My professor defined a flat module as follows:
$M$ is flat if $M \otimes_R(.)$ is exact.
Then she gave us this example for a module that is not flat:
For $n \in \mathbb Z, n \geq 2.$ the following sequence $0 \rightarrow \mathbb Z \xrightarrow{f} \mathbb Z \xrightarrow{g} \mathbb Z /n \mathbb Z \rightarrow 0$ with $f: 1 \mapsto n$ and $g$ surjective is exact but $$0 \rightarrow \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z \xrightarrow{f \otimes 1} \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z \xrightarrow{g \otimes 1} \mathbb Z /n \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z \rightarrow 0$$ where we have $$t \otimes 1 \mapsto nt \otimes 1 \mapsto 0$$ and $nt \otimes 1 = 0$ is not exact. And she added, so $\mathbb Z /n \mathbb Z$ is not a flat $\mathbb Z-$module.
My question:
I know that $\mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z = \mathbb Z /n \mathbb Z$ and $\mathbb Z /n \mathbb Z \otimes_{\mathbb Z} \mathbb Z /n \mathbb Z = \mathbb Z /n \mathbb Z.$ Also, I know that tensoring is right exact, we proved that before, so why the second sequence she gave is not left exact?
Any explanation to why my professor said that is appreciated.
The crucial observation was already pointed out in the comments by the user Alex Wertheim. I will just add an answer spelling it out.
We first note that $\mathbb Z\otimes_{\mathbb Z}\mathbb Z/n\mathbb Z\cong\mathbb Z/n\mathbb Z$ as usual. Along this identification the tensored map $f\otimes1$ becomes
$$\mathbb Z/n\mathbb Z\to\mathbb Z/n\mathbb Z,\,\overline x\mapsto\overline{nx}=\overline0$$
i.e. the zero map as $n\equiv0\mod n$. This map is clearly not injective for $n\ge2$ and hence the tensored sequence not left exact.