Let $A$ be an integral domain and, for any $A$-module $M$, let $T(M)$ be the torsion submodule of $M$.

Is it possible to have $0\ne T(M)\ne M$ for an indecomposable $A$-module $M$?

In such a case, $A$ would not be a principal ideal domains: see p. 36 of Kaplansky's book Infinite abelian groups.

Edit: The ring $A$ would not even be a Dedekind domain by Theorem 10 in Kaplansky's article Modules over Dedekind rings and valuation rings.

  • $\begingroup$ Let $A$ be a Noetherian local domain which its completion isn't a domain. Maybe the completion is neither torsion nor torsion-free.(I am not sure about this) $\endgroup$ Sep 17, 2019 at 15:41
  • 1
    $\begingroup$ See my answer, math.stackexchange.com/questions/3353989/… $\endgroup$
    – Mohan
    Sep 17, 2019 at 18:05
  • $\begingroup$ @Mohan - Sorry for answering so late! It took me an awful lot of time to find a (hopefully correct) proof of the indecomposabilty. (I never doubted that you were right, but I wanted to find a proof by myself.) Your example looks incredibly ingenious to me! It also answers this question. (So it answers at least 3 MSE questions!) - Would you like to post an answer? (Otherwise I can write a community wiki answer giving you due credit.) $\endgroup$ Sep 18, 2019 at 12:20
  • 2
    $\begingroup$ Thank you for your kind words. The example is not ingenious at all, if you think about it. You need a non-split exact sequence $0\to T\to M\to N\to 0$, where $T$ is torsion and $N$ torsion free. The simplest comes from $A=k[x,y]$ using $T=k[x,y]/(x,y)$, $N=(x,y)$, since $Ext^1(N,T)\neq 0$. $\endgroup$
    – Mohan
    Sep 18, 2019 at 13:24

1 Answer 1


As explained by Mohan in the comments, the answer is Yes.

Indeed, let $K$ be a field, let $x$ and $y$ be indeterminates, set $A:=K[x,y]$, and denote abusively by $K$ the $A$-module $A/(x,y)$. As we have $$ \operatorname{Ext}_A^1((x,y),K)\simeq\operatorname{Ext}_A^2(K,K)\simeq K, $$ there is a non-split exact sequence $$ 0\to K\to M\xrightarrow\pi(x,y)\to0. $$ Clearly $K$ is the torsion module $T(M)$ of $M$, and $M$ is neither torsion nor torsion-free. It suffices to check that $M$ is indecomposable.

Let $M_1$ and $M_2$ be two submodules of $M$ such that $M=M_1\oplus M_2$. We have $$ K=T(M)=T(M_1)\oplus T(M_2), $$ and thus $K$ is contained in $M_1$ or $M_2$. Say that $K$ is contained in $M_1$. The exact sequence being non-split, $K$ is a proper submodule of $M_1$. We get $(x,y)=\pi(M_1)\oplus\pi(M_2)$ and $\pi(M_1)\ne0$. The fact that $(x,y)$ is indecomposable implies $\pi(M_1)=(x,y)$ and thus $M_1=M$, showing that $M$ is indecomposable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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