Is $R/N(R)$ a faithfully flat $R$-module?

Solution 1:

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).