Vtgyjfy

Linked to my comment there.

Let $M$ be a torsion-free module over a domain $R$. Is $Motimes_R M$ torsion-free ? If not, a counterexample would be appreciated.

Here is what I believe is the argument Mohan had in mind:

Let $k$ be a field and $x$ and $y$ indeterminates. Set
$$
R:=k[x,y],quad M:=(x,y)subset R.
$$
As we have
$$
xy (xotimes y-yotimes x)=x^2otimes y^2-x^2otimes y^2=0
$$
in $Motimes_RM$, it suffices to check $xotimes yne yotimes x$ (again in $Motimes_RM$).

To do that we equip $k$ with an $R$-module structure by letting $x$ and $y$ act by zero, and we note that the map $Motimes_RMto k$,
$$
(fotimes g)mapstofrac{partial f}{partial x}(0,0) frac{partial g}{partial y}(0,0)
$$
is well-defined, $R$-linear and sends $xotimes y$ to $1$ and $yotimes x$ to $0$.

Following the suggestions of Mohan and Ben, I put here an example of torsionfree (and even torsionless) module $M$ over a domain $R$ such that its tensor square $Motimes_R M$ has torsion. If, of course, somebody has a shorter proof/construction, I'll be delighted to accept his/her answer.

Let $k$ be a field and $x,y$ two commuting variables; set $R=k[x,y]$. As a $k$-algebra, $R$ is graded by the total degree
$$
R_n=span_k{x^py^q|p+q=n}
$$
Consider the ideal $M:=R_{geq 1}=oplus_{ngeq 1}R_n=(x)+(y)=(x,y)$. As a $k$ vector space $M$ is graded by the preceding definition (the range of degrees is $mathbb{N}_{geq 1}$). We first look at the tensor square
$Motimes_k M$ and its graded structure given by
$$
(Motimes_k M)_n=oplus_{p+q=n}R_potimes_k R_q .
$$
(this time, the range of degrees is $mathbb{N}_{geq 2}$) and we consider the quotient
$$
N={(Motimes_k M)}/{(Motimes_k M)_{geq 3}}
$$
and, denoting by $s:(Motimes_k M)to N$ the canonical quotient mapping, we see that $N$ has $k$-dimension 4 with basis $B={e_{ab}}_{a,bin {x,y}}$ with $e_{ab}=s(aotimes_k b)$. Now, one checks easily that the $k$-bilinear map
begin{align*}
M times M &to N, \
(u,v) &mapsto Phi_k(uotimes_k v)=s(uotimes_k v)
end{align*}
is, in fact, $R$-bilinear (check it with monomials $x^py^q$). Hence it gives rise to a linear map $Phi_R : M otimes_R M to N$. This suffices to prove that
$t=xotimes_R y-yotimes_R xnot=0$ as $Phi_R(t)=e_{xy}-e_{yx}$ (note that $tin Motimes_R M$). But now
$$
x.t=x^2otimes_R y-xyotimes_R x=xotimes_R xy-xotimes_R yx=0
$$

which shows the claim.

Late edit Here, tensor products are thought and constructed as in
Bourbaki Algebra II § 3 eq. (1). More precisely, if $M$ (resp. $N$)
is a $B$-module on the right (resp. $N$ is a $B$-module on the left),
$Motimes_B N$ is a the solution of the universal problem with
$f, (Mtimes N)to G$ bi-additive and satisfying,
for $xin M, yin N, lambdain B$
$$
f(xlambda,y)=f(x,lambda y)
$$
but, if $M,N$ have other structures commuting with the $B$-actions, the
tensor product $Motimes_B N$ automatically inherits these properties. For example, here, as $R$ is commutative, all $R$-modules are $R-R$-bimodules.