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A module is decomposable if it is the direct sum of two modules. The process of splitting summands off of a decomposable module does not need to terminate, so infinitely generated modules do not typically split into sums of indecomposable modules.

But they do over certain rings, and I wonder which kinds of rings. Clearly, fields are okay, but even rings as simple as $\mathbb Z$ are not: an infinite product of $\mathbb Z$ is not free.

On the other hand, if we look at nonnegatively graded, connected $k$-algebras and their categories of nonnegatively graded modules, there seem to be more examples. I think some Zorn yoga shows that any nonnegatively graded module over $k[t]$, with $t$ in positive degree, splits as a sum of cyclic modules.

Are there more examples? What about, let's say, graded modules over the Steenrod algebra?

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    $\begingroup$ This doesn't address the graded question, but Warfield proved that the commutative rings for which every module is a direct sum of indecomposable modules are precisely the Artinian principal ideal rings. $\endgroup$
    – Jeremy Rickard
    Dec 17, 2021 at 12:27
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    $\begingroup$ It is proved in Birge Zimmermann-Huisgen, Rings whose right modules are direct sums of indecomposable modules, Proc. Amer. Math. Soc. 77 (1979), no. 2, 191-197, that every right R-module is a direct sum of indecomposables if and only if R is right pure semisimple, meaning that all pure-exact sequences of right R-modules are split. This is true if R is an artinian ring of finite representation type, and according to the pure semisimplicity conjecture, R must be such a ring.. $\endgroup$
    – wcb
    Dec 17, 2021 at 12:47
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    $\begingroup$ Krull-Schmidt category maybe a useful search term for you $\endgroup$
    – Chris McDaniel
    Dec 17, 2021 at 17:15

2 Answers 2

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In the book Spectra and the Steenrod Algebra, Margolis proves the following (Theorem 21 in chapter 11): if $A$ is a graded connected algebra over a finite field and if $M$ is an $A$-module which is finite-dimensional in each degree, then $M$ decomposes uniquely as a direct sum of indecomposables. In particular this applies for such modules over the Steenrod algebra.

Edit: Margolis also has some related results for the Steenrod algebra $A$: Proposition 13 in Chapter 13 says that any bounded below module $M$ over $A$ has a unique expression of the form $F \oplus N$ where $F$ is free and $N$ has no free summands. He also points out (p. 202) that even the existence part of this may fail if $M$ is not bounded below, for example if $M = \prod_{j \in \mathbb{Z}} \Sigma^j A$.

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  • $\begingroup$ My modules are bounded below, but not of finite type. Thanks for the pointer though, I’ll have a look at Margolis’s book for relevant results. $\endgroup$
    – Tilman
    Dec 18, 2021 at 19:01
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Regarding the question about the Steenrod algebra, it is not true that every non-negatively graded module for the $\text{mod }2$ Steenrod algebra is a direct sum of indecomposable modules. I haven't checked the $\text{mod }p$ Steenrod algebra for odd $p$, but I would be astonished if something similar didn't work.

First, note that not every (ungraded) $\mathbb{F}_2[x]$-module is a direct sum of indecomposables. This follows from more general results, since $\mathbb{F}_2[x]$ is not Artinian, or (more directly) at least one of the proofs of the corresponding fact about a countably infinite product of copies of $\mathbb{Z}$ also shows that a countably infinite product of copies of $\mathbb{F}_2[x]$ is not a direct sum of indecomposable $\mathbb{F}_2[x]$-modules.

Next, I will describe a full exact embedding of the category of $\mathbb{F}_2[x]$-modules into the category of non-negatively graded modules for the Steenrod algebra. So applying this to the example above will give a non-negatively graded module for the Steenrod algebra that is not a direct sum of indecomposables.

Let $V$ be a $\mathbb{F}_2[x]$-module. I will contruct a graded module $\bigoplus_{n\geq0}V_n$ for the Steenrod algebra with $$V_n=\begin{cases}V&(0\leq n\leq3)\\ 0&\text{(otherwise).} \end{cases}$$

The only Steenrod squares $\text{Sq}^i$ that can act nontrivially are for $i\leq3$, so the only Adem relations that will need to be checked are $\text{Sq}^1\text{Sq}^1=0$ and $\text{Sq}^1\text{Sq}^2=\text{Sq}^3$.

When $i<j$ I'll write $S_{i,j}$ for $\text{Sq}^{j-i}$ considered as a map $V_i\to V_j$. Setting $$S_{0,1}=S_{2,3}=S_{1,3}=\text{id}_V,$$ $$S_{1,2}=0,$$ $$S_{0,3}=S_{0,2}=x,$$ the Adem relations are satisfied, and so we have constructed the required embedding.

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  • $\begingroup$ There's something perplexing to me about your argument. I'll write $A$ for the Steenrod algebra. In your graded $A$-module $V$, the Steenrod squares $Sq^n$ act trivially for all $n>3$. So the action of $A$ on $V$ factors through the projection $A \rightarrow A/I$, where $I$ is the ideal generated by all homogeneous elements of degree $>3$. But $A/I$ is certainly Artin. It is also gr-local, i.e., has a unique maximal homogeneous ideal. Isn't it already known that the infinite product of (degree 0, i.e., not suspended) copies of a gr-local Artin ring remains free in the graded module category? $\endgroup$
    – user164898
    Dec 19, 2021 at 18:38
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    $\begingroup$ I’m not just taking an infinite product of copies of $A/I$. The structure maps of my module involve $x$: I’m not taking an infinite product of copies of anything Artinian, but of something that, as an $\mathbb{F}_2$ module, is a direct sum of four copies of $\mathbb{F}_2[x]$. $\endgroup$
    – Jeremy Rickard
    Dec 19, 2021 at 18:56
  • $\begingroup$ Okay, thanks for your reply. $\endgroup$
    – user164898
    Dec 19, 2021 at 20:11
  • $\begingroup$ Nice construction! $\endgroup$
    – Tilman
    Dec 20, 2021 at 12:38

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