Questions on the construction of a Noetherian domain that is not an N-1 ring In this wiki article there is an example I'm trying to understand.  The same example seems to appear in two articles, actually.
The relevant definition is

An $N$-$1$ ring is an integral domain $A$ whose integral closure in its quotient field is a finitely generated $A$ module

The example is

If $R$ is the subring of the polynomial ring $k[x_{1},x_{2},...]$ in infinitely many generators generated by the squares and cubes of all generators, and $S$ is obtained from $R$ by adjoining inverses to all elements not in any of the ideals generated by some $x_{n}$, then $S$ is a 1-dimensional Noetherian domain that is not an $N$-$1$ ring.

First off, I think it makes sense that $R=k[\{x_i^2\mid i\in\mathbb N\}\cup\{x_i^3\mid i\in\mathbb N\}]$.  Already it seems like its integral closure should be $R_0=k[\{x_i\mid i\in\mathbb N\}]$ which would seem to be not a finitely generated $R$ module.  So this is already a domain that isn't N-1, right?
My second question is what is meant by the rest of the construction. Let $M_1$ be the multiplicative subset of $R_0$ that is the intersection of complements of ideals of the form $(x_i)$, and let $M_2=M_1\cap R$.  Am I interpreting this correctly that $S=RM_2^{-1}$? If I'm reading that right, then how do I see that it makes $S$ $1$-dimensional and Noetherian?
$R$ itself does not look Noetherian, since you can generate an infinite ascending chain whose terms are sums of the first $n$ ideals of the form $(x_i^2, x_i^3)$. That also seems to show $R$'s Krull dimension is infinite.  I'm not very clear on how the localization step fixes this in $S$.
 A: I agree that $R$ is already not N-1: the integral closure is $k[x_1,\cdots]$ which is not finitely generated over $R$, as any finite collection of generating elements would only be able to express finitely many of the $x_i$.
I agree with your interpretation that $S=M_2^{-1}R$. To see that $S$ is one-dimensional, the intuition is that localizing at $M_2$ removes any subvariety of $\operatorname{Spec} R$ which is contained in two or more of the subvarieties $V(x_i^2,x_i^3)$, leaving you with only these sets as proper subvarieties, which gives dimension one. Indeed, if $I\subset R$ is an ideal which contains both $I_i=(x_i^2,x_i^3)$ and $I_j$ for some $i\neq j$, then it intersects $M_2$ and $M_2^{-1}I=S$. So every prime ideal of $S$ comes from a prime ideal of $R$ which contains at most one $I_i$, and it is not so hard to show that such an ideal must actually be $I_i$ or $0$. Therefore the only prime ideals of $S$ are $0$ and $M_2^{-1}I_i$, and as the only nontrivial inclusions among these are $0\subset M_2^{-1}I_i$, $S$ is of dimension one. Further, as any ring where any prime ideal is finitely generated is noetherian and all of these ideals are finitely generated, we have that $S$ is noetherian.
(For what it's worth, I agree with your arguments in the final paragraph too.)
