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Let $K$ be a field and $K(X)$ be the field of its rational functions.

Now let $\phi \in K(X)$ be a rational function such that $K(\phi) \neq K(X)$.

Now, since $\phi$ is transcendental over $K$, $K(\phi)$ is isomorphic to $K(X)$.

Is this a correct example of a field being isomorphic to its subfield?

Are there any other examples?

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8  
Yes. But there are easier examples, like $K(X_2, X_3, \ldots) \subsetneq K(X_1, X_2, \ldots)$... –  Zhen Lin Feb 26 '12 at 13:14
4  
@Zhen: Do you think yours is easier than $K(X^2) \subset K(X)$? –  GEdgar Feb 26 '12 at 14:41
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1 Answer

up vote 4 down vote accepted

If $K_1$ and $K_2$ are algebraically closed fields of the same uncountable cardinality and of the same characteristic, then they also have the same transcendence degree over their prime field, and are therefore isomorphic.

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1  
You mean "cardinality", right? –  Martin Brandenburg Feb 26 '12 at 14:12
    
Yes, thank you. I changed "cofinality" to "cardinality". –  g.castro Feb 26 '12 at 16:31
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