Let $k$ be a field, and let $E$ and $F$ be fields extending $k$, both contained in some single extension of $k$. If $E$ and $F$ are finitely generated (as fields) over $k$, must $E\cap F$ also be finitely generated? If not, is there a simple counterexample?



When are intersections of finitely generated field extensions finitely generated?
As Brian Conrad remarked above, subextensions of finitely generated extensions are also finitely generated. Here is a prove. I wish there would be a simpler one!


* If $L/K$ is a finitely generated field extension and $L'$ an intermediate field, then $L'/K$ is also finitely generated.


Proof: Since $tr.deg\_K(L) = tr.deg\_{L'}(L) + tr.deg\_K(L')$ is finite, the same is true for $tr.deg\_K(L')$. Choose a transcendence basis $B'$ of $L'/K$. Replacing $K$ by $K(B')$, we may asume that $L'/K$ is algebraic.


Now let $B$ be a transcendence basis of $L/K$. Then $L/K(B)$ is algebraic and a finitely generated field extension, thus finite. Let $C \subseteq L'$ be linearly independent over $K$. If we knew that $B$ is also algebraically independent over $L'$, we could conclude that $C$ is linearly independant over $K[B]$ and thus over $K(B)$. This implies $|L':K| \leq |L : K(B)| < \infty$. Thus it remains to prove:


* Let $L/L'/K$ be a tower of fields such that $L'/K$ is algebraic. Let $B \subseteq L$ be algebraically independent over $K$. Then $B$ is also algebraically independent over $L'$.


Proof: Since algebraically independence is of finite character, we may assume that $B$ is finite. Since $L'(B) / K(B)$ is algebraic, we have


$tr.deg\_{L'}(L'(B)) = tr.deg\_K(K(B)) + tr.deg\_{K(B)}(L'(B)) = |B|$


Since $B$ generated $L'(B)/L'$, some subset of $B$ is a transcendence basis of $L'(B)/L'$, but this has cardinality $|B|$. Thus $B$ is itsself this basis.


