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.

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- Dec 3rd 2011, 10:20 AMsorv1986Is R-P closed under addition?
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. - Dec 3rd 2011, 02:56 PMDrexel28Re: Is R-P closed under addition?
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.

- Dec 5th 2011, 09:29 PMsorv1986Re: Is R-P closed under addition?
- Dec 5th 2011, 09:37 PMDrexel28Re: Is R-P closed under addition?