Is the polynomial $x^2+1$ reducible in $\mathbb Z_7[x]$?

Question

Is the polynomial $x^2+1$ reducible in $\mathbb Z_7[x]$?

I think no, because $f(0)\ne0$; similarly, $f(1),f(2),f(3),f(4),f(5)$ and $f(6)\ne0$.

Is it correct?

Answer 1

Yes, you are correct, but only because the given polynomial is quadratic, which implies that it is irreducible iff it has no linear factors.

Checking for linear factors will also work for proving irreducibility of cubics, but not quartics and higher, where they may factor into irreducible polynomials of degree at least 2.

Answer 2

Your reasoning is correct but another way to look at it is that reducibility would imply that -1 is a square $\pmod{7}.$ However, this is not true as $7 \equiv 3 \pmod{4}.$