When Rees algebra is Noetherian? 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 $T$?
 A: 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.
