2
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • 8
    $\begingroup$ 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). $\endgroup$
    – Alex Degtyarev
    Commented Jan 16, 2015 at 0:30
  • $\begingroup$ @AlexDegtyarev, that works. Thanks. $\endgroup$
    – user44143
    Commented Jan 16, 2015 at 0:45
  • $\begingroup$ I don't see how a negative answer to this would give a negative answer to the linked question. $\endgroup$
    – Eric Wofsey
    Commented Jan 16, 2015 at 1:37
  • $\begingroup$ @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 $\{x,y\}$. $\endgroup$
    – user44143
    Commented Jan 16, 2015 at 1:53
  • $\begingroup$ 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. $\endgroup$
    – Emil Jeřábek
    Commented Jan 17, 2015 at 21:44

0

Reset to default

You must log in to answer this question.