I'm new to this medium, but I'm quite stuck with an exercise so hopefully someone here can help me.
This is the exercise: Let $I$ be an ideal in a Noetherian ring $A$, and assume that for every $i\in I$ there exists an $n_i$ such that $i^{n_i} = 0$. Show that there is an $n$ such that $i^n = 0$ for every $i\in I$.
I thought about this: $A$ is Noetherian so $I$ is finitely generated. That means, there exist $i_1, \ldots, i_m$ such that all of the elements in $I$ are linear combinations of $i_1, \ldots, i_m$. Now is it possible to take $n = n_1n_2\cdots n_m$?
I was thinking, maybe I have to use something like this: $(a + b)^p = a^p + b^p$. This holds in a field of characteristic $p$. If this is true, then indeed $(a_1i_1 + \ldots + a_mi_m)^n = 0 + \ldots + 0 = 0.$