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Assume that $R$ is a commutative Noetherian ring with $1$ and $\{I_k\}_{k\in\mathbb{N}}$ is a family of ideals in $R$ s.t. $I_k I_j\subset I_{k+j}$. Then we can form the ring $T=R+I_1 X+ I_2 X^2+\cdots$ (subring of $R[X]$). I was wondering if the following statement is true:

$T$ is Noetherian if and only if there exists an ideal $J\subset R$ s.t. $I_k=J^k$ for sufficiently large $k$.

I think I proved that if almost all $I_k$ are of the form $J^k$, then $T$ is Noetherian, but I have no clue about the proof in the other direction (or counterexample). If $T$ is Noetherian, then $\sum_{m_1+\dots+m_k=n}I_{m_1}\cdots I_{m_k}=I_n$ for almost all $n$, so maybe this should be the correct characterization of a Noetherian $T$?

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  • $\begingroup$ T is called the Rees algebra associated to the given family of ideals. $\endgroup$ – user26857 Apr 11 '20 at 10:00
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No, this is not true. One good source of examples comes from symbolic powers of ideals. For more information on symbolic powers I would recommend either these notes or this survey paper. For the purposes of producing a counterexample, I'll work in a more special case to make things simpler. Let $R=k[x_1,\dots,x_n]$ and let $I$ be a sqaurefree monomial ideal. Then $I$ is a radical ideal which has a primary decomposition of the form $I=\bigcap_i Q_i$ where the $Q_i$ are prime ideals generated by variables of $R$. The $n$th symbolic power can then be defined (in this setting) as $I^{(n)}=\bigcap_i Q^{n}_i$. Then we have

$$I^{(p)}I^{(q)}=\bigcap_i Q^p_i \cdot \bigcap_jQ^q_j \subseteq \bigcap_{i,j} Q_i^pQ_j^q \subseteq \bigcap_i Q_i^{p+q}=I^{(p+q)}.$$

However, it is well known that the symbolic Rees algebra $\mathcal{R}_s(t)=\bigoplus_{n \ge 0} I^{(n)}t^n \subseteq R[t]$ is Noetherian (see this paper). Disclaimer: it is critical that $I$ be a monomial ideal for this to be true as it is very much not true in general; this paper of Paul Roberts has the first known counterexample.

It is sometimes the case that symbolic powers agree with ordinary powers, but it's quite rare, and, in particular, if we take, for example, $R=k[x,y,z]$ and $I=(xy,xz,yz)$, then $I^{(n)} \ne I^n$ for any $n \ge 2$. One can check this by hand, or appeal to something like Proposition 1.8 here. But, as $I^n \subseteq I^{(n)}$ for every $n$, the radical of $I^{(n)}$ is $I$. Thus if there is a $J$ with $I^{(n)}=J^n$, then $J$ must also have radical $I$, in particular, is contained in $I$. But then $J^n \subseteq I^n$ so this would force $I^n=I^{(n)}$.

For the correct characterization of Noetherian filtrations (which you are close to in the second paragraph of your question), see, for example, the top of page $2$ in this paper which can also guide you to references for the original result.

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    $\begingroup$ Thank you kindly for really detailed answer with many references (+1). $\endgroup$ – Shingle Apr 11 '20 at 8:46
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    $\begingroup$ I have one question concerning statements from reference papers that you posted. In definition of filtration by ideals it is assumed that sequence of ideals is not increasing. I think this is not needed for $T$ to be a subring. What is the reason behind putting such restriction? $\endgroup$ – Shingle Apr 11 '20 at 13:56
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    $\begingroup$ @Shingle Inspired by the collection of powers of an ideal, there are three conditions that one might want to impose on a collection of ideals $\{I_n\}_{n \ge 0}$: (1) That $I_0=R$. (2) That $I_m \cdot I_n \subseteq I_{m+n}$. (3) That $I_{m+1} \subseteq I_m$. Rings satisfying (1) and (2) are called graded systems or graded families, precisely because the Rees ring of the system is a graded $R$-algebra. Collections satisfying all three are called (descending) filtrations. $\endgroup$ – metalspringpro Apr 11 '20 at 23:52
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    $\begingroup$ @Shingle Many things are better when we have a filtration instead of just a graded family. For instance, we have access to other blow up algebras such as the associated graded ring $G(E)=\bigoplus_{n \ge 0} I_n/I_{n+1}$ of the filtration. Yet another reason is that a filtration defines a topology on $R$ and there's a whole theory around filtered modules etc. Many families we care the most about, e.g. powers, symbolic powers, integral closure of powers, etc. are filtrations so it makes sense to study this notion abstractly. But there are certainly reasons to consider general graded systems too. $\endgroup$ – metalspringpro Apr 11 '20 at 23:53
  • $\begingroup$ Thank you again and sorry for such late reply. $\endgroup$ – Shingle Apr 18 '20 at 14:30

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