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I would like to find an example of a ring $R$ and a $R$-module $M$, which can't be written as a direct sum of indecomposable submodules, i.e. $$ M \not \cong \bigoplus\limits_{i \in I} M_i$$ for all set $\{M_i \;\vert\; i \in I \}$ of indecomposable submodules.

In that case, I know that $M$ should be neither noetherian nor artinian, but I wasn't able to find such an example. Any help would be appreciated!

user26857
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Watson
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  • Related: https://math.stackexchange.com/questions/130511/noetherian-module-implies-finite-direct-sum-of-indecomposables – Watson Jun 12 '18 at 16:14

1 Answers1

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A good way to find examples like this is to look at infinite products. For instance, let $k$ be a field (or more generally, any ring with no nontrivial idempotent elements), let $T$ be an infinite set, let $R=k^T$ (a product of copies of $k$ indexed by $T$), and let $M=R$. It is not too hard to show that any direct summand of $M$ is of the form $k^S$ for some subset $S\subseteq T$ (to show this, use the fact that any direct summand of a ring as a module over itself is generated by an idempotent). Thus the only indecomposable direct summands of $M$ are those of the form $k^S$ when $S$ is a singleton. But the direct sum of all of these is just the infinite direct sum $k^{\oplus T}$, which is the proper submodule of $M$ consisting only of elements which are $0$ on all but finitely many coordinates. Thus $M$ is not a direct sum of indecomposable submodules.

Here's another example, which shows you don't have to be working over some big complicated ring. Take $R=\mathbb{Z}$ and let $M=\mathbb{Z}^T$ for any infinite set $T$. I claim that $M$ is not a direct sum of indecomposable submodules. Since $M$ is not free, it suffices to show that any indecomposable submodule of $M$ is isomorphic to $\mathbb{Z}$. To show this, note that if $N\subseteq M$ is any nonzero submodule, then for some $t\in T$ the projection $p_t:N\to\mathbb{Z}$ onto the $t$th factor of the product is nonzero. Thus $p_t(N)=n\mathbb{Z}$ for some nonzero $n\in\mathbb{Z}$, and since $n\mathbb{Z}$ is free the surjection $p_t:N\to n\mathbb{Z}$ splits and gives a direct summand of $N$ isomorphic to $n\mathbb{Z}\cong \mathbb{Z}$. So if $N$ is indecomposable, it must be isomorphic to $\mathbb{Z}$.

Community
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Eric Wofsey
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  • I was thinking that taking that if, say $R=k^T$ with $T$ countable is the vectorspace of functions from $T$ to $k$, then by Zorns lemma there is a basis $B$ so $R\cong \oplus_{i\in B} k$ right? (And the basis is then necessarily uncountable). Doesn't that contradict your argument? – Myself Dec 14 '15 at 19:33
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    I think your isomorphism is a $k$-vector spaces isomorphism, not a $R$-modules isomorphism. Is it right ? – Watson Dec 14 '15 at 19:39
  • @Myself: As Watson said, that is only an isomorphism of $k$-vector spaces, not of $R$-modules. Most of the copies of $k$ in that direct sum representation are not $R$-submodules of $M$. – Eric Wofsey Dec 14 '15 at 19:41
  • Ok thanks! I was indeed thinking about $R=k$ instead of $R=k^T$! – Myself Dec 14 '15 at 19:42