Let $k$ be any field and $K:=k(X_1,X_2,\dots)$ be the field of rational functions over $k$ in countably many variables. Now $K$ has the interesting property that it is isomorphic to a transcendental extension of itself namely $K\cong K(X)$.

Are there any other examples of this phenomenon or is the following true?

When $K$ is a field that is isomorphic to a transcendental extension of itself, then there is some field $k$ s.t. $K\cong k(X_1,X_2,\dots)$.


As written, no.

$\mathbb C$ is isomorphic to $\overline{\mathbb C(X)}$ which is a transcendental extension, but $\mathbb C$ is not isomorphic to anything of the form $k(X_1,X_2,\ldots)$ because the latter is not algebraically closed ($X_1$ does not have a square root, for example).

  • $\begingroup$ Why is $\mathbb{C}$ isomorphic to $\overline{\mathbb{C}(X)}$? $\endgroup$ – Unit Jul 18 '18 at 22:05
  • 2
    $\begingroup$ @Unit because they are both algebraically closed fields of the same characteristic and transcendence degree over their prime field. Any two such fields are isomorphic. $\endgroup$ – Achilles Jul 18 '18 at 23:24
  • $\begingroup$ I see that by considering the theory consisting of the diagram of $K(X_1,X_2,\dots)$ and the theory of $K$ we get a field $F$ which is an extension of $K(X_1,X_2,\dots)$ and elementarily equivalent to $K$. But how can i make $F$ actually isomorphic to $K$? $\endgroup$ – Achilles Jul 18 '18 at 23:29
  • $\begingroup$ @Achilles: On further inspection, the argument I thought I had in mind for the last paragraph was terminally confused. All I'd get would be an $F$ elementarily extending $K$ such that $F$ is isomorphic to an extension of $F(X_1,X_1,\ldots)$, which is probably too far removed from the question to be interesting. I have removed the claim. $\endgroup$ – Henning Makholm Jul 19 '18 at 1:04

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