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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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    @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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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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    dim(M)= dim( Supp (M)) as an algebraic variety in Spec(R)2014-08-31