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The Statement

I suspect the following proposition is well known, but I found no reference.

Proposition If $A$ is a principal ideal domain, if $I$ is a nonzero ideal of $A$, and if $M$ is an $A/I$-module, then $M$ is a direct sum of finitely generated submodules.

The Proof

Proof (sketch) By the Chinese Remainder Theorem we can assume that $I$ is of the form $(p^n)$ where $p$ is a prime element of $A$ and $n$ a positive integer. Let $K$ be the residue field $A/(p)$, and let $S\subset A$ be system of representatives of the classes mod $(p)$ such that $0\in S$. We shall prove:

(a) There is a subset $\widetilde B=\bigsqcup_{i=0}^{n-1}\widetilde B_i$ of $M$ such that each element of $M$ can be written in a unique way as $$ \sum\ s(i,\widetilde b_j)\ p^i\ \widetilde b_j $$ where the sum runs over those pairs $(i,\widetilde b_j)$ with $0\le i\le j < n$, $\widetilde b_j\in B_j$, the $(s(i,\widetilde b_j))$ forming a finitely supported family of elements of $S$.

Clearly (a) implies the proposition. We sketch a proof of (a).

The multiplication by $p$ induces $K$-linear surjections $$ M/pM\xrightarrow{f_1}pM/p^2M\xrightarrow{f_2}\cdots\xrightarrow{f_{n-2}}p^{n-2}M/p^{n-1}M\xrightarrow{f_{n-1}}p^{n-1}M. $$ Pick a $K$-basis $B_0$ of $\operatorname{Ker} f_1$,

$\bullet$ complete $B_0$ to a $K$-basis $B_0\sqcup B_1$ of $\operatorname{Ker}(f_2\circ f_1)$,

$\bullet$ complete $B_0\sqcup B_1$ to a $K$-basis $B_0\sqcup B_1\sqcup B_2$ of $\operatorname{Ker}(f_3\circ f_2\circ f_1)$,

$\bullet\ \cdots$,

$\bullet$ complete $B_0\sqcup\cdots\sqcup B_{n-3}$ to a $K$-basis $B_0\sqcup\cdots\sqcup B_{n-2}$ of $\operatorname{Ker}(f_{n-1}\circ\cdots\circ f_1)$, and

$\bullet$ complete $B_0\sqcup\cdots\sqcup B_{n-2}$ to a $K$-basis $B_0\sqcup\cdots\sqcup B_{n-1}$ of $M/pM$.

Let $\widetilde b_j\in M$ be a lift of $b_j\in B_j\subset M/pM$, and let $\widetilde B_j\subset M$ be the set of all $\widetilde b_j$.

We claim that

(b) the $\widetilde B_j$ satisfy (a).


(c) the map $f_{k-1}\circ\cdots\circ f_1$ induces a $K$-linear isomorphism $$ \frac{M/pM}{\operatorname{Ker}(f_{k-1}\circ\cdots\circ f_1)}\simeq\frac{p^kM}{p^{k+1}M}\quad. $$ Let $\sigma_k:p^kM\to p^kM/p^{k+1}M$ be the canonical projection. It is easy to see that (c) implies

(d) $\sigma_k\left(p^k\left(\widetilde B_k\sqcup\cdots\sqcup\widetilde B_{n-1}\right)\right)$ is a $K$-basis of $p^kM/p^{k+1}M$.

Taking (d) for granted, we prove the uniqueness of the family $(s(i,b_j))$. (As often a close look at the proof of uniqueness suggests a proof of existence.)

Assume by contradiction we have distinct families $(s(i,b_j))$ and $(t(i,b_j))$ as above such that $$ \sum\ s(i,b_j)\ p^i\ \widetilde b_j=\sum\ t(i,b_j)\ p^i\ \widetilde b_j. $$ Let $k$ be the least nonnegative integer for which there is a $b_j$ with $s(k,b_j)\ne t(k,b_j)$. The congruence $$ \sum\ s(k,b_j)\ p^k\ \widetilde b_j\equiv\sum\ t(k,b_j)\ p^k\ \widetilde b_j\bmod p^{k+1}, $$ where the sums run over $k\le j < n$ and $b_j\in B_j$, contradicts (d). QED

The Question

My question has five parts:

(e) Is the above proposition true?

