Let $k$ be a field. I was trying to construct an example of a subalgebra of $k[x,y]$ which is not finitely generated. I came up with this example $$S=\{f(x,y)\in k[x,y]:f\text{ does not have a linear term}\}$$ i.e. $$f(x,y)=a_0+b_1x^2+b_2xy+b_3y^2+\dots+c_ky^m.$$ Next, I want to show that $S$ is not finitely generated. My idea is the following:
I can consider a grading on $k[x,y]=k\oplus k\langle x,y\rangle\oplus k\langle x^2,xy,y^2\rangle\oplus\dots$ where $$S=k\oplus k\langle x^2,xy,y^2\rangle\oplus\dots.$$ Then, if I assume that $S$ is finitely generated i.e. $S=\langle s_1,\dots,s_n\rangle$ with $s_i\in S$, then I want to show $\{s_1,\dots,s_n\}\subset k\langle x^2,xy,y^2\rangle$. My intuition is that we cannot obtain an element of the lower degree as a product and a linear combination of linear independent generators of the same degree and the generators of the higher degree. In other words, I want to show that $S$ can be generated by the elements from $\langle x^2,xy,y^2\rangle$.
So, if I can show that $S$ is actually generated by $\{x^2,xy,y^2\}$, then I will obtain a contradiction that as $y^3\not\in k[x^2,xy,y^2]$. Or, do I just need to consider a general case that each $s_i$ is a $k$-linear combination of $\{x^2,xy,y^2\}$ to obtain a contradiction? Am I on the right track? Is there a faster way of doing this?
I will appreciate your comments, and I will fix my mistakes by editing this post correspond