# Closed but not rational points of a cubic

In Mumford's Red Book of Varieties and Schemes, page 102, he gave the example of the closed but not rational points of the cubic $y^2=x^3-x$ on the real field : I have some difficulty to recover by elementary methods the figure he traced. Especially, he seems to imply that these closed points (where their residue field is the complex field) formed a set homeomorphic to the region $y^2>x^3-x$ in the real plane (that is why he depicted a cylinder in the projective plane). Can somebody give me a simple explanation ? (I suppose the maximal ideals of the spectrum of the algebra defined by the cubic have to be parametrized the right way ?)

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I am looking at Mumford's book, p. 101-102, but I am not sure what your question is. An elliptic curve $E$ over $\mathbb{C}$ is a torus, and Mumford depicts $E(\mathbb{C})$ modulo conjugation in $\mathbb{C}/\mathbb{R}$ by a cylinder. I am not sure where are you seeing $y^2>x^3-x$. – Álvaro Lozano-Robledo Aug 21 '11 at 14:49
@Alvaro Thank you for your comment. I understood what Mumford did but I wanted an elementary proof using the tools of the chapter : in the plane $y^2>x^3-x$ is the cylinder before identifying with the point at infinity $X=0$ right ? So how can I recover it with using the maximal ideals, like he did page 101 with the circle ? – brunoh Aug 21 '11 at 16:21

Ok, I believe I know what the problem is. I think you are taking Example L in Mumford's book (p.100-101) too literally, and you are thinking that the trick that is used in Example L is something more general that one can use to any equation... I don't think this is the case, and you will have to do something slightly different for the cubic equation if you want to parametrize the non-rational points in terms of maximal ideals of the function field.

Let me first explain what is happening in Example L. Take $k_0=\mathbb{R}$ with algebraic closure $k=\mathbb{C}$, let
$$X_0 = \operatorname{Spec}{(\mathbb{R}[x,y]/(x^2+y^2-1))},$$ and $X=X_0\times_{k_0} k$. Then $X$ is an affine conic with $2$ points at infinity. As in Mumford's book, you can visualize $X$ as a sphere in $\mathbb{C}^2$ and the equator, $X\cap \mathbb{R}^2$, is a circle of radius $1$. The rest of the points are not rational (not defined over $\mathbb{R}$), and they correspond to the two (open) hemispheres of the sphere. Up to conjugation, $(a+bi,c+di)\in X$ and $(a-bi,c-di)\in X$ are identified. Thus, conjugation identifies a point in one hemisphere with another point in the other hemisphere. Hence, we could pick the points in one of the hemispheres (say the north hemisphere) as representatives of the non-rational points. The northern open hemisphere is in 1-1 correspondence with the interior of $x^2+y^2=1$, hence the non-rational points up to conjugation are parametrized by $\{(\alpha,\beta) : \alpha^2+\beta^2<1\}$.

How do we write this correspondence in terms of maximal ideals? We need a family of maximal ideals $\mathcal{M}$ of $\mathbb{R}[x,y]$, such that the locus of each $\mathcal{M}$ is precisely a pair of conjugate points (defined over $\mathbb{C}$) in $X$. One way to do this is to intersect the circle $X$ with a line defined over $\mathbb{R}$, such that the points of intersection are not rational (real). Let $\alpha$ and $\beta$ be reals and let $L$ be the line $\alpha x + \beta y = 1$ (we can write any line in $\mathbb{R}^2$ like this). If you try to find the intersection of $X/\mathbb{C} : x^2+y^2=1$ and $L$, you will find that the condition for both points to be complex is precisely that $\alpha^2+\beta^2<1$ (the points of intersection are always complex conjugates because $L$ is defined over $\mathbb{R}$). On the other hand, if you pick two points $P=(a+bi,c+di)$ and $Q=(a-bi,c-di)$ in $X$, and find the line $L=\overline{PQ} : \gamma x + \delta y =1$, you can show that $\gamma,\delta\in\mathbb{R}$ and also $\gamma^2+\delta^2<1$. Hence, the pairs of complex conjugate points in $X$ are parametrized by the intersection of $X$ and arbitrary lines $\alpha x + \beta y =1$ with $\alpha^2+\beta^2<1$. In terms of maximal ideals, the pairs of conjugate points are parametrized by maximal ideals $\mathcal{M}_{\alpha,\beta} = (x^2+y^2-1,\alpha x +\beta y -1)$, with $\alpha^2+\beta^2<1$.

