# Problem with Ring $\mathbb{Z}_p[i]$ and integral domains

Let $$\Bbb Z_p[i]:=\{a+bi\;:\; a,b \in \Bbb Z_p\,\,,\,\, i^2 = -1\}$$

-(a)Show that if $p$ is not prime, then $\mathbb{Z}_p[i]$ is not an integral domain.

-(b)Assume $p$ is prime. Show that every nonzero element in $\mathbb{Z}_p[i]$ is a unit if and only if $x^2+y^2$ is not equal to $0$ ($\bmod p$) for any pair of elements $x$ and $y$ in $\mathbb{Z}_p$.

(a)I think that I can prove the first part of this assignment. Let $p$ be not prime. Then there exist $x,y$ such that $p=xy$, where $1<x<p$ and $1<y<p$. Then $(x+0i)(y+0i)=xy=0$ in $\mathbb{Z}_p[i]$. Thus $(x+0i)(y+0i)=0$ in $\mathbb{Z}_p[i]$. Since none of $x+0i$ and $y+0i$ is equal to $0$ in $\mathbb{Z}_p[i]$, we have $\mathbb{Z}_p[i]$ is not an integral domain.

However, I don't know how to continue from here.

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(b) Is wrong - if $x=y=0$ in $\mathbb Z_p$ then $x^2+y^2=0$. So you want this theorem for all$x,y$ where both are not zero. Alternatively, you could say, "for all $x,y\in\mathbb Z_p, x^2+y^2=0 \iff x=y=0$" –  Thomas Andrews Sep 21 '12 at 12:42

Note that $(a+bi)(a-bi)=a^2+b^2$. If $a^2+b^2\equiv0\pmod p$, then $a+bi$ is not a unit. And if $a^2+b^2$ is not zero modulo $p$, then it's invertible modulo $p$, so $a+bi$ is a unit.
Hint  unit $\rm\alpha\in \Bbb Z_p[{\it i}]\iff$ unit $\alpha\bar \alpha\in \Bbb Z_p \iff \alpha\bar \alpha\not\equiv0,\:$ since $\rm\:\alpha\:|\:1\iff \alpha\bar\alpha\:|\:1\:$ via norms.
Remark $\$ The point of expressing it this way (vs. prior answer) is that it explicitly highlights how the norm, being a multiplicative homomorphism, must preserve purely multiplicative properties, such as the property of being a unit. That is the essence of the matter here (and throughout number theory), so it is well-worth explicit emphasis.