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) $\endgroup$ – Mohammad Bagheri Sep 17 '19 at 15:41
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    $\begingroup$ See my answer, math.stackexchange.com/questions/3353989/… $\endgroup$ – Mohan Sep 17 '19 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.) $\endgroup$ – Pierre-Yves Gaillard Sep 18 '19 at 12:20
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    $\begingroup$ 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 '19 at 13:24

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.


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