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R is an integral domain and P is a prime ideal.

Show R\P (R complement P or R-P) is a multiplicative set.

-Well since R is an integral domain it contains 1.

-{0} would be a prime ideal, and that was removed (is this too much to assume)

I'm not sure how to show multiplication is closed. My idea is that since we removed all prime ideals, so we have a, b in R and ab in R, we are only left with subsets where multiplication is closed.

2)

Local rings have only one maximal ideal

Could this be prove as such:

Suppose an ideal J that contains all non units. Now suppose another maximal ideal I in J and "a" (a non-unit) in J. This means that I must either be the ring R or J again. Since J is maximal, I=J and a is in J.