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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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- Are bimodules over a commutative ring always modules?
- An ideal that is maximal among non-finitely generated ideals is prime.
- $k]$ is UFD for $k$ field

Please feel freely answering my question. Thanks.

- Radical ideal of $(x,y^2)$
- Embedding of free $R$-algebras
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- Normalization of $k$
- Question about fields and quotients of polynomial rings
- Flatness and intersection of ideals
- Calculating $Spec(\mathbb{C}/\langle x^2 \rangle)$
- Does every Noetherian domain have finitely many height 1 prime ideals?
- How to prove every radical ideal is a finite intersection of prime ideals?
- Generators for the intersection of two ideals

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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