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

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

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Related question: mathoverflow.net/questions/63745/… – Felipe Voloch Jun 8 '12 at 16:50
I remember one can use weighted projective space to get a smooth model fast and easy. But someone more knowledgable will have to confirm/explain this. – Dror Speiser Jun 8 '12 at 17:30

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.

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Anyway, an interesting tidbit I just learned, which I'm going to explain here mostly for my own benefit (though I'd appreciate any clarifying comments), is why the closure of the image of the curve $C_1 : y^2 = x^6 + 1$ in $\mathbb{P}^4$ has two points at infinity, whereas the curve $C_0 : y^2 = x^5 + 1$ in $\mathbb{P}^4$ has only one point at infinity.

From here on let $C_i$ denote the affine curve given by the equations above over $K$. Let $C_i'$ be the smooth complete model of $C_i$, ie the closure of the image of the map to $\mathbb{P}^4$.

If we consider the projection onto $x$ map $x : C_i \rightarrow \mathbb{P}^1$, we get an inclusion of function fields $K(x)\subseteq K(x,y)$ of index 2. The smooth complete model also has the same function field $K(x,y)$, so the number of points it has at infinity is just the number of preimages of $\infty$ under the projection $x$, extended to $C_i'$. Since $x$ is a double cover, this is either 1 or 2.

Now, the trick here is to note that at least for nonsingular curves (for example the ring of integers of a number field), points correspond to valuation rings of their function field, so here, we are looking at valuations of $K(x,y) = K(C_i')$ lying above the valuation ring $K(1/x)$ of $K(x) = K(\mathbb{P}^1)$. Now, noting that $$K(x,y)\otimes_{K(x)} K((1/x)) = \bigoplus_{v\mid \frac{1}{x}} K(x,y)_v$$ we see that there are two valuations if and only if $K((1/x))(y) = K((1/x))$.

Indeed, in our case $y = \sqrt{x^d + 1}$, and the square root exists in $K((1/x))$ if and only if $d$ is even.

This is pretty awesome, especially since you can't otherwise detect the behavior of the curve $C_i$ at infinity without first finding a normalization.

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Ugh, so apparently I am missing a relation, namely $X_3^2X_0 - X_2^3$. This shows that the only point at infinity is $(0,0,0,1,0)$. You'd think that 4 relations would be enough for a curve in 3-space...

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It is best to edit your question and not to use the answer box for further observations. – Chandan Singh Dalawat Jun 8 '12 at 16:36