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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 \} $?

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  • $\begingroup$ You got your $AB$ wrong. Note that if $r \in R$ and $a \in A$, then $r a \in A$. So what you have written is the ideal $A + B$. Instead, $A B$ is the ideal spanned by the products $a b$, for $a \in A$ and $b \in B$. Similarly, $I^{m}$ is the ideal spanned by all $m$-fold products $x_{1} \cdots x_{m}$, for $x_{i} \in I$. $\endgroup$ – Andreas Caranti Jan 26 '16 at 21:20
  • $\begingroup$ Nope, your definition is still wrong. $AB=\sum a_ib_i$, not product. $\endgroup$ – Thomas Andrews Jan 26 '16 at 21:24
  • $\begingroup$ More like a sum in the edited definition of $A B$. $\endgroup$ – Andreas Caranti Jan 26 '16 at 21:24
  • $\begingroup$ Thank you for the corrections, and for the clarification about $I^m$, any hint for the exercise? $\endgroup$ – HaroldF Jan 26 '16 at 21:24
  • $\begingroup$ Your rings have a unity, right? $\endgroup$ – Andreas Caranti Jan 26 '16 at 21:25
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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 $1\in I^m + J^m$.

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  • $\begingroup$ if I expand $1= (i+j)^{2m-1}= i^{2m-1}+j^{2m-1}+$"terms in $i^kj^{2m-1-k}"$ Now I have that the terms in $i^kj^{2m-1-k}$ lies in $IJ$ (and also their sum) but then? $\endgroup$ – HaroldF Jan 26 '16 at 21:37

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