Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following statement is an exercise in point set topology: If $E \to X$ is a covering with nonempty finite fibers and $X$ is compact, then also $E$ is compact. Now Grothendieck generalized covering theory so that in particular separable field extensions may be regarded as coverings.

Question: What is the corresponding statement in field theory?

share|cite|improve this question
I thought field extensions were embeddings and group extensions were coverings? – PEV Mar 6 '11 at 15:16
Branderburg: I think the corrispondence stated by Grothendieck is not so deep to allow questions such yours. In specific the corrispondence should be something like: – Giovanni De Gaetano May 3 '11 at 22:00
"The category of finite coverings over a (connected) topological space $X$ is equivalent to the category of finite sets under the action of the pro-fundamental group of the space. The category of finite separable algebras (not fields) over a given field $k$ is anti-equivalent (not equivalent) to the category of finite sets under the action of the pro-Galois group of the field." – Giovanni De Gaetano May 4 '11 at 7:39
In this sense, fixed a topological space $X$ and a field $k$ such that the pro-fundamental group of $X$ and the pro-Galois group of $k$ are isomorphic, the category of finite coverings of $X$ is anti-equivalent to the category of the finite separable algebras over $k$. This anti-equivalence conserve notions such as the connection, but I don't see how to translate the notion of compactness in the algebraic context. I'll think about it. – Giovanni De Gaetano May 4 '11 at 7:39
I apologize for the long comment, I hope to transform it in an answer in some time. – Giovanni De Gaetano May 4 '11 at 7:39
up vote 2 down vote accepted

To answer Martin's questions in his comments to Rayleigh's answer:

Fix a scheme $S$ and a proper morphism $f:X\to Y$ of $S$-schemes. Suppose that $Y$ is proper over $S$. If $f$ is proper, then $X$ is proper over $S$. This is simply because proper morphisms are stable under composition.

How does one prove that proper morphisms are stable under composition? One simply proves this property for finite type morphisms of schemes and separated morphisms. Then you're done.

If you stick to fields, all morphisms are separated so to prove that proper morphisms are stable under composition in this case, you simply have to prove that if you have a tower $K\subset L\subset M$ of finite degree field extensions, then $K\subset M$ is of finite degree. This is an easy fact. In conclusion, the proof of the statement for fields is easy and the statement itself doesn't give any nontrivial information in the case of fields.

I think Rayleigh added the additional hypotheses of "finite etale" to his statement, because he wanted to mimic the set-up of a "covering".

share|cite|improve this answer

Fix a scheme $S$.

Let $f:X\longrightarrow Y$ be a finite etale morphism of $S$-schemes, with $Y$ an integral $S$-scheme.

Note that $f$ is proper.

Therefore, if $Y$ is proper over $S$, we have that $X$ is also proper over $S$.

This is (I believe) the analogue of the statement in your question. (Take $S$ to be the spectrum of a field.)

share|cite|improve this answer
I see. Do we need at all that $f$ is étale? – Martin Brandenburg Oct 11 '11 at 15:14
Also, is there any nontrivial statement about fields (considering the original question) which we can derive from this? – Martin Brandenburg Oct 11 '11 at 15:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.