I have been working on the following problem, from an old p-set of a Galois theory course I found online:
Let $F = \mathbb{F}_2(t)$, the field of rational functions on $\mathbb{F}_2$, and let $f(x) = x^3 + tx + t$. Prove that $f(x)$ is irreducible over $F$, has nonzero square discriminant, yet its splitting field $L$ has degree $6$ over $F$.
I have been able to prove irreducibility and the discriminant is, with the formula for cubics without quadratic term, $-4t^3-27t^2 = t^2$, a nonzero square.
The first step is adjoin $r$ satisfying $f(r) = 0$ by setting $F(r):= F[r]/(f(r))$, so that now over $F(r)$ $f$ has one root.
$x^3+tx+t = (x+r)(x^2+rx+(r^2+t))$
To show the splitting field has degree $6$, it's enough to show that over $F(r)$ the quadratic $x^2+rx+(r^2+t)$ has no roots. This would make the next root to be adjoined to have a quadratic minimal polynomial and we can use the tower law.
I am stuck here. I don't know how to show the quadratic is irreducible over $F(r)$. Would someone have a hint at this?