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Let $(R,\mathfrak{m})$ be a local ring over a field. Suppose the ring has flat functions, i.e. $\mathfrak{m}^\infty\neq\{0\}$. (The prototype is of course $C^\infty(\Bbb{R}^p,0)$, or a quotient of it, by some finitely generated ideal.)

  1. Is there a notion of the 'infinite-radical' of an ideal, e.g. $\sqrt[\infty]{\mathfrak{m}^\infty}$ ? I would like in the 'geometric' case $\sqrt[\infty]{\mathfrak{m}^\infty}$ to be the defining ideal of the set $V(\mathfrak{m}^\infty)$. For example, for the ring $R=k[[\underline{y}]]\otimes C^{\infty}(\Bbb{R}^p,0)$ we have $\mathfrak{m}=(\underline{x},\underline{y})$, here $\underline{x}$ are the local coordinates on $(\Bbb{R}^p,0)$. Therefore $\mathfrak{m}^\infty=(\underline{x})^\infty$. Thus $\sqrt[\infty]{\mathfrak{m}^\infty}=(\underline{x})\neq\mathfrak{m}$.

  2. What is the definition of the height of $\mathfrak{m}^\infty$? For some rings $\mathfrak{m}^\infty$ is a prime ideal, (e.g. if the $\mathfrak{m}$-adic completion is a domain), is there some general characterization? Any general results about the prime sub-ideals $\mathfrak{p}\subset\mathfrak{m}^\infty$? At least for the case of $C^\infty(\Bbb{R}^p,0)$?

  3. Is there any place summarizing the known results about rings with the 'flat functions'? (Maybe extending some properties of flat elements of $C^\infty(\Bbb{R}^p,0)$ to more general rings? Some analogue of Tougeron's book for more general rings?)

I have asked this on matheoverflow, no replies :(

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  • $\begingroup$ Is 1. really what you want ? I mean, for usual radicals, you have $I\subset \sqrt{I^n}$. It seems you want something else because in your example $\sqrt[\infty]{\mathfrak{m}^\infty}\subset\mathfrak{m}$. My guess would be to put $x\in\sqrt[\infty]{I}\Leftrightarrow (x)^\infty\subset I$. But this definition does not behave as you want with respect to $V$... $\endgroup$
    – Roland
    Commented Oct 18, 2015 at 9:50
  • $\begingroup$ A agree that $\sqrt[\infty]{\mathfrak{m}^\infty}\subset\mathfrak{m}$ looks strange. But I would like to have set theoretically: $V(\sqrt[\infty]{\mathfrak{m}^\infty})=V(\mathfrak{m}^\infty)$. $\endgroup$ Commented Oct 18, 2015 at 10:03
  • $\begingroup$ Yep I understand, but I don't know, interesting question though (+1). I just think that you also want a precise definition of $V$ and $I$ (defining ideal) in this context, because otherwise the usual radical should work. $\endgroup$
    – Roland
    Commented Oct 18, 2015 at 10:21

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