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Let $R$ be a commutative ring with unity, and $M$ an $R$-module. Assume that $I$ and $J$ are finitely generated ideals and $K$ another ideal of $R$. Let $\textbf{x}$ be a sequence of generators of $I$. Denote by "$\check{C}.grade_A (I, M)$", the "$\check{C}$ech grade" of $I$ with respect to $M$, and by $H^i_{\textbf{x}}(M)$ the $i$-th cohomology of $\check{C}ech$ complex of $M$ with respect to $\textbf{x}$. The definitions are:
$$\check{C}.grade_A (I, M):= \inf \{i\in N_0| H^i_{\textbf{x}} (M) \neq 0\}.$$ and $$\check{C}.grade_A (K, M) := \sup \{ \check{C}.grade_A (I, M)| I\text{ is a finitely generated ideal contained in }K\}.$$ It is a fact that if $\sqrt I =\sqrt J $ then $H^i_I(M)=H^i_J(M)$, for all i.

Questions: Assuming $\sqrt I =\sqrt K $,
1- can one deduce that $\check{C}.grade_A (I, M)=\check{C}.grade_A (K, M)?$
(if not:)
2- can one deduce that $\check{C}.grade_A (I, R)=\check{C}.grade_A (K, R)?$

Thank you.

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  • $\begingroup$ What do you mean (In the last line before the questions, and in the first question) with $H_I^i(M)$? $\endgroup$
    – Fred Rohrer
    Commented Jan 7, 2016 at 12:20
  • $\begingroup$ $I$ is finitely generated, say by $\textbf{x}$. then $H_I^i(M):=H^i_{\textbf{x}}(M)$; it is well-defined $\endgroup$
    – user 1
    Commented Jan 7, 2016 at 14:54

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