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

  • $\begingroup$ Sorry, ignore my previous comment. I misread your question twice and ignored "finitely generated" so none of what I wrote makes any sense. $\endgroup$ – Rudy the Reindeer Nov 1 '12 at 16:45

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.)

  • $\begingroup$ could you add more details? I don't understand why this is a counterexample $\endgroup$ – Chris Nov 1 '12 at 17:20

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