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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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    I only remember this convention because if $R$ is a domain then using $R/\operatorname{Ann}_R(I)$ is really silly.2012-01-07
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    @Leon: Eisenbud comments on the non-equivalence of these definitions in his book: see page 226. Is your textbook silent on the matter?2012-01-07
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    @Zhen: Thanks for the reference! I use [Grillet](http://www.amazon.com/Abstract-Algebra-Graduate-Texts-Mathematics/dp/1441924507/ref=sr_1_1?s=books&ie=UTF8&qid=1325957420&sr=1-1), [Greuel&Pfister](http://www.amazon.com/Singular-Introduction-Commutative-Algebra/dp/3540735410/ref=sr_1_1?s=books&ie=UTF8&qid=1325957458&sr=1-1), [Kemper](http://www.amazon.com/Course-Commutative-Algebra-Graduate-Mathematics/dp/3642035442/ref=sr_1_4?s=books&ie=UTF8&qid=1325957458&sr=1-4). There is very little said on the matter so far.2012-01-07

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