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 $n-1=\dbinom{2n-1}{2}$.


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.

  • $\begingroup$ 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

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