Currently, I am studying torsion group. Reading the lecture notes, I found this theorem
Let $R$ be an integral domain and $Q:= \operatorname{Frac}(R)$ with $K=Q/R$, then $\operatorname{Tor}_1^R(K,A) \cong tA$ for all $R$-module $A$.
I figure out from the stackexchange question Why field of fractions is flat?, if $A$ is a torion $R$-module, then $tA=A$, so is $tA$ stands for torion of $A$? How one can define that?
As far as I know, the definition of my $\operatorname{Tor}_n^R(A,B) = H_n(P_A \otimes_R B) = H_n(A \otimes_R P_B)$ which is a homology of tensor products between one objects and the projective resolution
You have an exact sequence $0\to R\to Q\to Q/R\to 0$, so tensoring with $A$ gives an exact sequence $$Tor_1(Q,A)\to Tor_1(Q/R,A)\to R\otimes_R A\cong A\to Q\otimes_R A.$$ However, since $Q$ is a flat $R$-module, we have $Tor_1(Q,A)=0$, so $$Tor_1(Q,A)=\ker(A\to Q\otimes_RA).$$ It is easy to see that this is exactly the torsion elements in $A$, as Thomas mentions.