Problem:
Let $R = k[x_i]_{i \in \mathbb{N}}$ be the polynomial ring over a field $k$ in infinitely many variables. Let $I = \langle x_1, x_2^2, x_3^3, \ldots \rangle$ be an ideal of $R$.
Find:
(a) all the prime ideals of $R/I$.b) Show that $R/I$ is not noetherian.
Attempt: For b), I think I found the infinite ascending chain $$ \langle x_1 \rangle + I \subset \langle x_1, x_2 \rangle + I \subset \ldots $$ which shows $R/I$ is not noetherian.
For a), I was not sure. How does one solve these kinds of problems? I was thinking that all ideals of the form $$ \langle x_1 \rangle + I, \qquad \langle x_2 \rangle + I, \ldots $$ are surely prime. For if $$fg \in \langle x_i \rangle + I $$ then at least $f$ or $g$ must be divided by $x_i$ I think.
Help is appreciated.
Any prime ideal of the quotient ring $R/I$ can be seen as a prime ideal of $R$ containing $I$. Let $\mathfrak p$ be such a prime ideal. As an ideal of $R$, it must contain every element $x_k^k$ for $k\in \mathbb N$. Because the ideal $\mathfrak p$ is prime, it means that it actually contains every single variable $x_k$.
Whence, there is only one prime ideal in $R/I$, which can be seen as the ideal generated by all the variables in $R$. Note that it is indeed a prime (and even maximal) ideal since the quotient of $R$ by it is no other than $k$.
Because this prime ideal is not finitely generated, the quotient ring $R/I$ is not Noetherian.