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I have a curve $(x(\theta),y(\theta))$ in $\mathbb{C}^2$, where $x(\theta)$ is described as $$x(\theta) = (k-1)\cos(\theta) + \cos((k-1)\theta) + i[(k-1)\sin(\theta)- \sin((k-1)\theta)]$$ and $y(\theta)$ is just the conjugate of $x(\theta)$.

This curve is a rational curve, so there is a polynomial $P(x,y)$ such that the polynomial is 0 on the curve. There is a minimal such polynomial w.r.t. to degree, that satisfies this, which is not the 0 polynomial.

How do I find this minimal degree?

For example, the curve has rotational symmetry $2\pi/k$ so this tells me that the degree must be a multiple of $k$, or something very similar, but I need something that gives me a lower bound.

EDIT: For each $k$ I have a candidate for $P$, but it is tedious to express this $P$. However, I do not think it is too hard to calculate the degree of each $P$. Now, if this degree matches the minimal degree above, then I know that $P$ is in fact the smallest polynomial. Thus, for a fixed $k$ I can of course use variable elimination, and compare the result to my candidate $P$, but how do I prove that these are the same for all $k$?


Turns out the solution was to start at the other end. I had a certain discriminant $P$ in mind, that was the Zariski closure to this hypocycloid. There was not a nice formula for the discriminant explicitly, but I managed to get a very nice parametrization of the set where the discriminant vanish. Restricting this discriminant to the subspace $x$ is the conjugate of $y$ gives exactly the hypocycloid above.

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up vote 1 down vote accepted

You write the cosines in terms of the variable $z\exp(i \theta)$ in the usual way, then write $x = f(z), y = g(z),$ and eliminate $z$ from this pair of equations by computing the resultant (the minimal polynomial will be a factor of the resultant, so you will need to factorize and check all the factors). For more on the subject, see

Ideals, Varieties, and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra (Undergraduate Texts in Mathematics) by Cox, Little, and O'Shea.

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Yes, I am aware that I can do a variable elimination for a fixed k, by Groebner bases or similar. However, I need to find (and prove) a formula depending on k. GB only does this for a fixed k, (and I assume the same is for the resultant method), see my edits to the question. – Per Alexandersson Nov 10 '11 at 8:55
@Paxinum: I figured that you wanted a general answer, but once you try a few small cases (which should be easy) you would have a conjectured form for the x-y dependence, which might be easy to prove. This is presumably easier than actually thinking. – Igor Rivin Nov 10 '11 at 10:37
This actually lead me in the right direction, although I could not use it directly; The resultant argument was just what I needed. – Per Alexandersson Nov 10 '11 at 20:57

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