TITLE: Hilbert polynomial of disjoint union of lines in $\Bbb{P}^3$
QUESTION [8 upvotes]: Let $X$ be the disjoint union of the two lines in $\Bbb{P}^3$ given by $Z(x,y)$ and $Z(z,w)$. Letting $R = k[x,y,z,w]$, I have computed the following free resolution for the homogeneous coordinate ring  $S(X)$:
$$0 \to R(-4) \stackrel{\varphi_0}{\longrightarrow} R(-3)^4 \stackrel{\varphi_1}{\longrightarrow} R(-2)^4 \stackrel{\varphi_2}{\longrightarrow} R \longrightarrow S(X) \to 0.$$
$\varphi_0$ is the map that sends $r \in R(-4)$ to the $4$ - tuple  $(ry,-rz,rw,-rx)$, $\varphi_1$ is multiplication by the matrix $$\left( \begin{array}{cc} z & y & 0 & 0 \\ -w & 0 &y & 0 \\ 0& 0& -x & w \\ 0 & -x & 0 &-z \end{array}\right)$$
and $\varphi_2$ is multiplication by the row matrix matrix $\left(\begin{array}{cccc} xw & xz & yw & yz\end{array}\right)$. Note I have used that $I(X)= (x,y) \cap (z,w) = (xw,xz,yw,yz)$. From the free resolution above I then computed the Hilbert polynomial $H_X(t)$ for $X$:
$$H_X(t) = 2(t+1).$$


My questions are:

Is the free resolution for $S(X)$ above a minimal resolution for $S(X)$? If not how can I find the minimal one? 
What is the significance of the Hilbert polynomial in this case, for example how does it encode information about the dimension of $X$? What if I have a general projective variety?

REPLY [3 votes]: The answer to the first question is yes, the resolution that you computed is the minimal free resolution of the ideal.  A computer algebra system like Macaulay 2 or CoCoA will generate the minimal free resolution.  However, in this particular case your ideal is a monomial ideal generated by quadrics, so we can think of it as the edge ideal of the four-cycle (the graph with vertices $\{x,y,z,w\}$ and edge set $\{xw,wy,yz,zw\}$). See the PhD thesis of S. Jacques for the formula for the graded Betti numbers of all edge ideals of cycles (especially Theorem 7.6.28).