Does there exist an example of a module $X$ over some ring $R$ together with submodules $T_i$ such that:

  • $X$ is projective,
  • $X$ splits as an internal direct sum $X\cong T_1\oplus T_2\oplus \ldots \oplus T_n\oplus S_n$ (with some $S_n$) for every $n$,
  • $X$ does not split off the infinite direct sum $\bigoplus_{i=1}^\infty T_i$,
  • $R$ is hereditary.

Remark: I also don't know the answer without the last condition, so this would already be interesting, though for my specific application I definitely need all conditions.

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    $\begingroup$ In the second condition, do you just want an abstract isomorphism for each $n$, or do you want the sum of the submodules actually to be a direct summand? $\endgroup$ – Jeremy Rickard Mar 12 '20 at 22:00
  • $\begingroup$ I meant an internal direct sum, as in the answer below, added it to the question -- thanks $\endgroup$ – nikola karabatic Mar 13 '20 at 7:31

Here is an example if we interpret all direct sums as internal direct sums.

Example. Let $R$ be a discrete valuation ring with uniformiser $\pi$ and fraction field $K$. Let $X = R^{(\mathbf N)}$, and let $T_i$ be the free rank $1$ submodule with basis $\pi e_{i+1}-e_i$. Then the natural map $$\bigoplus_{i=1}^n T_i \to X$$ is injective with image $T_{\leq n} = \operatorname{span}(\pi e_2 - e_1, \ldots, \pi e_{n+1} - e_n)$, because the latter clearly has rank $n$. Moreover, $$S_n = \bigoplus_{i > n} Re_i \subseteq X$$ is a complement of $T_{\leq n}$: one easily sees that $S_n \cap T_{\leq n} = 0$, and they span $X$ because $e_n = \pi \cdot e_{n+1} - (\pi e_{n+1} - e_n)$, etcetera. But if $T = \bigoplus_{i \in \mathbf N} T_i = \sum_i T_i \subseteq X$, then \begin{align*} X/T &\stackrel\sim\to K\\ e_i &\mapsto \pi^{-i}. \end{align*} This surjection does not split because $X$ has no infinitely divisible elements. $\square$

What's going on is that we wrote $K$ as a filtered colimit of surjections $S_n \twoheadrightarrow S_{n+1}$ of free modules: $$K = \underset{\substack{\longrightarrow \\ n}}{\operatorname{colim}}\ S_n.$$ Each $X \twoheadrightarrow S_n$ has a splitting $S_n \hookrightarrow X$, but $X \twoheadrightarrow K$ does not.

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    $\begingroup$ Just a comment that this nice construction also works for $R=\mathbb{Z}$. Let $0\to T\to X\to\mathbb{Q}\to0$ be a free resolution of $\mathbb{Q}$ and write $T$ as a direct sum $T=\bigoplus_iT_i$ of copies of $\mathbb{Z}$. $\endgroup$ – Jeremy Rickard Mar 13 '20 at 10:40
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    $\begingroup$ @JeremyRickard absolutely! That's actually the example I started with, but the DVR case was easier. $\endgroup$ – R. van Dobben de Bruyn Mar 13 '20 at 17:20

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