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I've seen two definitions of Krull dimension of a module $M$ over a (commutative) ring $R$, and their equivalence does not seem obvious:

Matsumura on page 31 of his book Commutative Ring Theory defines it as

$\dim M=\dim R/\operatorname{Ann}(M)=$ maximal length of a chain of primes in $V(\operatorname{Ann}(M)).$

Enochs and Jenda on page 54 of Relative Homological Algebra define it as

$\dim M=\dim {\rm Supp}(M)=$ maximal length of a chain of primes in ${\rm Supp}(M).$

I guess this "maximal length" is the same for two sets above, but what's the proof? Otherwise how are two definitions equivalent?

PS: I already know that $\mathrm{Supp}(M)\subseteq V(\operatorname{Ann}(M))$ and that both definitions are equivalent for finitely generated modules.

rschwieb
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QED
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2 Answers2

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These definitions are not the same in general, if $M$ is not f.g.

Consider the module $\mathbb Q_p/\mathbb Z_p$ over $\mathbb Z_p$. Its annihilator is $0$, so the first definition gives dimension $1$. On the other hand, its support is the closed point of Spec $\mathbb Z_p$, and so the second definition gives dimension $0$.

If you are reading an article that applies the notion of dimension in the non-f.g. context, then you will either have to look and see if the author defines their terms, or else determine from the context (e.g. how they argue) which definition is in use.

Matt E
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Another example (of non-finitely generated module) showing the two definitions of dimension give different results is the $\mathbb Z$-module $M=\oplus_{n\ge1}\mathbb Z/2^n\mathbb Z$. (Maybe there are users who find hardly understood MattE's answer using the $p$-adics.)

We have $\operatorname{Ann}M=(0)$, and therefore $\dim M=1$, if one uses the first definition of dimension.

On the other side, $\operatorname{Supp}M=\{2\mathbb Z\}$, and then $\dim M=0$, if one uses the second definition.

user26857
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