Small Question on the Tor functor Suppose that I have an $A$ - module $N$ with $A$ commutative and I take a projective resolution of $N$:
$$\ldots \rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow N \rightarrow 0.$$
Suppose $M$ is some other $A$ - module. Now why is it the case that 
$$\ldots \rightarrow P_2 \otimes_A M \rightarrow P_1 \otimes_A M \rightarrow P_0 \otimes_A M \rightarrow N \otimes_A M \rightarrow 0 $$
is not exact? I know that the tensor product is not in general left exact. However if the projective resolution is an infinite one then there is no "left" so why should the sequence above not be exact?
There has to be some problem with my understanding for then we always have $\textrm{Tor}_i^A(M,N) = 0$ for all $i$.
 A: The exactness of a long exact sequence is equivalent to the exactness of lots of short exact sequences. Precisely, if:
$$\cdots\to X\stackrel{f}{\to}Y\stackrel{g}{\to}Z\to\cdots$$
is part of a long exact sequence, then the short sequence:
$$0\to\operatorname{im}{f}\stackrel{\iota}{\to}Y\stackrel{g}{\to}{\operatorname{im}{g}}\to0$$
is exact, where $\iota$ is the inclusion. Conversely, if each such short sequence (one for each object of the long sequence) is exact, then the long sequence is exact. As the tensor product may fail to take one of the short exact sequences to a short exact sequence, due to the failure of left exactness, it also may not preserve the exactness of the long sequence.
A: Example: $A = k[x]/x^2$, $k$ some field, $M = N = k$, the trivial module.
A resolution of $N$ is given by 
$$ \cdots A \to A \to A \to A \to k $$
where the maps $A\to A$ are multiplication by $x$, and the map $A \to k$ sends $1$ to $1$ and $x$ to $0$.
I claim all the maps $A\otimes _A k \to A \otimes _A k$ are zero.  This is true because $1 \otimes _A 1 \mapsto x \otimes _A 1 = 1 \otimes _A x\cdot 1 = 1\otimes _A 0 = 0$.  The new sequence fails to be exact since it consists of non-zero modules, but all of the maps (except the last) are zero.
