19
$\begingroup$

Is there a field $F$ such that $F \cong F(X,Y)$ as fields, but $F \not \cong F(X)$ as fields?

I know only an example of a field $F$ such that $F$ isomorphic to $F(x,y)$ : this is something like $F=k(x_0,x_1,\dots)$. But in this case we have $F \cong F(x)$.

This is related to this question on MO. However, it doesn't follow obviously from $R \not \cong R[x]$ that $\text{Frac}(R) \not \cong \text{Frac}(R)(X)$ (I'm not even sure that the given $R$ is an integral domain...).

These questions could be relevant: (1) ; (2).

Thank you very much!

$\endgroup$
7
  • $\begingroup$ One could ask about a generalization: let $(k_i)_{i \in I \cup J}$ be distinct cardinals, is there a field $F$ such that $F \cong F(X_s, s \in k_i)$ for $i \in I$ and $F \not \cong F(X_s, s \in k_j)$ for $j \in J$. $\endgroup$
    – Watson
    Commented Dec 26, 2016 at 21:49
  • $\begingroup$ It would be also nice to have an example of a field $F$ such that $F[x] \not \cong F \cong F[x,y]$ (and the same generalization as above can be done), or a ring $R$ such that $R \not \cong R[x,y]$ but $R[x] \cong R[x,y,z]$. $\endgroup$
    – Watson
    Commented Dec 26, 2016 at 22:04
  • $\begingroup$ Writing $\phi:F(X,Y)\to F$ we must have $F=\phi(F)(A,B)=\phi^2(F)(A,B,C,D)=\cdots$ in which one might hope that $F=(\bigcap \phi^n(F))(A,B,C,D,\cdots)$ but I see no way to prove that. $\endgroup$
    – anon
    Commented Dec 27, 2016 at 15:35
  • $\begingroup$ In my above comment, I wanted to say "of an integral domain $F$" ($F \cong F[x,y]$ isn't possible for a field $F$, since $x$ isn't invertible). Notice that there is no field $F$ such that $F(x,y) \not \cong F \cong F(x)$, because we would have $F \cong F(x) \cong F(y) \cong F(x)(y) = F(x,y)$. $\endgroup$
    – Watson
    Commented Jan 3, 2017 at 19:30
  • $\begingroup$ Cross-posted: mathoverflow.net/questions/258647 $\endgroup$
    – Watson
    Commented Jan 3, 2017 at 19:30

0

Reset to default

You must log in to answer this question.