Quadratic subfields of Quartic extensions

Question

In this article, on page 18, the author describes quadratic subfields of quartic extensions. However, the proofs are somewhat obscure (at least to me). Let $L/K$ be a quartic extension containing a quadratic extension $M$, $K$ not characteristic $2$. The aim is to prove that $L=K(\theta)$, where $\theta$ has minimum polynomial over $K$ of the form $X^4 + AX^2 + B$. $M=K(\sqrt{a})$ with $a \in K^{\times}$, obviously $a$ not square in $K$. Also $L=M(\sqrt{\theta})$, with $\theta \in M^{\times}$, obviously $\theta$ not square in $M$. The author puts it immediately like this: $L=M(\sqrt{u + v\sqrt{a}})$, $u,v \in K$, $v \neq 0$ (does something similar in characteristic $2$). The fact that $v \neq 0$ is key, because then $L=M(\sqrt{u + v\sqrt{a}})=K(\sqrt{u + v\sqrt{a}})$. Without justification, to me, it seems that the author uses what he is about to prove as a result. So, why $L=M(\sqrt{u + v\sqrt{a}})$, $u,v \in K$, $v \neq 0$ ? As counterexample, where this is not immediately obvious, consider e.g. $\mathbb{Q}(\sqrt{2})(\sqrt{-1})$.

Answer 1

Right, the case with $v=0$ must be settled. I keep all the previous notations. If $v=0$, one has $L=K(\sqrt \theta)=K(\sqrt a, \sqrt u)$. The case $K(\sqrt a)=K(\sqrt u)$, which corresponds to $a$ mod ${K^*}^2 = u$ mod ${K^*}^2 $, leads to a contradiction between the degrees. In the other case, $L/K$ is a biquadratic extension, i.e. a galois extension with group $G\cong C_2\times C_2$, hence contains exactly 3 quadratic subextensions which are $K(\sqrt a),K(\sqrt u), K(\sqrt {ua})$: this is an immediate consequence of Kummer theory, but can also be shown "by hand".

Let us introduce the sum $s=\sqrt a + \sqrt u\in L$. Since $s^2=a+u+2\sqrt a\sqrt u$ does not belong to $K$, the degree of $s$ over $K$ must be 4, and its minimal polynomial has the form $X^4+AX^2+B$. This can be checked by computing $s^4$ and eliminating $\sqrt a\sqrt u$. So the announced property holds even if $v=0$. I guess the author neglected this abelian case because he implicitly considered it as well known in the abelian case.