Let $V=V(Y^2-X^2(X+1))\subset\mathbb{A}^2$, e $\overline{X}, \overline {Y}$ the residues of $X,Y$ in $A(V)$ its coordinate ring; let $z= \dfrac{\overline{Y}}{\overline{X}}\in K(V)$. Find the pole sets of $z$ and of $z^2$.
I know that $z^2$ has no poles, it is polynomial. I know that points of the form $(0,y)$ where $y\neq 0$ are pole, and points of the form $(x,0)$, where $x\neq 0$ $x\neq −1$. I do not know if these are all poles. There are other poles ?
The pole set of $z$ is $\{ (0,0)\}$.
Let $J_z = \{ G\in k[X,Y]| \overline G z \in A(V) \}$.
Claim that: $J_z = (X,Y).$
$J_z \subset (X,Y)$ is trivial. To see that $(X,Y) \subset J_z$, it suffices to check that $\overline X z = \overline Y \in A(V)$ and $\overline Y z = {\overline {X^2(X+1)}}/{\overline X} = \overline {X(X+1)} \in A(V)$.
Then the pole set of $z$ is $V(J_z) = \{ (0,0)\}$.