Hilbert Polynomials and Schemes

Here I collect a set of computations of hilbert polynomials of various varieties.

The union of two skew Lines in $\mb P^3$

Lemma:

All schemes which are the union of two skew lines in $\mb P^3$ differ by an automorphism from $X = V(x, y)\cup V(z, w)$, which is a union of skew lines.

Proof. Recall that two lines are skew in $\mb P^3$ if and only if they do not intersect; note that $V(x, y)\cap V(z, w) = V(x, y, z, w) = \emptyset$, and so $X$ is the union of skew lines. Let $\ell_1$ and $\ell_2$ be skew lines, given by $V(P_1, Q_1)$ and $V(P_2, Q_2)$ respectively, for linear polynomials $P_1, Q_1, P_2, Q_2$. The lines $\ell_1$ and $\ell_2$ are skew if and only if they do not intersect, which holds if and only if their defining ideals sum to the irrelevent ideal, i.e. iff they generate the degree one peice of $k[x, y, z, w]$. By a dimension argument, this occurs if and only if the set $S = \{P_1, P_2, Q_1, Q_2\}$ is linearly independent. Then the matrix define by the coeficients of the $P_i$ and $Q_i$ is invertible, and thus (by the universal property of free algebras) so is the induced automorphism on $k[x, y, z, w]$. Moreover, that automorphism sends $x\to P_1, y\to P_2, z\to Q_1, w\to Q_2$, and the result follows by the functoriality of the proj construction in linear automorphisms.

Proposition:

The hilbert polynomial of any union of two skew lines $Y$ is $P(d) = 2d+2$.

Proof. Since the hilbert polynomial is invariant under automorphisms of $\mb P^n$, it suffices to compute using $X = V(x, y)\cup V(z, w)$ (by Lemma 1). Our scheme is then given by the ideal $I= (x, y)(z, w) = (xz, xw, yz, yw)$. If $R = k[x, y, z, w]$ and $S = R/I$ (where for each we denote the naturally induced grading by a subscript, so $\bullet_i$ is the $i$th peice of $\bullet$), it will suffice to compute $\dim_kS_i = \dim_kR_i-\dim_kI_i$. Combinatorics gives us that, since monomials generate $R_i$ as a vector space,

as this is the number of unique monic monomials in four variables of degree $d$. Moreover, a monomial $x^{c_1}y^{c_2}z^{c_3}w^{c_4}$ lies in $I$ if and only if both $c_1+c_2$ and $c_3+c_4$ is nonzero (recall that no $c_i$ can be negative). The number of monic monomials where this does not occur is the number of monic monnomials in $x, y$ plus the number of monic monomials in $y, z$, or exactly twice the number of monic monomials in two variables. Subtracting this from the total yields the number of monic monomials in $I_i$

Then

and we are done.