# Some generating set of modules

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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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.

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.

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.