# Composition length of surjective inverse limits

Let $M$ be a left module over some ring $R$ and suppose that $M$ is an inverse limit of a family of modules $M_i$ with $i\in I$. We suppose also that the maps of the inverse limit $\pi_i:M\to M_i$ are all surjective.

Is it true that the composition length of $M$ is the supremum of the composition lengths of the $M_i$?

Note. By composition length I mean a function that takes values in $\mathbb N\cup \{\infty\}$, where $\infty$ is just a symbol, I do not want to distinguish among infinities.

Idea. My idea for a proof is to use the fact that the composition length of a module is just the length of the lattice of its submodules. Now, it is known that the length of a lattice which is the union of a directed family of sublattices is the supremum of the lengths of such sublattices. Now I am wandering if the dual lattice of the lattice of submodules of $M$ is the union of the directed family of the dual lattices of the submodules of the $M_i$. If this is true, one should conclude by showing that the length of a lattice equals the length of its dual... Do you think this can be arranged to work?

Remark. If the proof works like this, I think there will be no difficulty in arranging the whole thing in a Grothendieck category and not only in a category of modules. Am I right?

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After thinking for a bit more I see that my question has a positive answer but the idea I was suggesting for a proof badly fails. I'll edit later with a proof.

EDIT: Let me start fixing some notation. We let $(I,\leq)$ be a directed set and, for all $j\leq i$ in $I$ we let $\pi_{ji}:M_i\to M_j$. Furthermore we denote by $M=\underleftarrow\lim_IM_i$ the inverse limit of the system $(M_i,\pi_{ji})_{j\leq i\in I}$ and we denote by $\pi_i:M\to M_i$ the maps of the limit. By assumption, all the $\pi_i$ are epi so, one deduces that all the transition maps $\pi_{ji}$ are epi as well.

Now, the fact that $\ell(M)\geq \sup_{i\in I}\ell(M_i)$ follows by the fact that $\ell$ is monotone on quotients. To prove the converse inequality we distinguish two cases:

$\bullet$ if $\ell(M)=n<\infty$, then the monotone map $f:I\to \mathbb N$ such that $f(i)=\ell(M_i)$ has bounded image. Let $\bar n$ be the maximum of $f(I)$, one can show that $f^{-1}(\bar n)$ is a cofinal subset of $I$. Furthermore, an epimorphism between modules of finite length is necessarily an isomorphism and so $M_i\cong M_j$ for all $i,j\in f^{-1}(\bar n)$. In particular, there exists $i\in I$ such that $M=M_i$ and we can conclude.

$\bullet$ if $\ell(M)=\infty$, then we can consider a monotone map $f:I\to \mathbb N\cup\{\infty\}$. If the image of $f$ is a finite subset of $\mathbb N$, then one concludes as in the previous case that $\ell(M)<\infty$, yielding a contradiction. On the other hand, if $f(I)$ is not a finite set or it contains $\infty$, then one can easily conclude that $\sup_I\ell(M_i)=\infty$, as desired. \\\

Let me conclude suggesting a counterexample to the idea I was suggesting in my question. In particular, it is true that the dual lattice of submodules of each $M_i$ can be identified with sublattices of the dual lattice of submodules of $M$. Anyway, their union can be very far from being the whole lattice. For example, the inverse limit of the modules $\mathbb Z/p^n$ with $n\in\mathbb N$ and the obvious transition maps, is the group of $p$-adic integers. This group is not countable (so it is not countably generated) and so its lattice of subgroups is not countable. Anyway, the union of the lattice of subgroups of the $\mathbb Z/p^n$ is certainly countable.

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