Home > Uncategorized > Torus as zero of a polynomial

Torus as zero of a polynomial

Find the least positive integer n such that there exists a polynomial P(x,y,z) of degree n such that P(x,y,z)=0 is a torus.

Hint: In how many points can a line intersect a torus?

My proof: It is easy to find a torus which is zero of a polynomial of degree four. There is an example in the given link.

Suppose now that P has degree less or equal to 3. There is a line which intersects the torus in four points. Denote it a+bt,\ t \in \Bbb{R}. Then you can define a one variable polynomial Q(t)=P(a_1+b_1t,a_2+b_2t,a_3+b_3t), and this polynomial has degree at most 3. But since the line intersects the torus in four points, and the torus is the set of zeros for P it follows that Q must have four roots. Since its degree it is smaller or equal to 3, this can only happen if Q is the zero polynomial, which is a contradiction to the fact that P is the set of zeros of a torus.

Categories: Uncategorized
  1. João Carlos M. A. Soares
    March 25, 2012 at 9:21 pm

    I don’t know of you have a more elegant argument in mind, but here a sketch of a possible answer:

    (1) I’ll take the torus to be in standard position, the other dimensions being irrelevant.

    (2) Given the torus’ symmetries, each monomial must have even degree; furthermore, the same applies to the individual exponents of each variable.

    (3) Now, considering the horizontal lines in a vertical plane containing the z-axis, we see that they intersect the torus at four distinct point for z in (-c,c), for some c > 0; at two points when z = +/-c, and zero otherwise.

    (4) Given this, consider a polynomial P(r,z), of fourth degree on r, and whose coefficients are quadratic polynomials in z, such that P has four real distinct roots for z in (-c,c), two roots with multiplicity two, when z = +/-c, and no real roots otherwise.

    (5) Finally, let r = (x^2 + y^2)^(1/2); given (2), the result will still be a polynomial, whose degree will be 4.

    (6) The degree cannot be smaller than four, because the only other possibilities are 0 and 2.

    I hope I didn’t let something obvious pass :-).

    • March 26, 2012 at 9:47 am

      Thank you for your comment and for your solution. I posted a solution myself. 🙂

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