Tensor square of a torsion-free module ($R$ domain)

Question

Linked to my comment there.

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

Answer 1

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 $$ xy\ (x\otimes y-y\otimes x)=x^2\otimes y^2-x^2\otimes y^2=0 $$ in $M\otimes_RM$, it suffices to check $x\otimes y\ne y\otimes x$ (again in $M\otimes_RM$).

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 $M\otimes_RM\to k$, $$ (f\otimes g)\mapsto\frac{\partial f}{\partial x}(0,0)\ \frac{\partial g}{\partial y}(0,0) $$ is well-defined, $R$-linear and sends $x\otimes y$ to $1$ and $y\otimes x$ to $0$.

Answer 2

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 $M\otimes_R M$ 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 $M:=R_{\geq 1}=\oplus_{n\geq 1}R_n=(x)+(y)=(x,y)$. 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 $M\otimes_k M$ and its graded structure given by $$ (M\otimes_k M)_n=\oplus_{p+q=n}R_p\otimes_k R_q\ . $$ (this time, the range of degrees is $\mathbb{N}_{\geq 2}$) and we consider the quotient $$ N={(M\otimes_k M)}/{(M\otimes_k M)_{\geq 3}} $$ and, denoting by $s:(M\otimes_k M)\to N$ the canonical quotient mapping, we see that $N$ has $k$-dimension 4 with basis $B=\{e_{ab}\}_{a,b\in \{x,y\}}$ with $e_{ab}=s(a\otimes_k b)$. Now, one checks easily that the $k$-bilinear map \begin{align*} M \times M &\to N, \\ (u,v) &\mapsto \Phi_k(u\otimes_k v)=s(u\otimes_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 $t=x\otimes_R y-y\otimes_R x\not=0$ as $\Phi_R(t)=e_{xy}-e_{yx}$ (note that $t\in M\otimes_R M$). But now $$ x.t=x^2\otimes_R y-xy\otimes_R x=x\otimes_R xy-x\otimes_R yx=0 $$
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), $M\otimes_B N$ is a the solution of the universal problem with $f, (M\times N)\to G$ bi-additive and satisfying, for $x\in M,\ y\in N,\ \lambda\in B$ $$ f(x\lambda,y)=f(x,\lambda y) $$ but, if $M,N$ have other structures commuting with the $B$-actions, the tensor product $M\otimes_B N$ automatically inherits these properties. For example, here, as $R$ is commutative, all $R$-modules are $R-R$-bimodules.