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 $D_+(X_0)\subseteq\text{Proj } k[X_0,\ldots,X_4]$.


  • $\begingroup$ Related question: mathoverflow.net/questions/63745/… $\endgroup$ Jun 8, 2012 at 16:50
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    $\begingroup$ 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. $\endgroup$ Jun 8, 2012 at 17:30

3 Answers 3


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.


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.


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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    $\begingroup$ It is best to edit your question and not to use the answer box for further observations. $\endgroup$ Jun 8, 2012 at 16:36

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