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In commutative algebra, for a module $M$ over a (possibly unital) commutative ring $R$, when is the number $\mu_R(M)$ well-defined?

For example, if $R$ is a local ring, then (by Nakayama Lemma and elementary linear algebra), any minimal generating set has the same number of elements.

-Rashid

Rashid Malik
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    I think that the question in your first sentence is not exactly what you mean to ask. For any module $M$, the minimal number of generators (finite or infinite) is well-defined thanks to well-ordering. Do you mean to ask about rings such that for every module $M$, every minimal generating set of $M$ has the same cardinality? – Manny Reyes Sep 16 '14 at 18:42
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    Minimal generating sets of modules are also discussed here. Moreover, in this article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that $\Bbb R$ has no minimal generating set as field (or $\Bbb Q$-algebra, i.e. as ring). – Watson Oct 05 '16 at 19:49

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Suppose $R$ is not local (and $R$ has a $1$). Pick distinct maximal ideals $m_1 \ne m_2$. As $m_1 + m_2 = R$, there exists $a \in m_1$, $b \in m_2$ with $a + b = 1$, so $\{1\}$ and $\{a, b\}$ are generating sets of $R$ of different sizes, with no proper subset generating $R$ (note: $Ra \subseteq m_1 \ne R$).

Thus only local rings have the property that you desire.

zcn
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  • @RashidMalik: As pointed out by Manny Reyes above, for any (f.g.) module $M$, there is always an integer $n$ such that any generating set of $M$ has size at least $n$. The example here just shows that there can exist generating sets of size $> n$, such that no proper subset generates – zcn Sep 19 '14 at 18:34
  • @RashidMalik: That is certainly a valid definition (whether or not a definition is "correct" depends on other things, such as usefulness). With this definition, it is true that for any ring $R$, $\mu_R(R) = 1$. – zcn Sep 21 '14 at 20:18
  • @RashidMalik: This definition is easy to come up with, but because of undesirable properties (like the counterexample I've given here), you may not find it in many books. If this answers your question, please consider accepting the answer (by clicking the check mark to the left) – zcn Sep 22 '14 at 18:12