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

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.

• 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? – Jeremy Rickard Mar 12 '20 at 22:00
• I meant an internal direct sum, as in the answer below, added it to the question -- thanks – nikola karabatic Mar 13 '20 at 7:31

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.
• 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 $\mathbb{Z}$. – Jeremy Rickard Mar 13 '20 at 10:40