How to show that $I = \langle x^2+2\rangle \subseteq \mathbb{Z}[x]$ is a prime ideal WITHOUT proving that $x^2+2$ is irreducible?

Question

How to show that $I = \langle x^2+2\rangle \subseteq \mathbb{Z}[x]$ is a prime ideal WITHOUT proving that $x^2+2$ is irreducible in $\mathbb{Z}[x]$?

I know how to show it by showing that it is irreducible, but how do I show it is prime without showing that? Do I assume it's not prime and get a contradiction? Like, take $a,b\in \mathbb{Z}[x]$ such that $ab\in \langle2+x^2\rangle. $ I want to show that $a\in\langle 2+x^2\rangle$ or $b\in \langle 2+x^2\rangle$ (i.e. either $a$ or $b$ is of the form $r(x^2+2), r\in\mathbb{Z}[x]$). I'll use contradiction. Assume $a\not\in\langle x^2+2\rangle$ and $b\not\in \langle x^2+2\rangle.$ Then $a$ and $b$ are not of the form $r(x^2+2),r\in\mathbb{Z}[x].$ I know we can't use the division algorithm since $\mathbb{Z}[x]$ is not a euclidean domain.

Answer 1

We assume $a(x)b(x)=(x^2+2)g(x)$. Let $x=\sqrt{-2}$. Then $a(\sqrt{-2})b(\sqrt{-2})=0$, so $a(\sqrt{-2})=0$ or $b(\sqrt{-2})=0$. Assume $a(\sqrt{-2})=0$ (a similar argument will work if we assume $b(\sqrt{-2})=0$). Then also $a(-\sqrt{-2})=0$ [this part takes a little justification]. So $a(x)$ is a multiple of both $x-\sqrt{-2}$ and $x+\sqrt{-2}$, hence of $(x-\sqrt{-2})(x+\sqrt{-2})$, which is $x^2+2$, done.

Answer 2

Consider the ring homomorphism $f:\mathbb Z[x] \to \mathbb C$ induced by $x \mapsto \sqrt{2}\,i$. Then $\mathbb Z[x]/\ker f$ is certainly a domain and so $\ker f$ is a prime ideal. Prove that $\ker f = \langle x^2+2\rangle$.