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.
 A: 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. 
A: 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.
