Determining if a quotient of $\Bbb Q[t_1, t_2, \ldots]$ is Noetherian Let $L = \Bbb Q[t_1, t_2, \ldots]$ (polynomial ring in infinitely many variables). Let $I$ be the ideal of $L$ generated by $t_1^2$ and $t_i - t_{i+1}^2$ for all $i$. I am allowed to assume that $t_1 \notin I$.
The question is, is $R = L/I$ Noetherian?
I have no idea how to attack this problem. $L$ itself is very clearly not Noetherian, and $I$ isn't finitely generated. The thing I'm allowed to assume gives me that for all $i$, $t_i \notin I$ since if $t_2 \in I$ then $t_1 - t_2^2 + t_2^2 \in I$, so $t_1 \in I$. And repeat to get $t_i \notin I$ for all $i$.
Does this perhaps give me an infinite chain of ideals in $R$:
$$Rt_1 < Rt_1 + Rt_2 < Rt_1 + Rt_2 + Rt_3 <\ldots$$
This seems too easy and must not work, I guess because maybe they're not distinct because some stuff ends up in $I$.
I can't understand what $R$ really looks like or how its ideals behave, any help would be appreciated.
 A: This ring is not Noetherian. To see this quickly, consider the left-shift map
$$\tilde\pi:L\rightarrow L$$
defined by taking $x_1$ to $0$ and $x_{i+1}$ to $x_i$. Observe that $\tilde\pi(I)\subseteq I$ by looking at the generators of $I$, therefore this descends to a map $\pi:R\rightarrow R$.
Now, observe that
$$\pi^n(t_{n})=0$$
$$\pi^n(t_{1+n})=t_{1}$$
Since you can assume $t_1\not\in I$, these equations witness that the inclusions
$$\ker \pi < \ker \pi^2 < \ker \pi^3 <\ldots$$
are all strict, thus we have an infinite ascending chain of ideals.
A little more work, noting that $\tilde \pi(I)=I$, shows that $\ker \pi^n$ lifts to an ideal of $L$ given by $I+\ker \tilde \pi^n$, so $\ker \pi^n$ is actually $Rt_1+\ldots+Rt_n$, so this is exactly the chain you suggested.

The intuition to take here is that $t_{i+1}$ is basically defined to be a square root of $t_i$, and $t_1$ is defined to be a square root of $0$. These relations have the property that we can simply "shift" the whole sequence backwards while preserving the relations that hold between the generators of $L$.
You could also write this ring out explicitly as the direct limit of a sequence of rings, where you define $L_i=\mathbb Q[x]/(x^{2^i})$ and define attaching maps $f_i:L_i\rightarrow L_{i+1}$ with $f_i(x)=x^2$. This gives a more or less explicit form for the ring and makes it fairly easy to work with ideals directly.
