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a trivial question perhaps. Given the projective variety defined by the condition $$ rank \left( \begin{array}{lll} x_0 & x_1 & x_2 \\ x_1 & x_2 & x_3 \end{array} \right) \leq 1.$$ So it is given as the intersection of three quadrics: $x_0x_2 -x_1^2=0, x_1x_3-x_2^2=0$ and $x_0 x_3 -x_1 x_2=0$. How to show that it can't be given as the intersection of just two quadrics? Thanks

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  • $\begingroup$ You forgot the dollar signs, I guess! $\endgroup$ – Gigili Jun 3 '12 at 11:36
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Your curve has degree 3, since it has parametric representation $[u^3:u^2v:uv^2:v^3]$.
It cannot be described as the intersection of any two distinct surfaces $S,T\subset \mathbb P^3$ of degrees $s$ and $t$ whatsoever.
Indeed by Bézout's theorem the intersection would have degree $st$ and if you want $st=3$ you must have $s=1$ (say).
But this means that your curve would be included in the plane $S$ with equation $ a_0x_0+a_1x_1+a_2x_2+a_3x_3=0$ which is impossible, since it would imply that $ a_0u^3+a_1u^2v+a_2uv^2+a_3v^3=0$ for all $[u:v]\in \mathbb P^1$, an absurd statement.

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  • $\begingroup$ do you know of a solution which doesn't require Bézout's theorem? $\endgroup$ – Marco Flores Oct 3 '16 at 3:29
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This is the standard example of a twisted cubic. Consider $x_0 x_3-x_1 x_2=0$ and intersect this quadric with either of the other two and you will get a line and a twisted cubic. The third equation is needed to get rid of the line.

It comes down to algebra manipulation.

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    $\begingroup$ I know that the intersection of two quadrics of the above form cannot give the cubic right, but I meant can we prove the intersection of {\it any} two quadrics cannot give us the required curve. $\endgroup$ – user17090 Jun 3 '12 at 11:53

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