Proving that the ring $\mathbb Z[i]/\langle{p}\rangle$ is a field with $p^2$ elements.

Question

Can someone help me with this problem?

Let $p\in\mathbb{Z}$ be a prime of the form $4k+3$. Prove that the ring $\mathbb Z[i]/\langle{p}\rangle$ is a field with $p^2$ elements.

Answer 1

Hint : You can show that : $$\mathbb Z[i]/\langle{p}\rangle \cong\mathbb Z[x]/\langle x^2+1,p\rangle\cong (\mathbb Z/p\mathbb Z)[i]$$

Answer 2

Step 1: Why is it a field? Notice that $2=1+1$ and every number that is $1\pmod{4}$ is the sum of two squares. If $p=2$ or if $p=4k+1$ then the conclusion is false!

Step 2: Prove it has at most $p^2$ elements, because every number in $\mathbb{Z}[i]$ is equivalent to a number in the set $S:=\{a+bi:a,b\in\{0,1,\ldots, p-1\}\}$.

Step 3: Show that no two elements of $S$ get mapped to the same place by the map $\varphi:x\to x +\langle p\rangle$, which means that every element of $S$ is a unique element of $\mathbb{Z}[i]/\langle p\rangle$, and so that set has at least $p^2$ elements.

Thus it follows that it is a field with exactly $p^2$ elements.

If any of these steps prove problematic, do comment with what you've tried and I'll give you some advice.