Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
It's also interesting to ask whether minimal generating sets have the same size. – Dylan Moreland Jun 22 '12 at 1:26
up vote 2 down vote accepted

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.

share|cite|improve this answer
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
@ArturoMagidin Would a minimal generating set exist for modules which are not finitely generated? I know it may not exist. Does it never exist? – Swapnil Tripathi Jan 26 at 13:30

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.

share|cite|improve this answer

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.

share|cite|improve this answer

protected by user26857 Nov 26 '15 at 12:51

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.