, say o(R) = ab with (a, b) = 1, then A = { r I ar = 0 } and B = { r I br = 0 } are two ideals of R such that A4pB and BgA which contradicts *(k). Thus o(R) =pc for some prime p. We assume k > 1 and choose xCR, xEER2. Then the subring (R2, x) of R generated by R2 and x is an ideal properly containing R2 whence (R2, x) =R. This gives R=(R2, x)a=(R,+l, Xa). Taking s=k-1, k-2, ... , we find that R is the image of zI [z] (I the ring of rational integers, z an indeterminate) by the homomorphism 4 which sends z into x. Now we claim that pxER2. Indeed, otherwise we should have (R2, px) =R whence there exists an integer s such that x-spxER2 which gives xk 1 = spxk-l = . . . = sapaxk-1 =0 whence Rk-1 = (0), a contradiction. Thus px = a2x2+a3x+ * * * +ak_lxk-1 with the ai rational integers, so that if we putf(z) =pz-a2Z2-a3z3. . .*ak-1Zk-1, the ideal (Zk, f(Z)) is contained in the kernel of 4. Conversely, every element of the kernel of 4 is congruent modulo (Zk, f(Z)) to a polynomial of the form blz+ * * * +bk_lzk-l with 0<?bi<p. If b1z0, then bix is in R2 which is impossible. Similarly each bi = 0 for 1 < i < k-I so that the kernel of q5 is (zk,f(z)), i.e. R--zI[Z]/(Zk,f(Z)). Conversely let J be any ideal of zI [z]/(Zk, f(z)) where f(z) =pz -a2z2 * * *-ak_lzk-l with 0< ai <p and let z denote the coset of z. Every element of J has the form b,2+ * * * +bk+lk-l with the bi