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Let $E$ be a field and $F$ be a subfield of $E$ which is isomorphic to $E$. Then is $F$ equal to $E$? It seems to be clear but I couldn't prove it. Could you please explain this statement?

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  • $\begingroup$ I find it interesting to mention that, on top of the great examples given below, $\mathbb{C}$ also has a proper subfield which is isomorphic to it (assuming the axiom of choice) $\endgroup$ – 35T41 Jun 30 '17 at 17:46
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Counterexample: $E$ is the field of rational functions in infinitely many variables $x_1,x_2,\dots$, $F$ is the subfield of functions that depend only on $x_2,x_3,\dots$.

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  • $\begingroup$ @2000 I don't think the question restricted the claim to fields which have "only" one subfield isomorphic to them (otherwise take the subfield to be the field itself...), but rather that if a proper subfield is isom. to the whole field then both are one and the same. David's counterexample works pretty fine. $\endgroup$ – DonAntonio Mar 31 '16 at 20:15
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Counterexample

Consider $\mathbb{R}$ and consider the fields of rational functions $\mathbb{R}(x^2)$ and $\mathbb{R}(x)$. They are isomorphic but $\mathbb{R}(x^2) \subset \mathbb{R}(x)$.

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