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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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  • $\begingroup$ I only remember this convention because if $R$ is a domain then using $R/\operatorname{Ann}_R(I)$ is really silly. $\endgroup$ – Dylan Moreland Jan 7 '12 at 2:42
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    $\begingroup$ @Leon: Eisenbud comments on the non-equivalence of these definitions in his book: see page 226. Is your textbook silent on the matter? $\endgroup$ – Zhen Lin Jan 7 '12 at 2:56
  • $\begingroup$ @Zhen: Thanks for the reference! I use Grillet, Greuel&Pfister, Kemper. There is very little said on the matter so far. $\endgroup$ – Leo Jan 7 '12 at 20:30
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They are not equivalent, but this never (?) leads to confusion.

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 $I=(x)$.

Of course, since you had to ask it is clear that that never I used above needs some qualification. But as soon as you get used to the funny convention the confusion disappears. Perhaps one should say that ït never leads again to confusion after one has unconfused oneself... Commutative algebra is fun!

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    $\begingroup$ 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. $\endgroup$ – Leo Jan 7 '12 at 2:42
  • $\begingroup$ dim(M)= dim( Supp (M)) as an algebraic variety in Spec(R) $\endgroup$ – kiranovalobas Aug 31 '14 at 20:03

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