Let $(R,\mathfrak{m},k)$ a local commutative and with unity ring such that $char(k) = 0$ and $I$ an ideal of $R$. I need to prove that $I^m = \sum_{y\in I}(y^m)$ for every $m\geq1$. Well, I already have some idea how I'm going to prove it, but I'm having some troubles
It's well known the following result from Basic Abstract Algebra
Let $(R,\mathfrak{m},k)$ a local ring such that $char(k) = 0$, then $n.1_R$ is unity in $R$ for every $n\in \mathbf{N}$
Using this fact, we can prove, for example, the case $m = 2$, because since $2ab = (a+b)^2 - a^2 - b^2$, then $ab = (2.1_R)^{-1}[(a+b)^2 - a^1 - b^2] \in \sum_{y\in I}(y^2).$
The case $m=3$, it's also simple, but a little more complicated: Using that $6ab^2 = (a+b)^3 - (a-b)^3 - 2b^3$ and calculating $(a+b+c)^3$, we can show that $abc \in \sum_{y\in I}(y^3)$.
I'm having difficult to prove the general case. Does someone know how to generalize or know some reference which proves this result or something like it?
I found the answer on the article: Ideals generated by powers of elements, by D.D. Anderson, Kent R. Knopp and Rebecca L. Lewis.