1
$\begingroup$

We work in the category of algebraic varieties over some algebraically closed field $k$.

By infinite dimensional variety I mean a filtration: $$ V_0\subset V_1\subset V_2\subset\ldots $$ where each $V_i$ is a closed subvariety of $V_{i+1}$.

For any affine variety X (reduced and irreducible), is it possible to give its coordinate ring $k[X]$ a structure of infinite dimensional variety in some canonical way?

This is trivially true for $X=\mathbb{A}^n$. More generally, I was trying to use the Noether's normalization lemma or look at the coordinate ring locally at some smooth point, but with no luck.

$\endgroup$
  • 2
    $\begingroup$ If you want the filtration to be part of the structure of the variety, then I don't see why $\mathbb{A}^n$ has a canonical filtration. On the other hand, the coordinate ring of a variety is a countable dimensional $k$-vector space, so a suitable filtration by finite dimensional affine spaces always exists. $\endgroup$ – S. Carnahan Dec 7 '13 at 13:33
  • $\begingroup$ In the case of $\mathbb{A}^n$ I was thinking about the filtration given by the degree function. I know it's not canonical in any way, so you're totally right. $\endgroup$ – Tomasz Lenarcik Dec 7 '13 at 13:51
  • $\begingroup$ If you want to use degree in any way, doesn't that mean that the ideal of the variety should be defined by homogeneous polynomials ? That would allow you to "descend" the degree function. $\endgroup$ – aginensky Dec 7 '13 at 15:09
  • $\begingroup$ I agree with Scott. If you don't ask for any kind of naturality then this trivially exists, and if you do ask for any kind of naturality then it's unclear if there are any examples of positive dimension (I also don't even know what you would do with this). One thing you could do is weaken a filtration to a filtered colimit and then again this trivially exists, since every vector space is the filtered colimit of its finite-dimensional subspaces, so is naturally an ind-variety. $\endgroup$ – Qiaochu Yuan Dec 8 '13 at 19:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.