# Is a module an inverse limit of finitely generated modules?

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

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

## 1 Answer

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.)

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