Class of finitely generated extensions is distinguished.

Question

I have to prove that $C$, the class of finitely generated extensions, is distinguished. I know that I have to prove that:

$i.$ If $K \subset F \subset E$ is a tower of fields, then: $E/K \in C$ if and only if $E/F,F/K \in C$.
$ii.$ If $E/K \in C$ and $F/K$ is any extension such that $E, F$ are contained in a common field, then $EF/F \in C$.

The problem is, I don't know how to start. I assume that $ E / K \in C $, then $E=K(\alpha_1,...,\alpha_n)$, but I don't see how to prove that the two intermediate extensions are fg. Any hint?

Answer 1

i. given $E/F/K$

• If $E/F,F/K$ are f.g. then so is $E/K$.

• If $E/K$ is f.g. then so is $E/F$.

  That $F/K$ is f.g. is a bit less obvious:

  Consider $F/M/K$ where $M/K$ is purely transcendental and $F/M$ is algebraic.

  Then repeat: $E/L/M$ where $L/M$ is purely transcendental and $E/L$ is algebraic.

  The trick is that $[F:M]=[LF:LM]$.

  $E/K$ is f.g. thus so is $E/L$,

  $E/L$ is a f.g. algebraic extension gives that $[E:L]$ is finite. Thus, $[LF:LM]$ and $[F:M]$ are finite.

  $trdeg(M/K)\le trdeg(E/K)<\infty$. Therefore $F/M,M/K$ are f.g. and so is $F/K$.

ii. $E=K(a_1,\ldots,a_n)$ gives that $FE=FK(a_1,\ldots,a_n)=F(a_1,\ldots,a_n)$.