# A field has only one isomorphic subfield to itself?

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?

• 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) – 35T41 Jun 30 '17 at 17:46

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 $x_2,x_3,\dots$.
Consider $\mathbb{R}$ and consider the fields of rational functions $\mathbb{R}(x^2)$ and $\mathbb{R}(x)$. They are isomorphic but $\mathbb{R}(x^2) \subset \mathbb{R}(x)$.