Is $\mathbb Z[i]/(q) \cong \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$? Yes/No I think  this answer  is wrong.
Let $q$ be a prime congruent to $3$ mod $4$, prove the quotient ring $\mathbb{Z}[i]/(q)$ is a field with $q^2$ elements
Answer
It is written that

I think computing the quotient (as sugggested in the other answer) is the easiest way to solve this. Nevertheless here is an alternative:
If you are fine about $\mathbb Z[i]/(q)$ being a field, you can find the number of elements by looking only at the abelian group structure.
As an $\mathbb Z[i]$ is generated by $1,i$, hence so is $G := \mathbb Z[i]/(q)$. $1$ and $i$ have order $q$ in $\mathbb Z[i]/(q)$. So we have two subgroups $A = \langle 1 \rangle, B =\langle i \rangle$ of order $q$ with $G=AB$ and $A \cap B=0$ (since they have both prime order, they are either equal or have trivial intersection), which implies $G = A \times B \cong \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$.

Why is this answer wrong ?
Here  $A \cap  B = q$  but in answer it is written that $A \cap B=0$ that is totally wrong
Also, $\gcd(q,q)=q \neq 1 \implies \mathbb Z[i]/(q) \not \simeq \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$
$\mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$ is not integral  domain
My thinking:
$\mathbb Z[i]/(q) \not \simeq \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$
Because take  $f:\mathbb Z[i] \to \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$  defined by $(a+bi) \to (a\mod q,b\mod q)$ is not  a homomorphism with  kernal $(q)$
 A: If the isomorphism is an isomorphism of additive groups then the statement is true. Not necessarily if this is supposed to be an isomorphism of rings though, depending on $q \bmod 4$.
Since we are in a PID The ideal $(q)$ is prime iff $q$ is prime iff $q$ is irreducible iff $(q)$ is maximal etc.
This is known to happen iff $q \equiv 3 \bmod 4$. Otherwise $q = (x+iy)(x-iy)$ for some integers $x$ and $y$.
When we quotient by $(q)$ this will give a field in the case where $(q)$ is maximal, and will fail to be an integral domain (it will have zero divisors) in the case where $q$ is reducible.
There is only one unique field of order $q^2$, and its multiplicative group is cyclic of order $q^2-1$ (we exclude only zero)
Edit - Thanks for the MathJax tip. Now that I slightly know what I'm doing I'll add a little more:
Additionally, if we want to see that there are exactly $q^2$ many elements in the quotient, observe that a set of (additive) coset representatives could be $\{ a+bi \in \mathbb{Z}[i] : 0 \leq a,b < q \} $, as any element of $\mathbb{Z}[i]$ differs from an element of this set by something in $q\mathbb{Z}[i]$, whereas elements of this set never differ from each other by something in $q\mathbb{Z}[i]$.
