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Let $R$ be a regular local ring and $M$ an $R$-module (not necessarily finite), then the injective dimension $\operatorname{id}_R(M)$ of $M$ is finite. When $M$ is finitely generated, we have $\dim(M)\leq \operatorname{id}_R(M)$.

When $M$ is an arbitrary $R$-module, does the relation $\dim(M)\leq \operatorname{id}_R(M)$ still hold?

(Here $\dim(M)$ denotes the supremum of the lengths of all chains of prime ideals in $\operatorname{Supp}(M)$.)

user26857
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nick
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  • See http://math.stackexchange.com/questions/623055/finite-injective-dimension and http://mathoverflow.net/questions/127379/does-there-exist-a-noetherian-ring-of-finite-injective-dimension-but-higher-krul. – Dietrich Burde Aug 20 '14 at 08:35
  • @DietrichBurde Unfortunately your links aren't helpful. – user26857 Aug 20 '14 at 21:04
  • @user26857, for the krull dimension of a module, I adopt the definition given in page 103 in Brodmann and Sharp's book "Local cohomology: an algebraic introduction with geometric applications". When $M$ is finitely generated, this definition is just $dim(R/ann(M))$. – nick Aug 21 '14 at 04:24
  • @nick I was used to consider as a definition of a module $\dim M=\dim R/\operatorname{Ann}(M)$ and this can differ from "the supremum of lengths of chains of prime ideal in $\operatorname{Supp}(M)$" when $M$ is not finitely generated. (Wikipedia also agrees with the first definition.) – user26857 Aug 22 '14 at 13:28

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