# polynomial-torus

I am wondering if it is possible to prove (or come up with an explicit example) that there is a polynomial $f ( x,y,z )$ of degree 8 such that the set $f ( x,y,z )=0$ is a union of two torri? Any specific proof or example of such polynomial? Thanks

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take a Torus, for example $$f_{R,r}(x,y,z) := (x^2+y^2+z^2+R^2-r^2)^2-4R^2(x^2+y^2)=0,$$ Now $f(x,y,z) = f_{2,1}(x,y,z) \cdot f_{2,1}(x,y,z-27)$ is a polynomial which zeroset is a union of two tori.

AB, martini.

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For example, $g(x,y,z) = (x^2+y^2+z^2+3)^2-16(x^2+y^2)$ is the equation of one torus centred at the origin, and is a polynomial of degree $4$. So $g(x,y,z) g(x,y,z+s) = 0$ is the equation of two tori with their centres separated by by $s$ on the $z$ axis, and has degree $8$.

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A torus can be defined in $\mathbb{R}^3$ as the zero locus of a quartic polynomial. For a simple example that is radially symmetric about the $z$-axis, the Pythagorean theorem and some geometry tells us that

$$\left(R-\sqrt{x^2+y^2}\right)^2+z^2=r^2$$

is the equation of the torus with major radius $R$ and tube radius $r$ (these terms aren't standard or anything). Some algebra (multiply out, isolate the sqrt, square, etc.) allows us to write this as

$$\left(x^2+y^2+z^2+R^2-r^2\right)^2-4R^2(x^2+y^2)=0.$$

This is a fourth-degree polynomial set to zero. Multiply two equations of this form and we're done.

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