Quotient of polynomial ring flat $R$-algebra Let $R$ be a Noetherian ring and let $P(T) \in R[T]$ be a monic polynomial. Let $A:= R[T]/(P(T))$. 
(1) Is $A$ a flat $R$-algebra and why? (About this, I think that's true, below I give an argument and would like to know if the argument is correct.)
(2) Assume $P(T)$ is not necessarily monic. What are  necessary and sufficient conditions to decide if $A=R[T]/(P(T))$ is flat $R$-algebra? Which role play the assumption that $R$ is Noetherian?
I think that I have elaborated a proof shows (1): Since we assumed $P(T)$ to be monic, $P(T)=T^{n}+c_{n-1}T^{n-1}+\cdots +c_{2}T^{2}+c_{1}T+c_{0}$ and $A$ is then a free $A$-module with basis $T^{0}=1, T^1, T^2,..., T^{n-1}$. We know free modules are flat. This seemingsly shows (1). Is the proof ok? 
 A: Let $f\in R[X]$, and let $c(f)$ denote the ideal generated by the coefficients of $f$. ($c(f)$ is called the content of $f$.) Then 

$A=R[X]/(f)$ is $R$-flat iff $c(f)$ is generated by an idempotent. 

"$\Rightarrow$" Tensor the exact sequence of $R$-modules $$0\to(f)\to R[X]\to A\to 0$$ 
by $R/c(f)$ and get 
$$0\to(f)/c(f)(f)\to (R/c(f))[X]\to A/c(f)A\to 0.$$
But $A/c(f)A=R[X]/c(f)R[X]$ since $(f)\subseteq c(f)R[X]$, hence $(f)=c(f)(f)$ and from $f=c(f)fg$ we get $c(f)=c(f)^2$.
"$\Leftarrow$" Let $P$ be a prime ideal of $R[X]$ containing $f$. We prove that $A_P$ is $R_p$-flat, where $p=P\cap R$. Note that we have $A_P=R[X]_P/fR[X]_P$.
Since $c(f)=Re$ with $e=e^2$, and $R=Re\oplus R(1-e)$, we may assume that $c(f)=R$. Then $f$ is a non-zerodivisor, and we have an exact sequence
$$0\to R[X]_P\stackrel{f\cdot}\to R[X]_P\to A_P\to0.$$
Since $c(f)\nsubseteq p$, $f$ has not all the coefficients in $p$, and thus $f$ is a non-zerodivisor modulo $p$. It follows that the forgoing exact sequence remains exact after tensorizing by $R/p$, and thus $\mathrm{Tor}_1^R(R/p,A_P)=0$ proving that $A_P$ is $R_p$-flat.
