# Is R-P closed under addition?

• Dec 3rd 2011, 10:20 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.
• Dec 3rd 2011, 02:56 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 \$\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 PM
sorv1986
Re: Is R-P closed under addition?
Quote:

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?
• Dec 5th 2011, 09:37 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 \$\displaystyle 3\$ nor \$\displaystyle -3\$ is divisible by \$\displaystyle 5\$, but how about their sum?