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$K$ is a field of characteristic $0$. Let $M$ be an extension of $\mathbb Q$ with $[M : \mathbb Q] = 4$. How do you prove that there exists a field $L$ such that $\mathbb Q ⊂ L ⊂ M$ and $[L : \mathbb Q] = 2 $ if and only if there exists $u ∈ M$ such that $ M = \mathbb Q(u)$ and the minimal polynomial of $ u$ is of the form $x^4 + ax^2 + b$ with $a, b ∈ \mathbb Q$

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Since $[M:L]=2$, $M=L(\sqrt{m})$ for some $m\in L$. Since $[L:\mathbb{Q}]=2$ we have $$m^2+am+b=0$$ let $u=\sqrt{m}$. Then $$u^4+au^2+b=0$$

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