How can I understand a hypocycloid as an ideal in a polynomial ring?

Question

Today I was reading On teaching mathematics by V. I. Arnol’d and came across the following quote.
"Rephrasing the famous words on the electron and atom, it can be said that a hypocycloid is as inexhaustible as an ideal in a polynomial ring. But teaching ideals to students who have never seen a hypocycloid is as ridiculous as teaching addition of fractions to children who have never cut (at least mentally) a cake or an apple into equal parts. No wonder that the children will prefer to add a numerator to a numerator and a denominator to a denominator."

I know hypocycloids as a fascinating geometric object and it is not hard to connect this to group theory and number theory via roots of unity. Also they are related with special unitary group and possibly related with octonions. But never thought and never seen how they related to ideals of a polynomial ring, though this might be obvious to someone who see the picture from the correct angle.

Can somebody explain me this connection? And, what is the analogous picture for Epicycloids and ordinary Cycloids?

(I am not sure about the suitable tags for this question. You are welcome to edit them as necessary.)

Answer 1

The key idea here is that a hypocycloid is actually an algebraic curve - it can be represented as the zero locus of a polynomial equation $f(x,y)=0$. Let's demonstrate how: if we parameterize our hypocycloid as $$ x(\theta) = r(k-1)\cos\theta + r\cos((k-1)\theta) $$ $$ y(\theta) = r(k-1)\sin\theta - r\sin((k-1)\theta) $$ then we can expand $\cos((k-1)\theta)$ and $\sin((k-1)\theta)$ as polynomials $p(\cos\theta,\sin\theta)$, $q(\cos\theta,\sin\theta)$ in $\cos\theta$ and $\sin\theta$ via de Moivre's formulas. Now we can write $\cos\theta =u$ and $\sin\theta=v$ and via the relation $u^2+v^2=1$, we can eliminate $u,v$ from the system of equations $$x=r(k-1)u+rp(u,v),$$ $$y=r(k-1)v-rq(u,v),$$ $$u^2+v^2=1$$ to get a polynomial equation in just $x$ and $y$.

For example, if $r=1$ and $k=4$, then we get the following equation for an asteroid (computation via Sage):

$$ x^6 + 3x^4y^2 - 48x^4 + 3x^2y^4 + 336x^2y^2 + 768x^2 + y^6 - 48y^4 + 768y^2 - 4096$$

and we see via graphing this equation (see Desmos, for instance) that it gives us the exact same points.

Expressing our hypocycloid as the vanishing locus of a polynomial puts us in the territory of algebraic geometry. As the hypocycloid is the vanishing locus of a polynomial, it is a closed subset in the Zariski topology on the plane. Via the nullstellensatz, we have a bijection between such sets and radical ideals of the coordinate ring of the plane, $k[x,y]$. This means that a hypocycloid is exactly equivalent to the data of the ideal cutting it out, which explains the connection Arnol'd was referring to.

As for the followup questions about epicycloids, the same procedure will work to show that they're also algebraic curves, and thus they're also equivalent to the data of the ideal of polynomial functions which vanish on them. (Plain cycloids don't quite work here - they can't be algebraic curves because they have infinite intersection with another algebraic curve without being equal. Specifically, the line traced by the center of the circle is algebraic and has infinitely many intersections with the cycloid.)