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Is there a field which is the union of finitely many proper subfields?

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No. A field is a vector space over any subfield, and here we get to the old problem of whether a vector space can be a union of finitely many proper subspaces - the answer is no unless the space is finite (finitely many elements), which is trivial for fields. I will elaborate on this in a detailed answer soon. –  darij grinberg Sep 22 '10 at 12:38
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Whether or not it's self-explanatory, it seems to me that it would be appropriate to include the question in the body of the question, along perhaps with any motivation you have for asking it. –  JBL Sep 22 '10 at 12:42
    
@darij: What about finite fields? –  Harry Gindi Sep 22 '10 at 13:20
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Deleted partition of unity tag, put question in body -- still could use some motivation from heiko. –  Cam McLeman Sep 22 '10 at 13:44
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Dear darij: The link with subspaces of a vector space seems not so obvious (though intuitively appealing) since in contrast with group-index there are examples (even finitely generated in char. 0) of fields $K$ and (necessarily proper) subfields $k$ and $k'$ such that $[K:k]$ and $[K:k']$ are finite but $[K:k \cap k']$ is infinite (hint: there are infinite groups generated by a pair of finite subgroups). I haven't given the matter any real thought, but anyway it may require an idea (perhaps tiny) to make the argument work. –  BCnrd Sep 22 '10 at 15:50
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3 Answers 3

There are 3 cases.

Case 1. The field is finite. Then, as Charles Matthews pointed out, the primitive element theorem does the job.

Case 2: The intersection of the subfields is infinite. This is covered by darij grinberg's comment. The field is a vector space over this intersection and subfields are subspaces, but a vector space over an infinite field cannot be a union of finitely many proper subspaces.

Case 3: The field is infinite but the intersection of the subfields is finite. Then everything is a vector space over $\mathbb F_p$ where $p$ is the characteristic of the field. Let us forget about fields and do linear algebra over $\mathbb F_p$. Let me call a subfield (more generally, an affine subspace) big if its codimension over $\mathbb F_p$ is finite, and small otherwise. We only need two basic facts: Let $X$ be an affine space over $\mathbb F_p$ and $Y,Z$ affine subspaces with nonempty intersection, then

(1) if $Y$ and $Z$ are big, then $Y\cap Z$ is also big;

(2) if $Y$ is big and $Z$ is small, then $Y\cap Z$ is small in $Y$.

(This and everything below does not actually depend on the base field.)

If all our subfields are big, then so is their intersection. Then it is infinite and we are in Case 2. So there must be small subfields in our union. Let us just remove them and show that the union will remain the same (so we are in Case 2 again):

Lemma. Suppose that $X$, an affine space over $\mathbb F_p$, is covered by a finite collection of subspaces. Then the big subspaces from the collection also cover $X$.

Proof (pretty standard). Use induction in the number of subspaces. Suppose there is a point $a\in X$ that does not belong to the union of the big subspaces. Let $Y$ be one of the big subspaces (if there are any) and $Y'$ be the subspace parallel to $Y$ containing $a$. Consider the covering of $Y'$ by its intersections with our subspaces. By the basic fact (2), $a$ is not covered by the big intersections in $Y$. And at least one of the subspaces (namely $Y$) does not produce an intersection, so we can apply the inductive hypothesis in $Y'$.

It remains to consider the case when all subspaces are small. In this case, they just cannot cover $X$. Indeed, let $Y$ be one of them, $H$ any hyperplane containing $Y$ and $H'\ne H$ any hyperplane parallel to $H$. Replace $X$ by $H'$ and proceed by induction as above.

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See A Bialynicki-Birula, J Browkin, A Schinzel, On the representation of fields as finite unions of subfields, Colloq Math 7 (1959) 31-32, MR 22 #2601. At the risk of violating copyright, here's the review, by S W Golomb:

This article contains a short proof of the fact that no field can be represented as a finite union of proper subfields. A counterexample is exhibited to the corresponding assertion for integral domains.

EDIT: The paper may be available at http://matwbn.icm.edu.pl/ksiazki/cm/cm7/cm717.pdf

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The reason it fails in the case of finite fields is the primitive element theorem (over the prime subfield, even). This is not quite an argument in the general case because of possible inseparability. Homework?

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It does take some of the structure theory of fields to handle the finite case, I think. I too am wondering if this is a homework problem, since it sounds familiar. –  Jim Humphreys Sep 22 '10 at 14:40
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Jim, by "structure of finite fields" I suppose you mean that the lattice of subfields matches that of the divisors of the field degree over the prime field. Amusingly, the proof that the sum of the sizes of the proper subfields (very crude idea!) manages to be strictly smaller than the size of the ambient field also comes up in the crude Mobius-inversion proof that there's an irreducible polynomial of each degree over fields of prime size. –  BCnrd Sep 22 '10 at 15:44
    
The proof that the multiplicative group is cyclic gives something stronger, but that is an inversion proof too (Serre's Course). –  Charles Matthews Sep 22 '10 at 15:51
    
I left "structure theory of fields" vague, since there seem to be several ways to approach the specific question when the field is finite. –  Jim Humphreys Sep 22 '10 at 16:52
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