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.

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1. 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. 🙂