Quotient ring of Gaussian integers $\mathbb{Z}[i]/(a+bi)$ when $a$ and $b$ are NOT coprime

Solution 1:

The best approach is to recall that $\mathbb{Z}[i]$ is a PID (Principal Ideal Domain), which is in fact an Euclidean domain with respect to its usual norm.

Once you notice this, you will realize that your approach using the Chinese Remainder Theorem is the correct one. The only problem is that you are factoring over $\mathbb{Z}$ instead that over $\mathbb{Z}[i]$. In this way, take $z\in\mathbb{Z}[i]$, factor it over $\mathbb{Z}[i]$ as $\prod q_k^{r_k}$ and you will obtain by the CRT that $$\mathbb{Z}[i]/(z)\cong \prod_k\mathbb{Z}[i]/(q_k^{r_k})$$For example, in your example with $13(2+3i)$, write it as $(2+3i)^2(2-3i)$ and so you obtain $$\mathbb{Z}[i]/(13(2+3i))\cong \mathbb{Z}[i]/(2+3i)^2\times \mathbb{Z}[i]/(2-3i)$$

Now, the only problem is studying which are the primes of $\mathbb{Z}[i]$ and determining the structure of $\mathbb{Z}[i]/(q^r)$ for $q$ prime in $\mathbb{Z}[i]$. The first question is can be answered using the fact that $z$ is prime in $\mathbb{Z}[i]$ iff $\mathbb{Z}[i]/(z)$ is a field (you have worked which primes of $\mathbb{Z}$ are primes of $\mathbb{Z}[i]$ without noticing and I let the proof to you), so we obtain:

The primes of $\mathbb{Z}[i]$ are of the form:

1. $(1+i)$. Up to multiplication by units, $1+i$ is the only prime associated to $2$.
2. $p\in \mathbb{Z}$ prime integer with $p \equiv 3$ (mod 4). Up to multiplication by units, $p$ is the only prime of this form for a given integer prime $p \equiv 3$ (mod 4).
3. $q=(x+iy)\in\mathbb{Z}[i]$ with $q\overline{q}$ prime integer. Up to multiplication by units, $q$ and $\overline{q}=(x-iy)$ are the only primes of this form for a given integer prime $q\overline{q}\equiv 1$ (mod 4).

Once this is known, one should determine the structure of $\mathbb{Z}[i]/(q^n)$ for each one of these primes. We only have to distinguish three cases (we just use the the isomorphism theorems):

1. $q=1+i$. When $r=2s$ is even, we reduce to $$\mathbb{Z}/(2^s)[X]/(X^2+1)$$ that can be realized as the matrix subalgebra of $M_2(\mathbb{Z}/(2^s))$ given by $$\mathbb{Z}/(2^s)\left[\begin{pmatrix}0&-1\\1&0\end{pmatrix}\right]$$ Note that for $s=1$, this is just $\mathbb{Z}/(4)$. When $r=2s-1$ is odd, we reduce to $$\mathbb{Z}/(2^s)[X]/(X^2+1,2^{s-1}(X+1))$$ which the better realization I can think of is a quotient of the subalgebra of $M_2(\mathbb{Z}/(2^s))$ $$\mathbb{Z}/(2^s)\left[\begin{pmatrix}0&-1\\0&0\end{pmatrix}\right]$$ by the ideal generate by $$\begin{pmatrix}2^{s-1}&2^{s-1}\\2^{s-1}&2^{s-1}\end{pmatrix}$$

2. $q=p$ is an integer prime. Then as noted by you, we reduce to $$\mathbb{Z}/(p^r)[X]/(X^2+1)$$ A concrete realization can be obtained by considering the matrix subalgebra of $M_2(\mathbb{Z}/(p^r))$ given by $$\mathbb{Z}/(p^r)\left[\begin{pmatrix}0&-1\\1&0\end{pmatrix}\right]$$

3. $q=a+bi$ is not an integer prime. In this case, it should be noted that for $(a+bi)^n=a_n+b_ni$, $a_n$ and $b_n$ should be coprime because otherwise it would be divisible by a prime non equivalent to $a+bi$ violating the unique factorization. Hence $$\mathbb{Z}[i]/(q^r)\cong \mathbb{Z}/((q\overline{q})^r)$$ in this case by your cited result.

This settles the question in general and in a completely satisfactory way in many case. In your considered example, we obtain $\mathbb{Z}/(13)\times \mathbb{Z}/(13^2)$. However, maybe there are better presentations for some of the above cases. In any case, the strategy I give works for any PID.