Why does a root extension has a Galois closure?

Question

The following are a definition and a lemma (Lemma 14.38 in Dummit's "Abstract Algebra").

Definition. An element $\alpha$ can be expressed by radicals or solved for in terms of radicals if $\alpha$ is an element of a field $K$ which can be obtained by a succession of simple radical extensions \begin{equation} F=K_0\subsetneq K_1\subsetneq\cdots\subsetneq K_i\subsetneq K_{i+1}\subsetneq\cdots\subsetneq K_s=K \end{equation} where $K_{i+1}=K_i(\sqrt[n_i]{a_i})$ for some $a_i\in K_i$ and $n_i\in\mathbb N^*$, $i=0,1,\ldots,s-1$. Here $\sqrt[n_i]{a_i}$ denotes some root of the polynomial $x^{n_i}-a_i$. Such a field $K$ will be called a root extension of $F$.

Lemma. Let $F$ be a field of characteristic $0$. If $\alpha$ is contained in a root extension $K$ as in the definition above, then $\alpha$ is contained in a root extension which is Galois over $F$ and where each extension $K_{i+1}/K_i$ is cyclic.

The proof starts with this:

Proof: Let $L$ be the Galois closure of $K$ over $F$. ...

But how do I know that there is a Galois closure $L$? I think the proof that there is a Galois closure $L$ should go like this.

1. All $K_{i+1}/K_i$ is a finite extension.
2. All $K_{i+1}/K_i$ is a separable extension.
3. Thus $K/F$ is a finite separable extension, since finiteness and separability of extensions are transitive.
4. All finite separable extension has a Galois closure, so $K/F$ has a Galois closure.

I couldn't prove 2. If $K_{i+1}$ is a splitting field of the minimal polynomial $m_{\sqrt[n_i]{a_i},K_i}(x)$, then $K_{i+1}/K_i$ is Galois since it is a splitting field of a separable polynomial, so it is separable. But I don't know if it really is a splitting field.

Answer 1

If $K$ is a separable algebraic extension of a field $F$, then its Galois closure is the smallest extension field, in terms of inclusion, which contains $K$ and is Galois over $F$.

If $K=F(\alpha)$ where $\alpha$ has irreducible polynomial $f$ over $F$, then the Galois closure of $K$ is the splitting field of $f$ over $F$.

If you have a finite extension in characteristic zero, then it is automatically separable. Hence there exists a Galois closure.