Is there a better way to find the polynomial equation for this curve?

Consider the curve in $\mathbb{R}^2$ defined by the equation $$ x^{1/3} + y^{1/3} + (xy)^{1/3} = 1, $$ where $x^{1/3}$ denotes the real cube root of $x$, etc.

Since the equation above involves only algebraic operations, this curve ought to be a branch of an algebraic curve, i.e. a curve defined by a polynomial equation involving $x$ and $y$. I know of only one algorithm to find this equation:

1. Compute the minimum polynomial of $x^{1/3} + y^{1/3} + (xy)^{1/3} - 1$ over $\mathbb{Q}(x,y)$. Since $\mathbb{Q}(x^{1/3},y^{1/3})$ has degree $9$ over $\mathbb{Q}(x,y)$, the minimum polynomial $p \in \mathbb{Q}(x,y)[t]$ presumably has degree $9$, so this will involve computing and then row reducing a $9\times 10$ matrix over $\mathbb{Q}(x,y)$.

2. The "constant term" of $p$ will be a rational function in $x$ and $y$ whose numerator is a polynomial $q(x,y)$. Then the given curve is a branch of the algebraic curve defined by the equation $q(x,y)=0$.

Is there some better way to do this?

Solution 1:

Let $\varepsilon=e^{2\pi i/3}$. Your equation will be $$ \prod_{k=0}^2 \prod_{\ell=0}^2 \Big( \varepsilon^k x^{1/3} + \varepsilon^\ell y^{1/3} + \varepsilon^{k+\ell} (xy)^{1/3} -1 \Big) =0. $$

Solution 2:

Let $K = \mathbf{Q}(\zeta_3)$. You could just directly take the norm of the element $x^{1/3} + y^{1/3} + x^{1/3}y^{1/3} - 1$ from $K(x^{1/3}, y^{1/3})$ down to $K(x, y)$. This norm is computed as the product of 9 Galois conjugates of your element and will be an element of $\mathbf{Q}(x, y)$. Should be very easy to do in any computer algebra system which can work with 3rd roots of 1.

But this is just exactly the same as what you suggest!

Solution 3:

Two automated approaches. With Gröbner basis: see the first entry of this result. With resultants: first compute this and then this.