If $R$ is integrally closed in $S$, then $R[x]$ is integrally closed in $S[x]$. First reduce to the case where $R$ is Noetherian? 
Let $R \subset S$ be rings. If $R$ is integrally closed in $S$, then $R[x]$ is integrally closed in $S[x]$.


The first hint for this exercise says:


"First reduce to the case where $R$ is Noetherian by passing to a subring finitely generated over $\mathbb Z$."


I'm not really sure what is going on here.

Is the exercise considering the cases where (i) $R$ is Noetherian and where (ii) $R$ is not Noetherian?; and the Noetherian case implies the non-Noetherian case somehow?
What does it mean to "pass to a subring" finitely generated over $\mathbb Z$? What does this accomplish?
Can someone explain how to make sense of this?
 A: To answer your questions:

Is the exercise considering the cases where (i)  is Noetherian and where (ii)  is not Noetherian?; and the Noetherian case implies the non-Noetherian case somehow?

The exercise is indeed claiming that the Noetherian case implies the non-Noetherian case.

What does it mean to "pass to a subring" finitely generated over $\Bbb{Z}$?

Let $T$ be a ring. By definition, a finitely generated $T$ algebra is a quotient of $T[x_1,\dots, x_n]$ for some $n.$ We may form a subring of $R$ which is finitely generated over $\Bbb{Z}$ by starting with $\Bbb{Z}[x_1,\dots, x_n]$ and choosing $n$ elements $r_1,\dots, r_n\in R.$ Then, there is a unique ring map $$\phi : \Bbb{Z}[x_1,\dots, x_n]\to R$$ sending $x_i\mapsto r_i,$ and the image of this is a subring isomorphic to $\Bbb{Z}[x_1,\dots, x_n]/\ker\phi$ (which we may denote $\Bbb{Z}[r_1,\dots, r_n]$). As such, it is finitely generated over $\Bbb{Z}.$

What does this accomplish?

Well, the point is that if $T$ is Noetherian, then $T[x]$ is also Noetherian. Moreover, if $T$ is Noetherian and $I$ is an ideal of $T,$ then $T/I$ is Noetherian. Together, these facts imply a ring finitely generated over a Noetherian ring is Noetherian ring. In particular, the subring of $R$ you want to consider is Noetherian.

How do we use all of this in the context of this problem?

To show that $R[x]$ is integrally closed in $S[x],$ we need to show that if we have a monic polynomial $f\in R[x][t],$ and $f(\alpha(x)) = 0$ for $\alpha(x)\in S[x],$ then $\alpha(x)\in R[x].$ Write $$f(t) = \sum_{i = 0}^n p_i(x)t^i,$$ where each $p_i\in R[x]$ and $p_n(x) = 1.$ Each of these polynomials can be written
$$
p_i(x) = \sum_{j = 0}^{n_i} r_{i,j} x^j.
$$
Since there are finitely many polynomials $p_i$, there are finitely many coefficients $r_{i,j},$ so $R' := \Bbb{Z}[\{r_{i,j}\}_{i,j}]$ is a subring of $R$ finitely generated over $\Bbb{Z}.$ It suffices to prove that $R'[x]$ is integrally closed in $S[x]$, so we have shown that the Noetherian case implies the non-Noetherian case!
Well really, it suffices to prove that all subrings of $R[x]$ of the form $\Bbb{Z}[\{r_{i,j}\}_{i,j}][x]$ are integrally closed in $S[x],$ as the coefficients of any monic polynomial $f\in R[x][t]$ lives in some subring $R_f[x]$ of this form. If $f$ has a root $\alpha\in S[x]$ and the particular $R_f[x]$ is integrally closed in $S[x],$ then $\alpha\in R_f[x]$ by the definition of being integrally closed, and as such $\alpha\in R[x].$
