M a finitely generated module over a commutative ring A. I can't think of an example of two maximal linearly independent subsets of M having different cardinality. I know that they all have the same cardinality if A is integral domain. Any suggestions are welcome!

1$\begingroup$ kwan, I believe that this question is based upon a misunderstanding of Georges' answer here mathoverflow.net/questions/29993/rankofamodule. When A is a domain, yes, the quantity you're talking about is welldefined and equal to rank. See Bruns and Herzog Section 1.4 for more in this case. When A is not a domain, this is not a very interesting quantity, since it does not obey the usual rules from linear algebra. E.g. let R=k[x,y]/(xy). Then R/(x) and R/(y) contain no lin.indep. elements (each is killed by an element of R), but R/(x)\oplus R/(y) contains (1,1), which is lin.indep. $\endgroup$ – Graham Leuschke Jun 30 '10 at 17:39
I found an old paper by Lazarus (Les familles libres maximales d'un module ontelles le meme cardinal?, Pub. Sem. Math. Rennes 4 (1973), 112) which contains the the following result: Let A be a commutative ring with unit and M an Amodule. In the following situations, maximal linearly independent subsets of M have the same cardinality:
If M is a free Amodule of infinite rank.
If A is reduced and has only finitely many minimal primes (e.g. integral domain, reduced Noetherian ring)
If A is Noetherian and M is a free Amodule.
If A is Noetherian and M is a submodule of a free Amodule of finite rank.
If A is Noetherian and M has an infinite linearly independent subset.
If A is Noetherian and M is a submodule of a flat Amodule.
If A is Artin local and the zero ideal $(0)\subset A$ is irreducible.
And the examples given in the paper of modules not satisfying this same cardinality property are highly nontrivial.

$\begingroup$ Dear kwan: Here is how I understand your answer. Lazarus proves the existence of nonequipotent maximal linearly independent subsets of modules over commutative rings. Is this interpretation of your wording correct? $\endgroup$ – PierreYves Gaillard Jul 3 '10 at 16:11

1$\begingroup$ Dear PierreYves, he actually constructs two nonequipotent maximal linearly independent subsets of a module M over a commutative ring A which do not satisfy the conditions listed in the answer. $\endgroup$ – ashpool Jul 4 '10 at 0:11

4$\begingroup$ Dear Victor, he gives one example of a module over a Noetherian ring which has maximal linearly independent subsets $S_{1}$ and $S_{2}$ where $S_{1}=1$ and $S_{2}=2$. $\endgroup$ – ashpool Jul 4 '10 at 0:17

1$\begingroup$ This is very interesting! I wasn't able to find the paper online, would you mind adding this example to your answer? $\endgroup$ – Victor Protsak Jul 4 '10 at 1:39

3$\begingroup$ Here is the example: $A=k[x,y]/(x,y)^{2}$. Then the $A$module $A^{3}/((\bar{x},0,\bar{x}),(\bar{y},\bar{x},0))$ has a maximal linearly independent singleton { (1,0,0) } and a linearly independent subset { (0,1,0), (0,0,1) }. $\endgroup$ – ashpool Jul 4 '10 at 15:39
If you consider rings that are not necessarily commutative, here's an example: let $V$ be a countable dimensional vector space over a field $F$, and let $A$ be the ring of all endomorphisms of $A$. I claim that $A\cong A\oplus A$ (as left $A$modules); if so, then using (and iterating) this isomorphism you can find maximal linearly independent subsets of any finite cardinality.
To see that $A\cong A\oplus A$, it suffices to exhibit a twoelement $A$basis for $A$. Let $e_1,e_2,\ldots$ be a basis for $V$. Let $f_1\in A$ be the endomorphism that maps $e_2,e_4,e_6,\ldots$ to $e_1,e_2,e_3,\ldots$, respectively, and maps every oddindexed basis element to $0$; let $f_2\in A$ be the endomorphism that maps $e_1,e_3,e_5,\ldots$ to $e_1,e_2,e_3,\ldots$, and maps the evenindexed basis elements to $0$. Then $f_1,f_2$ spans $A$: if $\varphi \in A$, then we can write $\varphi$ as $\varphi=gf_1+hf_2$, where $g(e_i)=\varphi(e_{2i})$ and $h(e_j)=\varphi(e_{2j1})$. To see that $f_1$ and $f_2$ are $A$linearly independent, suppose that $af_1+bf_2=0$; evaluating at the odd indexed $e_i$ shows that $b(e_j)=0$ for all $j$, and evaluating at the even indexed $e_i$ shows $a(e_j)=0$ for all $j$. Thus, $f_1,f_2$ is also a basis for $A$, which gives an isomorphism $A\cong A\oplus A$. Being bases, they are certainly maximal linearly independent sets.

$\begingroup$ Thanks! I was considering commutative ring, though. $\endgroup$ – ashpool Jun 30 '10 at 16:33

1$\begingroup$ Duh (to myself); you did tag it "commutative algebra." I believe there can be no examples in the commutative case, because commutative rings have IBN (invariant basis number). $\endgroup$ – Arturo Magidin Jun 30 '10 at 16:43

1$\begingroup$ That rules out free modules (IBN means that the rank of a free module is welldefined), but not arbitrary modules. $\endgroup$ – Graham Leuschke Jun 30 '10 at 17:18

$\begingroup$ @Arturo: are you sure? So far as I know, the property IBN pertains to free modules only. (But maybe the arguments carry over to this context; I haven't thought it through.) $\endgroup$ – Pete L. Clark Jun 30 '10 at 17:21

$\begingroup$ No, I'm not sure; indeed, IBN applies only to free modules, which is why I only "believe" this to be the case in general for commutative rings. Intuitively, I'm thinking that if I had maximal l.i. sets $x_1,\ldots,x_n$ and $y_1,\ldots,y_m$, some nonzero multiple of each $y_i$ would lie in $\langle x_1,\ldots,x_n\rangle$ (which would be free), and viceversa; and then one could try to move this completly into a free setting and invoke IBN. But I run into technical problems (e.g,I think it is not in general true that if $y_1,\ldots,y_m$ is l.i., and $a\neq 0$, then so is $ay_1,\ldots,ay_m$). $\endgroup$ – Arturo Magidin Jun 30 '10 at 17:32