Can an indecomposable module be neither torsion nor torsion-free?

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.

• 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) Sep 17, 2019 at 15:41
• See my answer, math.stackexchange.com/questions/3353989/… Sep 17, 2019 at 18:05
• @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.) Sep 18, 2019 at 12:20
• 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 $Ext^1(N,T)\neq 0$. Sep 18, 2019 at 13:24

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.