Noether normalisation $A=\mathbb{C}[x,y]/(f)$ where $f=(x-a)y^2-(x-b)$ find a transcendence element

Question

Noether normalisation $A=\mathbb{C}[x,y]/(f)$ where $f=(x-a)y^2-(x-a)$ $a , b \in \mathbb{C}$ find $z \in A$. transcendence over $\mathbb{C}$ such that $A$ is integral over $\mathbb{C}[z]$

any suggestions ?

Answer 1

This is probably not the canonical solution, maybe someone else has a more elegant solution, but until then let $$z=y^2-(x-a)$$

then $$z^2=y^4-2(x-a)y^2+(x-a)^2$$ and on multiplying by $(x-a)^2$ we have

$$(x-a)^2z^2=(x-b)^4-2(x-a)(x-b)^2+(x-a)^4$$ which is easily seen to be a monic polynomial in $x$. So $x$ is integral over $A[z]$. If we now note that $x=y^2-z+a$ and substitute in the above equation we get a monic polynomial for $y$, and $y$ is also integral.

This choice of $z$ follows from one proof of the Noether Normalization theorem.