# Blow up and hyperelliptic curves.

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 $$Bl_{\mathcal{O}}(C’’)=\pi^{-1}[C’’]_0\cup_{\psi} \pi^{-1}[C’’]_1$$.

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.)