## Torus as zero of a polynomial

Find the least positive integer such that there exists a polynomial of degree such that 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 has degree less or equal to . There is a line which intersects the torus in four points. Denote it . Then you can define a one variable polynomial , and this polynomial has degree at most . But since the line intersects the torus in four points, and the torus is the set of zeros for it follows that must have four roots. Since its degree it is smaller or equal to , this can only happen if is the zero polynomial, which is a contradiction to the fact that is the set of zeros of a torus.

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 :-).

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