Let $R$ be a ring, $M$ is a $R$-module. Then the Krull dimension of $M$ is defined by $\dim (R/\operatorname{Ann}M)$.

I can understand the definition of an algebra in a intuitive way, since the definition by chain of prime ideals agrees with the transcendental degree.

So, why dimension of module $M$ should be $\dim (R/\operatorname{Ann}M)$?

Please feel freely answering my question. Thanks.

  • $\begingroup$ Are you asking for an intuitive reason for defining $\mbox{dim} R / \mbox{Ann} M$ to be the (Krull) dimension of a module? $\endgroup$ – Elchanan Solomon Mar 26 '12 at 3:37
  • $\begingroup$ @IsaacSolomon : Thanks, it's exactly my question. may be I need more information to understand it. $\endgroup$ – Arsenaler Mar 26 '12 at 13:58

Let $M$ be a finitely generated module over $R$.
Its support $\operatorname{ supp}(M) \subset \operatorname {Spec}(R)$ is the set of prime ideals $\mathfrak p$ such that the stalk $M_{\mathfrak p}$ satisfies $M_{\mathfrak p}\neq 0$ or, equivalently thanks to Nakayama, that the fiber $M_{\mathfrak p}\otimes _R\kappa (\mathfrak p)$ is $\neq 0$.

It is then quite reasonable to say that the dimension of $\operatorname{ supp}(M)$ is some measure of the size of $M$, since outside of $\operatorname{ supp}(M)$ the fibers of $M$ are zero and on $\operatorname{ supp}(M)$ they are not zero, so that on $\operatorname{ supp}(M)$ the module $M$ behaves a bit like a vector bundle (and the associated sheaf $\tilde M$ is a vector bundle if $M$ is projective).

So we define $\operatorname {dim }(M) =\operatorname {dim}(\operatorname{ supp}(M))$.

And since it is easy to see that $\operatorname{ supp}(M))=V(\operatorname{ Ann}M)$ we arrive at
$$ \operatorname {dim }(M) =\operatorname {dim}(\operatorname{ supp}(M))= \operatorname {dim}(V(\operatorname{ Ann}M))= \operatorname {dim}(A/\operatorname{ Ann}M) $$
Summing up, we could say that the genuine dimension of $M$ is $\operatorname {dim}(\operatorname{ supp}(M))$ and that the formula $\operatorname {dim}(M)=\operatorname {dim}(A/\operatorname{ Ann}M)$ is just a technical device for computing it.

I have just remembered that there are two fantastic pictures of $M$ and $\operatorname{ supp}(M)$ in Miles Reid's Undergraduate Commutative Algebra: page 98 and right at the beginning of the book, as a frontispiece.
These illustrations are among the cleverest and most helpful I have ever seen in a mathematics book.

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  • $\begingroup$ Dear Georges, I am sorry but what do you mean by $\kappa(\mathfrak{p})$? $\endgroup$ – Arsenaler Mar 26 '12 at 14:00
  • $\begingroup$ Dear @msbaber, it is the ring of fractions $Frac(R/\mathfrak p)$ , which is also isomorphic to $R_\mathfrak p/\mathfrak p \cdot R_\mathfrak p$. $\endgroup$ – Georges Elencwajg Mar 26 '12 at 17:19
  • $\begingroup$ Dear Georges, may be I need to study more to understand your answer. Thank you very much. $\endgroup$ – Arsenaler Mar 29 '12 at 9:38
  • $\begingroup$ Dear @msnaber, in that case I warmly recommend Miles Reid's book, which is extremely user-friendly and probably the most elementary introduction to commutative algebra. Do not hesitate to post new questions if you are stuck. $\endgroup$ – Georges Elencwajg Mar 29 '12 at 17:39
  • $\begingroup$ Dear @GeorgesElencwajg, Could you please tell me what is a fiber of a module ? I know the fiber product of 2 maps but I do not know what is a fiber of a module $\endgroup$ – Arsenaler Apr 10 '12 at 4:07

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