Multiplication in a quotient ring of the Gaussian integers

Question

In order to prove that I={a+bi: a and b are even} is not a maximal ideal of Z[i], we need to prove that Z[i]/I is not a field. However, I am stuck at the step (1+i+I)*(1+i+I)=0 + I which indicates it has zero divisor. Can anyone help me understand how the above result has come about?

Answer 1

Are you confused as to why $(1+i+I)^2=I$, or do you not understand why it's true while knowing that you need to prove it? The existence of a zero divisor shows that $Z[i]/I$ cannot be a field, since fields are integral domains, so the fact that you stated completes the proof.

If you're confused about why $(1+i+I)^2=0$, recall that for all $c, d \in Z[i]$, $(c+I) + (d+I) = (c+d) + I$ and $(c+I)(d+I)=(cd+I)$.