Define the $k$-algebra $R=k[x,y]/(x^3-y^2)$. Fix some nonzero $m,n\in\mathbb{N}$. Set $s=x^my^n+(x^3-y^2)\in R$, and let $S=k[s]$. I want to show that $R$ is a free $S$-module with rank $2m+3n$.
My thoughts: one can show there is a $k$-algebra isomorphism $\psi:R\stackrel{\simeq}{\longrightarrow} k[t^2,t^3]$ given by $\psi(p+(x^3-y^2))\mapsto p(t^2,t^3)$. Therefore, it suffices to prove that $k[t^2,t^3]$ is a free $\psi(S)$-module of rank $2m+3n$. Since $\psi(s)=(t^2)^m(t^3)^n=t^{2m+3n}$, we have $\psi(S)=k[\psi(s)]=k[t^{2m+3n}]$. Therefore we need to find a basis of $2m+3n$ elements for $k[t^2,t^3]$ as a module over the subring $k[t^{2m+3n}]$. However, I have no idea how to come up with such a basis. Any suggestions?
Best Answer
You have a good start. Let me remove some unnecessary notation. Let $r\geq 2,$ and notice that it suffices to prove that $k[t^2,t^3]$ is free of rank $r$ over $k[t^r].$
Now, notice that $$k[t^2,t^3] = k + kt^2 + kt^3 + kt^4 + kt^5 + \cdots,$$ and that $$k[t^r] = k + kt^r + kt^{2r} + kt^{3r} + \cdots.$$ Let's pick off the terms of $k[t^r]$ from $k[t^2,t^3]$: we get $$k[t^2,t^3] = k[t^r] + kt^2+kt^3 + \dots + kt^{r-1} + kt^{r+1} + \cdots.$$ In fact, we can repeat this process: the $kt^2 + kt^{2+r} + kt^{2+2r} + \cdots$ gives us another copy of $k[t^r],$ as $$k[t^r] t^2 = kt^2 + kt^{2+r} + kt^{2+2r} + \cdots.$$ Applying the same logic to each residue class modulo $r,$ we find $$ k[t^2,t^3] = k[t^r] + k[t^r]t^2 + k[t^r]t^3 + k[t^r] t^4 + \dots + k[t^r]t^{r-1} + k[t^r]t^{r+1}. $$
So, your desired basis is $\{1, t^2, t^3,t^4,\dots, t^{r-1}, t^{r+1}\}$ (if $r = 2,$ the basis is $\{1,t^3\}$). Can you fill in the remaining details?