A subring of $k[x,y]$ is not Noetherian even if $k$ is finite It was an exercise in Basic Algebra by Cohn:

If $k$ is infinite field then the subring of $k[x,y]$ given by $k+xk[x,y]$ is not Noetherian.

Proof: Consider following chain of ideals in $k+xk[x,y]$:
$$(xy) \subset (xy,xy^2) \subset (xy,xy^2,xy^3)\subset \cdots$$
Claim: this chain of ideal in $k+xk[x,y]$ is strictly ascending. 
For example, lets show second inclusion is strict. For this, if $xy^3$ is in the ideal $(xy,xy^2)$ then 
$$xy^3=xyf_1 + xy^2f_2 \hskip5mm \mbox{ for some } f_1,f_2\in k+xk[x,y].$$
We cancel $x,y$ from both sides and let $f_1=a+xg_1$ and $f_2=b+xg_2$ where $a,b\in k$ and $g_1,g_2\in k[x,y]$. Then we have 
$$y^2=(a+xg_1) + y(b+xg_2).$$
Rewrite this in way 
$$y^2-(a+by)=x(g_1+yg_2)$$
So this is valid in $k[x,y]$. Now left side is in $k[y]$ and right side is in $xk[x,y]$ (i.e. ideal generated by $x$ in $k[x,y]$). But $k[y]\cap xk[x,y]=0$, so last equation gives contradiction. So given ring is not Noetherian.
Question. Is the proof correct? In this proof, I not used $k$ is infinite. Therefore was unsure about answer. (Cohn may have different argument in the case $k$ is infinite; so what could be other way to prove non-Noetherian of ring if $k$ is infinite?)
 A: The proof is quite correct but a little bit sloppy("For example...").It´s an elegant argument though. I just write it out a little bit: For $n\in\mathbb{N}$ the inclusion $$(xy,xy^2,...,xy^{n-1})\subset(xy,xy^2,...,xy^{n-1},xy^n)$$
is a real inclusion, i.e. the chain doesn´t become stationary and thus the subring $k+xk[x,y]$ is not Noetherian. Assume to the contrary that $$xy^n\in(xy,xy^2,...,xy^{n-1}).$$ Then we have $$xy^n=xyf_1+xy^2f_2+...xy^{n-1}f_{n-1}$$ where $f_j=a_j+xg_j,a_j\in k,g_j\in k[x,y]$ for $j=1,...,n-1$. Thus dividing by $xy$ yields $$y^{n-1}=f_1+yf_2+...+y^{n-2}f_{n-1}=$$ $$a_1+xg_1+ya_2+xyg_2+...+y^{n-2}a_{n-1}+xy^{n-2}g_{n-1}$$ and rewriting this yields $$y^{n-1}-(a_1+ya_2+...+y^{n-2}a_{n-1})=x(g_1+yg_2+...+y^{n-2}g_{n-1})$$ where the left hand side is in $k[y]$ and the right hand side is in $xk[x,y]$.Because $k[y]\cap xk[x,y]=0$ both sides of this equation have to be $0$ which cannot be since the left hand side is of degree $n-1$ in $k[y]$: contradiction. And I really think this proof holds if $k$ is finite too.
