$\newcommand{\BA}{\mathbb{A}}$ $\newcommand{\BQ}{\mathbb{Q}}$ $\newcommand{\disc}{\operatorname{disc}}$ Consider the number field $K=\BQ[\sqrt{-23}, \alpha]$ where $\alpha$ is a root of $x^3-x+1$. How can I show that $K/\BQ[\sqrt{-23}]$, or more generally $K/\BQ$, is Galois?
I know that:
$\disc(\BA \cap \BQ[\sqrt{-23}]) = \disc(\BA \cap \BQ[\alpha]) = -23$.
$\BQ[\sqrt{-23}]/\BQ$ is degree 2 and Galois. $\BQ[\alpha]/\BQ$ is degree 3 and not Galois.
$\BA \cap \BQ[\sqrt{-23}]$ has an integral basis $\{1, \sqrt{-23}\}$ and $\BA \cap \BQ[\alpha]$ has an integral basis $\{1, \alpha, \alpha^2\}$.
We need to show that the other two roots of $x^3-x+1$ belong to $K$. Somehow the fact that $\disc(\BA \cap \BQ[\alpha]) = -23$ implies this is the "right" polynomial to adjoin a root of.
(The further point of the exercise is to show that $L/K$ is unramified, similar to this question, and thus to conclude that $K$ is the Hilbert class field of $\BQ[\sqrt{-23}]$. However I am asking a more basic question: I do not know why $K/\BQ[\sqrt{-23}]$ is even Galois.)
Consider the splitting field $K$ of $x^3-x+1$. Certainly $K$ is Galois over $\Bbb Q$ and indeed over any subfield. The polynomial is irreducible, and its discriminant is $-23$ which is not a square. Therefore the Galois group is $S_3$ and $|K:\Bbb Q|=6$. The quadratic subextension $L$ is generated by the square root of the discriminant, so $L=\Bbb Q(\sqrt{-23})\subseteq K$. As $|K:L|=3$ then $K=L(\alpha)$ where $\alpha$ is a zero of $x^3-x+1$.