When are intersections of finitely generated field extensions finitely generated? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:36:41Z http://mathoverflow.net/feeds/question/21086 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21086/when-are-intersections-of-finitely-generated-field-extensions-finitely-generated When are intersections of finitely generated field extensions finitely generated? SJR 2010-04-12T09:34:13Z 2010-04-12T12:05:23Z <p>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?</p> http://mathoverflow.net/questions/21086/when-are-intersections-of-finitely-generated-field-extensions-finitely-generated/21093#21093 Answer by Martin Brandenburg for When are intersections of finitely generated field extensions finitely generated? Martin Brandenburg 2010-04-12T12:02:50Z 2010-04-12T12:02:50Z <p>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!</p> <ul> <li>If $L/K$ is a finitely generated field extension and $L'$ an intermediate field, then $L'/K$ is also finitely generated.</li> </ul> <p>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.</p> <p>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 $|L':K| \leq |L : K(B)| &lt; \infty$. Thus it remains to prove:</p> <ul> <li>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 $L'$.</li> </ul> <p>Proof: Since algebraically independence is of finite character, we may assume that $B$ is finite. Since $L'(B) / K(B)$ is algebraic, we have</p> <p>$tr.deg_{L'}(L'(B)) = tr.deg_K(K(B)) + tr.deg_{K(B)}(L'(B)) = |B|$</p> <p>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.</p>