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.

Every module is the direct limit of finitely generated modules. Is it true that every module is the inverse limit of finitely generated modules?

share|improve this question
    
Sorry, ignore my previous comment. I misread your question twice and ignored "finitely generated" so none of what I wrote makes any sense. –  Matt N. Nov 1 '12 at 16:45

1 Answer 1

up vote 5 down vote accepted

No. Consider $\mathbb{Q}$ or any other nonzero divisible group. Every homomorphism from $\mathbb{Q}$ into a finitely generated abelian group is zero: subgroups of finitely generated groups are finitely generated, quotients of divisible groups are divisible, and the only finitely generated divisible group is $0$.

More details: (using Wikipedia's notation)

Suppose towards a contradiction that $(X_i, f_{ij})$ is an inverse system of finitely generated abelian groups and $\pi_i \colon \mathbb Q \to X_i$ are the projections identifying $\mathbb{Q}$ as the inverse limit $\mathbb{Q} = \varprojlim\nolimits_i X_i$, in particular $f_{ij}\pi_j = \pi_i$. We know that $\pi_i = 0$ .

Consider the maps $\psi_i = \pi_i \colon \mathbb{Q} \to X_i$. Both $u = 1_{\mathbb{Q}}$ and $u' = 0$ are homomorphisms $\mathbb{Q} \to \mathbb{Q}$ such that $\psi_i = \pi_i u$ and $\psi_i = \pi_i u'$, hence the uniqueness statement in the universal property of the inverse limit implies $u = u'$, so $1_\mathbb{Q} = 0$. Nonsense.

(More generally, you can try to show that any map into $\mathbb{Q}$ would have to be zero.)

share|improve this answer
    
could you add more details? I don't understand why this is a counterexample –  Chris Nov 1 '12 at 17:20

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.