Can you help me solve iv) in the following problem?
I believe I'm supposed to use the results from i), ii) and iii).
But why can't I just say that $R/I$ is a finite ring from ii), and since $\varphi$ is surjective we have $\varphi(R/I)=R/J$ which must be finite ring, when $R/I $ is finite?
On
I'll assume you have already done (i) and (ii).
If $S$ is a subring of $R$ and $J$ is an ideal of $R$, then $J\cap S$ is an ideal of $S$.
In your case $S=\mathbb{Z}$, so $J\cap\mathbb{Z}=n\mathbb{Z}$ for some $n\ge0$.
Since $J\ne0$, take $a+bi\in J$, $a+bi\ne0$; then $(a+bi)(a-bi)=a^2+b^2\in J\cap\mathbb{Z}$ and $a^2+b^2\ne0$. Hence $n\ne0$. Let $I=nR$. Then $n\in I\cap(J\cap\mathbb{Z})$, so $n\in J$, which implies $I=nR\subseteq J$.
Now apply (ii) and (i).
To elaborate on my comment. We have an ideal $J \subset R = \mathbb{Z}[i]$, and we wish to construct another ideal $I \subset J$ of the form $I = nR$ for some $n \in \mathbb{Z}$.
We know $\mathbb{Z} \cap J$ is a nonzero ideal of $\mathbb{Z}$, hence is of the form $(n)$ for some $n \neq 0$.
By the above $n\mathbb{Z} \subset J$ so we know that $I := nR \subset J$ (you should convince yourself of this using the axioms of an ideal). But then we can use the fact that $R/I$ is finite and surjects onto $R/J$.
Note the difference with your comment - $J \cap \mathbb{Z}$ is not an ideal of $R$, you have to throw in some extra stuff.