Blow up and hyperelliptic curves.

Question

Let $C'$ be a non singular affine curve $y^2=x^5+3$ over $\mathbb{C}$. $C'^\#$ be its projective closure : $Y^2Z^3=X^5+3Z^5$. It has singular point at $\mathcal{O}'=(0:1:0)$.

On the other hand, let $C$ be a hyperelliptic curve whose affine part is $C’$ i.e. $C=C'\cup_{\phi} \tilde{C'}$ (gluing curve) where $\tilde{C'}$ is an affine curve defined by $y^2=x(1+3x^5)$ and $\phi:C'\dashrightarrow \tilde{C'}$ is a rational map such that $(x,y)\mapsto (1/x,y/x^3)$.

Question: Now, let $C’’$ be a affine part of $C'^\#$ defined by $Y=1$ (i.e. $C’’:z_{2/1}^3=x_{0/1}^5+3z_{2/1}^5$ where $x_{0/1}:=X/Y, z_{2/1}:=Z/Y$), I want to check the blow up of $C’’$ at $\mathcal{O}=(0,0)$ will be $C$. How do I do?

My computation: Take a blow up of the affine plane $\mathbb{A}_{(x_{0/1},z_{2/1})}$ of its coordinate is $(x_{0/1},z_{2/1})$:

$\pi:Bl_{\mathcal{O}}(\mathbb{A}_{(x_{0/1},z_{2/1})})\to\mathbb{A}_{(x_{0/1},z_{2/1})}$

Now let the coordinates of affine parts of $Bl_{\mathcal{O}}(\mathbb{A}_{(x_{0/1},z_{2/1})})$ are $\mathbb{A}_{(x_{0/1},u)}$ and $\mathbb{A}_{(v,z_{2/1})}$ (then $z _{2/1} =ux _{0/1}, v=1/u$). Then the strict transforms of $C’’$ in $\mathbb{A}_{(x_{0/1},u)}$, $\mathbb{A}_{(v,z_{2/1})}$ are $\pi^{-1}[C’’]_0: u^3=x_{0/1}^2+3u^5x_{0/1}^2$, $\pi^{-1}[C’’]_1: 1=v^5z_{2/1}^2+3z_{2/1}^2$.

There exists a rational map $\psi:\pi^{-1}[C’’]_0\dashrightarrow \pi^{-1}[C’’]_1; (x_{0/1},u)\mapsto (ux_{0/1},1/u)$ and we have the blow up $Bl_{\mathcal{O}}(C’’)=\pi^{-1}[C’’]_0\cup_{\psi} \pi^{-1}[C’’]_1$.

Answer 1

Let's prove this using Hartshorne's exercise IV.2.2, which shows that hyperelliptic curves are uniquely determined by the branch points of their hyperelliptic cover (up to automorphisms of $\mathbb{P}^1$).

It's easy to see that projection to the $x$-axis gives a hyperelliptic cover $C \to \mathbb{P}^1$ branched over $-3, -3\omega,... -3\omega^4, \infty \in \mathbb{P}^1$ where $\omega$ is a primitive fifth root of unity. We will construct a hyperelliptic cover $\tilde{C} \to \mathbb{P}^1$ramified over the same points, where $\tilde{C}$ is the resolved curve obtained from blowing up $\mathcal{O}$. This will prove the desired result.

To this end, let $\pi: C'^\# \dashrightarrow \mathbb{P}^1$ be the map defined by taking the affine projection of $C'$ onto the $x$-axis. (Explitly, take the composition $C' \to \mathbb{A}_y^1 \hookrightarrow \mathbb{P}^1$. This first map surjects to $\mathbb{A}^1$, so the composition only misses the point at $\infty$.) Looking at the equation of $C'$, we see that $\pi$ is branched exactly over the aforementioned points of $\mathbb{P}^1$, except for $\infty$.

We also note that this map is defined everywhere except $\mathcal{O}$. Indeed, it is defined everwhere in $C'$ we can describe the locus of points of $C'^\#$ not in $C'$ as the vanishing set of $z$ on $C'^\#$. Plugging this into the equation forces $x$ to be zero as well, so we're only left with $\mathcal{O}$.

Now, since $C’^\#$ has a single tangent direction at $\mathcal{O}$ it follows that there is only one point of $\tilde{C}$ lying over it. (cf. the remark below)

Now since $\tilde{C}$ is a nonsingular projective curve, the rational map $\tilde{C} \dashrightarrow \mathbb{P}^1$ extends to a surjective double cover $\tilde{C} \to \mathbb{P}^1$. The only possible point mapping to $\infty$ then is the unique point of $\tilde{C}$ lying over $\mathcal{O}$. Hence, $\infty$ is our sixth branch point, so $\tilde{C} \to \mathbb{P}^1$ and $C \to \mathbb{P}^1$ have exactly the same branch points. The mentioned exercise then gives us an isomorphism $\tilde{C} \cong C$.

Remark: I want to quickly explain why $\tilde{C}$ only has one point lying over $\mathcal{O}$. If $f(x,y) = f_{r} + \dots + f_d$ is the homogeneous decomposition of some $f \in \mathbb{C}[x,y]$, we can factor the smallest homogeneous part $f_r = (a_1 x - b_1 y)(a_2 x - b_2 y) \cdots (a_r x - b_r y)$. These factors are called the tangent directions of $f$ at the origin. Now, when we blow up the origin, the tangent directions through the origin correspond to points of the exceptional curve (via $ax - by \leftrightarrow [a: b] \in E \cong \mathbb{P}^1$), so the intersection of $\tilde{C}$ with the exceptional curve consists of only one point. This is the unique point of $\tilde{C}$ lying over $\mathcal{O}$. (If you haven't seen this, I would encourage you to justify this correspondence algebraically.)