In Mumford's Red Book of Varieties and Schemes, page 102, he gave the example of the closed but not rational points (that is to say points having residue field the complex field and not the real field) 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 formed the region $y^2>x^3−x$ in the real plane (which looks like the cylinder he pictured 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 ?)

  • $\begingroup$ Perhaps to make this question clearer, I should precise that I am looking at a way to parametrize the maximal ideals of the real algebra determined by the cubic in order to get something that looks in the projective plane like a cylinder based on the two connected components of the cubic in the real plane. This what I understood when Munford said he will leave "the details to the reader", having explained in the previous example L page 101 such a method that worked well with the real circle (he then got a disk in the plane). Unfortunately I have some problems to work out these details ... $\endgroup$ – brunoh Aug 22 '11 at 14:07
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    $\begingroup$ Your comment was very helpful, and your expectation is correct if you meant that you're looking for a cylinder that looks like a 1-meter section of a pipe of diameter $1/(2\pi)$. Remember that the complex picture is a torus gotten by taking the quotient of the plane by a square lattice, say of width and height 1, and that the closed points over $\mathbb{R}$ are what you get by taking the quotient by the equivalence relation of complex conjugacy. In this latter relation, not only the points on the real axis but those $1/2$ unit above are self-conjugate. $\endgroup$ – Lubin Aug 22 '11 at 17:33
  • $\begingroup$ @Lubin Thank you for your comment, it gave me something new to think about ! Especially considering that I am stuck because of the following reasoning : I looked at the maximal ideals in the algebra determined by the cubic, and I found them to be like $(x^2-px-q,ax+by-1)$, with relations between $(a,b)$ and $(p,q)$ to be sure they contain $(y^2-x^3-x)$. If I want their residue field to be like the complex field and not the real, I have to make $p^2+4q<0$, and be sure that $a$ and $b$ stay real. I obtained in the $(p,q)$ plane something like a parabola, not like a cylinder in the projective ! $\endgroup$ – brunoh Aug 22 '11 at 22:37

I don't know what Mumford had in mind, but here (in some detail) is a down-to-earth way to topologically identify this space with a cylinder.

Let $C$ be our projective cubic curve with affine equation $y^2=x^3-x$. We're considering complex conjugate pairs of points on $C$, that is, pairs $\{(x,y), (\bar x,\bar y)\}$ of solutions of $y^2=x^3-x$. While those points are not real, the line joining them is real: there are real numbers $a,b,c$, not all zero, such that the line $l_{a,b,c}: aX + bY + c = 0$ passes through $(x,y)$ and $(\bar x, \bar y)$, and the coefficient vector $(a,b,c)$ is determined uniquely up to multiplication by a nonzero scalar. That is, $(a:b:c)$ is a well-defined point in the "dual projective plane" ${\bf P}^*$ of lines on the projective plane with coordinates $(x:y:1)$ where $C$ lives. Now these points $(x,y)$ and $(\bar x, \bar y)$ are on $l_{a,b,c} \cap C$, which contains three points in all, so there is a third point $(x_0,y_0) =: p_0$, necessarily real. Conversely, any line $l$ meets $C$ in at least one real point, and if there is only one such point (and $l$ is not tangent to $C$ at that point) then the other two points of $l \cap C$ constitute a closed-but-not-rational point of $C$.

That is,

the space we're looking for is homeomorphic with the subset, call it $S$, of ${\bf P}^*$ consisting of lines whose real intersection with $C$, with multiplicity, has size $1$

as opposed to size $3$.

One way to describe $S$ is to start from $p_0 = (x_0,y_0)$. It is geometrically clear that this point must be on the infinite component of $C$, call it $C_0$: the other component $C_1$ is a closed curve in the affine plane ${\bf R}^2$, so any line meets it with even total multiplicity. Given $p_0$, the lines through $p_0$ constitute a real projective line, which is topologically a circle and the lines through $p_0$ that meet $C_0$ in two other points $q,q'$ constitute the union of two closed arcs, one for lines where $q,q' \in C_0$ and the other for lines where $q,q' \in C_1$. [The boundary points correspond to the four points $q$ whose tangent passes through $p$, which are the solutions of $2q=-p$ in the group law of $C$.] So the lines through $p_0$ in $S$ constitute two open intervals. Now the subtlety is that when $p_0$ goes around the closed curve $C_0$, these two intervals switch as each of the boundary points makes a complete cycle around $C_0$ or $C_1$, so we must traverse $C_0$ twice to traverse our cylinder once. In effect we're getting a Möbius band cut down the middle, which is indeed a cylinder (with a "full twist", true, but that is an artifact of the embedding in three-dimensional space that we use to visualize $S$).

