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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 $\dim (R/\operatorname{Ann}M)$?

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- Suggestions for further topics in Commutative Algebra

Please feel freely answering my question. Thanks.

- Non-Noetherian rings with an ideal not containing a product of prime ideals
- Infinite linear independent family in a finitely generated $A$-module
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- Noetherian and Artinian modules
- If $\{M_i\}_{i \in I}$ is a family of $R$-modules free, then the product $\prod_{i \in I}M_i$ is free?
- Showing the polynomials form a Gröbner basis
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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.

**Edit**

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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