Let $R$ be a domain. Then $\operatorname{Tor}_n^R(A,B)$ is a torsion module I have some problem to understanding the proof of this problem. This theorem is on page $414$ introduction to homological algebra Rotman. The theorem says: 

If $R$ is a domain, then $\operatorname{Tor}_n^R(A,B)$ is a torsion module for all $A$, $B$ and $\forall n\ge 1$.

for proof we use the $$0\to {tB}\to B \to B/tB\to0$$
gives exactness of $$ Tor_1^R(A,tB)\to Tor_1^R(A,B)\to Tor_1^R(A,B/tB).$$the flanking terms are torsion, thus $Tor_1^R(A.B)$ is torsion(?!) and says the proof by dimension shifting(?)
I have thought about this, but I don't know how to use dimension shifting!
Can you help please?
thank you
 A: Lemma 7.14, directly before Theorem 7.15, and the first step in the proof of 7.15 show why they flanking terms are torsion.
Suppose $A \xrightarrow{f} B \xrightarrow{g} C$ is an exact sequence of $R$-modules with $A$ and $C$ torsion. Let $x \in B$. If $x \in \ker g$, then by exactness there is $a \in A$ with $f(a) = x$. Since $A$ is torsion, there is $r\neq 0$ with $ra = 0$. So, $0 = rf(a) = rx$ implying $x$ is a torsion element. If $x \not\in \ker g$, then $g(x)\neq 0$ in the torsion module $C$. So, there is some $r\neq 0$ with $rx \in \ker g$. Then, by exactness again there is $a \in A$ with $f(a) = rx$. Repeat the process for $a \in A$ and you get some $s\neq 0$ in $R$ with $0= sf(a) = (sr)x$. Since $R$ is a domain, $sr \neq 0$ and so $x$ is a torsion element again. This implies $B$ is torsion.
For the inductive step we assume $Tor_{n-1}^R(A,B)$ is torsion for all $R$-modules $A,B$ and $n > 1$.  Then, let $\Omega(tB)$ and $\Omega(B/tB)$ be first syzygies of $tB$ and $B/tB$ respectively. Dimension shifting says that $Tor_n^R(A,tB) \cong Tor_{n-1}^R(A,\Omega(tB))$ and $Tor_n^R(A,B/tB) \cong Tor_{n-1}^R(A,\Omega(B/tB))$. Combine this with the exact sequence you wrote above (but in degree $n$ instead) to get another exact sequence whose flanking terms are torsion by induction.
