I am not sure about this problem.
Let $K/F$ be a finite separable extension and let $\widetilde{K}/F$ be a normal closure of $K/F$. Is $\widetilde{K}/F$ necessarily separable?
I tried considering $\alpha \in \widetilde{K} \backslash K$ and assume it is inseparable over $F$. Then I considered $m(x)$ that is satisfied by $\alpha$, and $\widetilde{K}$ contains all roots of $m(x)$. Then I am not sure how to proceed. Can anyone help?
On
Not sure if this helps but I will give it a shot.
Since $K/F$ is a finite separable extension, we can write $K = F(\alpha_{1},\alpha_{2},\dots,\alpha_{n})$ with each $\alpha_{i}$ separable over $F$. Now, let $p_{i}(t) \in F[t]$ be the minimal polynomial of $\alpha_{i}$ for all $i$. Since $\widetilde{K}/F$ is normal, and $\alpha_{i} \in \widetilde{K}$ for all $i$, hence $\widetilde{K}$ is a splitting field for $p_{i}(t)$ for all $i$, with each $p_{i}(t)$ separable over $F$. Also, since $\widetilde{K}/F$ is a normal closure and by definition the smallest field containing $K$ and having $\widetilde{K}/F$ normal, thus any $\beta \in \widetilde{K} \backslash K$ that is separable will be a root of one of the $p_{i}(t)$ for some $i$. If not, such an $\beta$ will force $\widetilde{K}$ to contain all roots of its minimal polynomial $p'(t)$. However, the field $E = F(S)$, where $S$ = {all roots of $p_{i}(t)$}, will be a normal extension of $F$ and will contain $K$, contradicting the normal closure property of $\widetilde{K}$. Thus, the separable elements of $\widetilde{K}$ will be the roots of $p_{i}(t)$ which are contained in $\widetilde{K}$, and hence any element in $\widetilde{K}$ is separable.
Since the extension is separable and finite, it is generated by a primitive element $x$. By definition of separable, the roots of the minimal polynomial $P$ of $x$ are distinct so the splitting field of $P$ is a separable extension and the normal closure of $K$. The fact that splitting field of a separable polynomial is separable is shown here.
Splitting field of a separable polynomial is separable