Characterization of factor rings of Gaussian integers It was given in class as a example that $$\displaystyle {\mathbb{Z}[i]} \,/\,{(1+i)\mathbb{Z}[i]}\simeq \mathbb{Z}_2,$$
Then I was wondering if all the factor rings of $\mathbb{Z}[i]$ can be written in the form of $\mathbb{Z}_m$ (classification of factor rings of $\mathbb{Z}[i]$). Also I am not sure about if all the ideals of $\mathbb{Z}[i]$ are in the form of $r\cdot\mathbb{Z}[i]$, where $r$ is any element in $\mathbb{Z}[i]$.
Thanks in advance.
 A: $1+i$ is somewhat exceptional in this case since it is the only ramified prime. We have $(2)=(1+i)^2$ as ideals. However we can see your equation as $\mathbb{Z}/(2)=\mathbb{Z}[i]/(1+i)$. In general primes of the form $p=4n+1$ will factor into two distinct primes, $\mathfrak{p}_1$ and $\mathfrak{p}_2$ and we will have $ \mathbb{Z}/(p)=\mathbb{Z}[i]/\mathfrak{p}_i$. On the other hand primes $p=4n+3$ will remain prime and $\mathbb{Z}[i]/(p)$ will be a field extension of $\mathbb{Z}/(p)$ of degree two, called $\mathbb{F}_{p^2}$. All ideals in this ring are principal.
A: The ring $\mathbf{Z}[i] / 5\mathbf{Z}[i]$ is not of the form $\mathbf{Z} / m \mathbf{Z}$: it is, in fact, isomorphic to $\mathbf{Z}/5 \mathbf{Z} \times \mathbf{Z}/5 \mathbf{Z}$. One such isomorphism sends $i \to (2, -2)$.
The ring $\mathbf{Z}[i] / 2 \mathbf{Z}[i]$ is isomorphic to $(\mathbf{Z}/2\mathbf{Z})[x] / x^2$. One such isomorphism sends $i \to 1+x$.
The ring $\mathbf{Z}[i] / 3 \mathbf{Z}[i]$ is isomorphic to $(\mathbf{Z}/3\mathbf{Z})[t] / (t^2 + 1)$, a finite field of $9$ elements. One such isomorphism sends $i \to t$.
It is true that $\mathbf{Z}[i]$ is a principal ideal domain: every ideal can indeed be written in the form $r \mathbf{Z}[i]$.
I believe all of the quotients by primary ideals are of the following forms:


*

*$\mathbf{Z}[x] / (2^n, x^2)$

*$\mathbf{Z}[x] / (2^n, 2^{n-1} x, x^2)$

*$\mathbf{Z} / p^n$ where $p \equiv 1 \bmod 4$

*$\mathbf{Z}[x] / (p^n, x^2 + 1)$ where $p \equiv 3 \bmod 4 $


and so all nontrivial quotients of $\mathbf{Z}[i]$ are a product of finitely many rings of the above form, as determined from the prime factorization of the ideal you are modding out by (I am using the fact that $\mathbf{Z}[i]$ is a Dedekind domain when I say this). The product can only have one factor over $2$ and only one factor over primes that are $3 \bmod 4$, and can only have two factors over primes that are $1 \bmod 4$
