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Let $A$ be a commutative noetherian ring with two ideals $I,J$ such that $\sqrt{I}=\sqrt{J}$. Does there always exist integers $p,q,r$ such that

$$ I^p \subset J^q \subset I^r? $$

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Are the ideals $I=\langle x_n y_n^n, \,n\geq1\rangle$ and $I=\langle x_n^{n}y_n, \,n\geq1\rangle$ of the polynomial ring on the variables $\{x_n,y_n:n\geq1\}$ a counterexample? – Mariano Suárez-Alvarez Apr 18 '11 at 22:29
@Mariano, if you have infinitely many indeterminates, is the ring Noetherian? – Gerry Myerson Apr 19 '11 at 1:22
@Mariano: A polynomial ring in countably many variables is not noetherian, though... – Arturo Magidin Apr 19 '11 at 1:22
Ah! I missed that :) – Mariano Suárez-Alvarez Apr 19 '11 at 1:27
up vote 5 down vote accepted

Since $A$ is noetherian, $I$ and $J$ are finitely generated, say $$\begin{align*} I &= (a_1,\ldots,a_n)\\ J &= (b_1,\ldots,b_m). \end{align*}$$

Note that $J^r$ is generated by all elements of the form $$b_1^{\beta_1}\cdots b_m^{\beta_m}$$ where $\beta_i\geq 0$ are integers and $\beta_1+\cdots+\beta_m=r$.

Since $J\subseteq \sqrt{J}=\sqrt{I}$, for each $b_j$ there exists $k_j$ such that $b_j^{k_j}\in I$. Letting $k=\max\{k_1,\ldots,k_m\}$, we have that $b_j^k\in I$ for $j=1,\ldots,m$. Letting $q=m(k-1)+1$, we conclude that every generator of $J^q$ described as above lies in $I$, so $J^q\subseteq I$.

Now notice that $\sqrt{J^q} = \sqrt{J}$, so applying the argument to $I$ and $J^q$ we conclude that there exists $p$ such that $I^p\subseteq J^q$.

Thus, if $A$ is commutative noetherian, and $I$ and $J$ are ideals such that $\sqrt{I}=\sqrt{J}$, then there exist integers $p$ and $q$ such that $I^p\subseteq J^q \subseteq I$.

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HINT $\ $ Every ideal contains a power of its radical in a Noetherian ring. This basic property is familiar to anyone who has studied primary decomposition of ideals in Noetherian rings. The proof can be abstracted out from the proof in Arturo's answer.

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