Proving I^n+J^m=R for Coprime I,J in a Ring R

[Galileo]

I want to show that if I and J are coprime ideals of a ring R, so I+J=R, then for any positive numbers m and n we also have $I^n+I^m=R$.

I thought the easiest way to do it was to show that $1 \in I^n+J^m$ given that there exist $i\in I$ and $j\in J$ such that $i+j=1$. But I haven't had much luck yet. Any hint would be appreciated.

 

[matt grime]

I'm going to hope that R is commutative, for then consider (i+j) raised to some power to do with m and n but larger than both.

 

[Galileo]

Yes, R is commutative. I forgot to mention that.
I already had tried expanding (i+j) to some power. For example m+n or mn:
$$1=(i+j)^{m+n}=(i+j)^m(i+j)^n=\sum_{k=0}^m {m \choose k}i^{m-k}j^k\sum_{k=0}^{n}{n \choose k}i^{n-k}j^k=\sum_{k=0}^{m+n}i^{m+n-k}j^k$$

$$1=(i+j)^{mn}=\sum_{k=0}^{mn}{mn \choose k}i^{mn-k}j^k=\left(\sum_{k=0}^m {m \choose k}i^{m-k}j^k\right)^n$$
I can't see where that leads me. I understand that if I can write 1 as $i^n x+j^m y$ whatever x and y are, then I`m done.

 

[Galileo]

I think I've got it.

$$1=(i+j)^{m+n}=\sum_{k=0}^{m+n}{m+n \choose k}i^{m+n-k}j^k$$
$$\sum_{k=0}^{m}{m+n \choose k}i^{m+n-k}j^k +\sum_{k=m+1}^{m+n}{m+n \choose k}i^{m+n-k}j^k=$$
$$i^n\left(\sum_{k=0}^{m}{m+n \choose k}i^{m-k}j^k\right) +j^m\left(\sum_{k=0}^{n-1}{m+n \choose m+1+k}i^{n-1-k}j^{1+k}\right)$$