I'm beginning to learn about Hilbert polynomials and I'm trying to find it for the variety $X=Z(x^2+y^2+z^2+w^2)\subset \mathbb{P}^3$.

I know that the leader term must be of the form $\frac{2}{2!}t^2$, since $\dim X=2$ and the degree of $x^2+y^2+z^2+w^2$ is $2$. I tried to look at the vector spaces $S(X)_m$ explicitly, for example, for $m=2$:

$$S(X)_2=(x^2,y^2,z^2,w^2, xy,xz,xw,yz,yw,zw)/(x^2+y^2+z^2+w^2)$$

Since $\overline{w}^2=-\overline{x}^2-\overline{y}^2-\overline{z}^2$, then $\{\overline{x}^2, \overline{y}^2,\overline{z}^2, \overline{x}\overline{y},\overline{x}\overline{z},\overline{x}\overline{w}, \overline{y}\overline{z}, \overline{y}\overline{w}, \overline{z}\overline{w}\}$ is a $k$-basis for $S(X)_2$ therefore $\dim_k S(X)_2=9$.

But I don't know where to go from here. What is the idea?

  • 1
    $\begingroup$ The Hilbert series is $\frac{1+t}{(1-t)^3}$, so the $h$-vector is $(1,\ 1)$. Then the Hilbert polynomial is $P_0+P_1$, where $P_j(X)={{X+d-j-1}\choose{d-1}}$ and $d=\dim X$. $\endgroup$ – user26857 Nov 19 '17 at 20:02
  • $\begingroup$ What is an $h$ vector? How did you find all these information? I've looked up the definition of Hilbert series but it didn't clarify much $\endgroup$ – rmdmc89 Nov 19 '17 at 20:49
  • $\begingroup$ This is a standard terminology. Look in Bruns and Herzog. $\endgroup$ – user26857 Nov 19 '17 at 22:02

Let $h_X$ be the Hilbert function of $X=V_{+}(w^2+x^2+y^2+z^2)$, let $S(X)=\mathbb{K}[w,x,y,z]_{\displaystyle/(w^2+x^2+y^2+z^2)}$; by definition $$ h_X:d\in\mathbb{Z}\to\dim_{\mathbb{K}}S(X)_d\in\mathbb{N}_{\geq0}, $$ and by hypothesis: $$ h_X(d)=\begin{cases} 0\iff d<0\\ 1\iff d=0\\ 4\iff d=1\\ \vdots \end{cases}. $$ For $d\geq2$ a base $B_d$ of $S(X)_d$ as $\mathbb{K}$-vector space is $$ \left\{\overline{x^ay^bz^c}\in S(X)\mid a,b,c\in\{0,1,\dots,d\},\,a+b+c=d\right\}\cup\left\{\overline{wx^ay^bz^c}\in S(X)\mid a,b,c\in\{0,1,\dots,d\},\,a+b+c=d-1\right\}; $$ the size of $B_d$ is $$ \binom{3+(d-1)}{d}+\binom{3+(d-2)}{d-1}=\binom{2+d}{d}+\binom{1+d}{d-1}=\frac{(d+2)!}{d!2!}+\frac{(d+1)!}{(d-1)!2!}=\frac{(d+1)!}{(d-1)!2!}\left(\frac{d+2}{d}+1\right)=\frac{(d+1)d}{2!}\frac{2d+2}{d}=(d+1)^2; $$ summerize: $$ h_X(d)=\begin{cases} 0\iff d<0\\ (d+1)^2\iff d\geq0 \end{cases}; $$ that is Hilbert polynomial $\chi_X(d)$ of $X$ is $(d+1)^2$.


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