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Let $k$ be an algebraic closed field with character not equal to 2, $a,b,c\in k$ be distinct numbers, and consider the curve $C: y^2=(x-a)(x-b)(x-c)$. Let $P=(a,0),P_{\infty}$ for the point at infinity. Then $div(x-a)=2(P)-2(P_{\infty})$. Here, the curve $C$ really means projective curve $C: y^2z=(x-az)(x-bz)(x-cz)$, so I guess $div(x-a)$ here should mean $div(\frac{x}{z}-a)$.

My problem is how to compute $ord_{P_{\infty}}(\frac{x}{z}-a)$. I konw the local ring at $P_{\infty}$ is $(k[x,z]/(z-(x-az)(x-bz)(x-cz)))_{(x,z)}$, and it is a regular local ring. But I cannot find the uniformizer and prove $ord_{P_{\infty}}(\frac{x}{z}-a)=2$.

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Dear Li Zhan, How do you know that your curve is regular at $P_{\infty}$? If you can actually prove this, then your proof will provide a uniformizer. More precisely, your proof will show that there is a unique tangent line to the curve at $P_{\infty}$. Then, the equation of any other line passing through $P_{\infty}$ will be a uniformizer. Regards, – Matt E Sep 28 '11 at 4:04
Thank you, I can find the uniformizer. But I don't know why at smooth point, non-tangent line forms a uniformizer, could you point out some reference for that result? – Li Zhan Sep 28 '11 at 23:43
Dear Li Zhan, If $\ell = 0$ is the equation of the line, then passing through the point means that $\ell$ (thought of as a function on the curve) vanishes at the point. Passing through the point and being tangent to the point means that the function $\ell$ vanishes with order $\geq 2$ at the point. Passing through the point but being non-tangent at the point means that $\ell$ vanishes to exact order $1$ at the point, i.e. is a uniformizer. Regards, – Matt E Sep 29 '11 at 1:46
up vote 2 down vote accepted

One way to compute the order is to look at the degree of the map and the fiber. Since the generic fiber of the map $x: C \to \mathbb{P}^1$ has degree two, but the fiber over the point at $\infty$ has cardinality one, the map $x$ must be doubly ramified there (by a general fact about compact Riemann surfaces*): in other words, the meromorphic function $x$ has a double pole at the point at infinity. This implies that $x-a$ has a double pole there for any $a$.

(You might, similarly, try showing that $y$ has a triple pole at the point at infinity.)

*This fact is the following: if $X, Y$ are compact Riemann surfaces and $f$ a nonconstant holomorphic map, then the sum $\sum_{x \in f^{-1}(y)} e_f(x)$ is constant (the degree) where $e_f(x)$ denotes the ramification of $f$ at $x$ ( 1 if unramified).

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Thank you very much! – Li Zhan Sep 28 '11 at 23:46

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