Given $I,J$ ideals of $R$, show that $\forall m \geq 1; I^m + J^m =R$

Question

Given $R$ a commutative ring and $I,J$ ideals of $R$, such that $I+J=R$, show that $I^m+J^m=R, \forall m \geq 1$

My problem is that I don't know what is the meaning of $I^m$ and in literature I didn't find anything. I know that, given $A,B$ ideals, then $AB= \{ \sum_{i=1}^n a_ib_i |a_i \in A, b_i \in B, n \in \mathbb{N} \}$, but with this notation $I^2=II=I$, so what is the meaning of $I^m$? It could be $I^m = \{ x^m | x \in I \} $?

Answer 1

The definition of $I^m$ is recursive: $I^1=I$ and $I^{m+1}=II^m$, where the right side is the product of ideals.

Hint: Note that $1=i+j$ for some $i\in I,j\in J$. Then expand $1=(i+j)^{2m-1}$ to see that $1\in I^m + J^m$.