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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2$\begingroup$ 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 $K$. $\endgroup$– ACLCommented Nov 23, 2012 at 11:20
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2 Answers
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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.
answered Nov 23, 2012 at 10:02
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3$\begingroup$ 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. $\endgroup$ Commented Nov 23, 2012 at 10:23
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$\begingroup$ Yeah, you are right. I don't know why I took 2 variables! $\endgroup$ Commented Nov 23, 2012 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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$\begingroup$ You seem to be asking another question here, but you don't write all the hypothesis. Could you please clarify? $\endgroup$ Commented Nov 23, 2012 at 10:28
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1$\begingroup$ Moreover, this should appear in the body of the question or in a comment, not as an answer. $\endgroup$ Commented Nov 23, 2012 at 10:30
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$\begingroup$ O，I am a freshman in this website.Lots of thing need to learn. $\endgroup$– bo.guCommented Nov 23, 2012 at 10:37