How do I show that the ideal $(2)$ of $\Bbb{Z}/4\Bbb{Z}$ is not flat? 
Let $R=\Bbb{Z}/4\Bbb{Z}$ and let me take the ideal $(2)\subset R$. I want to show that $(2)$ is not flat in $R$.

We only had the following definition:

A module $M\subset R$ is flat if $M\otimes_R-$ is left exact.

Using this definition I want to show it.
My idea was the following. Since we know that tensoring is already right exact we only need to worry about left exactness. The problem is that I don't see what left exact sequence I need to consider such that after tensoring it isn't left exact anymore. Could someone give me a hint?
 A: Assuming that you intend to look at modules over the ring $R = \mathbb{Z}/4\mathbb{Z}$, consider the short exact sequence
$$
0 \to \mathbb{Z}/2\mathbb{Z} 
\to \mathbb{Z}/4\mathbb{Z} 
\to \mathbb{Z}/2\mathbb{Z} 
\to 0, 
$$
where the left map is multiplication by $2$, embedding $\mathbb{Z}/2\mathbb{Z}$ into the regular module $\mathbb{Z}/4\mathbb{Z}$ as the submodule, i.e. ideal, $(2)$. Can you see how tensoring with $\mathbb{Z}/2\mathbb{Z}$ destroys injectivity?
After tensoring with $\mathbb{Z}/2\mathbb{Z}$, the map
\begin{align}
\mathbb{Z}/2\mathbb{Z} &\otimes \mathbb{Z}/2\mathbb{Z} 
&&\;\longrightarrow\; 
&\mathbb{Z}/2\mathbb{Z} &\otimes \mathbb{Z}/4\mathbb{Z} \\
x &\otimes y &&\;\longmapsto\; &x &\otimes 2y
\end{align}
sends
$$
1 \otimes 1 \;\longmapsto\; 1 \otimes 2 
= 2 \otimes 1 = 0 \otimes 1 = 0 \otimes 0, 
$$
i.e., it's the $0$ map, which is clearly not injective.
Here it is all summarized in a commutative diagram of $\mathbb{Z}/4\mathbb{Z}$-modules, where the vertical maps are all isomorphisms and the numbers labeling the other arrows are where the class of $1$ or $1 \otimes 1$ is sent (which determines maps since they're cyclic modules):
$$
\require{AMScd}
\begin{CD}
0 @>>> 
\mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} 
@>{1 \otimes 2}>> 
\mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/4\mathbb{Z} 
@>{1 \otimes 1}>> 
\mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}
@>>> 0 \\
@. @V{\sim\,}VV @V{\sim\,}VV @V{\sim\,}VV @. \\
0 @>>> 
\mathbb{Z}/2\mathbb{Z} @>{\smash[t]{0}}>> 
\mathbb{Z}/2\mathbb{Z} @>{\smash[t]{1}}>>
\mathbb{Z}/2\mathbb{Z} @>>> 0
\end{CD}
$$
