Pole set of rational function defined on a variety The problem: Let $V = V(y^2-x^2(x+1))$, and let $\overline{x}, \overline{y}$ denote the $I(V)$-residues of $x$ and $y$ in the coordinate ring $\Gamma(V)$. Set $z=\overline{y}/\overline{x}$. Find the pole sets for $z$ and $z^2$.
My progress: Since $I(V) = (y^2-x^2(x+1))$ (the polynomial $y^2-x^2(x+1)$ is irreducible), we have $\overline{y}^2 = \overline{x}^2(\overline{x} + \overline{1})$ in $\Gamma(V)$, so $$z=\overline{y}/\overline{x} = (\overline{x}/\overline{y})(\overline{x}+\overline{1})$$ and $$z^2 = \overline{y}^2/\overline{x}^2 = \overline{x} + \overline{1}.$$
Hence $z^2$ has no poles since one representation is a polynomial, and $z$ has no poles at $(x,y)$ if either $x\neq 0$ or $y\neq 0$. So I have concluded that the only possible pole for $z$ is $(0,0)$. However, I am unsure of how to check this point. 
I would appreciate any hints.
 A: I will write for simplicity $x, y$ instead of $\bar{x}, \bar{y}$. 
If $z$ is defined at $(0, 0)$, then there exist $b(x,y), a(x,y)\in \Gamma(V)$ such that $b(x,y)z=a(x,y)$ and $b(0,0)\ne 0$. Equivalently, $by=xa$. 
Because of the relation $y^2=x^2(x+1)$, we see that any element of $\Gamma(V)$ can be written in a unique way as $f(x)+g(x)y$. So 
$$(b_0(x)+b_1(x)y)y=x(a_0(x)+a_1(x)y), \quad b_0(0)\ne 0.$$ 
So
$$ b_0(x)y+b_1(x)x^2(x+1)=xa_0(x)+xa_1(x)y.$$
By the uniqueness of the decomposition, $b_0(x)=xa_1(x)$, hence $b_0(0)=0$. Contradiction. 
So $z$ can't be defined at $(0,0)$. 
A: The curve (more-or-less) is $ y = \pm \sqrt {x^2(x+1)}$ an elliptic curve.  The function 
\[ z =  \frac{y}{x} = \frac{\sqrt {x^2(x+1)}}{x} \approx \sqrt{x+1} \quad\text{ and }\quad z^2 = x+1\]
So, there shouldn't be any poles, even at $(0,0)$.  Certainly, away from the origin this logic carries through.
Since this is algebraic geometry, I suspect $z$ has a pole at the origin - since the division does not carry though - while $z^2$ does not have a pole.
