A module $B$ is flat if Tor $= 0$ From Weibel's "An Introduction to Homological Algebra":
Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A. 
It seems to me that the obvious way to do this would be to use the definition:
Tor$_i^R(A,B) = $H$_i(P.\otimes B)$ where $P.$ is a projective resolution of A.
We need to show that for an exact sequence:
$...\rightarrow A_{n+1} \rightarrow A_{n} \rightarrow A_{n-1} \rightarrow ...$ ,
$...\rightarrow A_{n+1} \otimes B \rightarrow A_{n} \otimes B \rightarrow A_{n-1}  \otimes B \rightarrow ...$ is exact.
Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.
Any help is appreciated.
 A: We want to show that $\_ \otimes B$ is exact. Now Lets take for this a short exact sequence $$0 \to X \to Y \to Z \to 0$$
Now lets apply the functors $\mathrm{Tor}^i(\_,B)$ to that sequence. Now this gives by the definition of $\mathrm{Tor}^i(\_,B)$ as a homological functor a long exact sequence $$ ... \to \mathrm{Tor}^1(X,B) \to \mathrm{Tor}^1(Y,B) \to \mathrm{Tor}^1(Z,B) \to\mathrm{Tor}^0(X,B) \to \mathrm{Tor}^0(Y,B) \to \mathrm{Tor}^0(Z,B) \to 0  $$
Now  since we have a natural isomorphism $\mathrm{Tor}^0(Z,B) \cong Z\otimes B$ we may rewrite the top sequence as:
$$ \mathrm{Tor}^1(X,B) \to \mathrm{Tor}^1(Y,B) \to \mathrm{Tor}^1(Z,B) \to X\otimes B \to Y\otimes B \to Z \otimes B \to 0  $$
But since $ \mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0\to X\otimes B \to Y\otimes B \to Z \otimes B \to 0  $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $\mathrm{Tor}$-terms vanish)
