about flat $R$-modules with different conditions

Question

Need to know if the following statements are true or false:

1. Let $S$ be a multiplicative set in $R$. Then $(S^{-1}R)[x,y,z]$ is always a flat $R$-module.

2. Let $I,J \subset R$ be proper ideals of $R$ such that $I+J=R$. Then $R/I$ is always flat $R$-module.

3. Let $R[[x]]$ be the formal power series ring over $R$. Then $R[[x]]/\left<\sum^{\infty}_{i=2} x^i\right>$ is always flat $R$-module.

4. $R/N(R)$ is always a flat $R$-module. ($N(R)$ the nilradical). False. Seen here. Sure?

Answer 1

I assume you are working over commutative rings.

1. This is correct. $S^{-1}R$ is flat over $R$ (localizations are flat) and if $A\to B\to C$ are ring homomorphisms with $A\to B$ and $B\to C$ are flat, then so is $A\to C$. Since $S^{-1}R\to S^{-1}[x,y,z]$ is flat, you are done.

2. This is false. Take $R=\mathbb{Z}$, $I=pR, J=qR$ where $p,q$ are distinct primes.

3. This is correct. $f=\sum_{n=2}^{\infty} x^n=x^2u$ where $u=\sum_{n=0}^{\infty} x^n$. Easy to check that $u$ is a unit. So, $R[[x]]/(f)=R[[x]]/(x^2)$ which is free module over $R$ with basis $1,x$.

4. This is false. For example, let $R=\mathbb{Z}/(p^2)$, $p$ a prime. Then $R/N(R)=R/(p)$ is not flat over $R$.