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.
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?