TITLE: If $k\subset R\subset k[x]$, then $R$ is Noetherian?
QUESTION [9 upvotes]: Is there a way to prove that any subring $R$ of the polynomial ring over a field $k$ such that $k\subset R$ is Noetherian without appealing to integral extensions, Eakin-Nagata, etc.?
The reason I ask is because I found this as an exercise 15.1.9 in Dummit and Foote, which well precedes any discussion of integral extensions. 
It says to show that if $y\in R\setminus k$, then $k[x]$ is f.g. as a $k[y]$-module. Why is that? Since $k[y]$ is Noetherian, $k[x]$ is a Noetherian $k[y]$-module, but how does this show $R$ is Noetherian?

REPLY [5 votes]: Let $y \in k[x]$ be a non-constant polynomial.
Claim: $k[x]$ as a $k[y]$-module is generated by the polynomials $1,x,\ldots ,x^{d-1}$, where $d=\deg (y)$.
Proof: let $f\in k[x]$ be a polynomial of degree $n\geq d$. Let $e$ be the maximal integer such that $n-de\geq 0$. By the division-with-remainder algorithm one has $f=qy^e+r$, with $\deg (q)<d$ and $\deg (r)<de$ or $r=0$. Applying the same procedure to $r$ if necessary after finitely many steps leads to a linear combination $f=a_0+a_1x+a_2x^2+\cdots +a_{d-1}x^{d-1}$ with $a_j\in k[y]$. 
If $y \in R$, then every ideal $I$ of $R$ is a submodule of the $k[y]$-module $k[x]$. If the latter is noetherian, then $I$ is noetherian too. In particular it is finitely generated as a $k[y]$- module and thus as an $R$-module.