Two non isomorphic rings with $\mathbb{Z}_5 \times \mathbb{Z}_5 $ as their additive group.

Question

I am trying to figure out a method for finding two non isomorphic rings of which their additive groups are isomorphic to $\mathbb{Z}_5\times \mathbb{Z}_5$

Answer 1

1. Try a cartesian product of two (obvious) rings.
2. Try a finite field of an appropriate order.

To make sure those are not isomorphic, consider the number of invertible elements.

Answer 2

There's a way to do this without knowledge of finite fields, but it assumes that you do NOT require rings to have a multiplicative identity. For one, take $\mathbb{Z}_5 \times \mathbb{Z}_5$ and give it the usual multiplication of congruence classes (the direct product of the "usual" ring $\mathbb{Z}_5$ against itself). Next, take the same underlying group but give it the trivial multiplication $ab=0$ for all $a, b \in \mathbb{Z}_5 \times \mathbb{Z}_5$. This is a ring, not iso to the first one (that one had an identity, this one doesn't).

Answer 3

More generally, let $K$ be a field and consider the rings $$ A = K \times K, \qquad B = K[x]/(x^2) $$ Then their additive groups are both isomorphic to $K \times K$ (even as $K$-vector spaces), but $A$ is not isomorphic to $B$ because $B$ has non-zero nilpotent elements while $A$ does not.

Moreover, if there is an irreducible quadratic polynomial $p \in K[x]$, then $C=K[x]/(p)$ is a field and so is not isomorphic to either $A$ or $B$, which have nontrivial zero divisors.

Thus, for $K=\mathbb Z_5$, we get three non-isomorphic rings with isomorphic additive groups.