Abstract
PROOF. Assume S----Z2, $ A~ ~ A1 is a member of R, and let ~ be an additive endomorphism of R. We will prove y must be a ring endomorphism of $. The map y will send A1 to A1 and will send Z~ ~ A2 to Z2, @ A 2 since mx = 0 implies that my(x) ~ O. It will then suffice to show that y restricted to Z~ ~B A~ is a ring endomorphism. Therefore we will assume A 1 = 0. Products in S are defined by xy = 2n-ix A 2n-ly where the symbol a A b is defined to be a if a = b and 0 ff a ---- 0 or b = 0. It is undefined otherwise. From the definition of a A b ~(a A b) = ~(a) A ~(b) whenever a A b is defined. Therefore
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