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I know that $I:=\{a+ib \in \mathbb Z[i] : 2\mid a-b\}$ is a maximal ideal of $\mathbb Z[i]$. My question is: what is the number of elements of $\mathbb Z[i] /I $?

I am totally stuck. Please help .

(Since $\mathbb Z[i]$ is a euclidean domain, $I$ must be generated by some element $\alpha \in \mathbb Z[i]$, then $|\mathbb Z[i] /I|=|\alpha|^2$ but I have no idea what this $\alpha$ should be here)

user26857
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1 Answers1

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If $z=a+bi$ does not fulfill $2\mid a-b$, then $z-1$ does. Hence the order of the quotient is $2$. (The generator of the ideal is $1+i$).

kahen
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MooS
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