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



Smooth Models of Hyperelliptic Curves (+ a concrete question)
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.


