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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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How do you define basis ? Isn't it a minimal generating set? –  Joel Cohen Nov 7 '11 at 2:34
    
@JoelCohen Sorry, I missed that definition. I edited it in. –  Bruno Stonek Nov 7 '11 at 2:35
    
From this page, it seems P.M. Cohn proved that for commutative rings, this is the case (that is, that a free module of rank $n$ cannot be generated by fewer than $k$ elements. –  Arturo Magidin Nov 7 '11 at 2:56
    
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: math.stackexchange.com/questions/20178/… ! Because of the different formulation, I hadn't found it before asking the question; I wouldn't have asked the question if I had ;P –  Bruno Stonek Nov 7 '11 at 18:55

3 Answers 3

up vote 6 down vote accepted

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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Thank you for your answer. Could you elaborate, though, on yout last line? Also, what if $r$ is infinite? –  Bruno Stonek Nov 7 '11 at 12:44
    
Dear Bruno, in the last line I have used the fact that if you have a surjective linear map from a vector space (over a field!) of dimension $s$ onto a vector space of dimension $r$, then $s\geq r$. The proof works also in the case of infinitely many generators: I'll write an edit instead of cramming more mathematics into this comment. –  Georges Elencwajg Nov 7 '11 at 13:49
    
Thank you for the clarification; I was misunderstanding the argument above, I thought it used that $k^s$ and $k^r$ were finite dimensional. Please, correct me if I'm wrong: what you have done is to "extend scalars" from $R$ to $k$, right? $k$ is a $k$-module and hence an $R$-module through restriction of scalars with the canonical map $R\to k$; and we have that $k\otimes_R R^n$ is a $k$-module (extension of scalars). –  Bruno Stonek Nov 7 '11 at 16:45
    
Dear Bruno, there is nothing to correct: you are absolutely right. The mathematical precise way to extend scalars is by using tensor products, and this preserves surjections.(In elementary linear algebra courses, teachers extend scalars from $\mathbb R$ to $\mathbb C$ by ad hoc constructions, in order to force matrices to have eigenvalues, etc., but the general procedure is to use tensor products) –  Georges Elencwajg Nov 7 '11 at 16:51
    
Great! Yes, I'm aware of the restriction-extension of scalars constructions, but I'm not yet familiar with their applications. What an useful and elegant application! Thank you once again. –  Bruno Stonek Nov 7 '11 at 16:56

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! –  Bruno Stonek Nov 7 '11 at 16:15

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