$p$ is a prime in $\mathbb{Z}[i]$ if and only if $-1$ is not a square modulo $p$, direct proof using $\mathbb{Z}[x]$?

Question

I'm trying to show directly that $p$ is prime in $\mathbb{Z}[i]$ if and only if $-1$ is not a square modulo $p$, using $\mathbb{Z}[x]$.

I see how to prove the result using the description of prime Gaussian integers (and using loads of intermediate results) but I read it can be proved directly using polynomials with integer coefficients and I don't see how...

I guess the idea is to prove that $<p>$ is a prime ideal but I get stuck on the actual procedure. Could you help me, I'm confused? Thank you very much!

Answer 1

First note that $\mathbb Z[i]/(p) \cong \mathbb Z[x]/(p,x^2+1)$ because $\mathbb Z[i] \cong \mathbb Z [x]/(x^2+1)$ and you can use the correspondence between ideals given by the isomorphism theorem: $(p)$ in $\mathbb Z[i]$ corresponds to $(p,x^2+1)$ in $\mathbb Z[x]$.

@Abellan has proved one direction. Here is the other direction:

If $−1$ is a square mod $p$, then the ideal $(p,x^2+1)$ is not prime in $\mathbb Z [x]$

Write $a^2=-1+pk$. Then $a^2+1 \in J=(p,x^2+1)$ and so $(x-a)(x+a)=x^2-a^2=(x^2+1)-(a^2+1) \in J$. But neither $x-a$ nor $x+a$ are in $J$.

Answer 2

I'm not sure but:

$\implies$ Supposing p is a prime number, $\mathbb{Z}_p$ is a field. Now we consider $\mathbb{Z}_p[X]$ and the polynomial $x^{2}+1$ which is irreducible since $i\notin$$\mathbb{Z}_p$. We can conclude $\mathbb{Z}_p[i]$ is a field extension of $\mathbb{Z}_p$ and $\mathbb{Z}_p[i]$$\cong$$\mathbb{Z}[i]/p$