# Over which (graded) rings are all modules decomposable into indecomposables?

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?

• 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. Dec 17, 2021 at 12:27
• 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..
– wcb
Dec 17, 2021 at 12:47
• Krull-Schmidt category maybe a useful search term for you Dec 17, 2021 at 17:15

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 $$M = \prod_{j \in \mathbb{Z}} \Sigma^j A$$.

• 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. Dec 18, 2021 at 19:01

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 $$i I'll write $$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.

• 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?
– user164898
Dec 19, 2021 at 18:38
• 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 $\mathbb{F}_2[x]$. Dec 19, 2021 at 18:56