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?

  • 3
    $\begingroup$ Subextension inherits finite generation; this is a homework-level exercise. Please work on it by yourself to find a proof. (The version for $k$-algebras is of course a completely different story...) $\endgroup$ – BCnrd Apr 12 '10 at 10:17
  • 4
    $\begingroup$ But we should admit that it's a rather hard exercise. I've written it up here matheplanet.com/matheplanet/nuke/html/… (german). $\endgroup$ – Martin Brandenburg Apr 12 '10 at 11:33
  • 6
    $\begingroup$ But it's OK for some exercises to be hard. To succeed in research one needs the experience of struggling with hard exercises on one's own. $\endgroup$ – BCnrd Apr 12 '10 at 15:39

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.

| cite | improve this answer | |
  • 2
    $\begingroup$ Sigh. If students know they can search on MO for full solutions to many standard exercises, they'll never learn to figure stuff out for themself, which is an important part of learning ideas and training to do creative work later. Please keep this in mind before future posting of arguments like this. I'll just say that there is a more geometric approach if one thinks in terms of irreducible components and nilpotence on geometric fiber of a $K$-variety associated to $L$, but please try that one in private; no need to post it on MO. $\endgroup$ – BCnrd Apr 12 '10 at 17:09
  • 9
    $\begingroup$ I understand your objection. I posted the proof because I'm pretty sure that this exercise is even interesting and not straight-forward for some graduates. As for me, to be honest, I have no intuition for the proof. Perhaps you can provide some. $\endgroup$ – Martin Brandenburg Apr 12 '10 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.