Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In what classes of fields does CSB hold? That is to say, in what classes of fields is it true that if there exist embeddings $F\to K$ and $K\to F$ then $F$ and $K$ must be isomorphic?

I know this holds for algebraically closed fields, but all of the counter-examples I've seen are variations on the same idea ($F=\overline{\mathbb Q(x_0,x_1,x_2,...)}$ and $F(x)$).

Does CSB hold for fields of finite transcendence degree? What about for fields with no algebraically closed subfields?

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

A trivial example of a family that satisfy CSB is the set of completions $\{\mathbb{Q}_{v}\}_{v}$ of $\mathbb{Q}$. They satisfy the condition that you require about not containing any ACF, however their transcendence degree is big. This is a trivial example in the sense that even if one of such fields is embedding in the other then they are isomorphic. More generally if you take all locally compact fields the same should be true.

share|improve this answer
    
I guess you meant non-discrete locally compact fields. –  Ramiro de la Vega Feb 28 '12 at 21:40
    
@Ramiro: You are right, of course, I'm only thinking local fields. Saludos. –  Guillermo Mantilla Feb 29 '12 at 5:01
add comment

No. For any field $k$, there exists a pair of isogenous elliptic curves over $k$ that are not isomorphic. The dual isogeny yields suitable maps of function fields that are not isomorphisms. If your base field has finite transcendence degree and does not contain an algebraically closed subfield, then the same is true for the function fields.

Edit: I see that this answer is essentially identical to Robin Chapman's answer to a more general question, and to KConrad's answer to a slightly different question.

share|improve this answer
add comment

Your Answer

 
discard

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.