$p$ a prime satisfying $p \equiv 3 \mod 4 $. Then, the quotient field $ F_p [x] / (x^2 + 1)$ contains $\bar{x}$ that is a square root of -1

Question

I know that $x^2 + 1$ is irreducible in $F_p[x]$ if and only if $-1$ is not a square in $F_p$. Otherwise, $x^2 + 1$ could be factored out.

$-1$ not being a quadratic residue in $F_p$ is equivalent to saying that $p$ is a prime satisfying $p \equiv 3 \mod 4 $.

However, my book later writes:

Let $p$ be a prime satisfying $p \equiv 3 (\mod 4) $. Then the quotient field $ F_p [ x ] / (x^2 + 1)$ contains an element $\bar{x}$ that is a square root of -1.

But how is that so? Isn't $-1$ not a quadratic residue in $F_p$?

Answer 1

It is certainly true that $-1$ is not a square in $F_{p}$. We, however, are not in $F_{p}$. In our field, $x^2+1=0$, that is exactly what it means to mod out by it. Thus, $x$ (really the equivalence class of $x$) is the square root of $-1$.