Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $k$?

SOLUTION:

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.