I need to find the splitting field of $f =x^4 -t \in \mathbb{F}_p(t)[x]$, with $p$ prime, distinguishing the cases $p \equiv 1 \bmod 4$ and $p \equiv 3 \bmod 4$.
I solved it in the first case, not the second. Here's what I did:
$p \equiv 1 \bmod 4$. I recall $\mathbb{F}_p^*$ is cyclic, say it has generator $\alpha$ of order $p-1$. Then take a root $\zeta$ of $f$, say the image of $x$ in $\frac{\mathbb{F}_p(t)[x]}{(x^4 -t)}$, and it is easy to see that $\zeta\alpha^{\frac{p-1}{4}}, \zeta\alpha^{\frac{p-1}{4}\cdot 2}, \zeta\alpha^{\frac{p-1}{4}\cdot 3}$ are also roots of $f$, and they are all distinct thanks to $\alpha$ being a generator. Hence, $\mathbb{F}_p(t,\zeta)$ is the splitting field.
When $p \equiv 3 \bmod 4$, the $\frac{p-1}{4}$-trick does not work, and I noted that $\alpha^{\frac{p-1}{2}}$ is just $-1$. I feel in this case I'll have to attach two distinct roots $\zeta_1,\zeta_2$ so that all the roots are $\pm\zeta_1,\pm\zeta_2$. But I don't know how to prove it.
Thanks in advance for any help.