# Why is the twisted cubic given as maximal minors of a matrix?

Today I began reading Hartshorne, and was doing an exercise about the twisted cubic, parametrically given by $t \mapsto (t,t^2,t^3)$. According to Wikipedia, its (non-homogeneous) ideal is $(xz-y^2,y-z^2,x-yz)$. Toying a bit with these equations, I noticed that they are precisely the maximal minors of the matrix

$\begin{pmatrix}x &y &z \\ y& z& 1\end{pmatrix}$

Is there a reason for this? I tend to think of determinants as having to do only with linear equations of some sort, but the twisted cubic does not seem very "linear".

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There is a whole class of "determinantal varieties", the Segre embedding (used to define the product of two projective varieties) usually being the first one that is explicitly given in this way. – Dylan Moreland Jan 12 '12 at 20:49
Actually, are these equations in the ideal correct? The second generator, for example, seems off: $t^2 - t^6$ isn't usually zero. – Dylan Moreland Jan 21 '12 at 8:11
Hm, seems you're right. I just copied the homogeneous equation from Wikipedia (en.wikipedia.org/wiki/Twisted_cubic), and put $w=1$. – Fredrik Meyer Jan 21 '12 at 19:44