Let $F$ be a field of characteristic not 2. Let $a,b \in F$ and $K/F$ is a splitting field of the polynomial $(X^{2}-a)(X^{2}-b)$.Prove that the Galois group of $K/F$ is isomorphic to either $\{e\}$ or $\mathbb{Z}/ 2\mathbb{Z}$ o r $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Give an example in each case.
I'm kind of stuck on this problem. Any help will be appreciated. Thanks.
If $f(X) = (X^2 - a)(X^2 - b)$ then the splitting field $K$ of $f$ over $F$ is the smallest field extension in which $f(X)$ factorises completely into distinct linear factors. To do this, we need elements $\alpha, \beta$ such that $\alpha^2 = a, \beta^2 = b$, since then we will be able to factorise $f(X) = (X-\alpha)(X+\alpha)(X-\beta)(X+\beta)$. By minimality then, $K = F(\alpha, \beta)$.
The Galois group of $K/F$ is then the group of all automorphisms of $K$ that fix all the elements in $F$. So we have some choices:
If $\alpha \in F$ and $\beta \in F$ (i.e. $a, b$ are squares in $F$), then $K = F$. The only automorphism of $K$ that fixes all of $F$ is the one that fixes all of $K$, the identity. So here $Gal(K/F) = \{e\}$.
If only $\alpha \in F$, so $a$ is a square but $b$ is not, then $K = F(\beta)$. Suppose $\sigma : K \to K$ is an automorphism fixing all of $F$. Then $\sigma$ is uniquely determined by it's value at $\beta$, since every element of $K$ can be written using only field operations (which commute with $\sigma$), elements of $F$, and $\beta$. We also know $\sigma(b) = b$ since $b \in F$, so $\sigma(\beta^2) = b$. On the other hand, since $\sigma$ commutes with field operations, $\sigma(\beta)^2 = b$, and so $\sigma(\beta) = \pm \beta$ (why can $b$ not have any other square roots other than these 2 obvious ones?)
Choosing $\sigma(\beta) = \beta$ leaves us with just the identity automorphism. Choosing $\sigma(\beta) = -\beta$ gives us an automorphism of order 2: $\sigma(\sigma(x)) = x$ for all $x \in F$. The group of automorphisms in this case is $\{e, \sigma\} \cong \mathbb{Z}/2\mathbb{Z}$, where $\sigma: \beta \to -\beta$.
In the case $\alpha, \beta$ both not in $F$, a similar computation shows that the Galois group is either $\mathbb{Z}/2\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ (why is it not always the second one?)