MathOverflow is a question and answer site for professional mathematicians. 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

Let K and L are fields,L is a sub field of K,and L is isomorphic to K,whether can we get K=L?If true,how to prove? Thanks.

share|cite|improve this question

closed as too localized by Andreas Blass, Franz Lemmermeyer, Emil Jeřábek, Leonid Positselski, Andrés E. Caicedo Nov 26 '12 at 6:53

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Aakumadula and Peter Mueller already answered that the answer is no. And it is far from being yes, even if you assume that $K$ and $L$ are algebraically closed. For instance, algebraically closed fields of a given characteristic are characterized by their transcendence degree. So take an algebraically closed field $K$ with infinite transcendence degree, take a transcendence basis, remove one element and call $L$ the algebraic closure of the field generated by this smaller set. Then $L$ is a strict subfield of $K$, but is isomorphic to $K$. – ACL Nov 23 '12 at 11:20
up vote 1 down vote accepted

No. ${\mathbb C}(X^2,Y)=L$ is a subfield of $K={\mathbb C}(X,Y)$ where $X,Y$ are algebraically independent variables over $\mathbb C$. Hence $L$ is isomorphic to $K$ but not equal.

share|cite|improve this answer
What is $Y$ good for? If $F$ is any field, and $X$ a transcendental over $F$, then $L=F(X^2)$, $K=F(X)$ is an example. – Peter Mueller Nov 23 '12 at 10:23
Yeah, you are right. I don't know why I took 2 variables! – Venkataramana Nov 23 '12 at 10:32

If K and L are F-field extensions, K/F and L/F are both finite dimensional, and the isomorphism from K to L is an F-homomorphism, then the proof is easy, but the general case seems difficult.

share|cite|improve this answer
You seem to be asking another question here, but you don't write all the hypothesis. Could you please clarify? – François Brunault Nov 23 '12 at 10:28
Moreover, this should appear in the body of the question or in a comment, not as an answer. – François Brunault Nov 23 '12 at 10:30
O,I am a freshman in this website.Lots of thing need to learn. – Nov 23 '12 at 10:37

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