This is exercise 13.15 in Harris' book "A First Course...".

Let $X$ be the twisted cubic with ideal $I(X) = (XZ-Y^2,YW-Z^2,XZ-YW).$ Let $S(X)$ denote the homogeneous coordinate ring of $X$ and $S$ the homogeneous coordinate ring $k[X,Y,Z,W]$ of $\mathbb{P}^3$ (this is Harris' notation).

The goal is to compute the Hilbert polynomial of the twisted cubic by producing a minimal free resolution of $S(X)$.

I know the polynomial is supposed to be $h_X(m) = 3m+1$, so I know I am constructing the wrong resolution. Here is what I have:

The polynomials generating the ideal of $X$ are all degree 2. We thus have an exact sequence of graded $S$-modules $$ S(-2)^3 \to S \to S(X) \to 0$$ where $S(d)$ denotes the twist of $S$ by the integer $d$. We know the kernel of the first map is generated by the relations that $XZ-Y^2$, $YW-Z^2$, $XW-YZ$ satisfy. I think I am having trouble producing these relations. For convenience put $$F_1 = XZ-Y^2 \qquad F_2=YW-Z^2 \qquad F_3 = XZ-YW.$$ The only relations I can think of are the tautological ones $F_iF_j = F_jF_i$, i.e. the matrices $A=(F_2,-F_1,0)$, $B=(F_3,0,-F_1)$, $C=(0,F_3,-F2)$, thought of as maps $S^3 \to S^3$. Following Harris, this would give the exact sequence $$ S(-4)^3 \to S(-2)^3 \to S \to S(X) \to 0.$$ Then the matrices $A$, $B$, and $C$ satisfy the relation $-F_3A + F_2B = F_1C$, i.e. the matrix $(-F_3,F_2,F_1)$. This is the only relation I can think of, so I guess the kernel is a free module, so we have the exact sequence $$0 \to S(-6) \to S(-4)^3 \to S(-2)^3 \to S \to S(X) \to 0.$$

By p. 170 in Harris we should have $$h_X(m) = {m+3 \choose 3} - 3 {m+3-2 \choose 3} + 3 {m+3-4 \choose 3} - {m+3-6 \choose 3} = 8.$$

This is obviously wrong, so my free resolution is wrong.

  • 2
    $\begingroup$ I think you have the wrong generators for $I(X)$; Harris in Ex. 1.10 says $I(X) = (XZ-Y^2,YW-Z^2,XW-YZ)$. Correcting this, and numbering the $F_i$ as you did, we have $YF_2 = WF_1 + ZF_3$ and $YF_3 = ZF_1 + XF_2$, which should give a different exact sequence. $\endgroup$ – Takumi Murayama Jul 10 '15 at 4:10
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    $\begingroup$ @TakumiMurayama Thanks! That did it. (I should say I computed the new sequence and it gives the correct result) $\endgroup$ – Kopper Jul 10 '15 at 4:25
  • $\begingroup$ @Kopper You could post the solution as an answer to your post. =) $\endgroup$ – Pedro Tamaroff Jul 10 '15 at 4:48

Per Takumi's comment, I had the wrong generators for $I(X)$. Indeed $I(X) = (XZ-Y^2, YW-Z^2, XW-YZ) = (F_1, F_2, F_3)$ and the relations are generated by $WF_1 + YF_2 - ZF_3$ and $ZF_1 + XF_2 - YF_3$, i.e. the matrices $(W,Y,-Z)$, $(Z,X,-Y)$. (Observe, for example, that $(F_2, -F_1, 0) = Y(W,Y,-Z) - Z(Z,X,-Y)$).

Since $(W,Y,-Z)$ and $(Z,X,-Y)$ are independent, the new exact sequence is $$ 0 \to S(-3)^2 \to S(-2)^3 \to S \to S(X) \to 0.$$

We obtain $$h_X(m) = {m+3 \choose 3} - 3 {m+3-2 \choose 3} + 2{m+3 - 3 \choose 3} = 3m+1$$ as desired.


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