If $...F_{2}\rightarrow F_{1}\rightarrow F_{0}\rightarrow N\rightarrow 0$ is a free solution of $N$ as $R$-module(R is a commutative ring),then $Tor^{i}(M,N)$ is the homology at $M\otimes F_{i}$ of the complex $M\otimes F_{i+1}\rightarrow M\otimes F_{i}\rightarrow M\otimes F_{i-1}$. I am very confused about two things:
1:$Tor^{0}(M,N)\cong M\otimes N$.I think since $Tor$ is right exact ,so $M\otimes F_{1}\rightarrow M\otimes F_{0}\rightarrow M\otimes N\rightarrow 0$ is exact ,$Tor^{0}(M,N)$ should be $0$.
2 Why we can also choose the free solution of $M$ to compute the $Tor^{i}(M,N)$?
Can you help me?
$\text{Tor}_k(M,N)$ is the homology of the sequence $$\cdots\to M\otimes F_{k+1}\to M\otimes F_k\to\cdots\to M\otimes F_1\to M\otimes F_0\to 0$$ at the $k$-th position. So $\text{Tor}_0(M,N)$ is the cokernel of $$M\otimes F_1\to M\otimes F_0$$ which is precisely $M\otimes N$ by right-exactness.
There are various tricks to showing that $\text{Tor}$ can be computed by resolving either argument, or more generally that $\text{Tor}_n^R(M,N)\cong\text{Tor}_n^R(N,M)$. Each text on homological algebra will give its favourite method. One way is to resolve both arguments and consider the homology of the "double complex" got by tensoring both resolutions together.