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. $\mathbb{Q}(\sqrt{2})(\sqrt{-1})$.
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Do you agree that $M=K(\sqrt a), a\in K^,\notin {K^}^2$, and $L=M(\sqrt\theta), \theta \in M^*$ with similar properties ? I recall the proof for $L$, which is a priori of the form $M(x)$, where $x$ is a root of a quadratic polynomial $\in L[X]$, so actually $L=M(\sqrt \theta)$, where $\theta $ is the discriminant. But $\theta \in M$ is of the form $u+v\sqrt a$, with $u,v\in K$. If $v=0$, one would have $L=M(\sqrt u)=M(\sqrt a)=M$, contradiction. – nguyen quang do Dec 10 '19 at 21:31
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Not true. $v$ being not zero is crucial in $M(\sqrt{u})$ being equal to $M(\sqrt{a})$ in your example. That's just the problem I was looking to adress. – dario Dec 12 '19 at 17:54
1 Answers
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.
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