That's Example L. Now you should try to do something similar for Example M, where $k_0=\mathbb{R}$ with algebraic closure $k=\mathbb{C}$,
$$X_0 = \operatorname{Spec}{(\mathbb{R}[x,y]/(y^2 - x(x^2-1)))},$$ and $X=X_0\times_{k_0} k$, as before. An elliptic curve, such as $X: y^2=x(x^2-1)$, is a complex torus when considered over $\mathbb{C}$. More concretely, you can find a lattice $\Lambda$ such that $X \cong \mathbb{C}/\Lambda$. The real locus of $X$ has two connected components, which correspond to the intersection of the real plane $\mathbb{R}^2$ with the torus $X$ in $\mathbb{C}^2$, so the intersection would look like as two circles in $\mathbb{R}^2$ (except that one of the points in one circle is the point at infinity of $X_0$). Complex conjugation identifies points in the torus, and we can choose half an open torus as a set of representatives of the non-rational points on $X$. This is loosely depicted in Mumford's book as a cylinder, whose ends are the two connected components of the real locus. To make this rigorous, you have to go as we did above: find a maximal ideal whose locus is precisely a pair of conjugate points on $X$. One hint: if you intersect a line with a cubic, you will have either 3 real points, or one real point and a pair of complex conjugate points (at least, that's the case in projective space)... I hope this helps, and good luck!

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I believe you have perfectly understood my problem ! Thank you for your detailed and precise analysis. I have given you an upvote for your kind efforts ! So here is where I am a bit stuck : the maximal ideals I found look like $(x^2−px−q,ax+by−1)$ with special relations between $(a,b)$ and $(p,q)$ to ensure that the complex conjugates that are solutions of $x^2−px−q$ corresponds to an y given by the cubic equation. So I get a set of two parameters $(p,q)$ with $p^2+4q<0$ to ensure the $x$ is not a real solution. It does not look like a cylinder to me in the projective plane ! – brunoh Aug 22 '11 at 10:53
So I looked more closely at the necessary relations between $(a,b)$ and $(p,q)$ and I found that $a$ has to be real solution of a second degree equation with discriminant $q(p^2−(q−1)^2)$ that has to be positive and then to get also $b$ real I have to force $(1+a^2∗q)/(pq)$ to be positive. In the end I only got something like a parabola in the real $(p,q)$ plane, not quite a cylinder in the projective. Where did I do wrong ? – brunoh Aug 22 '11 at 10:56

The question has been answered on MathOverflow. The right way to parametrize the maximal ideals was just to consider those containing $(ax+by-1)$ and impose that the intersection of this line with the cubic has only one point (so that a discriminant involving parameters $(a,b)$ is negative). Then one finds a cubic region looking like the cylinder I was looking for in the projective. The flaw of my approach was that I was parametrizing by projecting on the wrong real plane $(p,q)$ instead of $(a,b)$ because I found the equations easier being of degree 2. But then I could not grasp the right picture (because I crushed the component corresponding to the second coordinate $y$ therefore seeing only a disk in the projective) ! Silly me ...

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You should add a link to the MO answer. – Zhen Lin Aug 24 '11 at 11:47
@Zhen You are perfectly right. Sorry and here it is : mathoverflow.net/questions/73326/… – brunoh Aug 24 '11 at 11:50