Prove that $\Bbb{Z}[i]/J$ is not isomorphic to $\Bbb{Z}$ 
If $J$ is a prime ideal, then prove that $\Bbb{Z}[i]/J$ is not isomorphic to $\Bbb{Z}$.

My attempt: Since $\mathbb{Z}[i]$ is a PID, then $J$ is a maximal ideal, then  $\mathbb{Z}[i]/J$ is a field. Hence the result follows. I just want some alternative approaches.
 A: The ideal $J=0$ clearly does not work, so we can assume that there exists an element $a+bi\neq0$ in $J$. But then the integer $m=(a+bi)(a-bi)=a^2+b^2>0$ is an element of $J$. Therefore the additive order of $1$ in the quotient $\Bbb{Z}[i]/J$ is a factor of $m$. Most notably it is finite. Therefore the quotient ring is not isomorphic to $\Bbb{Z}$.

Looks like I didn't need the assumption that $J$ is a prime ideal :-)
A: Suppose $\Bbb{Z}[i]/J$ is isomorphic to $\Bbb{Z}$ for some ideal $J\subset\Bbb{Z}[i]$. Since $\Bbb{Z}[i]$ is a PID there are $a,b\in\Bbb{Z}$ such that $J=(a+bi)$. Then  $(a+bi)(a-bi)=a^2+b^2\in J\cap\Bbb{Z}$, from which it follows that $a^2+b^2=0$. But then $a+bi=0$ so $\Bbb{Z}[i]/J$ is not isomorphic to $\Bbb{Z}$, a contradiction.
A: Consider the unique ring homomorphism $\chi\colon\mathbb{Z}\to\mathbb{Z}[i]/J$, $n\mapsto n+J$. 
If $J\ne\{0\}$, $\chi$ is not an isomorphism: it is not injective, as $x\in J$, $x\ne0$, implies $x\bar{x}\in\ker\chi$.
If $J=\{0\}$, $\chi$ is not an isomorphism: it is not surjective, as $i\notin\chi(\mathbb{Z})$.
