Is a inverse limit of indecomposable again indecomposable?

Question

In truth, I do not need in the general case.

Let $R$ be a ring (associative, with unit element) and let $\mathrm{Comp}(R)$ the category of complexes over $\mathrm{Mod}\ R$.

If $\mathbb{N}$ is the natural numbers with the usual partial order and $\{X_i^\bullet, \varphi_{i,j}\colon X_j^\bullet \to X_i^\bullet\}$ is an inverse system over $\mathbb{N}$ such that $X_i^\bullet \in \mathrm{Comp}(R)$ is indecomposable, for all $i \in \mathbb{N}$, so is indecomposable $\varprojlim X_i^\bullet$?

Edit. Of course, I'm assuming $\varphi_{i,j} \neq 0$ for all $j\geq i \in \mathbb{N}$.

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Remark. A category $\mathcal{C}$ is complete if $\varprojlim A_i$ exists in $\mathcal{C}$ for every inverse system $\{A_i , \varphi_{i,j}\}$ in $\mathcal{C}$. The categories $\mathcal{C} = \mathrm{Mod} \ R$ and $\mathcal{C} = \mathrm{Comp}(R)$ are examples of complete categories.

Answer 1

Suppose $M$ and $M'$ are two $R$-modules with a common submodule $N$ that has a descending chain of submodules $$N=N_0\supseteq N_1\supseteq N_2\supseteq\dots$$ with zero intersection.

Then if $X^\bullet_i$ is the two term complex $$\dots\to 0\to N_i\to M\oplus M'\to 0\to\dots,$$ where the differential embeds $N_i$ diagonally into $M\oplus M'$, the inverse limit of the $X^\bullet_i$ is the complex $$\dots\to 0\to 0\to M\oplus M'\to 0\to\dots,$$ which is decomposable, but it's not too hard to construct cases where all the complexes $X^\bullet_i$ are indecomposable.

For example, for complexes of abelian groups, take $M$ to be the $p$-adic integers $\mathbb{Z}_p$, and $M'=\mathbb{Z}_q$ for primes $p\neq q$, and take $N=\mathbb{Z}$ with the chain of subgroups $N_i=2^i\mathbb{Z}$.