Proof of irreducibility of $x^2+1$ over $\mathbb{Z}_p[x]$

Question

I am trying to prove that $x^2+1$ is irreducible over $\mathbb{Z}_p[x]$. In order to do that, let's say it's not irreducible, then we have:

$x^2+1=(x+a)(x+b) \rightarrow x^2+(a+b)x+ab=x^2+1\mod p$, thus we have:

$a=-b \mod p$, $ab=1\mod p \rightarrow b^2=-1\mod p$

Now, can I say that the last equation doesn't have any answer in $\mathbb{Z}[x]_p$? If yes, how? Because the last equation implies that $b^2 = p-1\mod p$, and I am not sure how to prove there is no $b$ exist to satisfy this equation

Answer 1

Let's set $p=5$.

$x^2 +1 \equiv x^2 - 4 \equiv (x-2)(x+2) \pmod 5$.

What's really going on? You have to out find when $-1$ is a quadratic residue modulo $p$. You should read really carefully this interesting page on Wikipedia about Legendre Symbol and its properties! https://en.wikipedia.org/wiki/Legendre_symbol.

Answer 2

It is not in general true that $x^2+1$ is irreducible in $\mathbb{Z}_{p}[x]$. For example, we have that $$x^2+1 \in \mathbb{Z}_{2}[x]$$ may be written as: $$(x+1)(x+1) \in \mathbb{Z}_{2}[x].$$

Answer 3

To add some details to Maffred's answer, there's actually an exact criterion for this, which is well known as part of the answer to a different question.

Theorem: (Fermat's Two Square Theorem) Let $p$ be a prime of $\mathbb{Z}$. The following are equivalent:

1. $\exists x\in\mathbb{Z}$ such that $p|x^2+1$
2. $p$ is not prime as an element of $\mathbb{Z}[i]$
3. $\exists a,b\in\mathbb{Z}$ such that $p=a^2+b^2$
4. $p=1\pmod{4}$ or $p=2$