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I know the rational normal curve as the image of a polynomial map \begin{gather} \phi:K\rightarrow K^n\\ \phi(t)=(t,t^2,\dots,t^n) \end{gather}

My question is proving the variety defined by the set of homogeneous quadratics obtained by taking all possible $2 \times 2$ sub-determinants of the matrix \begin{bmatrix} x_0 & x_1 & x_2& \dots &x_{n-1}\\ x_1 & x_2& x_3& \dots &x_{n} \end{bmatrix} is equivalent to the parametric representation.

What I have done so far was look at a special case \begin{bmatrix} x_0 & x_1 & x_2\\ x_1 & x_2& x_3 \end{bmatrix} and the map $$\phi(t)=(t,t^2,t^3,t^4).$$

Clearly when I take the sub-determinants the ideal of $\varphi(K)$ is $$I =\langle x_0x_2-x_1^2,x_0x_3-x_1x_2,x_1x_3-x_2^2\rangle.$$ We can see that the image of the map is contained in the ideal generated by the sub determinants, but I am not sure how to argue that the image contains the ideal.

Maybe helpful: this is in I.V.A. by Cox, Little and O'Shea as a part of a problem in 8.4, problem 11, part d.

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    $\begingroup$ See this answer. (Let me know if need more details.) It's also the Proposition 6.1 in Eisenbud, The Geometry of Syzygies. $\endgroup$ – user26857 May 13 '16 at 14:27

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