Normal extension, why is $E/\mathbb F_p(t)$ normal?

216 Views Asked by At

I have a theorem that says that $E/K$ is a normal extension if $E$ is the splitting field of one separable polynomial $f \in F[X]$.

In an exercise, I have that the splitting field $E$ of $X^p-t$ is normal since $X^p-t$ splits over $E$, the extension $E/\mathbb F_p(t)$ is normal. But $X^p-t$ is not separable, since $X^p-t=(X-\sqrt[p]t)^p$, so the theorem doesn't hold... Any explanation?

Moreover, does $E=\mathbb F_p(t)(\sqrt[p]t)$?

1

There are 1 best solutions below

2
On

You have written the answer. The point is that $X^p-t$ has a unique root thus its splitting field is normal by definition.