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What is an example of a field $K$ which is of degree $2$ over two distinct subfields $E$ and $F$ respectively, but not algebraic over $E \cap F$?

mathcounterexamples.net
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  • Copying a comment from a now-deleted answer, to show there are examples with "algebraic" replaced with "finite": Let $K=\overline{\mathbb Q}$ and $E=K\cap\mathbb R$. There is an automorphism $\sigma$ of $K$ which sends $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$ since the two elements are conjugate, let $F=\sigma(E)$. Then $E,F$ are index 2 subfields of $K$, but their intersection cannot be a finite index subfield, because of Artin-Schreider. It is an algebraic extension, but not a finite one. – Wojowu Nov 25 '16 at 15:50
  • By the way, do you know whether the statement is actually true? In a comment to a deleted answer you mention Lang's "Algebra"; does it state this result or the one about finite extensions? – Wojowu Nov 25 '16 at 15:57
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    @Wojowu This is an exercise stated as I did. – mathcounterexamples.net Nov 25 '16 at 15:58

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Let $X$ be an indeterminate and consider $E=\mathbb C(X^2),F=\mathbb C(X^2+X),K=\mathbb C(X)$. Clearly $K$ is quadratic over both $E$ and $F$. We claim $E\cap F=\mathbb C$.

Suppose first $p(X^2)=q(X^2+X)$ for some rational functions $p,q$ over $\mathbb C$. Consider them as equal to a function $f(X)$ over $\mathbb C$ in a variable $X$. This equality tells that if $z$ is a zero (resp. a pole) of $f$, then $-z$ and $-1-z$ are also zeros (resp. poles) of $f$. Unless $f$ is constant, it has either a zero or a pole, and by repeatedly applying $z\mapsto -z,z\mapsto -1-z$ we find it has infinitely many of these. But this is impossible for rational function unless it's identically zero. It follows that $f$ is constant, so $p(X^2)=q(X^2+X)\in\mathbb C$.

Wojowu
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