= Solution
Put
$$
\delta=\sqrt c,
\qquad
\beta=\frac{\delta}{\alpha}.
$$
The four roots are $\alpha,-\alpha,\beta,-\beta$. If $\delta\in\mathbb Q(\alpha)$, then
$$
K=\mathbb Q(\alpha),
\qquad
[K:\mathbb Q]=4,
$$
because $g$ is irreducible. Hence the Galois group has order four and acts transitively, hence regularly, on the roots.
Let $\tau$ be the automorphism taking $\alpha$ to $\beta$. Since $\tau(\delta)=\pm\delta$,
$$
\tau^2(\alpha)
=\tau\left(\frac{\delta}{\alpha}\right)
=\frac{\tau(\delta)}{\beta}
=\begin{cases}
\alpha,&\tau(\delta)=\delta,\\
-\alpha,&\tau(\delta)=-\delta.
\end{cases}
$$
If $\tau(\delta)=\delta$, then $\tau$ is an involution. Together with the distinct involution taking $\alpha$ to $-\alpha$, it gives
$$
\boxed{\operatorname{Gal}(K/\mathbb Q)\cong C_2\times C_2.}
$$
In the ordered root set $(\alpha,-\alpha,\beta,-\beta)$ these generators act as double transpositions.
If $\tau(\delta)=-\delta$, then
$$
\alpha\longmapsto\beta\longmapsto-\alpha
\longmapsto-\beta\longmapsto\alpha,
$$
so $\tau$ has order four and
$$
\boxed{\operatorname{Gal}(K/\mathbb Q)\cong C_4.}
$$
This is the degree-four case in the <Galois group of an irreducible even quartic>.
Back to article page
