My general question is : given a hyperelliptic curve $y^2 = f(x)$ with disc$f(x) \ne 0$, is there a general formula for finding a smooth complete model of the curve?
Specifically, I want a smooth, complete model of the curve $C_0 : y^2 = x^5 + 1$ over some field $k$ of characteristic 0.
Silverman (Exercise 2.14, the Arithmetic of Elliptic Curves) seems to think that I can find such a model by considering the closure of the image of $C_0$ in $\mathbb{P}^4$ under the map $[1,x,x^2,x^3,y]$.
It seems to me that the closure of the image of this curve is the scheme $\text{Proj }k[X_0,X_1,X_2,X_3,X_4]/(X_2X_0 – X_1^2, X_3X_0^2 – X_1^3, X_3X_0 – X_1X_2, X_4^2-X_2X_3-X_0^2)$
But, you can quickly check that there are infinitely many points at infinity (ie $X_0 = 0$), namely the points $(0,0,1,a^2,a)$. However, this is impossible since the points at infinity are closed, and hence finite since the affine piece $C_0$ is embedded in $D_+(X_0)\subseteq\text{Proj } k[X_0,\ldots,X_4]$.
thanks
Best Answer
An alternative, and I think easier, way to get a smooth model for $y^2=f(x)$ is to glue together two non-singular affine curves. If $f(x)$ has degree $2d$ or $2d-1$, then glue $C:y^2=f(x)$ and $C':v^2=u^{2d}f(1/u)$ via $x=1/u$ and $y=v/u^d$. Note that the affine curves $C$ and $C'$ are both smooth. Also, the "points at infinity" on $C$ are the points where $u=0$, so there are two such points if $f$ has even degree and one such point if $f$ has odd degree.
I want to reiterate what Dalawat says. You generally should not answer your own question. Instead, edit the question if there are further things that you want to say.