Let $t$ be transcendental over $\mathbb{F}_3$.
I used the polynomial $f = x^3 - 1$ to prove that $\mathbb{F}_3(t)(\sqrt[3]t)$ a splitting field over $\mathbb{F}_3(t)$. This was convenient since $f = (x - \sqrt[3]{t})^3$ in $\mathbb{F}_3(t)(\sqrt[3]t)$.
Now I have to show whether the same applies to $\mathbb{F}_3(t)(\sqrt[4]t)$. I have the feeling this is not a splitting field over $\mathbb{F}_3(t)$, but to show this I somehow have to show that there is no polynomial in $\mathbb{F}_3(t)$ for which $\mathbb{F}_3(t)(\sqrt[4]t)$ is the smallest subset containing all of it's zero's.
Can anyone give me a hint?
A normal extension is stable - invariant under ambient automorphisms. This means in particular that it contains all its elements' conjugates. What are all of the roots of $X^4-t$?
The roots aren't just $\pm\sqrt[4]{t}$, any more than they would be with $\Bbb Q(t)$ instead of $\Bbb F_3(t)$. Notice that to get the roots with $\Bbb Q$, you need to invoke primitive $4$th roots of unity. Same idea with $\Bbb F_3$.