Return to Answer

When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves rather badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. However, it is much more restricted than what I knew about before reading the comments on this question. In particular, it follows from the Forster-Swan Theorem that when $M$ is projective of rank $n$ then $\operatorname{mg}(M) \in \{n,n+1\}$. (In an earlier version of this answer I knew only that $\operatorname{mg}(M) \leq 2n-1$ and "guessed" that it could be that large over suitable Dedekind domains. Not a terrible guess, perhaps, but not the most educated one either...)

When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves rather badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. However, it is much more restricted than what I knew about before reading the comments on this question. In particular, it follows from the Forster-Swan Theorem that when $M$ is projective of rank $n$ then $\operatorname{mg}(M) \in \{n,n+1\}$. (In an earlier version of this answer I knew only that $\operatorname{mg}(M) \leq 2n-1$ and "guessed" that it could be that large over suitable Dedekind domains. Not a terrible guess, perhaps, but not the most educated one either...)

When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves very badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. The structure theory shows that if $r(M) = n$ then $n \leq \operatorname{mg}(M) \leq 2n-1$ and I give examples in the exercises with $(r(M),\operatorname{mg}(M))$ equal to $(1,1)$, $(1,2)$, $(2,2)$ and $(2,3)$. I would guess that all values in the above range are possible assuming that $\operatorname{Pic} R$ is sufficiently large, but I haven't thought seriously about it.

When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves rather badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. However, it is much more restricted than what I knew about before reading the comments on this question. In particular, it follows from the Forster-Swan Theorem that when $M$ is projective of rank $n$ then $\operatorname{mg}(M) \in \{n,n+1\}$. (In an earlier version of this answer I knew only that $\operatorname{mg}(M) \leq 2n-1$ and "guessed" that it could be that large over suitable Dedekind domains. Not a terrible guess, perhaps, but not the most educated one either...)

Added: Following quid's comments I checked out Fuchs's text on infinite abelian groups, and indeed the rank of an arbitrary abelian group is defined in a way so as to take $p$-primary torsion into account. (I simply didn't know this was true.) Thus the rank defined there would be the sum over all the values of the rank function in the sense of 1). In fact, for a perfectly arbitrary module $M$ over any commutative ring $R$, we can define a function $r_M$ with domain $\operatorname{Spec} R$ and taking values in the cardinal numbers, and then the rank of an abelian group in Fuchs's sense is the cardinal sum over all the values of the rank function.

All this seems to indicate, even more than my answer, that different notions of rank proliferate.

Added: Following quid's comments I checked out Fuchs's text on infinite abelian groups, and indeed the rank of an arbitrary abelian group is defined in a way so as to take $p$-primary torsion into account. (I simply didn't know this was true.) Thus the rank defined there would be the sum over all the values of the rank function in the sense of 1).

All this seems to indicate, even more than my answer, that different notions of rank proliferate.

Added Later: Quid also suggests consideration of the quantity $\operatorname{mg}(M)$, the minimal number of generators of an $R$-module $M$: this is a cardinal invariant which is (clearly) finite if and only if $M$ is finitely generated. Let $R$ be a Dedekind domain with fraction field $K$. For a maximal ideal $\mathfrak{p}$ of $R$ and a finitely generated $R$-module $M$, let $M[\mathfrak{p}^{\infty}]$ be the submodule of $M$ consisting of elements annihilated by some power of $\mathfrak{p}$. Then $M[\mathfrak{p}^{\infty}]$ is a finitely generated torsion module over the DVR $R_{\mathfrak{p}}$ and is thus a direct sum of $\operatorname{mg}(M[\mathfrak{p}^{\infty}])$ copies of $R_{\mathfrak{p}}/\mathfrak{p}^k R_{\mathfrak{p}}$. Let us define $tr(M,\mathfrak{p})$ to be this number of copies. (When $M$ is infinitely generated I believe one should also count copies of $K_{\mathfrak{p}}/R_{\mathfrak{p}}$ in a certain sense in order to recover Fuchs's $p$-primary torsion rank in the $R = \mathbb{Z}$ case. Let me omit this for now.)

Now define

$R(M) = r_M((0)) + \sum_{\mathfrak{p} \in \operatorname{MaxSpec} R} tr(M,\mathfrak{p})$

When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves very badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. The structure theory shows that if $r(M) = n$ then $n \leq \operatorname{mg}(M) \leq 2n-1$ and I give examples in the exercises with $(r(M),\operatorname{mg}(M))$ equal to $(1,1)$, $(1,2)$, $(2,2)$ and $(2,3)$. I would guess that all values in the above range are possible assuming that $\operatorname{Pic} R$ is sufficiently large, but I haven't thought seriously about it.