# Equivalence of definitions of Krull dimension of a module

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.

• This is not an ambiguity: you are in the presence of two different definitions, but each of them is perfectly unambiguous. Commented Aug 22, 2013 at 18:33
• @ Mariano Suárez-Alvarez This is a real ambiguity. What's the meanning of $dim(M)$ when I'm studying a a more advanced book or an article?
– QED
Commented Aug 22, 2013 at 18:49
• I think this question is answered here although the original questions do not match. And if you ever get interested in Krull dimension in the noncommutative setting, there is an important version called the (Rentschler-Gabriel) Krull dimension. Commented Aug 22, 2013 at 19:38
• Ah, and seeing Matt E's solution below, I now see that Georges' answer over there is talking about a f.g. module :) Commented Aug 22, 2013 at 19:42

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 $0$.
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.