We can work over $\mathbb{C}$ for simplicity. Let $X$ be defined by the equation $y^2 = h(x)$ where $h(x)$ has odd degree and distinct roots. Let $\pi: X \to \mathbb{P}^1$ be the ramified double covering.

Just to make sure, the roots of $h(x)$ are ramification points because if $y = 0$, then we get each root having multiplicity two, correct? But this argument doesn't really shed light on why there is another ramification point outside of the roots when $h$ has odd degree.

Which point in $X$ maps to the point at $\infty$, and why is it a ramification point? Why isn't this a ramification point when $h(x)$ has even degree?

Edit: I would like an explanation without using the Riemann-Hurwitz formula since the text I am reading counts the ramification points to compute the genus of the curve.


Write $h(x) = a_{2g+1}x^{2g+1}+\ldots+a_1 x + a_0$. Remember that $X$ is really a projective curve, so let us switch to another affine chart with coordinates $(u,v) = (\frac{1}{x}, \frac{y}{x^{g+1}})$ where $X$ is represented by the equation \begin{align*} v^2 = \frac{y^2}{x^{2(g+1)}} &= \frac{a_{2g+1} x^{2g+1} + \ldots + a_1 x + a_0}{x^{2(g+1)}}\\ &= a_{2g+1} \frac{1}{x} + \ldots + a_1 \frac{1}{x^{2g}} + a_0 \frac{1}{x^{2g+1}} \\ &= a_{2g+1} u + \ldots + a_1 u^{2g} + a_0 u^{2g+1}. \end{align*} The projection map $X \to \mathbb{P}^1$ on this affine chart is given by $(u , v) \to u$. From this, we see that $X$ has a ramification point above $u=0$ (indeed, the projection map $X \to \mathbb{P}^1$ has a ramification point wherever it fails to be a double cover). This point $u =0$ corresponds to $x=\infty$ in the other chart.

What changes if $h(x)$ had even degree? If $h(x) = a_{2g+2} x^{2g+2} + \ldots + a_1 x + a_0$, then in the new coordinates $(u,v)$, the curve is represented by the equation $$ v^2 = a_{2g+2} + a_{2g+1} u + \ldots + a_1 u^{2g+1} + a_0 u^{2g}. $$ This polynomial in $u$ (on the right hand side of the above equation) does not have a root at $u=0$, hence the projection map will not have a ramification point at $x=\infty$.

Thanks to @Jyrki Lahtonen for pointing out the error in my answer.

  • $\begingroup$ Thank you! It's clear now. $\endgroup$ – Daniel Levine Dec 28 '15 at 21:05
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    $\begingroup$ Is there a good motivation for this choice of coordinates, perhaps in terms of twisting? I had to work quite hard for the same result... $\endgroup$ – Slade Dec 28 '15 at 21:56
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    $\begingroup$ I can explain how I first saw these coordinates come up (though if anyone knows other ways in which they arise I'd love to hear it!). When the degree of $h(x)$ is even, it comes from the map to $\mathbb{P}^1$: if we realize $\mathbb{P}^1$ as two copies of $\mathbb{A}^1$, with coordinates $x,u$ respectively, glued together by the rule $x = \frac{1}{u}$, then we can try to pull these coordinates back to the hyperelliptic curve, in some sense. Above the copy of $\mathbb{A}^1$ with coordinate $x$, the curve $X$ can be represented as $y^2 = h(x)$,... $\endgroup$ – msteve Dec 28 '15 at 22:06
  • $\begingroup$ ...and if we'd like it to be of a similar form e.g. $v^2 = g(u)$ in the new coordinates, then we're forced into making $v = \frac{y}{x^{g+1}}$ (or some slight variation thereof). In this manner, the curve $X$ is of the same form above the other copy of $\mathbb{A}^1$. $\endgroup$ – msteve Dec 28 '15 at 22:09

Let's clarify your definition of $X$. $y^2 = h(x)$ defines a nonsingular affine curve (we'll assume $h(0)\neq 0$), but the natural projectivization may be singular at infinity.

By convention, when we present a hyperelliptic curve as $y^2=h(x)$, we are resolving any singularities. Since birational nonsingular curves are isomorphic, this resolution, $\hat{X}$, is unique. Furthermore, the projection $(x,y)\mapsto y$ from $X$ onto $\mathbb{A}^1$ lifts uniquely to a map $\hat{X} \to \mathbb{P}^1$.

Now we look at the projectivization. If $h(x)$ has degree $k\geq 3$, then the projectivization has defining equation $Y^2 Z^{k-2} - h(X,Z)=0$, where $h(X,Z) = \prod_i (X - \alpha_i Z)$, $\alpha_i$ ranging over the roots of $h(x)$. The map to $\mathbb{P}^1$ is given by $[a:b:c] \mapsto [b:c]$.

Plugging in $Y=1$ gives us an affine curve $Z^{k-2} = h(X,Z)$, where the points mapping to $\infty$ are just those with $Z=0$, i.e. just the point $[0:1:0]$.

But this might not actually be a point of ramification, since we haven't resolved the singularity yet. If it resolves to one point, there is ramification. If it resolves to two points, there is no ramification.

I am not sure of the most direct way to count the points over a singularity (without invoking more advanced theory), but this can be done with a series of blow-ups, i.e. we can write the equation as $(Z/X)^{k-2} = X^2 h(Z/X)$, then as $(Z/X)^{k-4} = (X^2/Z)^2 h(Z/X)$, $(Z/X)^{k-6} = (X^3/Z^2)^2 h(Z/X)$, etc.

Finally, we either arrive at an equation $S = T^2 h(S)$ or $S^2 = T^2 h(S)$, depending on whether $k$ is odd or even. In the first case, we have resolved the singularity to a single point ($S=T=0$), so there is ramification. In the second case, we blow-up again to get the nonsingular $U^2 = h(S)$, and then $S=0$ has two solutions, so there is no ramification.

  • $\begingroup$ Okay, I finally got what you were saying. I was looking for an explicit calculation using charts, but I like this explanation too! $\endgroup$ – Daniel Levine Dec 28 '15 at 21:17
  • $\begingroup$ @dl1990 I don't see what isn't explicit, though I glossed over the blow-ups. msteve's answer is much shorter because of a clever change of coordinates, but we're still both basically blowing up charts containing $\infty$. $\endgroup$ – Slade Dec 28 '15 at 21:32
  • $\begingroup$ I just prefer seeing it through the change of coordinates since it makes it easier (for me at least) to compute divisors of rational functions. $\endgroup$ – Daniel Levine Dec 29 '15 at 0:05
  • $\begingroup$ @dl1990 Oh, no question that it's a much better approach. $\endgroup$ – Slade Dec 29 '15 at 0:22

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