# Thread: How would you prove this?

1. ## How would you prove this?

If I and J are ideals in a ring R that is commutative with identity, and I=(a) and J=(b), prove that IJ=(ab)

So if I=(a), then I is the set of all multiples of a, and likewise J is the set of all multiples of b. So IJ={i1j1+....+ikjk} with each i a multiple of a and each j a multiple of b, and so, each product a multiple of ab. So the sum of products that are multiples of ab is a multiple of ab?

Is that correct?

2. Originally Posted by zhupolongjoe
If I and J are ideals in a ring R that is commutative with identity, and I=(a) and J=(b), prove that IJ=(ab)

So if I=(a), then I is the set of all multiples of a, and likewise J is the set of all multiples of b. So IJ={i1j1+....+ikjk} with each i a multiple of a and each j a multiple of b, and so, each product a multiple of ab. So the sum of products that are multiples of ab is a multiple of ab?

Is that correct?
This shows that $IJ \subseteq (ab)$.

Now you need to show $(ab) \subseteq IJ$. Can you see how this is done?

3. Hm....so you start with all multiples of ab, that is, all numbers of the form nab for some integer n. No, sorry, not really sure how to go this direction.

4. Originally Posted by zhupolongjoe
Hm....so you start with all multiples of ab, that is, all numbers of the form nab for some integer n. No, sorry, not really sure how to go this direction.
For the other direction, you need to show $ab \in IJ$. Since $IJ$ is an ideal, it follows that $(ab) \subseteq IJ$.