Torus as algebraic set - minimal polynomial Let $T=S^1\times S^1$ be a torus. We know that it can be described by as the solution set to $x_1^2+x_2^2=R,\  x_3^2+x_4^2=r$. 
We can find a polynomial of degree 4 such that its set of zeros is $T$ :
$f=(x^2+y^2+z^2+R^2-r^2)^2-4R^2(x^2+y^2)$, where we used embedding in $\mathbb{R}^3$. 
How to prove that $f$ is a polynomial of minimal degree such that its set of zeros is $T$ ? 
 A: You know that $T$ is the zero set of $f$. So if there was a polynomial $g$ of smaller degree such that $T$ is the zero set of $g$, we would have $f=g^2$ or $f=g^4$. If $f$ were a square (or a forth power which also implies that it's a square), then for example $f(0,y,0)$ would also be a square. But we have $f(0,y,0)=(r - R - y) (r + R - y) (r - R + y) (r + R + y)$. 
These four factors are pairwise coprime because $0<r<R$.
The geometric intuition behind that argument is that you can find a line which intersects the torus in four points. Thus it can not be defined by a polynomial of degree less than four.
A: The last comment in Hans' answer is actually entirely sufficient to solve the problem; in particular one can show that if the desired set of zeros $T$ has that there exists a line passing through exactly $n$ points*, then the minimal polynomial having those zeros is of degree at least $T$. This is clear since the restriction of a polynomial of degree $d$ to a line is a polynomial of degree at most $d$ (since we can imagine parameterizing the line by a set of linear expressions of one variable for each coordinate in a single variable). A non-zero polynomial of degree at most $d$ has at most $d$ distinct roots (true of any integral domain). Since it must have $n$ distinct roots, we get $d\geq n$.
In this particular case, there are clearly lines intersecting the torus $4$ times. Thus any polynomial (other than the zero polynomial) which vanishes on the torus has degree at least $4$.
(*We require "exactly $n$ roots" to exclude the case where the restriction of the polynomial to that line is identically zero, where the argument would otherwise fail. We really only need "$n$ roots and some point which is not a root")
