# Subextension of a finitely generated extension of fields

If $E/K$ is a finitely generated field extension and $F$ is an intermediate field how can I prove that $F/K$ is finitely generated?

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Possible duplicate: math.stackexchange.com/questions/34424/… – Pierre-Yves Gaillard Mar 29 '12 at 9:40

A proof of the general case can be found in $\S 12.4$ of my field theory notes.

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An excellent, freely available reference. – Georges Elencwajg Mar 29 '12 at 13:47

The result is easy if you suppose that the big extension $E/K$ is algebraic ( equivalently: finite) and Benjamin has given you a proof.

But it is also true in complete generality: if $E/K$ is a field extension and if $E=K(x_1,...,x_n)$ is a finitely generated extension, then any intermediate field $K\subset F\subset E$ is also finitely generated .

The difficulty is that some or all of the $x_i$'s might be transcendental over $K$.
Even in the case of a purely transcendental extension $K\subset K(X_1,...,X_n)$ the situation is quite complicated and it is not true for $n\gt 1$ that $F$ must be purely transcendental too : this is the failure of Lüroth's theorem in higher dimensions.

A proof of finite generation of $F$ in the general non-algebraic case is surprisingly difficult to locate in the literature. The only reference I could find is Theorem 24.9 in Isaacs's book.

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Dear Georges: I found this MathOverflow answer by our friend Martin Brandenburg. – Pierre-Yves Gaillard Mar 29 '12 at 9:30

First of all if $E/K$ is finitely generated this means that $E = K(a_1,\ldots a_n)$ where the $a_i$ are algebraic over $K$. Since $F$ is an intermediate field, you have the containment

$$K \subseteq F \subseteq E.$$

We now need the result the following result:

If $E = K(a_1,\ldots a_n)$, then $[E:K]$ finite.

$\textbf{Proof:}$ Since $a_1$ is algebraic over $F$, $[K(a_1) : K]$ is finite. Since $a_2$ is algebraic over $K$, it is algebraic over $K(a_1)$ because $K \subset K(a_1)$. Hence $[K(a_1,a_2):K]$ is finite by the dimension counting formula. Continuing in this fashion we see that $[E:K] = [K(a_1, \ldots a_n):K]$ is finite, proving our claim.

Since $[E:K]$ is finite, the dimension counting formula implies that $[E:F]$ and $[F:K]$ are finite. In particular this means that $[F:K]$ is finite so that $F/K$ is finitely generated.

Alternatively if you know linear algebra, $F$ being an intermediate field is a $K$ - vector subspace of a finite dimensional $K$ - vector space $E$, hence is finite dimensional as well.

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This is not correct, finitely generated does not mean algebraic. – Gregory Grant Apr 20 '15 at 15:38