Let $k$ be a field, $f(x,y)\in k[x,y]$ be any polynomial, and consider the variable $x'=x-y^n$. Show that $k[x,y]=k[x',y]$, and that if $n$ is sufficiently large, then as a polynomial in $x'$ and $y$, $f$ is a scalar times a monic polynomial in $y$. Deduce that $k[x,y]/f$ is integral over its subring $k[x']$. Use this to show that dim $k[x,y]=2$.
This is Exercise 9.2(a) from Eisenbud's Commutative Algebra, but I don't quite understand it. I don't know why if $n$ is sufficiently large, then $f$ can be written as scalar times a monic in $y$ for any polynomial $f$. Also, I don't think $k[x']$ can always be a subring of $k[x,y]/f$, like when we take $f=y$, as we can easily check, the previous statement doesn't hold in this case.
It is required to show that if $n$ is sufficiently large, then as a polynomial in $x'$ and $y$, $f$ is (almost) monic in $y$, and this is easy enough: when replace $x$ by $x-y^n$ in a monomial $x^iy^j$ the only monomial in $y$ is $y^{ni+j}$. This could interfere with other monomials in $y$ of $f$, but for $n$ large enough it actually doesn't, and we get a power of $y$ times a non-zero element of $k$. Anyway, when consider the ideal generated by $f$ we may assume that $f$ is monic as a polynomial in $y$.
Then we have $k[x']\subset k[x',y]=k[x,y]\to k[x,y]/(f)$. You can check that the composition of these two morphisms is injective. In particular, $\dim k[x,y]/(f)=\dim k[x']=1$.