$\mathbb{Z}/n\mathbb{Z}$ is not flat On the flat module Wikipedia page, it's stated that $\mathbb{Z}/n\mathbb{Z}$ is not flat over $\mathbb{Z}$.  But I don't understand their explanation of why.  It is said that

$n:\mathbb{Z}\rightarrow \mathbb{Z}$ is injective, but tensored with $\mathbb{Z}/n\mathbb{Z}$ it is not.

Why does this imply that $\mathbb{Z}/n\mathbb{Z}$ is not flat?  What exact sequence is not preserved by $\oplus_\mathbb{Z}\mathbb{Z}/n\mathbb{Z}$, and why? I apologize if this is an elementary question, but I am new to flatness.  I may need to have it spelled out for me.
 A: Every monomorphism $f:A \to B$ induces an exact sequence $0 \to A \to B \to \text{coker}(f) \to 0$. As tensoring is right-exact, a module $M$ is flat iff tensoring with $M$ preserves monomorphisms. 
Now use that $\Bbb Z \otimes_{\Bbb  Z}\Bbb Z /n \Bbb Z = \Bbb Z/n \Bbb Z$ to see why the map  $x \to nx$ is not injective on the tensors.
A: Guessing that $n:\mathbb{Z}\to\mathbb{Z}$ is multiplication by $n$? If so, this is certainly injective, so starts an exact sequence
$$ 0 \to \mathbb{Z} \stackrel{n}{\to} \mathbb{Z} $$
and the kernel of the next map will have to be the image, $n\mathbb{Z}$, so we take the canonical map from $p:\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}$ (which is surjective) to get
$$ 0\to \mathbb{Z}\stackrel{n}{\to} \mathbb{Z} \stackrel{p}{\to} \mathbb{Z}/n\mathbb{Z}\to 0. $$
If $\mathbb{Z}/n\mathbb{Z}$ were flat, we could tensor and
$$ 0 \to \mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z}\to 0 $$
would have to be exact as well. But is it?
