Every ideal contains a power of it's radical in a Noetherian ring.

Question

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)

Answer 1

Here is an alternative without the hint the OP suggested. Suppouse that is not the case and $rad(I)^n\setminus I\not=\emptyset\:,\:\forall n$. Then,

$$\exists r_1 \in rad(I)\setminus I$$

There must be $r_2\in rad(I)^{2}\setminus I$ such that $r_2\not\in(r_1)$. In fact, if there weren't such an element, $rad(I)^{2}\setminus I\subseteq (r_1) $, but for $n$ sufficiently large $r_1^n\in I$, so $rad(I)^{2n} \subset I$ and so $rad(I)^{2n}\setminus I=\emptyset$, which we suppoused couldn't hold. Therefore we actually have an $r_2$:

$$(r_1)\subset(r_1)+(r_2)$$

Where the inclusion is proper. Now, there must be $r_3\in rad(I)^{3}\setminus I$, such that $r_3\not\in (r_1)+(r_2)$. Otherwise, $rad(I)^{3}\setminus I\subseteq (r_1)+(r_2)$, but for $n$ sufficiently large $r_1^n,r_2^n\in I$. Therefore, $rad(I)^{6n} \subset I $ which is absurd. Therefore, there is an $r_3$ which makes the inclusion proper:

$$(r_1)\subset(r_1)+(r_2)\subset (r_1)+(r_2)+(r_3)$$

Proceeding inductively, we have a sequence of strict ascending ideals which never stabilize in contradiction to $R$ being a Noetherian ring.

Answer 2

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 $I$.