Let $R$ be a commutative ring with unity and $M$ be a finitely generated free $R$-module.

Let $S$ be a finite subset of $M$ generating $M$ as $R$-module.

From this, can we say that $M$ has finite basis?

If $R$ is a field, it is the case because the maximal linearly independent set in $S$ gives rise to the basis of $M$.

But when $R$ is just commutative ring not being a field, we cannot apply this argument.

If $M$ also has finite basis, can we say further that the number of basis is equal or less than $|S|$?

  • $\begingroup$ I am not sure what the precise question is (maybe "not commutative ring" should be "not a field"?). Anyway, over a commutative ring $R$ with unity, in general, not every finitely generated $R$-module has a basis. What is true is that a finitely generated $R$-module which has a basis must have a finite basis. (Better yet, its given basis must be finite.) Moreover, if an $R$-module $M$ has a generating set with $g$ elements and a linearly independent set with $i$ elements, then $i \leq g$ unless $R = 0$. $\endgroup$ – darij grinberg Oct 12 '15 at 19:52
  • $\begingroup$ @darij grinberg, Oh, Sorry. I omitted the free condition. Here, $M$ is finitely generated free $R$-module. I modified my question a bit upon your reply. Please read it again! $\endgroup$ – Andrew Oct 12 '15 at 19:58

No, in general you can't extract a basis from a set of generators. The simplest counterexample is $R=\mathbb{Z}$ (the integers) and $M=\mathbb{Z}$. The subset $S=\{2,3\}$ generates $M$, because $1=-1\cdot2+1\cdot 3$, but no subset of $S$ is a set of generators.

It is however true that, if $M$ is a finitely generated free module over a commutative ring, then the rank (number of elements in a basis) is less than or equal to the number of elements in any finite set of generators.

How can we prove it? It's quite simple. Consider $M$ a finitely generated free module and $\mathfrak{m}$ a maximal ideal of $R$. Then $M/\mathfrak{m}M$ is a vector space over $R/\mathfrak{m}$. Any set of generators in $M$ projects to a set of generators of $M/\mathfrak{m}M$ and a basis projects to a basis.

Over non commutative ring, the notion of rank is not well defined for finitely generated free modules: there are non commutative rings $R$ such that, as $R$-modules, $R\cong R\oplus R$, for instance.

  • $\begingroup$ This is what I am looking for. Thank you for your illumination! $\endgroup$ – Andrew Oct 12 '15 at 20:11
  • $\begingroup$ Thinking on your reply, some question have come to my mind. Projecting a basis of $M$ to $M/mM$, there is some vanishing possibility. (I mean, a non-zero basis element of $M$ may zero in $M/mM$.) So, the cardinality of basis of $M/mM$ is equal or less than that of $M$. Did you consider this possobility? $\endgroup$ – Andrew Oct 12 '15 at 20:46
  • $\begingroup$ @Andrew No, it can't happen, because the module is free. $\endgroup$ – egreg Oct 12 '15 at 20:50
  • $\begingroup$ Ah, I see! Thank you again~ $\endgroup$ – Andrew Oct 12 '15 at 21:06

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