For a different kind of explicit picture of $S$, note that a real cubic polynomial has one real root (with multiplicity) if and only if its discriminant is negative. So we can describe $S$ by eliminating of the variables from $aX+bY+c=0$, substituting into $Y^2=X^3-X$, computing the discriminant $\Delta$ of the resulting cubic, and plotting the region $\Delta < 0$. For example, in the affine piece $b \neq 0$ of ${\bf P}^*$, we may set $b=1$, compute $Y = -(aX+c)$, find that $$ \Delta = -27c^4 - 4(ac)^3 + 30(ac)^2 + 4 a^5 c + 24 ac + a^4 + 4 $$ (I didn't promise it would be pretty), and ask www.wolframalpha.com

plot(-27*c^4-4*a^3*c^3+30*a^2*c^2+4*a^5*c+24*a*c+a^4+4 < 0)

to get a picture with two blue components that join up at infinity to form a topological cylinder:

alt text (source)

[The two visible cusps come from the inflection points where $p=q=q'$, which are real 3-torsion points on $C$; there's a third such singularity at infinity. This means that of the two boundary components of $S$ (it looks like four but they pair up at infinity) the one containing the cusps is $C_0$, and the other is $C_1$.] Try also

plot(-27*c^4-4*a^3*c^3-30*a^2*c^2+(24*a-4*a^5)*c+a^4-4 < 0)

for the picture arising from the curve $y^2=x^3+x$ with only one real component; this time it is a Möbius band embedded in ${\bf P}^*$ so that there boundary and the complement have only one component each:

alt text (source)

To connect this with the usual (but less elementary) picture of an elliptic curve over ${\bf C}$ as a complex torus: as Lubin noted, the complex locus of $C$ is isomorphic as a Riemann surface with ${\bf C} / L$ where $L$ is the Gaussian lattice ${\bf Z} + {\bf Z} i$; this is consistent with complex conjugation, and the real locus consists of the cosets mod $L$ of the complex numbers of integral or half-integral imaginary part, constituting the components $C_0$ and $C_1$ respectively. We're looking to identify the conjugate pairs $\{(z,\bar z)\} \bmod L$ with a cylinder; in terms of the group law the real point $p_0$ associated above to $\{(z,\bar z)\}$ is $-2 \phantom. {\rm Re}(z)$, which as before can only be on $C_0$ and goes around $C_0$ twice (and in the opposite direction, as it happens) as $z$ goes around the cylinder once.

  • $\begingroup$ [edited to correct the description of the space of closed-but-not-rational points on $y^2=x^3+x$: in that case it's a Möbius band, not a cylinder.] $\endgroup$ – Noam D. Elkies Aug 24 '11 at 3:56
  • $\begingroup$ @Noam Your answer is very clear and pedagogical : thank you very much for your time ! I especially loved the part about the analysis of the boundary components recovering $C_0$ and $C_1$ and the extension to the other cubic. Everything is clear for me now and I understood also why I got stuck trying to read to precisely the complicated equations that arouse ... $\endgroup$ – brunoh Aug 24 '11 at 7:48
  • $\begingroup$ @Noam And I also loved the explanation about the Moebius band aspect ... $\endgroup$ – brunoh Aug 24 '11 at 8:44
  • $\begingroup$ @Noam Because of your excellent answer, I also understood my mistake : what I was doing was not wrong but by using the parameters $(c,d)$ of the $(x^2-cx-d)$ in the maximal ideals corresponding to each real points $(x^2-cx-d,ax+by+1)$ I was projecting on the wrong slice of the torus in $\mathbb{C}/L$, therefore crushing in the real plane the $y$ component. I could therefore only saw a disk in the projective (the parabola I found). Silly me, right ? $\endgroup$ – brunoh Aug 24 '11 at 11:46

Starting always from the knowledge that a closed point over $\mathbb{R}$ is either an $\mathbb{R}$-rational point or a pair of conjugate $\mathbb{C}$-rational points, let's think of a complex point $P=(z,w)$ together with its conjugate $\overline P$, if need be, and call $z=a+bi$, $w=c+di$. Then your maximal ideal corresponding to $P$ is $(x^2-2ax+a^2+b^2,y^2-2cy+c^2+d^2)$, with special forms in case either $z$ or $w$ is real, for example $(x-a,y^2-2cy+c^2+d^2)$ in case $z$ is real but not $w$. Of course the condition that $w^2=z^3-z$ is expressed by a pair of equations in the real variables $a,b,c,d$, namely $a + c^2 - d^2 - a^3 + 3ab^2=0$ and $b + 2cd - 3a^2b + b^3=0$, which I haven't gotten much help from, even though I've stared at them long and hard hoping to verify your very interesting insight about the bordered region $S=\lbrace (x,y):y^2 \ge x^3-x\rbrace$. If you want a surface in $\mathbb{R}^3$ to look at, you can take the points $(a,c,b^2+d^2)$, subject to the two conditions above. Its intersection with the plane $(\*,\*,0)$ is just the locus of real-rational points, but its projection onto that plane is not your region $S$. Notice that conjugate points have the same image under this mapping, and nonconjugate points have different images.

The edition of Mumford's Red Book that I'm looking at does not support your guess about $S$, in my opinion, even though as a topological space with border, $S$ is exactly right. Perhaps there's a transcendental argument justifying your insight, using the $\wp$-function for the appropriate lattice.

  • $\begingroup$ @Lubin Thank you very much for your detailed explanation : I was reassured by the fact that the equations are indeed not easy to read and you are right by the suggesting I should just try to depict S topologically ! $\endgroup$ – brunoh Aug 24 '11 at 7:55

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