Gotta solve this problem:
If $K|F$ is a separable field extension, must its normal closure $N|F$ be separable?
Read another answer and when $K|F$ is finite, the answer is yes and the problem is already solved by using the primitive element theorem. Now this time $K|F$ is not necessarily finite, so I'm not quite sure if the result is still true.
What I thought is that if I take $\alpha \in N\setminus K$, I have two possibilities:
-$MinPol_F(\alpha)$ has other root $\alpha_1$ in $K$, so as $K|F$ is separable, and $MinPol_F(\alpha) = MinPol_F(\alpha_1)$ $\implies$ $MinPol_F(\alpha)$ is separable and so on the normal closure $N$ is separable.
-$MinPol_F(\alpha)$ doesn't have any roots in $K$.
In this second case, how can I tell if the polynomial $MinPol_F(\alpha)$ is separable? If it has no roots in $K$, I can't use the fact that $K|F$ is separable in any way... but neither I can say that ''we can't assure that this polynomial is separable so the normal closure is not necessarily separable".
Maybe this polynomial with no roots in K can't exist?? Or this is not a good approach?