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.



Projective module which splits off sequence of submodules, but not the sum
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.


