Ring Isomorphisms

Question

Why are $\mathbb{Z} [\sqrt{2}]$ and $\mathbb{Z} [ \sqrt {2} ] \times \mathbb{Z} [ \sqrt {2} ]$ not isomorphic to each other?

Put this question off for like a week while studying but I feel there is a very simple answer. I think it might have to do with commutativity.

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Answer 1

The former is a domain and the latter is not a domain.

Answer 2

Hint: $\Bbb Z[\sqrt{2}]$ is a domain.

Answer 3

If $\,R\ne 0\,$ then a ring hom from $\,R \times R\,$ to a domain maps the $\,3\,$ roots $\,(0,0),(1,1),(0,1)\,$ of $\,x^2 = x\,$ onto the $\,2\,$ roots $\,0,1\,$ in the domain, so the hom is not injective.