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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18$\begingroup$ 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 LeuschkeCommented Jun 30, 2010 at 1:41
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4$\begingroup$ 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. $\endgroup$– ashpoolCommented Jun 30, 2010 at 2:23
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8$\begingroup$ 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 $n \ne 0$. $\endgroup$– SashaCommented Jun 30, 2010 at 8:51
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1$\begingroup$ In my experience, "quotient field" and "field of fractions" are used interchangeably to mean the localization of a domain R at the set R0 . I have never seen either term used to mean a ring modulo a maximal ideal. I would say your use of "quotient field" is non-standard. The relationship is: if $P\in Spec(R)$, then the quotient field of the domain $R/P$ is the residue field of the local ring $R_P$. At any rate, as I said above, if you want to talk about the rank of a module $M$ over a domain $R$, you want the dimension $M\otimes_RK$, where $K$ is the quotient field of $R$. $\endgroup$– Keenan KidwellCommented Jun 30, 2010 at 16:19
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2$\begingroup$ I guess I used the term quotient field not as a standard terminology but as an adjective followed by a noun: $A/\mathfrak{m}$ is a field which is a quotient. I admit it is very nonstandard but I tend to associate the word quotient with "modding out" rather than "fraction." $\endgroup$– ashpoolCommented Jun 30, 2010 at 17:25
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1 Answer
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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.
Bibliography
Ischebeck and Rao have published a monograph Ideals and reality: projective modules and number of generators of ideals on exactly this theme
answered Jun 30, 2010 at 9:05