1. ## Is 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.

2. ## Re: Is R-P closed under addition?

Originally Posted by sorv1986
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.

3. ## Re: Is R-P closed under addition?

Originally Posted by Drexel28
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.

suppose i m taking two elements a,b from R-P.
then (a+P)+(b+P)=(a+b)+P.
now a+P≠P, b+P≠P.
then is it true that (a+b)+P ≠P?

4. ## Re: Is R-P closed under addition?

Originally Posted by sorv1986
suppose i m taking two elements a,b from R-P.
then (a+P)+(b+P)=(a+b)+P.
now a+P≠P, b+P≠P.
then is it true that (a+b)+P ≠P?
No. Neither $\displaystyle 3$ nor $\displaystyle -3$ is divisible by $\displaystyle 5$, but how about their sum?