# Show that there exists $n$ such that $i^n = 0$ for all $i$ in the ideal $I$

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 $(a_1i_1 + \ldots + a_mi_m)^n = 0 + \ldots + 0 = 0.$

-

Hint : Use multinomial theorem to see that the following is true $$(a_1i_1+\cdots +a_mi_m)^{n_1+\cdots +n_m}=0$$

-
What if $A$ is not commutative? –  N. S. Jul 5 '12 at 15:20
Since one of the tags is commutative-algebra, I think the OP probably just forgot to include the assumption that $A$ is in fact commutative. –  Derek Allums Jul 5 '12 at 15:33
Nice: +1, pritam. You can even take the smaller exponent $n_1+\cdots +n_m-m+1$ (but this is an essentially useless "amelioration"...) –  Georges Elencwajg Jul 5 '12 at 15:37
Thanks a lot! Well, the answer is, as usual, less complicated than I imagined. Why didn't I come up with this myself? ;) And excuse me, in the course I follow we always assume rings are commutative and have an identity so I forgot to mention it. –  Loes Jul 5 '12 at 16:00
You can also use induction by $m$ the number of generators of $I$, and just the binomial theorem, but this is basically the same proof. –  N. S. Jul 5 '12 at 16:53

Pritam's binomial approach is the easiest way to solve this problem in the commutative setting. In case you're interested though, it's also true in the noncommutative setting.

Levitzky's theorem says that any nil right ideal in a right Noetherian ring is nilpotent. This implies your conclusion (in fact, even more.) However this is not as elementary as the binomial proof in the commutative case :)

-