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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:

  1. Is the free resolution for $S(X)$ above a minimal resolution for $S(X)$? If not how can I find the minimal one?
  2. 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?
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@YACP Yes it's obvious, I spotted the typo before I read it fully. –  Loki Clock Jun 19 '13 at 14:48
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2 Answers

up vote 5 down vote accepted

The degree of the polynomial is one; this encodes the fact that your variety is of dimension $1$. The leading terms is $2$ (or, perhaps better, $2/1!$); this encodes the fact that your variety is of degree $2$ (a generic plane meets it in two points).

The meaning of the constant terms is a bit more subtle, but here is one way to think about it:

Suppose first instead you have two lines meeting in a point; equivalently, two co-planar lines. The two co-planar lines are a (degenerate) conic in the plane, and all planar conics have the same Hilbert polynomial, namely $2t+1$.

Now two disjoint lines have $1$ more point than two co-planar lines (because the two co-planar lines share a point in common), and this is reflected in the fact that the Hilbert polynomial is $2t + 2 = (2t+1) +1$; so that gives some significance to the constant term.

(This example is discussed somewhere in Hartshorne, maybe in the discussion of flat families near the end of Chapter III; if you degenerate you two skew lines into two lines that meet in a point in a flat family, you don't simply get two co-planar lines, but rather two co-planar lines with a non-reduced structure at the intersection point, reflecting the fact that the "extra point" can't disappear.)

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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).

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