(f) Is it well known?

(g) Do you know any reference?

(h) Is the above argument correct?

(i) Is there a better argument?

share|improve this question
Finitely generated $A/I$-modules "are" also finitely generated $A$-modules, hence direct sums of cyclic modules. So you have actually proven (if your proof is correct) that every $A/I$-module is a direct sum of cyclic modules. Is this true? What about $\prod_{\mathbb{N}} \mathbb{Z}/n\mathbb{Z}$ over the ring $\mathbb{Z}/n\mathbb{Z}$? Since this module is torsion-free, it could only be isomorphic to a direct sum of copies of $\mathbb{Z}/n\mathbb{Z}$. But this seems to contradict cardinalities ... –  Martin Brandenburg Jul 28 '14 at 10:58
Dear Martin, Thanks for your comment. Could you explain more precisely why "this seems to contradict cardinalities"? - Do you agree at least that, when $n$ is prime, $\prod_{\mathbb{N}}\mathbb{Z}/n\mathbb{Z}$ is isomorphic to a direct sum of copies of $\mathbb{Z}/n\mathbb{Z}$? –  Pierre-Yves Gaillard Jul 28 '14 at 11:11
@Pierre-YvesGaillard Don't worry, the proposition is true, and even for a much broader class of rings. –  rschwieb Jul 28 '14 at 13:29
@rschwieb - Thank you for your kind comment and your great answer! –  Pierre-Yves Gaillard Jul 28 '14 at 13:34

2 Answers 2

up vote 2 down vote accepted

Yes, this is very old work. In 1935, Koethe proved that the modules of Artinian principal ideal rings are all direct sums of cyclic submodules. Of course, your example falls into this category. In fact, all of the proper quotients of a principal ideal domain are Artinian principal ideal rings (in fact they are also self-injective, hence quasi-Frobenius.)

If you enjoy reading in German, you can look up

Köthe, Gottfried (1935), "Verallgemeinerte Abelsche Gruppen mit hyperkomplexem Operatorenring. (German)", Math. Z. 39: 31–44

Of course since then, there have been many proofs written in English :) My go-to references for serial rings are anything by Puninski, such as this article or even his textbook Serial rings.

Eisenbud and Griffith's article is also freely available: Serial rings

Rings for which all right modules are decomposable into direct sums of finitely generated modules are called right pure-semisimple rings. They are pretty well-studied, and the place I would begin studying them is Algebra, K-theory, Groups, and Education: On the Occasion of Hyman Bass's 65th birthday

share|improve this answer
Dear rschwieb: Thanks a lots for your terrific answer! Needless to say that I up-voted it. I'll try to digest it. I'm afraid my question is even much more naive than I feared... –  Pierre-Yves Gaillard Jul 28 '14 at 13:33
Dear rschwieb: Your wrote: "Rings for which all right modules are decomposable into sums of finitely generated modules are called right pure-semisimple rings." I think you meant "direct sum", not just "sum"... –  Pierre-Yves Gaillard Jul 28 '14 at 13:54
@Pierre-YvesGaillard Yes, that's an important omission to correct: I'll change it :) Obviously being a sum of f.g. submodules is not very interesting. –  rschwieb Jul 28 '14 at 14:02

This is a remark concerning points (e-g): your result is mentioned in the article


right at the beginning, but without proof. However some pointers are given to articles about certain non-commutative rings in which the problem under discussion seems to be treated.

share|improve this answer
Dear Hagen: Thank you very much for your nice answer! +1. For the other readers, here is a link to Warfield's article. I think Warfield's statement is stronger than mine, because it applies to all Artinian principal ideal rings, and I imagine this is a strictly larger class of rings. –  Pierre-Yves Gaillard Jul 28 '14 at 13:23

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