What's wrong with defining the rank of a finitely generated module over any (commutative) ring to be just the smallest number of generators? All books I know define rank only locally this way. But why not define it globally?

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    $\begingroup$ That's not the definition of rank in any book I know, it's the definition of minimal number of generators. Minimal number of generators is only defined locally because it's only well-defined locally -- you'd like every list of generators such that no element can be deleted to have the same length, and that fails (badly) for non-local rings. Rank is defined over domains (or, more complicatedly, as an n-tuple over for generically free modules over arbitrary rings) by relating it to vector-space dimension of a localization. $\endgroup$ – Graham Leuschke Jun 30 '10 at 1:41
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    $\begingroup$ If the ring is local, the minimal number of generators of a finitely generated module coincides with the rank (dimension of the vector space obtained by extending scalars to the residue field) by Nakayama's lemma. $\endgroup$ – ashpool Jun 30 '10 at 2:23
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    $\begingroup$ The minimal number of generators is not a good invariant. For example, it is not additive in exact sequences. E.g. let the ring be Z, and consider the short exact sequence $0 \to Z \to Z \to Z/nZ \to 0$ with $n \ne 0$. $\endgroup$ – Sasha Jun 30 '10 at 8:51
  • $\begingroup$ Don't extend scalars to the residue field; extend them to the quotient field. $\endgroup$ – Graham Leuschke Jun 30 '10 at 10:53
  • $\begingroup$ @kwan This is very much false. Consider $\mathbb{Z}_p$, the ring of $p$-adic integers. This is a discrete valuation ring. It's residue field is $\mathbb{F}_p$, the finite field of $p$ elements, but its quotient field is $\mathbb{Q}_p$. Note that, in this case, the residue field and quotient fields do not even have the same characteristic. The definition of rank you're thinking of is for modules over a domain, and it is the dimension of the extension of scalars to the quotient field of the domain. $\endgroup$ – Keenan Kidwell Jun 30 '10 at 13:08

Since your profile says you are interested in Algebraic Geometry, here are geometric considerations that might appeal to you.

Consider a projective module $P$ of finite type over a commutative ring $A$. It corresponds to a locally free sheaf $\mathcal F $ over $X=Spec(A)$. The rank of $\mathcal F $ at the prime ideal $\mathfrak p$ is that of the free $A_{\mathfrak p}$-module $\mathcal F_{\mathfrak p}$. The rank is then a locally constant function on $X$ and if $X$ is connected (this means that the only idempotents in $A$ are $0$ and $1$) it may be seen as an integer.

If $A$ is a domain, then $X$ is certainly connected and has a generic point $\eta$ whose local ring is the field of fractions $\mathcal O_\eta=K=Frac(A)$. The rank of $\mathcal F $ or of $P$ is then simply the dimension of the $K$ vector space $P\otimes_A K$.

Actually, if $A$ is a domain, this formula can be used to define the rank of any $A$-module $M$ (projective or not, finitely generated or not) : $rank(M)=dim_K ( M\otimes_A K) $ .
This is the definition given by Matsumura in his book Commutative Rings, page 84.
It corresponds to the maximum number of elements of $M$ which are linearly independent over $A$.

The minimum number of generators of $M$ (which started this discussion) is quite a different, but interesting invariant, which has been studied by Forster, Swan, Eisenbud, Evans,...
Geometrically it corresponds to the minimum numbers of global sections of $\tilde{M}$ which generate this sheaf at each point of $Spec(A)$.
Elementary example: Every non-zero ideal of a Dedekind domain is of rank one, can be generated by at most two elements and can be generated by one element iff it is principal. If the Dedekind domain is not a PID there always exist non free ideals which thus cannot be generated by less than two elements.

Ischebeck and Rao have published a monograph Ideals and reality: projective modules and number of generators of ideals on exactly this theme


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