Is there a field which is the union of finitely many proper subfields?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
5
2
|
|
|||||||||||||||||||||||
|
|
8
|
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. |
||
|
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
3
|
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 |
||
|
|
|
2
|
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? |
||||||||||||||
|
