How should one go about computing the dimension of the space of cubic polynomials over $\mathbb{P}^5$ which vanish on $\mathbb{P}^2$, where $\mathbb{P}^2$ sits inside $\mathbb{P}^5$ via Veronese embedding. (Here $\mathbb{P}^k$ is complex projective space). I am more interested in elementary approaches.
2026-03-30 17:03:20.1774890200
Dimension of the space of cubic polynomials over $\mathbb{P}^5$ which vanish on the Veronese surface.
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$\newcommand{\PP}{{\mathbb P}}$ $\newcommand{\Ohol}{{\mathcal O}}$ $\newcommand{\QQ}{{\mathbb Q}}$ $\newcommand{\sheaf}[1]{{\mathcal #1}}$
You can do the following calculation with Macaulay2. The idea is the following: One considers the sequence
$$0 \to \sheaf{I} \to \Ohol_{\PP^5} \to v_*\Ohol_{\PP^2} \to 0$$
where $v:\PP^2 \to \PP^5$ is the Veronese-imbedding. One applies $\bigoplus_d\Gamma(\PP^5,- \otimes \Ohol_{\PP^5}(d))$ and gets
$$0 \to I \to S \to R$$
with $S=\QQ[t_0,\ldots,t_5]$ and $R=\QQ[x,y,z]$ where $v:S \to R$ is given by $t_0 \mapsto x^2,\ldots,t_5 \mapsto z^2$.
Now the vanishing cubic polynomials are $I_3 = \Gamma(\PP^5,\sheaf{I}(3))$ which can be selected by the command basis(3,id1) where id1 is the ideal $I$ in Macaulay2 (line i13 below). One counts $28$ columns in the result matrix, which is therefore the dimension of the space of cubic polynomials vanishing on the veronese-surface. To double check, one computes basis(3,S/id1) (line i15), which gives $28$ too. As the cubic polynomials have dimension $\binom{5 + 3}{3} = 56$, our calculation must be right.