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If $k$ is any field, then the field $K=k(X_0,X_1,\dots)$ of rational functions in infinitely many variables satisfies $K(X)\cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_{n+1}$). My question is, does the converse hold? That is:

Suppose $K$ is a field such that $K\cong K(X)$. Must there exist a field $k$ such that $K\cong k(X_0,X_1,\dots)$?

(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)

Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $R\cong R[X]$ but $R\not\cong S[X_0,X_1,\dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.

In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.

Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $F\cong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)

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  • $\begingroup$ Might it be worth asking this interesting question on MathOverflow? $\endgroup$ – Watson Jul 23 '18 at 17:56
  • $\begingroup$ Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want. $\endgroup$ – Eric Wofsey Jul 23 '18 at 18:10
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    $\begingroup$ Do you know that the field of fractions of the rational group algebra of the Baer-Specker group is not a counterexample? Given the links in the question, I’m guessing you’ve thought about it. $\endgroup$ – Jeremy Rickard Jun 15 at 13:09
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    $\begingroup$ @JeremyRickard: That's a neat idea. I don't actually recall thinking through it before (I think the constructions I was trying were all countable). $\endgroup$ – Eric Wofsey Jun 15 at 14:50
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    $\begingroup$ It’s an example of something. Either it’s a counterexample for your question, or the rational group algebras of the nonisomorphic groups $\mathbb{Z}^\omega$ and $\mathbb{Z}^\omega\oplus\mathbb{Z}^{(\omega)}$ have isomorphic fields of fractions, answering my question. $\endgroup$ – Jeremy Rickard Jun 16 at 9:25

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