# Why does a root extension has a Galois closure?

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 $$$$F=K_0\subsetneq K_1\subsetneq\cdots\subsetneq K_i\subsetneq K_{i+1}\subsetneq\cdots\subsetneq K_s=K$$$$ where $$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.

• In characteristic zero, any finite extension is separable, so is contained in a Galois extension. Aug 9 '20 at 8:13

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 $$F$$.