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Let $R$ be a commutative ring with unity. Let $M$ be a free (unital) $R$-module.

Define a basis of $M$ as a generating, linearly independent set.

Define the rank of $M$ as the cardinality of a basis of $M$ (as we know commutative rings have IBN, so this is well defined).

A minimal generating set is a generating set with cardinality $\inf\{\#S:S\subset M, M=\langle S \rangle\}$.

Must a minimal generating set have cardinality the rank of $M$?

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    Georges' answer below made me realize that giving a generator with $s$ elements of a module $M$ is the same as giving an epimorphism $R^s\to M\to 0$. In light of this, my question is actually this one: http://math.stackexchange.com/questions/20178/given-a-commutative-ring-r-and-an-epimorphism-rm-to-rn-is-then-m-geq-n ! Because of the different formulation, I hadn't found it before asking the question; I wouldn't have asked the question if I had ;P2011-11-07

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Yes, a generating set of minimal cardinality must have cardinality $r=rank_R(M)$.
It suffices to show that for any generating set of $M$ with $s$ elements, we have $s\geq r$ .

Assume that $M=R^r$.
Our generating set gives rise to a surjective $R$-module morphism $R^s\to R^r\to 0 \quad (\star)$.
Let $\mathfrak m\subset R$ be a maximal ideal and tensor $(\star)$ with the field $k=R/\mathfrak m$ .
You get a $k$-linear map $k^s\to k^r \to 0 \quad$ which is still surjective by right-exactness of the tensor product.
Since $k$ is a field, this implies $s\geq r$.

Edit
Since Bruno just commented that he is also interested in the case of infinitely many generators, let me reassure him that the above reasoning remains true, with the obvious change from integers to cardinals and minor cosmetic adaptations in notation.
More precisely, we assume that $M=R^{(B)}$ where $B$ is a basis of $M$ of cardinality $card (B)=\beth$.
If $A$ is a generating set of $M$ of cardinality $card(A)=\aleph $, we have a surjective morphism $R^{(A )}\to R^{(B )} \to 0 \quad (\star)$ yielding once more by tensorization (still right-exact!):
$k^{(A )}\to k^{(B )} \to 0 $ and the conclusion is again $\aleph \geq \beth$.

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    Georges: just to let you know, here's a followup to this question, http://math.stackexchange.com/questions/80658/a-ring-with-ibn-which-admits-a-free-module-with-a-generator-with-less-elements-t2011-11-09
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In Some remarks on the invariant basis property, Topology 5 (1966), pp. 215-228, MR 33 #5676, P.M. Cohn proved that there are rings in which the notion of rank is well defined, but which admit free modules of rank $t$ that can be generated by fewer than $t$ elements. However, he also proved that this is impossible if the ring is Noetherian, Artinian, or commutative. I don't have access to that paper, so I don't know how easy it is to prove, but in your situation (commutative rings), the answer is therefore "yes".

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The statement in the Edit of Georges's answer also holds in the non-commutative case:

Let $R$ be an associative ring with $1$. If $M$ is an $R$-module, if $B$ is an infinite basis of $M$, and if $S\subset M$ is a generating subset, then we have $|S|\ge|B|,$ where, for any set $X$, the symbol $|X|$ denotes the cardinality of $|X|$.

In particular, if $C$ is another basis of $M$, then $|C|=|B|$.

Proof: For any $x\in M$ let $B_x$ be the finite set of those $b$ in $B$ such that the $b$-component of $x$ is nonzero. The fact that $S$ generates $M$ implies $ B=\bigcup_{s\in S}\ B_s, $ and thus $ |B|=\left|\ \bigcup_{s\in S}\ B_s\ \right|\ \le \left|\ \coprod_{s\in S}\ B_s\ \right|= \sum_{s\in S}\ \left|B_s\right|=|S|. $

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    Thank you, it's nice to know. It still puzzles me that facts regarding basis of free modules behave a little better in the infinite case!2011-11-07