In what follows we will always use this notation:

$R$ will be a commutative noetherian ring with unity, $X=\mathrm{Spec}\:R$, $f\colon X\rightarrow X$ a self-morphism of schemes, $\varphi\colon R\rightarrow R$ the ring endomorphism corresponding to $f$, and $Y=\mathrm{Spec}\:R/\mathfrak{a}$ a closed subscheme of $X$ defined by an ideal $\mathfrak{a}$ of $R$.

A closed subscheme $Y=\mathrm{Spec}\:R/\mathfrak{a}$ is called $f$-invariant if the restriction $f\restriction_Y$ gives a scheme self-morphism of $Y$ to itself. This means, in particular, that $\varphi(\mathfrak{a})\subset\mathfrak{a}$. We say $Y$ is completely $f$-invariant if 1) $Y$ is $f$-invariant and 2) $f^{-1}(Y)=Y$ as sets. It can be quickly checked that $Y$ is completely $f$-invariant if and only if $\mathrm{rad}\:(\mathfrak{a})=\mathrm{rad}\:(\varphi(\mathfrak{a})R)$.

Question. Is there an example of a local ring $(R,\mathfrak{m})$ and a self-morphism $f\colon X\rightarrow X$ with an $f$-invariant closed subscheme $Y=\mathrm{Spec}\:R/\mathfrak{a}$ satisfying:

a) The closed point $\{\mathfrak{m}\}$ is completely $f$-invariant (equivalently, $\varphi(\mathfrak{m})R$ is an $\mathfrak{m}$-primary ideal).

b) $\varphi^n(\mathfrak{m})R\subseteq\mathfrak{m}^2$ for $n\gg0$.

c) $Y$ is completely $f$-invariant (equivalently, $\mathrm{rad}\:(\mathfrak{a})=\mathrm{rad}\:(\varphi(\mathfrak{a})R)$).

d) $\varphi^n(\mathfrak{a})R\not\subset\mathfrak{a}^2$, for all $n$?

Or is there a proof showing a) $+$ b) $+$ c) $\Rightarrow$ $\varphi^n(\mathfrak{a})R\subset\mathfrak{a}^2$, for $n\gg0$?

Remark. To produce such examples, one can use $k$-linear endomorphisms of a ring of formal power series $k[[x_1,\ldots,x_n]]$ over a field $k$ by sending each $x_i$ to a power series $f_i$. Since this ring is regular, condition a) will imply that $\varphi$ is in fact flat. (cf. Theorem 23.1 in Matsumura's Commutative ring theory.) Even in this special case I would like to know whether a) $+$ b) $+$ c) $\Rightarrow$ $\varphi^n(\mathfrak{a})R\subset\mathfrak{a}^2$, for $n\gg0$?

What is this good for? Let $M$ be a finite $R$-module. In the above setting, one can use the decreasing sequence of ideals $\{\varphi^n(\mathfrak{a})R\}_{n\geq1}$ to define local cohomology-type objects (notation mine): $$\mathrm{H}^i_Y(X,f,M):=\varinjlim_n\:\mathrm{Ext}^i_R(R/\varphi^n(\mathfrak{a})R,M).$$ When the sequence $\{\varphi^n(\mathfrak{a})R\}_{n\geq1}$ defines the same topology as the $\mathfrak{a}$-adic topology on $R$, one gets the ordinary local cohomology modules $\mathrm{H}^i_{\mathfrak{a}}(M)$. This case has been studied in Local cohomology and pure morphisms, Illinois Journal of Math., (51), 287-298 (2007) by A. K. Singh and U. Walther and the Frobenius endomorphism in positive characteristic is a classic example of such an endomorphism. The case where $\{\varphi^n(\mathfrak{a})R\}_{n\geq1}$does not define the same topology as the $\mathfrak{a}$-adic topology has not been studied, as far as I know. However, even then one can obtain information about ordinary local cohomology $\mathrm{H}^i_{\mathfrak{a}}(M)$ from $\mathrm{H}^i_Y(X,f,M)$. For instance, when $Y$ is completely $f$-invariant, one can show isomorphisms ($R$ does not need to be local here) $$\mathrm{H}^i_{\mathfrak{a}}(M)\cong\varinjlim_n\:\mathrm{H}^i_{Y_n}(X,f,M)$$ where $Y_n$ is $\mathrm{Spec}\:R/\mathfrak{a}^n$. But I would like to see an example as I asked in the question, to see if it is even possible to have a good application.

  • 1
    $\begingroup$ This is indeed a very interesting question; I hope somebody has the answer! In a paper with Yongwei Yao (Criteria for flatness and injectivity, link.springer.com/article/10.1007%2Fs00209-011-0910-y; it's also on arXiv), we make use of this sort of notion, or rather the negation of it (that is, we assume c+the negation of d over subsets of Spec R). See 3.2-3.5 of that article if you're interested $\endgroup$ – Neil Epstein Feb 2 '13 at 18:45
  • $\begingroup$ Dear Neil: Thank you for the reference. $\endgroup$ – Mahdi Majidi-Zolbanin Feb 3 '13 at 17:23

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