about torsion and flat module

Question

Can someone explain me how (i) implies (ii) ? What is that lopping off term does ? And why at n=0 tensor is not exact?

Answer 1

(i) $\implies$ (ii) by definition, since any flat $A$-module $M$ will have $\operatorname{Tor}_i (M, N) = 0$ for all $i > 0$ and all $N$.

The Tor functors are the derived functors of the tensor product. Starting with $0 \to M' \to M \to M'' \to 0$ as ses of modules and $N$ is any module (over a fixed commutative ring $R$), then $M' \otimes N \to M \otimes N \to M'' \otimes N \to 0$ is exact.

Answer 2

$$\dots\longrightarrow P_2\otimes_A M\longrightarrow P_1\otimes_A M\longrightarrow P_0\otimes_A M\longrightarrow N\otimes_A M\longrightarrow 0$$ is an exact complex since $M$ is flat, so the lopped-off complex $$\dots\longrightarrow P_2\otimes_A M\longrightarrow P_1\otimes_A M\longrightarrow P_0\otimes_A M\longrightarrow 0$$ is exact in positive degree. Therefore the homology of this complex is $0$ in positive degree, i.e. by definition of the Tor functors, $$\operatorname{Tor}^A_n(M,N)=0\quad\forall n>0.$$

In degree $0$, we have the homology \begin{align} H_0&=\ker(P_0\otimes_A M\to 0)\big/\operatorname{Im}(P_1\otimes_A M\to P_0\otimes_A M)\\ &=P_0\otimes_A M\big/\operatorname{Im}(P_1\otimes_A M\to P_0\otimes_A M)\\ &\simeq N\otimes_A M. \end{align}