Need help to show $R/I$ is not necessarily flat over $R$ Let $R$ be a ring with unit and $I$ an ideal in $R$. I want to show that $R/I$ is need not be flat over $R$, but I do not know how to come up with a counter-example.
Any hint is appreciated.
 A: Analysis of the problem:
Suppose that  $R/I$ is flat over $R$.
Then tensoring  the short exact sequence $0\to I\to R$, by $R/I$ yields  a new exact sequence  $$0\to I\otimes_R R/I \to R\otimes_R R/I\quad (*)$$ Recalling the standard identification $M\otimes_R R/I\xrightarrow {\cong} M/IM:\tilde m\otimes \tilde  r\mapsto \overline  {rm}$ for any $R$-module $M$, we get from $(*)$ the injective map $$ 0\to I/I^2\to R/I:\tilde i  \mapsto \overline  {i} =\bar 0    \quad (**) $$ 
But  the morphism  $(**)$ is clearly the zero map.
It can only be injective if $I/I^2=0$ or equivalently if $I=I^2$.   So we have proved $$  R/I \; \text {flat}    \implies I=I^2              $$
Conclusion:
By contraposition, if $I\neq I^2$ the $R$-module $R/I$ is guaranteed to be non-flat.
So in a non formal  but very clear sense $R/I$ is practically never flat since an ideal is practically never equal to its square .
Here is a result ( a consequence of Nakayama's lemma) corroborating this informal statement :
Theorem:
 If $I$ is finitely generated and $I=I^2$, then $I=(i)$ for some idempotent $i=i^2\in R$
Corollary:
If $R$ is a noetherian domain and $0\subsetneq I\subsetneq R$  an ideal, then $R/I$ is not flat.
A: You could insist that $R$ be local, and then flatness is equivalent to projectiveness.
To thwart $R/I$ from being projective, you would just ensure that $I$ is not a summand of $R$. 
So there you have it, a blueprint to find an example. Any commutative local ring with an ideal which is not a summand will work. An obvious choice would be $\mathbb{Z}/(p^2)$ for a prime $p$.
A: Take the exact sequence 
$$ 0 \to \mathbb Z \xrightarrow{\cdot 2} \mathbb Z \to \mathbb Z / 2 \mathbb Z \to 0$$
and tensor with $\mathbb Z / 3 \mathbb Z$ to get 
$$ 0 \to \mathbb Z \otimes \mathbb Z / 3 \mathbb Z \xrightarrow{\cdot 2} \mathbb Z \otimes \mathbb Z / 3 \mathbb Z \to \mathbb Z / 2 \mathbb Z \otimes \mathbb Z / 3 \mathbb Z\to 0$$
which is isomorphic to 
$$ 0 \to \mathbb Z / 3 \mathbb Z \to  \mathbb Z / 3 \mathbb Z  \to 0 \to 0$$
which is no longer exact.
A: Consider $R = \mathbf Z$ and take the ideal $I = 2\mathbf Z$. Now take a look at the exact sequence $0\to 2\mathbf{Z}\to\mathbf{Z} \to \mathbf{Z}/2\mathbf{Z} \to 0$.
Can you see what happens when you tensor this exact sequence over $\mathbf{Z}$? You should be able to prove that the resulting sequence cannot be exact.
