When is the rank the biggest number for which $\Lambda^m(M) \neq 0$? I was doing some theory of Dedekind domains, and I found very useful to use the language of exterior algebra to prove the main results for finitely generated modules over Dedekind domains. I was, however, very frustrated to notice that the following theorem : 
$$
\forall k > \mathrm{rank}_A(M), \quad \Lambda_A^k(M) = 0
$$
does not hold as often as I'd like. I managed to prove something by myself, so I am not sure I am as general as possible (if as general as possible exists, maybe it depends on the context). The most general I got was as follows : 
Let $A$ be a (commutative with $1$) Noetherian integral domain (in particular, the rank of $M$ is defined as the dimension of the localization of $M$ at $(0)$). If $M$ is a finitely generated projective $A$-module of finite rank, then for all $m > \mathrm{rank}_A(M)$, $\Lambda_A^m(M) = 0$. 
Question : Does this fact hold under more general assumptions than $A$ Noetherian and $M$ finitely generated and projective? The projective assumption is the one which is most frustrating... note that in particular projective implies torsion-free ; the torsion-free assumption is of course necessary to stay in full generality. Or perhaps the statement I have is the wrong one ; what should I be looking for?
This is because torsion modules don't "behave so well" (as far as zero-ness is concerned) under $\Lambda(-)$ (in particular, they have rank zero and their exterior powers are not zero in general ; for example, if $\mathfrak a \trianglelefteq A$ is an ideal, then $\Lambda_A^n( (A/\mathfrak a)^n ) \supseteq \bigotimes_{i=1}^n \Lambda_A^1(A/\mathfrak a) \simeq A/\mathfrak a \neq 0$.
 A: Edit: (Short version over Noetherian Domain) Exactly when M is projective
One "extension" of your statement is that if $M$ has rank $r$, then  $\mbox{rank}\wedge^k M = {r \choose k}$.  In particular, when $k > r$ the module $\wedge^k M$ has rank $0$.  This holds in any commutative ring for any finite rank module.  (It comes considering $R^r \to M$, and the fact that tensor product commutes with exterior powers.)
Now, this is obviously is a weaker consequence than you'd like-- so let's see how it relates to the original statement.  Lets extend our notion of rank to every prime $P \in Spec \ R$, so that $\mbox{rank}_P(M) = \dim (M \otimes R_P/PR_P)$. (For Noetherian rings, this is the same as the min number of generators of $M_P$) We still have $$\mbox{rank}_P (\wedge^k M )= {\mbox{rank}_P \ M \choose k}.$$  
When our ring is Noetherian, Nakayama's lemma implies that if $\mbox{rank}_P (L) = 0$ then $R_P \otimes L = 0$.  And if this is true for all maximal ideals $P$, then $L$ must itself be zero.  
The punchline is as follows: If  $M$  such that $\mbox{rank}_0 \ M  = \mbox{rank}_P \  M$ for all maximal ideals $P$, the above equality will guarantee the ranks over every prime of the exterior powers $> \mbox{rank}_0 \  M$ are zero, hence $\wedge^k M = 0$.  Conversely, if $\wedge^k M = 0$ for $k = \mbox{rank}_0 M$, then the rank over every prime is zero, hence $\mbox{rank}_P \ M \leq \mbox{rank}_0 \ M$.  Nakayama shows that localizing can only decrease the rank, so $\mbox{rank}_P \ M = \mbox{rank}_0 \ M$.  
So in a Noetherian ring, $\forall k > \mbox{rank}_0 M, \wedge^k M = 0$ holds if and only if $\mbox{rank}_0 \ M = \mbox{rank}_P M$ for all primes $P$. 
This gives us some interpretations of what you've proved : Sufficiently small open subsets of $Spec D$ of a dedekind domain  principal ideal domains, so for any prime $P \in Spec D$ you can find an $f \in D$ so that $M_f$ is a module over a principal ideal domain, and so takes the form $M_f = D_f^r \oplus (D_f/p_1^{e_1}) \oplus \dots$.
Then the generic rank (the rank over $0$) is just $R$.  But the rank over $P D_f = p_1 D_f$ will be $r$ plus the number of copies of the $p_1$ torsion.  So if $M$ is torsion free, then we're set.  And this is necessary and sufficient.
But when things are two dimensional:  torsion free just doesn't cut it.  Take $R = k[x,y]$.  Then the module $M = (x,y)R$ is torsion free, but its generic rank 1, whereas its rank over $(x,y)$ is two.  In fact, this counterexample extends to any $\geq 2$ dimensional Noetherian domain.  
So you really need locally free (i.e. there exist elements $f_i$ such that $(f_i)R = R$ and $M_{f_i}$ free over $R_{f_i}$).  This is equivalent to projective or flat for a f.g. module over a Noetherian ring, but locally free is the easiest to work with.  
Locally free is sufficient, by a variant on the argument for torsion-free above. To see that it's necessary suppose that $M_P$ is not free.  Take a minimal generating set $\{n_i\}$.  Then the $n_i$ have a relation $\sum r_i n_i = 0$, so that when we localize at $0$, the $r_i$ become invertible so it takes fewer $n_i$ to generate the module, and so the rank decreases.
