4
$\begingroup$

Is there a nice description of all real algebraic curves of $\mathbb R^2$ which have the property that every such curve contains all zeroes (after the obvious identification of $\mathbb C$ with $\mathbb R^2$) of infinitely many monic rational polynomials of $\mathbb Q[z]$ having only simple roots.

Examples are the real line, the unit circle and there are many hyperelliptic curves (obtained by taking preimages of such curves with respect to suitable rational functions). (It is of course easy to construct also non-simple examples of such curves, eg. by considering all real lines determined by sets of zeroes of cyclotomic polynomials.)

I guess that there exists a curve $\mathcal C\subset \mathbb R^2\sim\mathbb C$ defined by a polynomial in $\mathbb Z[x,y]$ which contains the rootsets of only finitely many rational monic polynomials without multiple roots. Can someone exhibit such a curve?

Improve this question
$\endgroup$
5
  • $\begingroup$ I am not sure that I understood the question correctly: is the curve with equation $y=\pi$ an example? The real and imaginary parts of the roots of a polynomial with rational coefficients are algebraic numbers, and the only imaginary parts of points on the curve above are transcendental. $\endgroup$
    – damiano
    Commented Apr 30, 2010 at 13:09
  • $\begingroup$ No. I am asking for real curves in $\mathbb R^2$ containing all points $(x,y)$ for $z=x+iy$ root of a rational polynomial (and containing all such points for many polynomials). In particular, all points corresponding to roots of such polynomials have algebraic coordinates. A good example is the complex unit circle defined by $x^2+y^2-1$ where $z=x+iy\in\mathbb C$ which contains all roots of all cyclotomic polynomials. There are less trivial examples, eg. elliptic curves with only finitely many rational points. $\endgroup$
    – Roland Bacher
    Commented Apr 30, 2010 at 14:55
  • $\begingroup$ I guess that I was answering the last question you mentioned in your post (and Dror seems to have done the same!). So you are only interested in curves having the property you mention, not the ones lacking, right? $\endgroup$
    – damiano
    Commented Apr 30, 2010 at 16:28
  • $\begingroup$ @roland: can you please add an example of a hyperelliptic curve with the property? $\endgroup$
    – Dror Speiser
    Commented Apr 30, 2010 at 19:13
  • $\begingroup$ Yes, the elliptic curve $-x+x^3-2y^2+xy^2$ has this property in a non-trivial way (I did not check if it has infinitely many rational points, elliptic curves with infinitely may rational points have this property trivially of course). One can then apply the constructions mentionned above to get examples of higher genus. $\endgroup$
    – Roland Bacher
    Commented May 3, 2010 at 14:01

1 Answer 1

Reset to default
3
$\begingroup$

Yes. There exist many such curves:

For any rational monic polynomial without multiple roots if $x+iy$ is a root, then $x-iy$ is a root as well. So if your curve $C$ satisfies $y\not =0,\\ (x,y)\in C \rightarrow (x,-y)\not\in C$ then it cannot contain the rootset of any polynomial with complex roots. To finish, make the curve have finitely many real roots. For example: $$C: x-y=0$$

As for your first question, it seems hard to characterize all such curves. There two types of families I have thought of. Of course you can multiply curves that satisfy the property, so let us only talk about minimal curves with respect to the property.

  1. As you mention, the real line.

  2. Genus 0 or 1 curves with a rational point of infinite order and the required property (mentioned in comments) that $y$ appears only with even powers. This uses the simple fact that conjugate numbers in $\mathbb{Q}(\sqrt{-1})$ have minimal polynomial over the rationals. If we want to extend this to higher genus, there is the following question: Given a rational class in the jacobian of a curve, does it have infinitely many multiples such that $a(x)$, in its representation as a reduced divisor $[(a(x), b(x,y)]$, has only real roots? If so, then the property holds for all curves with a rational point on the jacobian of infinite order.

  3. The unit circle fits into the above. But as you mention, there is another way, which we can generalize: curves of the form $||f(z)||=1, z=x+iy$. The question is now: how many points in the union of all CM fields does this equation have? If I had to guess, infinitely many, so I think these probably satisfy the property.

Improve this answer
$\endgroup$
6
  • $\begingroup$ The (non-simple) curve given by $x^2-y^2$ falls into the trivial class of examples mentioned in the question. I know there are many such curves. I ask for specific examples which lack the property. $\endgroup$
    – Roland Bacher
    Commented Apr 30, 2010 at 14:57
  • 1
    $\begingroup$ I don't understand. Doesn't my example lack the property? The curve you mention is different. $\endgroup$
    – Dror Speiser
    Commented Apr 30, 2010 at 15:35
  • $\begingroup$ @roland-bacher: ok, so you want examples of curves whose points contain all the zeros of infinitely many monic irreducible polynomials with rational coefficients. Dror's argument shows that an irreducible such curve must contain only even or only odd powers of y. Thus either it is the curve y=0 or it must contain only even powers of y. Is this in line with the question that you want answered? $\endgroup$
    – damiano
    Commented Apr 30, 2010 at 16:55
  • $\begingroup$ @damiano: I was thinking about this earlier as well. Can you prove the statement about powers of $y$? I actually believe that almost all curves will fail to have the property. $\endgroup$
    – Dror Speiser
    Commented Apr 30, 2010 at 19:11
  • $\begingroup$ @Dror: let C be a curve with the required property. The set of points that we are interested in is infinite and hence it is dense in some irred component of C: replace C by this component. Thus, there is a Zariski dense set of pts (x,y) with (x,-y) in C. It follows that the symmetry $y \mapsto -y$ fixes C and if f(x,y) vanishes on C, then also f(x,-y) vanishes on C. We conclude that an equation defining C must involve only even or only odd powers of y. Having said that, I also suspect that the property is quite rare! $\endgroup$
    – damiano
    Commented May 1, 2010 at 12:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .