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Consider $\mathbf{Q}(x,y)$, the rational functions in $x$ and $y$, as a vector space over $\mathbf{Q}$.

Let $\sigma$ be the map interchanging $x$ and $y$. Is there a basis for $\mathbf{Q}(x,y)$ which is invariant under $\sigma$?

Motivation: a negative answer to this question would give a negative answer to that question too.

The obvious way to construct a basis for $\mathbf{Q}(x,y)$ starts from $\{$ monomials$\ /\ $irreducibles$\}$, but then we have to choose one of $1/(x-y)$ and $1/(y-x)$, and a basis can't symmetrically include both.

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    I believe that both invariant and skew-invariant subspaces are of countable dimension. Important is that they are equal, hence, a basis exist: $e_i\pm f_i$, where $e_i$ and $f_i$ are some bases for the invariant and skew-invariant parts, respectively (numbered somehow). – Alex Degtyarev Jan 16 '15 at 00:30
  • @AlexDegtyarev, that works. Thanks. –  Jan 16 '15 at 00:45
  • I don't see how a negative answer to this would give a negative answer to the linked question. – Eric Wofsey Jan 16 '15 at 01:37
  • @EricWofsey, suppose we have an asymmetric basis $B$ for $Q(x,y)$. Let $f_i(x,y)$ be a listing of the polynomials in $Q(x,y)$. Find the first $i$ such that $f_i(x,y) \in (B-\sigma B)\cup(\sigma B-B)$. If $f_i(x,y) \in B$, choose $x$; otherwise choose $y$. So from a basis for $Q(x,y)$ we would get a choice function on ${x,y}$. –  Jan 16 '15 at 01:53
  • On the other hand, if $k({x_n:n\in\mathbb N})$ had a symmetric (invariant under all permutations of the indeterminates) basis for every field $k$, this would give a positive answer to the linked question. – Emil Jeřábek Jan 17 '15 at 21:44
  • @EmilJeřábek, I hope to post an answer of that form soon. –  Jan 17 '15 at 22:50

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