Proving Containment in AB for Ideals A and B in a Commutative Ring with Unity

[e(ho0n3]

Homework Statement

If A and B are ideals of a commutative ring R with unity and A + B = R, show that A \cap B = AB.

The Attempt at a Solution

Showing AB \subseteq A \cap B is easy. I'm having trouble with containment in the other direction:

Let x \in A \cap B. Then x is in A and x is in B. To show that x belongs to AB, it suffices to show that 1 belongs to either A or B and so 1x or x1 belongs to AB. It seems to me that 1 isn't necessarily in A or B so this approach is unfruitful. Is there another decomposition of x into ab where a belongs to A and B belongs to B?

 

[matt grime]

If you show 1 is in A, then that means A=R. Clearly you don't want to do that. What information haven't you used? That A+B=R, ie there exist elements x in A and y in B such that x+y=1.

 

[e(ho0n3]

I have thought about that but it lead nowhere: Write 1 as a + b. Then x1 = xa + xb. Hmm...I see it know. Since R is commutative, xa = ax and so ax is in AB since x is in B. Similarly, xb is in AB. If AB is closed under addition, then surely xa + xb is in AB and so x is in AB. All that remains is to show that AB is closed under addition. Right?

 

[matt grime]

That is part of the *definition* of AB.

 

[e(ho0n3]

AB = {ab : a in A and b in B}. How does that definition demonstrate that AB is closed under addition?

 

[matt grime]

I think you should go and check your definition of the *ideal* AB.

 

[e(ho0n3]

Aha! There is an exercise where AB is defined. I had overlooked it. Thanks a lot.