Thread: 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.

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 is multiplicative. But additive? Take the simplest example possible, is additive? Probably not since neither nor are divisible by , but their sum is.

Originally Posted by Drexel28

Certainly not. You know by definition that is multiplicative. But additive? Take the simplest example possible, is additive? Probably not since neither nor are divisible by , 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?

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 nor is divisible by , but how about their sum?