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?
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): $$ \require{AMScd} \begin{CD} 0 @>>> \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} @>{1 \otimes 2}>> \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/4\mathbb{Z} @>{1 \otimes 1}>> \mathbb{Z}/2\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z} @>>> 0 \\ @. @V{\sim\,}VV @V{\sim\,}VV @V{\sim\,}VV @. \\ 0 @>>> \mathbb{Z}/2\mathbb{Z} @>{\smash[t]{0}}>> \mathbb{Z}/2\mathbb{Z} @>{\smash[t]{1}}>> \mathbb{Z}/2\mathbb{Z} @>>> 0 \end{CD} $$