Ring Isomorphism between integer ring and polynomial ring

Question

I want to check if $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})[x]/(x^2)$. To begin with, I know that both rings have the same cardinality, and I tried to explicitly map each element in $(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})$ to an element in $(\mathbb{Z}/2\mathbb{Z})[x]/(x^2)$. However, this didn't seem to work as the function $f$ I defined didn't satisfy $f(ab) = f(a)f(b)$ (due to $f[(0,1)(1,0)]$ not mapping to the right element). Is this enough to show that these rings aren't isomorphic?

Answer 1

As the comment on your post states: it is not enough to show that a particular map $f$ fails to be an isomorphism.

Typically, one shows that two rings (or more generally, two mathematical objects) fail to be isomorphic by showing that one ring has an "essential" property that the other ring does not have. In this case, $\Bbb Z_2[x]/(x^2)$ has a nilpotent element, but $\Bbb Z_2 \times \Bbb Z_2$ does not.

For a detailed proof, we might state the following:

Suppose (for the purpose of contradiction) that $f:\Bbb Z_2^2 \to \Bbb Z_2[x]/(x^2)$ is a ring isomorphism. There exists an element $a \in \Bbb Z_2^2$ such that $f(a) = x$. It follows that $$ f(a^2) = f(a)^2 = x^2 = 0. $$ Because $f$ is a ring isomorphism, we have $f(a^2) = 0 \implies a^2 = f^{-1}(0) = 0$. However, $a \in \Bbb Z_2^2$ can only satisfy $a^2 = 0$ if $a = 0$. Thus, we have $x = f(a) = f(0) = 0$, which is a contradiction.