We define $F:=\mathbb{F}_3(t)$, and let $F(\alpha)$ be the field extension created by adjoining $\alpha$ which is a root of $x^3 - t$. Is $F(\alpha)$ a normal extension of $F$?.
Question is from Galois Theory by David Cox.
I think it is no, but I'm not sure about my proof, which aims to show that $x^3-t$ is irreducible, and $F(\alpha)$ contains one but not all of its roots.
First, we factor $x^3 - t$, which is irreducible in $F[x]$.
$(x^3 - t) = (x - \alpha)(x^2 + \alpha x + t/ \alpha)$
Then we must show that $(x^2 + \alpha x + t/ \alpha)$ does not split in $F(\alpha)$.
Suppose $p(t)$ is a root of $(x^2 + \alpha x + t/ \alpha)$, so $p(t)^2 + \alpha p(t) + t/\alpha = 0$
If $p(t)$ has degree $>= 1$, then this is impossible, as LHS will have a $t^2$ or higher term. If $p(t)$ gas degree $0$, also impossible, since $t/\alpha$ can't be cancelled out. So $p(t)$ has no roots in $F(\alpha)$. Hence, $F(\alpha)$ is not normal.
Is this a valid solution? Also, is there a better proof of this? Because if $p(t)$ was more complicated I would be lost. Thanks!
Edit: I messed up the factorization. Writing $(x^2 + \alpha x + t/ \alpha) = (x - \beta)(x - \gamma)$ gives us $\alpha \beta\gamma = t$, which after multiplying both sides by $\alpha^2$, gives us $\beta \gamma = \alpha^2$. It also gives us $-\beta -\gamma = \alpha$. Combining, we have $\beta \gamma = (\beta + \gamma)^2$ which has solutions $\beta = \gamma = \alpha$. Hence the correct factorization is $x^3 - t = x^3 - \alpha^3 = (x - \alpha)^3$, which is actually kinda obvious.
Thus $F(\alpha)$ is a splitting field so it is normal (and finite).