Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|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
add comment

2 Answers

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|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
add comment

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|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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