Semi-algebraic sets

Question

The Astroid is a semi-algebraic curve. Since the points $(x,y)$ of the curve fulfill the equation $x^{2/3} + y^{2/3} = 1 $, and since they are points of said polynomial it is semi-algebraic. But my question is that enough? Since the coefficients of the polynomial are rational and I understand that a set is semi-algebraic if it is the finite union of finite intersections of polynomial systems with integer exponents.

Answer 1

It's possible to write down the explicit polynomial system if you want, but it's easier to use the fact that the projection of a semi-algebraic set is again semi-algebraic by Tarski-Seidenberg. Consider $\Bbb R^4$ with coordinates $x,y,z,w$ and the set $S$ cut out by the constraints $x^3=z^2$, $y^3=w^2$, and $z+w=1$. Then the projection of $S$ to the copy of $\Bbb R^2$ with coordinates $x,y$ exactly recovers your asteroid: the fiber of $S$ over $(x,y)$ is a single point when $x^{2/3}+y^{2/3}=1$ and empty otherwise.