We have the following setup: $K$ is a field of odd characteristic, $L$ is a biquadratic extension. We have already shown that $L = K( \sqrt{a}, \sqrt{b})$ for $a,b, a/b$ nonsquares in $K^\times$. The problem is the following:
Let $c \in L^\times$ be a nzonsquare, and let $E = L ( \sqrt{c})$. Prove that $E$ is Galois over $K$ if and only if for each $\sigma \in \text{Gal}(L/K)$ the ratio $\sigma(c)/c$ is a square in $L$.
I'm honestly not sure how to approach either direction. For the forward direction, I was hoping that $E/K$ being Galois would imply that $\tau(\sigma(c)/c)) = c/\sigma(c)$ for all $\tau \in \text{Gal}(L/K)$. But, I can't see where the hypothesis that $E/K$ is Galois comes in.
For the reverse direction, I was also hoping that I could somehow get the hypothesis that $\sigma(c)/c$ is a square to imply that $E$ is the splitting field of the minimal polynomial of $c$ over $K$. But, this also seems hard and I don't know how to approach it. The separable part of showing Galois-ness is already handled, since the composition of separable extensions is separable.
Any help would be much appreciated.
Best Answer
$\newcommand{\Q}{\mathbf{Q}}$
Let $G = \text{Gal}(L/K)$. Note that:
$\sigma(c)/c$ is a square in $L$ for all $\sigma \in G$
$\Longleftrightarrow$ $L(\sqrt{c}) = L(\sqrt{\sigma(c)})$ for all $\sigma \in G$
$\Longleftrightarrow$ $\sqrt{\sigma(c)} \in L(\sqrt{c})$ for all $\sigma \in G$
$\Longleftrightarrow$ $\pm \sqrt{\sigma(c)} \in L(\sqrt{c})$ for all $\sigma \in G$ and for all choices of $\pm$
This last condition looks a bit silly, but notice that the Galois conjugates of $\sqrt{c}$ are exactly the $\pm \sqrt{\sigma(c)}$ for $\sigma \in G$. Indeed, if $x$ is a conjugate of $\sqrt{c}$ then $x^2$ is a conjugate of $c$, so $x^2 = \sigma(c)$ and $x$ has the desired form. Conversely, if $\sigma \in G$ then we can lift it to an element of $\text{Gal}(\overline{K}/K)$ (or to the Galois group of the Galois closure of $E$ over $K$ if you prefer) where it now acts on $\sqrt{c}$, and $\sigma(\sqrt{c})^2 = \sigma(\sqrt{c}^2) = \sigma(c)$, so $\sigma(\sqrt{c})$ must be one of the square roots of $\sigma(c)$. Since $c$ is not a square, we know that its two square roots are conjugate to each other as well. This establishes the ``notice that'' from the start of this paragraph.
So if $E/K$ is Galois, then it certainly contains all the Galois conjugates of $\sqrt{c}$, so by the chain of equivalences above we are done. The converse is slightly more subtle since we need to account for the fact that $E$ might not be the splitting field of the minimal polynomial of $\sqrt{c}$ over $K$. (For example, what if $c \in K$?) But supposing that $E = L(\sqrt{c})$ contains all the Galois conjugates of $\sqrt{c}$, we can conclude that $E$ is the compositum of $L$ and the splitting field of the minimal polynomial of $\sqrt{c}$ over $K$, both of which are Galois.
As a final remark, notice that we never actually used that $L$ was biquadratic, only that it was Galois over $K$. Essentially this condition is asking if, from the point of view of $K$, the extension $L(\sqrt{c})$ is ``well-defined''. Let's think about $\Q$ for a second. $\Q$ can't tell $i$ and $-i$ apart, but $\Q(i) = \Q(-i)$ so that field is still Galois over $\Q$ because it contains all the conjugates of $i$. However, if we replace $i$ by $\sqrt[3]{2}$ (the real one), we see that $\Q(\sqrt[3]{2})$ doesn't contain the other Galois conjugates of $\sqrt[3]{2}$, and the fields generated by each of these conjugates are isomorphic but not literally equal as subfields of $\overline{\Q}$. Consequently, $\Q(\sqrt[3]{2})$ is not Galois over $\Q$. This is essentially what's happening in your problem as well, with $i$ or $\sqrt[3]{2}$ replaced by $\sqrt{c}$.