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

Question

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

The field portion I understand. $\mathbb{Z}[i]$ is a PID and because $q$ is congruent to 3 mod 4 it is irreducible thus prime. and hence the ideal $(q)$ is prime and therefore is maximal because $\mathbb{Z}[i]$ is a pid. Thus $\mathbb{Z}[i]/(q)$ is a field

I am not totally clear on the number of elements being $q^2$. I think perhaps the division algorithm would help but I am still not totally clear on how to use it in a case like this.

Answer 1

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 $G = A \times B \cong \mathbb Z/q\mathbb Z \times \mathbb Z/q\mathbb Z$.

Answer 2

You can get this field from $\mathbb{Z}[x]$ by first killing $x^2+1$ and then killing $q$, as you have shown above. But you could also get there by killing $q$ first, and then $x^2+1$. In this way, we get that the field is isomorphic to $F_p[x]$ mod $x^2+1$. How many elements does this field have?