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.
Best Answer
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.