$\text{Tor}_0^R(M,N)\simeq M\otimes_R N$

Question

I'm trying to understand why $\text{Tor}_0^R(M,N)\simeq M\otimes_R N$.

Let $\dots\to F_1\to F_0\to M\to 0$ be a free resolution of $M$. Then $$ F_1\to F_0\to M\to 0$$ is exact. Hence $$ F_1\otimes_R N\to F_0\otimes_R N\to M\otimes_R N\to 0$$ is exact. Call the left map $f_1$ and the right map $f_0$. Then $M\otimes N\simeq F_0\otimes N/\ker (f_0)\simeq F_0\otimes N/\text{im}(f_1)$.

But if I understand the definition correctly, to show the result we need that $M\otimes N\simeq H_0(\mathcal F)=\ker f_0/\text{im}(f_1)$. But why does $\ker f_0=F_0\otimes N$? If it does at all...

Answer 1

You have to trim your resolution to end with $F_1\otimes_R N\to F_0\otimes_R N\to 0$ before taking homology. With this fix your problem dissolves, and the explanation is contained in what you already explained.