Why does $\mathrm{Tor}_0^R(M,N)\cong M\otimes_R N$?

Several texts on commutative algebra (Eisenbud, Peter May's notes) state this fact as an obvious consequence of the fact that $(-\otimes_R N)$ is a right-exact functor, however, I think I'm missing their argument.

Let $$\cdots\to P_1\to P_0\to M\to 0$$ be the end of a projective resolution, so that $$\cdots\to P_1\otimes_R N\to P_0\otimes_R N\to M\otimes_R N\to 0$$ is a complex, and by right-exactness, the map $P_0\otimes_R N\to M\otimes_R N$ must be surjective. However, $P_1\otimes_R N\to P_0\otimes_R N$ need not be injective, so I don't see how to show that $$\mathrm{Tor}_0^R(M,N) = \mathrm{im}\left(P_1\otimes_R N\to P_0\otimes_R N \right)\big/\mathrm{ker}\left(P_0\otimes_R N\to M\otimes_R N \right)\cong M\otimes_R N.$$ What am I missing?

Solution 1:

As I mentioned in the comments, you need to consider the sequence $$\dots \to P_1\otimes N\to P_0\otimes N\to 0$$ Then $$\text{Tor}_0(M,N)=\frac{\ker P_0\otimes N\to 0}{\text{im} P_1\otimes N\to P_0\otimes N}=\frac{P_0\otimes N}{\alpha_1(P_1)\otimes N}$$ Where $\alpha_1:P_1\to P_0$ is the morphism from the projective resolution of $M$.