Given $f(x) = x^3 + x + 1$, is $\sqrt{-31}$ in $\mathbb{Q}[x]/(f)$? In the splitting field $K$ of $f$?

Question

Given $f(x) = x^3 + x + 1$, is $\sqrt{-31}$ in $\mathbb{Q}[x]/(f)$? In the splitting field $K$ of $f$?

This is a problem from Artin's Algebra I'm looking at for test prep.

Answer 1

Hint: $-31$ is the discriminant of the polynomial. So if $a_1, a_2, a_3$ are the roots of $f$, then:

$$\Delta = -31 = (a_1 - a_2)^2(a_1-a_3)^2(a_2-a_3)^2$$

So is $\sqrt{-31} \in K$?

Next, note that $\mathbb{Q}[x]/(f) \cong \mathbb{Q}[\alpha]$ where $\alpha$ is a root of $f$. Well, $[\mathbb{Q}[\alpha]:\mathbb{Q}] = 3$ and $[\mathbb{Q}[\sqrt{-31}]: \mathbb{Q}] = 2$.

If $\sqrt{-31} \in \mathbb{Q}[\alpha]$, then we have $\mathbb{Q} \subset \mathbb{Q}[\sqrt{-31}] \subset \mathbb{Q}[\alpha]$. Can you find a contradiction here?