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What are two polynomials $f,g \in \mathbb{C}[x,y,z]$ such that $$\{(x,y,z): f(x,y,z)=g(x,y,z)=0\}\;=\;\{(t^3,t^4,t^5): t \in \mathbb{C}\}$$ holds as an equality of subset of $\mathbb{C}^2$?

This answer, as I understand it, claims that there are indeed such polynomials. How can we find them?

Surely both polynomials need to be on the ideal of this curve, and thus $f(x,y,z)=p_1(xz-y^2)+p_2(yz-x^3)+p_3(z^2-x^2y)$, for some polynomials $p_i$ in $x,y,z$, and similarly for $g$.

Alas, we cannot simply equate coefficients of each monomial in the two equations, nor solve generally these equations so as to ensure their solutions will only be points on the curve. Is there a reasonable way to find $f, g$?

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    $\begingroup$ The OP asks for set-theoretic generators and indeed there are two equations which define the curve set theoretically. The curve is ideal theoretically defined by the $2\times 2$ minors of the matrix $\left(\begin{array}{ccc} x&y&z\\ y&z&x^2 \end{array}\right)$. The curve is set-theoretically defined by $xz-y^2$ and the determinant of the $3\times 3$ matrix $\left(\begin{array}{ccc} x&y&z\\ y&z&x^2\\ z&x^2&0\end{array}\right)$. $\endgroup$
    – Mohan
    Commented May 9, 2016 at 2:10
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    $\begingroup$ @Mohan Can you please explain or give a reference why the latter part is true? (it answers my question). $\endgroup$
    – Pachirisu
    Commented May 9, 2016 at 4:04
  • $\begingroup$ Why don't you just solve the two equations? For example, if $(a,b,c)$ is a solution to the two equations and $a=0$, we get from the first equation $b=0$ and from the last equation $c^3=0$. Next, look at $a\neq 0$. Then the first equation gives, $c=b^2/a$. Substitute in the second and see what you get. $\endgroup$
    – Mohan
    Commented May 9, 2016 at 14:22
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    $\begingroup$ @Mohan Well, the determinant is $-x^5+2x^2yz-z^3$, plugging in we get $a^4=b^3$, and writing $a=t^3$ we arrive at the required curve. But I'd love to know how did you find those polynomials, especially since the expression as a pretty determinant is non-obvious. $\endgroup$
    – Pachirisu
    Commented May 9, 2016 at 19:18
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    $\begingroup$ This is hard to explain, since the idea originated with Ferrand-Szpiro and the best I can say is, they have a neat way of `doubling' a variety, using the dualizing module. More elementarily, you may have noticed that both the equations are determinants of symmetric matrices and somehow that is what makes it work. And why the question is open for general curves in 3-space. $\endgroup$
    – Mohan
    Commented May 9, 2016 at 19:24

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