# Product Of Ideals - Why is the sum necessary?

1. Mar 22, 2010

### jd102684

So I'm learning about Ring Theory and have gotten to Ideals. My book tells me that the product of two ideals I and J or a ring R (standard * and + operations) is the set of all finite sums of elements of the form ij where i is in I and j is in J. I'm having trouble coming up with an example as to why we can't just define the product of I and J to be just the elements of the form ij (not any finite sum).

The definition without the sum seems to work for the rings i've tried (R being the integers Z, and ideals being nZ and mZ for different m and n, for example). I suspect it breaks down with matrices, or maybe polynomials, but i'm having trouble nailing down a specific example. Guidance would be much appreciated. Thanks!

- JD

2. Mar 22, 2010

### jbunniii

You could do that, but it wouldn't be all that useful because that set isn't an ideal. An ideal has to be closed under addition.

IJ is the smallest ideal containing all the elements of the form ij, and this forces finite sums of elements to be included as well.

Last edited: Mar 22, 2010
3. Mar 22, 2010

### jd102684

Right, I'm sorry, I guess I didn't exactly state my question. I'm trying to see an example where the way I defined IJ (without the sum) is NOT an ideal. I cant seem to work out an example where the summation is necessary to maintain closure, though I know there's probably plenty basic examples out there!

4. Mar 22, 2010

### jbunniii

Gotcha. I'll try to come up with an example. I think you may be right that you need to consider e.g. matrix rings.

5. Mar 22, 2010

### Hurkyl

Staff Emeritus
You don't have to leave the nice world of commutative algebra.

However, what you do need to do is leave the world of principal ideal domains. I claim the following:

If I or J is principal, then IJ = { ij | i in I and j in J }​