# Two ring theory problems

1. Feb 24, 2010

### 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.

2. Feb 24, 2010

### Hurkyl

Staff Emeritus
No, you removed the elements of P.

3. Feb 24, 2010

### tom.young84

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

4. Feb 24, 2010

### Tinyboss

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

5. Feb 24, 2010

### rasmhop

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?

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.

6. Feb 25, 2010

### 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.

7. Feb 25, 2010

### rasmhop

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.