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.

  • 2
    $\begingroup$ 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? $\endgroup$ Mar 12, 2020 at 22:00
  • $\begingroup$ I meant an internal direct sum, as in the answer below, added it to the question -- thanks $\endgroup$ Mar 13, 2020 at 7:31

1 Answer 1


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.

  • 2
    $\begingroup$ 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 $\mathbb{Z}$. $\endgroup$ Mar 13, 2020 at 10:40
  • 1
    $\begingroup$ @JeremyRickard absolutely! That's actually the example I started with, but the DVR case was easier. $\endgroup$ Mar 13, 2020 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.