Complete Intersection of a Veronese embedding as Five Quadrics

Question

Consider the Veronese embedding given by the linear system $|\mathcal{O}_{\mathbb{P}^2}(2)|$, $\phi:\mathbb{P}^2\to \mathbb{P}^5$. In exercise IV.5 of Beauville, Complex Algebraic Surfaces, we prove that that the image is in fact given by all the symmetric matrices of rank 1, considering points in $\mathbb{P}^5$ of the form: $$ \begin{pmatrix} z_0 & z_1& z_2 \\ z_1 & z_3 &z_4\\ z_2 & z_4 & z_5 \\ \end{pmatrix} $$ However, the same exercise asks to prove that this image (hence the set of all these matrices with rank 1) is the complete intersection of 5 quadrics. Setting all the determinants of the $2\times 2$ minors zero, I could find nine relations, but by symmetry I could drop 3 of them, so I get: $$\left\{ \begin{array}{l} z_0z_3-z_1^2=0 \\ z_0z_4-z_1z_2=0\\ z_3z_5 -z_4^2 =0 \\ z_1z_4-z_2z_3=0\\ z_1z_5-z_2z_4=0 \\ z_0z_5 -z_2^2=0 \end{array} \right.$$ But they are 6, not 5. Is Beauville's exercise wrong or we can really realize that just as complete intersection of 5 quadrics?

Answer 1

The book does not say complete intersection! It simply says: "Deduce that $V$ is the intersection of $5$ quadrics in $\check{Q}$." Note that $V$ (the image of the above map) is $2$-dimensional, while $\check{Q} \cong \mathbb{P}^5$ is $5$-dimensional. So $V$ has codimension $3$. Beauville asks you to find $5$ equations, not $3$. So we are not showing that $V$ is a complete intersection.

All the $6$ equations you listed are in the ideal of the image $V$. They are minimal degree ($I(V)$ contains no linear forms). And they are linearly independent, in fact no two of them share any monomial. So it is not possible for any subset of $5$ of them to generate $I(V)$!

Instead, look for $5$ of the equations that set-theoretically define $V$, i.e., $5$ quadrics that generate $I(V)$ up to radical. Hint: pick one of the quadrics and show that a power of it lies in the ideal generated by the other $5$.

For instance, $(z_0 z_4 - z_1 z_2)^2$ lies in the ideal generated by the other $5$ quadrics.