# 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.

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What is the definition of the poles on a singular variety ? –  user18119 Oct 29 '12 at 16:35
@QiL This is an exercise from Fulton. He says that a rational function $f$ is defined at $P\in V$ if $f=a/b$ for some $a,b\in \Gamma(V)$, where $b(P)\neq 0$. Then, the pole set of $f$ is the set of all points $P$ such that $f$ is not defined at $P$. It is discussed on pages 20-21 here: math.lsa.umich.edu/~wfulton/CurveBook.pdf –  Ekie Oct 29 '12 at 16:47
OK, then $(0, 0)$ is the unique pole of $z$. –  user18119 Oct 29 '12 at 17:03
But how do I know that $(0,0)$ is a pole at all? Wouldn't I have to check that $z$ is not defined at $(0,0)$ for any representation of $z$? –  Ekie Oct 29 '12 at 17:12

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 $(0,0)$.

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Thanks for this. That is a useful technique. –  Ekie Oct 30 '12 at 2:52

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.

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The problem is rather that taking square root does not carry. –  user18119 Oct 29 '12 at 20:46