# Is R-P closed under addition?

• December 3rd 2011, 11:20 AM
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.
• December 3rd 2011, 03:56 PM
Drexel28
Re: Is R-P closed under addition?
Quote:

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 $R-\mathfrak{p}$ is multiplicative. But additive? Take the simplest example possible, is $\mathbb{Z}-(3)$ additive? Probably not since neither $4$ nor $2$ are divisible by $3$, but their sum is.
• December 5th 2011, 10:29 PM
sorv1986
Re: Is R-P closed under addition?
Quote:

Originally Posted by Drexel28
Certainly not. You know by definition that $R-\mathfrak{p}$ is multiplicative. But additive? Take the simplest example possible, is $\mathbb{Z}-(3)$ additive? Probably not since neither $4$ nor $2$ are divisible by $3$, 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?
• December 5th 2011, 10:37 PM
Drexel28
Re: Is R-P closed under addition?
Quote:

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 $3$ nor $-3$ is divisible by $5$, but how about their sum?