suppose R be a commutative ring and P be a prime ideal of R. Then is S=R-P (the complement of P in R) closed under addition?
regards. thanks.
Certainly not. You know by definition that $\displaystyle R-\mathfrak{p}$ is multiplicative. But additive? Take the simplest example possible, is $\displaystyle \mathbb{Z}-(3)$ additive? Probably not since neither $\displaystyle 4$ nor $\displaystyle 2$ are divisible by $\displaystyle 3$, but their sum is.