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In a module, we know what a minimal generating set is. But, is it always true that such a set exists? If the module is finitely generated, is it possible?

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It's also interesting to ask whether minimal generating sets have the same size. – Dylan Moreland Jun 22 '12 at 1:26

3 Answers 3

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Not every module has minimal generating sets. As another example on the same vein as Hurkyl's, consider the Prüfer $p$-group as a $\mathbb{Z}$-module. A subset generates if and only if it contains elements of arbitrarily high order; but you can remove any finite subset of such a set (you can even remove infinite subsets) and still have a set with that property. Thus, no generating set is minimal: they all contains as proper subsets other generating sets.

It is also not true in general that two minimal generating sets, if they exist, will have the same size: $\mathbb{Z}$ has a minimal generating set (over itself) given by the single element $1$, but it also has a minimal generating set with two elements, $\{2,3\}$. And one with three elements: $\{6, 10, 15\}$. In fact, there is a minimal generating set with any finite number of elements.

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Thank. Magidin, is there the conditon that every generating set has same size? – Sang Cheol Lee Jun 22 '12 at 8:45
@SangCheolLee: It almost never happens; I suspect you would need every module to be free and for the ring to have IBN, but you may be able to get away with a bit less. – Arturo Magidin Jun 22 '12 at 14:41
@Artro Magudin: thank you vary much. I get solved my question thank to your answer thanks . – Sang Cheol Lee Jun 24 '12 at 9:24

Not every module has a minimal generating set. The $\mathbb{Z}$-module $\mathbb{Q}$, for example.

If a module is finitely generated, then the existence of a minimal generating set is easy to show: take any finite generating set and keep removing elements until you can't anymore.

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If M is a module over a ring (not necessarily commutative) which is not a finitely-generated module then every two minimal generating sets of M have the same cardinality (provided that at least a minimal generating set of M exists). This assertion (as stated in the above answers) in the finitely-generated case is not necessarily true. But, if the underlying ring is commutative we have then the following result: Every two bases of a free module have the same cardinality.

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