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Is there a ring homomorphism from $\mathbb{Z}[i]$ (Gaussian integers) onto $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$? or $\mathbb{Z}/4\mathbb{Z}$?

I know there are both 4 elements in those two rings. Not sure if they are onto homomorphisms though.

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A ring homomorphsism $A\to R\times S$ is just a pair of homomorphisms $A\to R$ and $A\to S$ (the two combine in a unique and obvious way). Any unitary ring morphism must map $1$ to $1$ (which does not imply it is a $1-1$ map ;-) so a morphism defined on $\def\Z{\Bbb Z}\Z[i]$ is determined by the image of$~i$, which must be a solution $x$ of $x^2+1=0$ in the destination ring. There are no such solutions in $\Z/4\Z$, which settles that case. In $\Z/2\Z$ there is a unique solution, so for a morphism $f:\Z[i]\to(\Z/2\Z)\times(\Z/2\Z)$ the two components $\Z[i]\to\Z/2\Z$ will be identical, and $f$ will fail to be surjective (it only maps to elements on the diagonal). So the answers are no and no.

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