TITLE: Every ideal contains a power of it's radical in a Noetherian ring.
QUESTION [6 upvotes]: Let $I$ be an ideal in a Noetherian ring $R$. Prove that there exists a positive integer $N$ such that $\text{rad}(I)^N ⊂ I$. [Hint:Let $\text{rad}(I)=⟨g_1,\ldots,g_k⟩$,and suppose $g_i^{n_i} \in I$.Use $N=n_1+n_2+\cdots+n_k]$.
Proof:
If $I$ is an ideal in a noetherian ring $R$, then if $R/\text{rad}(I)$ is finite dimensional over the field then so is $R/I$. 
Let $\text{rad}(I) = ⟨g_1,...,g_k⟩$. There exists $n_i$ such that $g_i^{n_i} ∈ I$ for all $i = 1,2,\ldots,k$. We have $g_i^n \in I$ for all $i = 1,2,...,k$
where n is the maximal of $n_i$. Thus, $(\sum x_ig_i)^{nk} ∈ I$ for all $x_1,\ldots,x_k \in R$. Therefore, $\text{rad}(I)^N ⊂ I$ for $N = nk$. (checking if my proof works)

REPLY [9 votes]: Let $I$ be an ideal in $R$, a Noetherian ring, and observe that $r(I) = \langle x_1, \ldots, x_k\rangle$ since $r(I)$ is an ideal of $R$ and every ideal in a Noetherian ring is finitely-generated.
Since $x_i \in r(I)$, there exists $n_i$ such that $x_i^{n_i} \in I$. Define $N = n_1 + \cdots + n_k$.
Observe that $r(I)^N$ is generated by elements of the form $x_1^{r_1} \cdots x_k^{r_k}$ where $r_1 + \cdots + r_k = N$ by the multinomial theorem. Observe that, for each such element, $r_i \geq n_i$ for some $i$, otherwise we contradict the fact that $r_1 + \cdots + r_k = N$.
This means that each $x_1^{r_1} \cdots x_k^{r_k} \in I$. Conclude that $r(I)^N \subseteq I$ since each of its generators is is in $I$.