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Learning about the Krull dimension of modules for the first time. The standard definition for an $A$-module $M$ seems to be $\dim(M)=\dim(A/ann(M))$. This definition seems to make sense for finitely generated modules, since in that case $Supp(M) = V(ann(M))$. To me it seemed that the information we are after is the topological dimension of $Supp(M)$.

Yet this breaksdown in the non-f.g. case. In particular, let $M=\oplus_{n \geq 1} \mathbb{Z}/3^n\mathbb{Z}$. Then $ann(M)=0$ so we should have that $\dim(M)=1$, yet $Supp(M)$ consists of a single point the ideal $(3)$. So its dimension as a topological space is $0$.

So what explains this discrepancy, are we not actually after the dimension of $Supp(M)$? Do we even care about Krull dimension in the non-finitely generated case?

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    $\begingroup$ Not really sure what your question is. Obviously, which definition is the "right" one to use in the non-f.g. case depends on what you want to use it for and what properties you want it to have. $\endgroup$ Oct 25, 2015 at 19:42
  • $\begingroup$ @EricWofsey I think that is a fair criticism, my question is a bit soft. I suppose what I was wondering is what motivates the standard definition, given that there is a nice geometric alternative with many of the same properties. Why in general does it seem to be the preferred definition? $\endgroup$
    – Bill Trok
    Oct 26, 2015 at 6:16
  • $\begingroup$ I wouldn't assume that the standard definition is necessarily "better". To large extent it is probably a historical accident that it became the most commonly stated definition, and it also has the advantage of being more elementary to define. $\endgroup$ Oct 26, 2015 at 6:20

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As you noticed, there are two different Krull dimensions for arbitrary modules which agree for finitely generated ones, but in general they can differ. (With respect to the question if we care about this in general, the answer is positive: the local cohomology modules are usually not finitely generated, but we care about their Krull dimension.)

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