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I was asked to do some computations involving a "twisted quartic" function $f : \mathbb{C} \to \mathbb{C}^4$ defined by $t \mapsto (t,t^2,t^3,t^4).$ However, I first know that I need to compute a basis (need not be a Groebner basis) for $I(V)$ where $V = f(\mathbb{C}) \subset \mathbb{C}^4.$

I know that the ideal of the twisted cubic is $\langle x_1^2-x_2,x_1^3-x_3 \rangle \subset \mathbb{C}[x_1,x_2,x_3]$. Moreover, I know that the twisted quartic is a $1$-dimensional curve in $\mathbb{C}^4$ which means that $I(V)$ should be generated by $3$ polynomials in $\mathbb{C}[x_1,x_2,x_3,x_4]$ but how do can I tell which three it should be? Will any three relating the components work? For example, $I(V) = \langle x_1x_3-x_2^2,x_1x_4-x_2x_3,x_2^2-x_4\rangle$?

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The most obvious thing seems to be to just rewrite the parameterization in terms of equations: $x_2 = x_1^2$, $x_3 = x_1^3$, $x_4 = x_1^4$.


The equations you suggest reduce (by eliminating $x_4$) to $x_1 x_3 = x_2^2$ and $x_2(x_1 x_2 - x_3) = 0$. So as well as the twisted quartic curve their vanishing locus also contains the crossed lines $x_1 x_3 = x_2 = x_4 = 0$.

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