4
$\begingroup$

One basic alg geom book convinced me that the genus of a complex curve given by $y^2=x^{2n}+a_{2n-1}x^{2n-1}+...+a_0$ is $n-1$.

Another textbook computes the genus of a singular variety given as $(y-b_1x)(y-b_2x)\cdot\cdot\cdot(y-b_{2n}x)$ to be $\dbinom{2n-1}{2}$ and then argues that, because genus is stable under coefficient change, that's going to be the genus of all curves of degree $2n$.

Naturally, I am confused about the implied equality $n-1=\dbinom{2n-1}{2}$.

$\endgroup$
2
$\begingroup$

These are only defining equations of the curves in a affine chart and they have different meanings.

The first one is the equation of a hyperelliptic curve. It is given as a double cover of the projective line (locally $(x,y) \mapsto x$). And this does not give you a plane curve in general.

For a projective plane curve $C$ of degree $d$ we have a relation between the degree and another invariant, the arithmetic genus $p_a(C)$. We have $p_a(C) = \frac{(d-1)(d-2)}{2}$. When $C$ is smooth $p_a(C)$ coincides with the genus of $C$.

Edit: Given $y^2 = f(x)$, $f$ a (generic) polynomial of degree $2g+1$ or $2g+2$ you can define a hyperelliptic curve taking $w^2 = z^{2g+2}f(1/z)$ and glue these curves identifying $$(z,w) = (1/x, y/x^{g+1})$$ this give you an abstract curve of genus $g$

We can also compactify the curve given by $y^2 = f(x)$ setting $$F(x,y,z) = z^{d}\left[(y/z)^2 - f(x/z) \right]$$where $d = \max\{2,\deg(f)\}$. This will give a curve in $\mathbb{P}^2$.

These two compactifications give us different curves.

Example: Consider $f(x) = x(x-1)(x-2)(x-3)(x-4)(x-5)$. First we construct the hyperelliptic curve $C$. We have two affine charts $$ U_1 = \{ (x,y) \mid y^2 = f(x) \} \, {\rm and}\, U_2 = \{ (z,w) \mid w^2 = z^6f(1/z) \} $$ Nothe that $z^6f(1/z) = (1-z)(1-2z)(1-3z)(1-4z)(1-5z)$. The curve $C$ is defined by

$$ C = (U_1 \sqcup U_2) /\sim $$ where $(x,y) \sim (z,w) \Leftrightarrow (z,w)= (1/x, y/x^{3})$. Since $f$ is squarefree it follows that $C$ is smooth.

We can define a map $\varphi\colon C \rightarrow \mathbb{P}^1$ by \begin{align*} \left.\varphi \right|_{U_1} \colon U_1\rightarrow \mathbb{C}, \, \left.\varphi \right|_{U_1}(x,y) = x \\ \left.\varphi \right|_{U_2} \colon U_2 \rightarrow \mathbb{C}, \, \left.\varphi \right|_{U_2}(z,w) = z \end{align*} Note that the gluing of two copies of $\mathbb{C}$ by $z=1/x$ gives $\mathbb{P}^1$ and this map is well defined.

Taking a preimage of a general point $p\in \mathbb{C}$ we see that $\varphi$ has degree two and is ramified at 6 points $(p,0)\in U_1$ such that $f(p)=0$. Using Riemann-Hurwitz we have that $C$ has genus 2.

Fact: there are no smooth plane curves of genus two.

Now lest build the plane curve $D$ of affine equation $y^2 = f(x)$. Homogeinizing we get $$ F(x,y,z) = z^6(y^2/z^2 - f(x/z)) = z^4y^2 - x(x-z)(x-2z)(x-3z)(x-4z)(x-5z) $$ And $D = \{(x:y:z) \in \mathbb{P}^2 \mid F(x,y,z) =0 \}$. We already have that $D\cap \{ z=1 \}$ is smooth. In $\{x=1\}$ we have that $D$ is given by $$ z^4y^2 - (1-z)(1-2z)(1-3z)(1-4z)(1-5z) = 0 $$ and it follows that $D\cap \{ x=1 \}$ is smooth. In $\{y=1\}$, however, $D$ is given by $$ z^4 - x(x-z)(x-2z)(x-3z)(x-4z)(x-5z) = 0 $$ which is singular (only) at the origin. Since $F$ has degree $6$, $D$ has arithmetic genus $\frac{(6-1)(6-2)}{2}=10$. Taking a resolution of singularities you can see that $D$ has geometric genus $2$. In fact, $C$ and $D$ are birational.

$\endgroup$
  • $\begingroup$ What do you mean by a plane curve? I don't understand the first equation doesn't give you a curve. $\endgroup$ – user99914 Sep 8 '18 at 16:51
  • $\begingroup$ It is a curve but not in the projective plane $\mathbb{P}^2$. See en.wikipedia.org/wiki/Hyperelliptic_curve and en.wikipedia.org/wiki/Algebraic_curve#Plane_projective_curves $\endgroup$ – Alan Muniz Sep 8 '18 at 17:19
  • $\begingroup$ I've included the two definitions. Tell me if it is clearer. $\endgroup$ – Alan Muniz Sep 8 '18 at 17:33
  • $\begingroup$ @AlanMuniz, one more question: why the map from the hyperelliptic curve to $P^2$ given by its equation in 2 variables ${x,y\in\mathbb{C}: y^2=x^{2n}+...+a_0}$ followed by compactification does not make the hyperelliptic curve a plane curve? $\endgroup$ – Michael Sep 10 '18 at 15:48
  • $\begingroup$ I'll try to include an example. $\endgroup$ – Alan Muniz Sep 15 '18 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.