Galois group of a biquadratic quartic

Solution 1:

There are two numbers $\alpha,\beta$ (in some algebraic closure of $K$) such that the identity $P=X^4+aX^2+b=(X^2-\alpha^2)(X^2-\beta^2)$ holds. Then the set $R$ of all roots of $P$ is $\lbrace \pm \alpha, \pm \beta\rbrace$. Since $P$ is irreducible, $G$ can be identified to a transitive subgroup of ${\mathfrak S}(R)$, the group of permutations of $R$.

Also, any $\sigma\in G$ obviously satisfies $\sigma(-\alpha)=-\sigma(\alpha)$ and $\sigma(-\beta)=-\sigma(\beta)$. The subgroup $H$ of permutations satisfying those two conditions consists of eight elements : in cycle notation,

$$ \begin{array}{lcl} H &=\lbrace& {\sf id},(\alpha,-\alpha)(\beta,-\beta),\\ & & (\alpha,\beta)(-\alpha,-\beta),(\alpha,-\beta)(-\alpha,\beta), \\ & & (\alpha,\beta,-\alpha,-\beta), (\alpha,-\beta,-\alpha,\beta), \\ & & (\alpha,-\alpha), (\beta,-\beta) \rbrace \end{array} $$

There exactly three transitive subgroups of $H$, namely $H$ itself and

$$ \begin{array}{lcl} H_1 &=&\lbrace {\sf id},(\alpha,-\alpha)(\beta,-\beta), (\alpha,\beta)(-\alpha,-\beta),(\alpha,-\beta)(-\alpha,\beta) \rbrace \\ H_2 &=& \lbrace {\sf id},(\alpha,-\alpha)(\beta,-\beta), (\alpha,\beta,-\alpha,-\beta), (\alpha,-\beta,-\alpha,\beta) \rbrace \end{array} $$

In case (a), $(\alpha\beta)^2=b$ is a square in $K$, so $\gamma_1=\alpha\beta$ is in $K$. Now if $\tau_1=(\alpha,-\beta,-\alpha,\beta)$, we have $\tau_1(\gamma_1)=-\gamma_1$, so $\tau_1\not\in G$ and this forces $G=H_1$.

In case (b), $(\alpha\beta(\alpha^2-\beta^2))^2=b(a^2-4b)$ is a square in $K$, so $\gamma_2=\alpha\beta(\alpha^2-\beta^2)$ is in $K$. Now if $\tau_2=(\alpha,\beta)(-\alpha,-\beta)$, we have $\tau_2(\gamma_2)=-\gamma_2$, so $\tau_2\not\in G$ and this forces $G=H_2$.

Finally, in case (c) we have $\gamma_1\not\in K$ and $\gamma_2\not\in K$. By the fundamental theorem of Galois theory, $\gamma_1$ is not fixed by all the elements of $G$, so $G\neq H_2$. Similarly $\gamma_2$ is not fixed by all the elements of $G$, so $G\neq H_1$. The only possibility left is then $G=H$.