the understanding of $Z[i]/(2)$

Question

I have to proof that $Z[i]/(2)$ is not a field. Am I right thinking that $Z[i]/(2)=\{0, 1, i, 1+i\}$? Thanks!

Answer 1

Yes you are right, in the sense that one can see what you mean. But this is not quite mathematically correct yet (what exactly is the induced structure on the set that you have written?).

Hint: a way to prove that $\Bbb Z[i]/(2)$ is a field is to prove that $(2)$ is maximal.

Answer 2

You can certainly observe that $(1+i+(2))^2=1+2i-1+(2)=(2)$, so the ring would have a nonzero nilpotent element, provided you show that $$ 1+i\notin(2) $$ However, if $2z=1+i$, then $|2z|^2=|1+i|^2$, that is, $2|z|^2=1$, a contradiction because $|z|^2$ is an integer.

In another way, you can observe that $2=(1+i)(1-i)$ and neither factor is invertible. Thus $2$ is not irreducible, hence not prime, so $(2)$ is not a prime ideal.

Answer 3

$1+i$ is a zero divisor. So it's not an integral domain, hence not a field.

Btw: $(2)\subsetneq (2,1+i)=(1+i)\subsetneq\mathbb Z[i]$, hence $(2)$ isn't maximal...

Answer 4

$\mathbb Z[i]\cong \mathbb Z[x]/(x^2+1) $, so $\mathbb Z[i]/(2) \cong \mathbb Z[x]/(2,x^2+1) \cong \mathbb Z_2[x]/(x^2+1)\cong \mathbb Z_2[x]/(x+1)(x-1)\cong \mathbb Z_2 \times \mathbb Z_2 $