Roots of $x^2+1$ over $\mathbb{Z}_7$?

Question

Prove that $x^2 + 1$ is irreducible in the ring of polynomials $\Bbb Z _7 [x]$ over the field $\Bbb Z _7$.

Is it enough to show that no single element of $\mathbb{Z}_7$ squared is equal to $-1 \pmod 7$? Or is there more to it than that?

Answer 1

Yes, it is enough, since $\mathbb{Z}_7$ is a field (thus $\mathbb{Z}_7[x]$ is a unique factorization domain), and $x^2+1$ has degree 2.

In fact, if $x^2+1$ were not irreducible, his factorization would be a product of two degree 1 polynomials $(x-a)(x-b)$ and $a,b \in\mathbb{Z}_7$ would be roots of $x^2+1$.

Just show $x^2+1$ has no roots in $\mathbb{Z}_7$ and you're done!