$k[h(x),y] \subseteq k[h(x),y] + \langle h(x),y \rangle \subseteq k[x,y]$

Question

Let $k$ be a field of characteristic zero. Let $h=h(T) \in k[T]$ with $\deg(h)=d \geq 2$ and $h(0)=0$ (namely, $h$ has zero constant term).

Consider the following chain of $k$-algebras: $$k \subseteq k[h(x),y] \subseteq k[h(x),y] + \langle h(x),y \rangle_{k[x,y]} \subseteq k[x,y]$$ where $\langle h(x),y \rangle_{k[x,y]}$ is the ideal in $k[x,y]$ generated by $h(x)$ and $y$.

Observe that $k[h(x),y] \subsetneq k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$, since, for example, $xy \in k[h(x),y] + \langle h(x),y \rangle_{k[x,y]} - k[h(x),y]$.

Recall the Artin–Tate Lemma: "Consider the inclusions of rings $R \subset B \subset A$. Suppose that $R$ is Noetherian, that $A$ is a finitely generated algebra over $R$ and that $A$ is a finitely generated module over $B$. Then $B$ is a finitely generated algebra over $R$".

Here $R=k[h(x),y]$, $B=k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$, $A=k[x,y]$. Here $A$ is moreover a finitely generated module over $R$ (with $d$ generators: $1,x,\dotsc,x^{d-1}$).

By the Artin–Tate Lemma, $k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$ is a finitely generated algebra over $k[h(x),y]$.

Could we find the algebra generators of $k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$ over $k[h(x),y]$? Are they just $yx,yx^2,\dotsc,yx^{d-1},yx^d$?

Same question with $h(0) \neq 0$.

Answer 1

You have the right idea. Every element of $B$ is of the form $r+ph+qy$ for some $r\in R$ and $p,q\in A$. Since $A$ is generated by $1,x,\ldots,x^{d-1}$ as an $R$-module, we see $$p=\sum_{i=0}^{d-1}p_ix^i$$ $$q=\sum_{i=0}^{d-1}q_ix^i$$ for some $p_i.q_i\in R$. But then $$r+ph+qy=r\cdot 1+\sum_{i=0}^{d-1}p_ihx^i+\sum_{i=0}^{d-1}q_iyx^i.$$ So, conclude that $B$ is generated by $1,h,hx,\ldots,hx^{d-1},y,yx,\ldots,yx^{d-1}$ as an $R$-module.