Let $U$ be a $\mathbb{Z}$-module with $\operatorname{Tor}_1^\mathbb{Z}(U,U)=0$. Does this imply that $U$ is torsion-free?
I know that if $\operatorname{Tor}_1^\mathbb{Z}(U,M)=0$ for every $\mathbb{Z}$-module $M$, then $U$ is called flat (by definition), but this is stronger than $\operatorname{Tor}_1^\mathbb{Z}(U,U)=0$. On the other hand, flat implies torsion-free. However, I have no idea how to attack the question.
I appreciate any help.
The functor $\operatorname{Tor}^\mathbb{Z}_1$ is left exact in each variable (by the long exact sequence for Tor and the fact that $\operatorname{Tor}^\mathbb{Z}_2=0$). So if $\operatorname{Tor}^\mathbb{Z}_1(U,U)=0$, then $\operatorname{Tor}^\mathbb{Z}_1(A,B)=0$ for any subgroups $A,B\subseteq U$. Now if $U$ is not torsion-free, let $x\in U$ be a nonzero torsion element and let $A$ and $B$ to both be the cyclic subgroup generated by $x$. Then $\operatorname{Tor}^\mathbb{Z}_1(A,B)\neq 0$ (since $\operatorname{Tor}^\mathbb{Z}_1(\mathbb{Z}/(n),\mathbb{Z}/(n))\cong\mathbb{Z}/(n)$ for all $n>0$), which is a contradiction. Thus if $\operatorname{Tor}^\mathbb{Z}_1(U,U)=0$, then $U$ is torsion-free.
More generally, if $M$ and $N$ are $\mathbb{Z}$-modules, then $\operatorname{Tor}^\mathbb{Z}_1(M,N)\neq 0$ iff there is a prime $p$ such that both $M$ and $N$ have elements of order $p$. See When is $\operatorname{Tor}_1 ^\mathbb{Z} (M,N) \neq 0$? for the details.