TITLE: Exactness of the Tensor Functor
QUESTION [13 upvotes]: This might turn out to be a very stupid question, so I apologize in advance, but it is confusing me a little bit.
I know in general that if $$M'\rightarrow M\rightarrow M''\rightarrow 0$$ is an exact sequence of $A-$modules, then $$M'\otimes_AN\rightarrow M\otimes_AN \rightarrow M''\otimes_AN \rightarrow 0$$ is an exact sequence for any $A-$module $N$.
If instead we are given an exact sequence of the form: 
$$M'''\rightarrow M'\rightarrow M\rightarrow M''\rightarrow 0$$ where $M'''$ is not the zero module (because I know that tensoring with a module doesn't necessarily take injective maps to injective maps), then I guess my question is why isn't the following an exact sequence as well?
$$M'''\otimes_A N\rightarrow M'\otimes_AN\rightarrow M\otimes_AN \rightarrow M''\otimes_AN \rightarrow 0$$ 
What is going morally wrong here? I guess i'm wondering where exactly is the proof of the first case failing to show that the second sequence is exact at $M'\otimes_AN$?
I would appreciate very much if anyone could clarify my confusion. I am probably just being stupid about this :/

REPLY [4 votes]: Some Counterexamples
There are many counterexamples to this statement - clearly, there has to be, for otherwise flat modules would be everywhere. And perhaps the best way to see why it fails is to examine some examples. Here is a class of examples that have two advantages: they are easy to compute with, and show something a little different than working over principal ideal domains (it's safe to ignore the parenthetical remarks):
Consider $\mathbb{Z}[x]$-the integer polynomial ring. Here is an example of an exact sequence of $\mathbb{Z}[x]$-modules:
$$
0\to \mathbb{Z}[x]\xrightarrow{\alpha} \mathbb{Z}[x]^2\xrightarrow{\beta}\mathbb{Z}[x]\xrightarrow{\gamma} \mathbb{Z}/2\to 0
$$
The maps are defined by

$\alpha(f) = (2f,xf)$ 
$\beta(f,g) = 2g - xf$ 
$\gamma$ the quotient map via the ideal $(2,x)$.

(This is just a special example of something called the Koszul complex given by the regular sequence $(2,x)$ - any two relatively prime elements generating a proper ideal will work, since $\mathbb{Z}[x]$ is a GCD domain.) Here are two examples to show why tensoring does not keep this exact:
Example 1
One can apply the functor $\mathbb{Z}/2\otimes_{\mathbb{Z}[x]}-$ to get the complex
$$
0\to \mathbb{Z}/2\xrightarrow{\alpha} \mathbb{Z}/2\oplus \mathbb{Z}/2\xrightarrow{\beta}\mathbb{Z}/2\to\mathbb{Z}/2\to 0
$$
Here, $\mathbb{Z}/2$ is a $\mathbb{Z}[x]$ module via the quotient map $\mathbb{Z}[x]\to \mathbb{Z}/2$, so $2$ and $x$ act trivially, and hence $\ker\beta = \mathbb{Z}/2\oplus\mathbb{Z}/2$ whereas $\operatorname{Im}\alpha = 0$. In other words, $\operatorname{Tor}_1^{\mathbb{Z}[x]}(\mathbb{Z}/2,\mathbb{Z}/2) = \mathbb{Z}/2\oplus\mathbb{Z}/2$. Note, to compute derived functors we "should have" chopped off the $\mathbb{Z}/2$ but that doens't matter since we are not computing the zeroth homology.
Example 2
Here is a more interesting example, still working with $\mathbb{Z}[x]$-modules: Consider the following $\mathbb{Z}[x]$-module: the underyling abelian group structure is given by $V = \mathbb{Z}/4\times\mathbb{Z}/4$. There is an endomorphism of $V$ as abelian groups given by the matrix
$$T=
\begin{pmatrix}
2 & 2\\
0 & 1
\end{pmatrix}
$$
Letting $x$ act via $T$ makes $V$ into a $\mathbb{Z}[x]$-module. Applying $V\otimes_{\mathbb{Z}[x]}-$ to the exact sequence gives a chain complex
$$
V\xrightarrow{\alpha} V^2\xrightarrow{\beta} V\to V\otimes_{\mathbb{Z}[x]}\mathbb{Z}/2\to 0
$$
Now $\beta([(1,0),(1,0)]) = (2,0) - (2,0) = 0$. However, $\alpha(a,b) = [ (2a,2b), (2a + 2b,b)]$. So the complex is not exact at $V^2$.
To Produce Your Own...
...find rings that have non-flat modules and modules with free resolutions of length at least two. Most of the time you should get a winner.