show that $$X^4-4X^2-21\in\mathbb{Q}[X]$$ is solvable by radicals.
$\mathrm{Def}$: Let $f(X)\in K[X]$ and let $\Sigma$ be a splitting field for $f(X)$ over $K$. We say $f(X)$ is solvable by radicals if $\exists\;\; M\;$ s.t $M\supseteq \Sigma$ and $M\supseteq K$ is a radical extension.
$\mathrm{Def (2)}$: $L\supset K$ is a rdical extension if there exists a chain of intermediate fields $K=K_{0}\subseteq K_{1}\subseteq K_{2}\subseteq....\subseteq K_{n}=L$ such that $K_{i+1}=K(\alpha_{i})$, where $\alpha_{i}^{r_{i}}\in K_{i}$ for $ \{1,...,n-1\}$
I have problems with understanding these concepts. I suspect just by following the above definitions that we first need to find a splitting field $\Sigma$ for $f(X)$ over $K$ and then show that $\Sigma$ is contained in a radical extension $L\supseteq K$.
Let $u=x^2$ then we get $u^2-4u-21=0$ with roots $x_{1}=-3$ and $x_{2}=7$
So the original equation splits in linear factors
$X^4-4X^2-21=(x-\sqrt{7})(x+\sqrt{7})(x-i\sqrt{3})(x+i\sqrt{3})$
Hence the splitting field $\Sigma=\mathbb{Q}(i,\sqrt{3},\sqrt{7})$ but as $\mathbb{Q}(\sqrt{3},\sqrt{7})=\mathbb{Q}(\sqrt{3}+\sqrt{7})$ we can simplify the splitting field to $\Sigma=\mathbb{Q}(i,\sqrt{3}+\sqrt{7})$
What should I do next? How to create (find) $M$ that will contain $\Sigma$? And how to show that $M\supseteq \mathbb{Q}$ is radical then? Any help appreciated.
In your description you actually give a solution in radicals, so you don't need any Galois theory to show that your equation is solvable in radicals.