TITLE: Is $R/N(R)$ a faithfully flat $R$-module?
QUESTION [10 upvotes]: I'm studying recently faithfully flat modules and I'd like to know the following: 

Is $R/N$ faithfully flat as $R$-module, where $R$ is a commutative ring with unit and $N$ is the ideal of nilpotent elements of $R$?

REPLY [13 votes]: If $I\subset R$ is an arbitrary  ideal, then $R/I$ can only be flat if $I=I^2$ :  this follows from tensoring  the exact sequence $0\to I\to R$ by $R/I$ and getting the exact sequence $0\to I\otimes _R R/I=I/I^2\stackrel {0}{\to} R/I$.   
This is a tool for proving that many quotients of a ring  are not flat over the ring:
 For example if $k$ is a field, then  the ring $R=k[\epsilon]=k[T]/(T^2)$ has as nilpotents $N=(\epsilon)$, and since $(\epsilon)^2=(0)\neq (\epsilon)$, the quotient $R/N$ is not flat over $R$, which answers your question negatively (we don't have flatness of $R/N$ over $R$, let alone faithful flatness).