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I'd like to understand a step in a counterexample of Reichardt and Lind to the Hasse principle. The example is given by the equation $2y^2=x^4-17z^4$ (1).

(1) has no rational solutions ($\neq (0,0,0)$), but it has non-zero solutions in $\mathbb{Q}_{p}$ for all $p$. The part that I don't get is the justification that the equation has solutions in $\mathbb{Q}_{p}$ for all $p$. First one can find explicit solutions for $p=2,17,\infty$. For the other values of $p$ the idea is to first show the existence of non-zero solutions over $\mathbb{F}_{p}$, and then use Hensel's lemma to lift these solutions. To show that (1) has solutions over $\mathbb{F}_{p}$ what I've seen people doing is using Hasse's bound on number of points over $\mathbb{F}_p$ of an elliptic curve. (see between minutes 15 and 30) or

The key point is the following observation: "The curve defined by the equation (1) has genus 1"

and thats precisely what I don't get, so my question is:

Question: Why is that a curve? and moreover why does it have genus 1?

Thank you.

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"genus 1 means it's an elliptic curve" that's true only if you have a point. "Any polynomial defines a curve as its set of solutions" that's definitely not true, in fact your example -which is intended to be over $\mathbb{R}$- is not a curve it is a 2-dimensional variety. The coincidence here is that as topological space you can identify it with $\mathbb{C}\cup \infty$, a 1-dimensional object over $\mathbb{C}$ with no holes. – user8317 Mar 19 '11 at 4:54
The above comment has no sense now that user quanta erased his/her comments. I'll recall his/her comments so mine makes sense. "genus 1 means that it is an elliptic curve..." and "yes it is a curve, any polynomial defines a curve e.g. $x^2 + y^2 + z^2=1$ which is a curve of genus 0" – user8317 Mar 20 '11 at 3:03
up vote 3 down vote accepted

I don't really know what you mean when you ask "Why is that a curve?". The question can be taken to mean either "I don't know what an algebraic curve is", or "Since there are three unknowns and two equations, it looks more like a surface to me". I am going to assume the latter.

So recall that a curve is most often defined either by a set of homogeneous equations in $n+1$ variables, in which case their solutions form a well defined subvariety of the projective space $\mathbf P^n$, or a set of arbitrary equations in $n$ variables, in which case their solutions are taken inside of affine space $\mathbf A^n$. We can go from one to the other by setting one of the variables equal to one and homogenizing respectively (but note that this process is not bijective).

So note first that our equation is not homogeneous. Based on the previous paragraph it would make most sense to interpret it as a subvariety of $\mathbf A^3$, but this is clearly not what they mean since that would give us an algebraic surface and not an algebraic curve.

Here are two ways to make sense of this:

  1. Set $z=1$, for example. Then we get the equation $2y^2 = x^4 - 17$ in two variables, and whenever we have a solution to this equation we also have a solution to the original one. This equation defines a plane curve in $\mathbf A^2$. By embedding $\mathbf A^2$ in a projective space and taking the closure we get a projective curve, and by desingularizing it we also get a smooth projective curve.

  2. Give the variable $y$ degree $2$, so that the equation is homogeneous with respect to a shifted degree. Then we can talk about the set of solutions inside of a weighted projective space. (Over $\mathbf C$ this can be defined as $\mathbf C^3 / \mathbf C^\ast$ where $\mathbf C^\ast$ acts by $\lambda \cdot (x_0,x_1,x_2) = (\lambda^2 x_0, \lambda x_1, \lambda x_2)$.)

Your second question - how do we see that this curve has genus one - can be determined in several ways. I am not sure what is the simplest way but here is one, using model (1) above and calculations in affine coordinates. Your curve has a degree 2 map to $\mathbf A^1$ given by projection to the $x$-coordinate, and this map will have branch points wherever $x^4-17 = 0$. The map can be extended to a map from the smooth projective version of the curve to $\mathbf P^1$ (this is a general fact about smooth curves, but you can also prove this by computing in explicit coordinates, see the Wikipedia article on hyperelliptic curves). Using the Riemann-Hurwitz formula one sees that:

  1. a map of degree two must have an even number of branch points, so the point $x = \infty$ can not be a branch point;

  2. a curve which has a degree two map to $\mathbf P^1$ with four branch points has genus one.

Oh, and please tell me which parts of this need elaborating.

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Dear Dan: First, this is a great answer thank you. I've in fact thought this along the lines of approach (1) so my conclusion is that saying that "the equation defines a curve" is wrong, however, as you point out, a point in the plain curve produces a point on the surface. Here is something that I'vent been able to do though. I'm trying to resolve the singularity $[0:1:0]$ of the projective closure $2y^{2}u^{2}=x^4-17u^4$ of the curve, but I'm having not luck at it. I'm not sure how to do blow ups of projective curves so I tried to resolve $2u^{2}=x^4-17u^4$ at $[0:0]$ but for some reason... – user8317 Mar 19 '11 at 5:21
... I did not get a smooth curve. I'm guessing that if I do the right calculations I'd get a cubic plane curve - a Weierstrass model - which is birational to $2y^{2}=x^4-17$. I imagine there are references that show how to get a cubic equation from an equation $y^2=f(x)$ where $f$ is a quartic polynomial with no repeated roots - I see from your answer why such equation exists- but imagine in practice there is some sort of algorithm that shows how to do it, and does not mention coverings, genus, etc... – user8317 Mar 19 '11 at 5:34
I'm trying to calculate the above example to see if I can derive such algorithm. If you know any reference for it that would be great. Thanks. – user8317 Mar 19 '11 at 5:35
In principle there is an algorithm for resolving any singularity of a curve: just keep blowing up and it will eventually disappear. But I would avoid doing this kind of explicit computation as far as possible so maybe it was misleading that I talked about blow-ups in my answer. For a curve of the type $y^2 = f(x)$ where $\deg f = 2g$ or $\deg f = 2g+1$, you can get a nonsingular model by gluing two affine pieces $y^2 = f(x)$ and $z^2 = w^{2g}f(1/w)$, which is a lot easier to compute with than whatever you would get by blowing up... – Dan Petersen Mar 19 '11 at 6:34
As for your algorithm, an elliptic curve is determined by the four points on $\mathbf P^1$ which are the branch points of the double cover (and the choice of one of them as the origin) and in the resulting equation $y^2 = f(x)$ the polynomial $f$ has degree three precisely if one of the four points is the point at infinity, and degree four otherwise. So you just need to pick an automorphism of $\mathbf P^1$ sending one of the roots to infinity and of course there are many such maps. This can be written as a change of coordinates, without mentioning geometry, but why would you want to do that? – Dan Petersen Mar 19 '11 at 6:41

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