I am working on the following problem:
If $K\subset L\subset M$ are fields and $M$ is finitely generated over $K$, prove that $L$ is finitely generated over $K$.
I've proved that you may reduce to the case where $L$ is algebraic over $K$. We want to bound the size of linearly independent subsets of linearly independent subsets of $L/K$. To do this, we may try considering a transcendence basis $S$ of $M/K$. Since $M$ is finitely generated, $M/K(S)$ should be finite dimensional. So, if a linearly independent subset $L/K$ stays linearly independent in $M/K(S)$, then we're okay. This would follow, if we could show that an algebraically independent subset remains algebraically independent over an algebraic extension, since then by standard Linear Algebra facts, I's be able to solve the problem. So, I was wondering how one might go about doing this (if it is indeed possible), or if there is a better approach to this problem.
To show the last claim (that a subset of $ L $ linearly independent over $ K $ remains linearly independent over $ K(S) $), note that given any linear dependence relation over $ K(S) $, you may first clear denominators to land in $ K[S] $, then look at the coefficient of the leading term (for example, using lexicographical ordering) to get a linear dependence relation over $ K $. Algebraically independent elements remain algebraically independent over algebraic extensions: say that some element $ s \in S $ was algebraically dependent to the remaining elements of $ S $ over some algebraic extension $ N/K $, then we may wlog assume that $ N/K $ is finite (there are only finitely many coefficients in any algebraic dependence relation) and then note that this implies that we have a tower $ K(S - \{s \}) \subset N(S - \{ s \}) \subset N(S) $, where each extension has finite degree, so the subextension $ K(S)/K(S - \{ s \}) $ is also of finite degree. This contradicts algebraic independence of $ S $ over $ K $.