Is R\P a Multiplicative Set in an Integral Domain with a Prime Ideal?

[tom.young84]

1)

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.

 

[Hurkyl]

My idea is that since we removed all prime ideals

No, you removed the elements of P.

 

[tom.young84]

Outside of the case with {0}, I'm not sure why this is a multiplicative set.

 

[Tinyboss]

(1) follows immediately from the definition of prime ideal.

 

[rasmhop]

1)

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.

As Hurkyl remarked you seem to have misunderstood the question. We haven't removed all prime ideals, just a single fixed one. You need to show that if a and b are elements of R\P, then ab is an element of R\P. That is assume a,b are elements of R that aren't in P. Now you need to show that ab can't possibly be in P. The easiest way is by contradiction. Suppose ab is in P. Then you have,
- a,b are not in P
- ab is in P
Can you see how this contradicts that P is a prime ideal?

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.

What is your definition of a local ring? It seems to be that all non-units in the ring form an ideal. If this is the case then this argument is correct.

 

[tom.young84]

For the first question, I understand your proof, but I don't understand why that answers the question. Your contradiction is that it violates the definition of a prime ideal. I don't understand why this demonstrates that A\P is closed under multiplication.

 

[rasmhop]

For the first question, I understand your proof, but I don't understand why that answers the question. Your contradiction is that it violates the definition of a prime ideal. I don't understand why this demonstrates that A\P is closed under multiplication.

We prove that if $a,b \in R\setminus P$ and $ab \in P$ we get a contradiction, so if $a,b \in R\setminus P$ we must have $ab \in R \setminus P$ which is exactly what it means for $R \setminus P$ to be closed under multiplication.