Solving I^m + J^n = R with Positive Integers m, n

[Pietjuh]

Suppose we have a commutative ring R and ideals I and J of R such that I + J = R. I have to show that there exist positive numbers m,n such that I^m + J^n = R.

I think the trick is just to show that I^m + J^m contains 1. Because I + J = R, I+J contains 1 so there exist i in I and j in J such that i + j = 1. Now I have to find a a^m in I^m and a b^n in J^n such that a^m + b^n = 1.
I tried a lot of things but none of them seemed to work :(

Can anyone give me a hint how to find these a^m and b^n ?

Thanks in advance

 

[matt grime]

Why isn't n=m=1 sufficient for your purpose?

It might help you to realize that a typical element in I^n is not of the form a^n for a in I, but a sum of n'th powers of elements in I