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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.

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

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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
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2 Answers 2

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.

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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
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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.

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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. –  bo.gu Nov 23 '12 at 10:37
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