# definition of Krull dimension of a module

Let $R$ be a commutative ring with $1$. We know that the Krull dimension of $R$ is by definition the length of the longest chain of prime ideals of $R$.

Now if $M$ is a $R$-module, the Krull dimension of $M$ is by definition $\dim(M):=\dim(R/\mathrm{Ann}_R(M))$. Since every ideal $I$ of $R$ is also a $R$-module, the Krull dimension of $I$ is $\dim(I)=\dim(R/\mathrm{Ann}_R(I))$.

However, in the literature, the Krull dimension of an ideal is $\dim(I):=\dim(R/I)$.

Are the two definitions equivalent?

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I only remember this convention because if $R$ is a domain then using $R/\operatorname{Ann}_R(I)$ is really silly. – Dylan Moreland Jan 7 '12 at 2:42
@Leon: Eisenbud comments on the non-equivalence of these definitions in his book: see page 226. Is your textbook silent on the matter? – Zhen Lin Jan 7 '12 at 2:56
@Zhen: Thanks for the reference! I use Grillet, Greuel&Pfister, Kemper. There is very little said on the matter so far. – Leon Jan 7 '12 at 20:30

Suppose for example that $R$ is a domain, so that $\mathrm{ann}_RI=0$. Then seeing $I$ as a module gives $\dim I=\dim R$, which the other definition gives $\dim I=\dim R/I$, which are usually different —for a boring example, take $R=k[x]$ and $I=(x)$.
Aha, a counterexample would be: $R=K[x,y], M=I=\langle y\rangle$. Then $dim(R/I)=dim(K[x])=1\neq2=dim(K[x,y]/0)=dim(R/\mathrm{Ann}_{R}(I))$. Hmm, is there any reason to define the Krull dimension of a module that way? I mean, it doesn't agree with the dimension of the ideal, nor with the vector space dimension when $R$ is a field? Does it at least agree with "longest chain of submodules of M"? I'm guessing not, since the case $M=_RR$ would mean $dim(R)=$longest chain of ideals, which is wrong, since we must consider only prime ideals. – Leon Jan 7 '12 at 2:42