# Is the algebraic subextension of a finitely generated field extension finitely generated?

This question is motivated by this other question (and its answer).

Suppose we have a field $F$, possibly imperfect. Consider the finitely generated field extension $F(a_1,\ldots,a_n)$. Is it always true that $K=F(a_1,\ldots,a_n)\cap F^{\textrm{alg}}$ is finitely generated?

The proof from 1 generalises to case when $F$ is perfect (or more generally when $F(a_1,\ldots,a_n)$ is separable over $F$, I guess), thanks to the primitive element theorem.

But what about the general case? What if the initial extension is inseparable?

• Is everything algebraic? If not, I think you can use counter-examples to the analog of Luroth theorem for greater transcendence degree. – user40276 Apr 11 '14 at 11:44
• @user40276: Of course it isn't. If everything was algebraic, the question would be completely trivial. – tomasz Apr 11 '14 at 22:09

Given an arbitrary field extension $F\subset G$, if $G$ is finitely generated (as a field) over $F$, then any intermediate field extension $F\subset K\subset G$ is also finitely generated over $F$.
Applying this to your situation (with $G=F(a_1,\cdots,a_n))$ you see immediately that your $K$ is both algebraic and finitely generated over $F$ so that actually it is even a finite-dimensional vector space over $F$.

The powerful theorem mentioned in the first sentence is unfortunately not as well-known as it should.
As often the best reference is Bourbaki: Algebra, Chapter 5, §15, Corollary 3, page 118.

• Interesting. Though it should be noted that in the version you linked, the corollary is on the page 224 (of the entire book, not just the chapter) and it is in paragraph 14, not 15. Also, the proof is remarkably simple. :) – tomasz Apr 13 '14 at 13:05
• Dear @tomasz, yes, the result is in §14: corrected. As to your other comment: Bourbaki has written a splendid chapter on field theory and His aim was not to arrive quickly or slowly at the result under discussion, but to develop the theory in a harmonious and natural way. Your last sentence confirms that He has succeeded. – Georges Elencwajg Apr 13 '14 at 17:17

Let $T$ be a transcendence basis of $F(a_1,\ldots ,a_n)/F$. Then $F(a_1,\ldots ,a_n)/F(T)$ is a finite extension and $F(T)$ and $K$ are linearly disjoint over $F$. Hence the cardinality of any $F$-linearly independent set of elements of $K$ is bounded from above by the degree of $F(a_1,\ldots ,a_n)/F(T)$.

If an algebraic extension $K/F$ of fields is not finite, then within $F$ there exists an infinite tower of proper finite extensions $K\subset K_1\subset K_2\subset \ldots$. Since $K_i$ is a $K$-sub vectorspace of $K_{i+1}$ a basis of $K_i$ can be extended to a basis of $K_{i+1}$. Thus the above infinite tower yields an infinite $K$-linearly independent set within $F$. By linear disjointness this set remains linearly independent over $F(T)$ - a contradiction.

• But how does the second sentence follow, exactly? That's pretty much exactly the point of this question... – tomasz Apr 12 '14 at 22:49