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In the note I'm reading there is the following theorem:

Let $X\subset \mathbb{P}^n$ be a projective variety, and $p_X$ is its Hilbert polynomial. If the dimension of $X$ is $n$, then the leading term of $p_X(m)$ is $$ \frac{\operatorname{deg}X}{n!}m^n $$

However, I tried to test it using the variety $X=\mathbb{V}(x_0)\cup\mathbb{V}(x_1,x_2)\subset\mathbb{P^2}$, and found some problems.

We have $\mathbb{I}(X)=\langle x_0\rangle\cap\langle x_1,x_2\rangle=\langle x_0x_1,x_0x_2\rangle$, thus $$ S(X)=k[x_0,x_1,x_2] / \mathbb{I}(X)\cong k[x_0]\oplus k[x_1,x_2] $$

Thus the Hilbert function is $$ h_X(m)=\operatorname{dim}_k S(X)_{m}=1+(m+1)=m+2 $$

So the Hilbert polynomial is $p_X(m)=m+2$, which indicates that the degree of $X$ is $1$, because the dimension of $X$, which is the maximal dimension of its irreducible component, is $1$.

However, the geometric definition of degree is the maximal number of intersections between $X$ and a hyperplane in general position with dimension equal to the codimension of $X$, in this case $1$. Since $X$ is just the union of $x_0$-axis and the $(x_1,x_2)$-plane, one can easily see the degree of it should be 2.

Could any tell me what goes wrong here?

Updated: following is my geometric intuition: The line $L\in\mathbb{P}^2$ intersects $X$ at two points: $L_1$ and $L_2$(which is the $x_0$-axis).

enter image description here

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  • $\begingroup$ I'm confused about what you are doing. You want a union of a line and a plane, but that won't fit in $\mathbb{P}^2$. $\endgroup$
    – hunter
    Nov 25, 2013 at 16:48
  • $\begingroup$ @hunter, 'a union of a line and a plane' is just a brief description. If you draw the picture of $X$ in $\mathbb{A}^3$, it looks like the union of a line and a plane, but indeed we should look at it in $\mathbb{P}^2$..hope you understand what I mean.. $\endgroup$
    – hxhxhx88
    Nov 25, 2013 at 16:52
  • $\begingroup$ So what you have is a line and a point not on that line. Then I disagree with your geometric intuition about what the intersection numbers should be. $\endgroup$
    – hunter
    Nov 25, 2013 at 16:54
  • $\begingroup$ (The problem lies in the ambiguity of "general position." The point is that lines that don't hit the point aren't sufficiently general. You can make this precise if you think of the set of all lines as being its own space -- the condition of passing through the point is Zariski closed and therefore not "general.") $\endgroup$
    – hunter
    Nov 25, 2013 at 16:56
  • $\begingroup$ (To be clear, the difference in our perspectives -- you think of the projective space as a quotient of $\mathbb{A}^3 \setminus \{0\}$ while i think of it as a completion of $\mathbb{A}^2$-- is not what's giving the wrong answer -- even calculating in $\mathbb{A}^3$, the generic plane through the origin misses your line. I just prefer to visualize things in the dimension they "are" if that makes any sense.) $\endgroup$
    – hunter
    Nov 25, 2013 at 16:59

1 Answer 1

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The calculation is: If $I_1 = <X_0>$ and $I_2 = <X_1, X_2>$ then we have $Z = Var(I)$ with $I = I_1 \cap I_2 = <X_0*X_1, X_0 * X_2>$ and $I_1 + I_2 = <X_0, X_1, X_2>$. To compute the homogenoues coordinate ring $S/I, S := k[X_0, X_1, X_2],$ we use the exact sequence

$$0 \longrightarrow S/I \longrightarrow S/I_1 \coprod S/I_2 \longrightarrow S/(I_1 + I_2) \longrightarrow 0$$

We have $S/(I_1 + I_2) = k$, hence $Hilb_{S/(I_1 + I_2)} = 0.$ Moreover $Hilb_{S/I_1}(t) = Hilb_{k[X_1,X_2]}(t) = t+1$. And $Hilb_{S/I_2}(t)=Hilb_{k[X_0]}(t) = 1$

Because the Hilbert polynomial is additive on exact sequences we get

$$Hilb_{S/I} = Hilb_{S/I_1} + Hilb_ {S/I_2} – Hilb_k$$

i.e.

$Hilb_{S/I}(t) = t+2-0 = t+2$ and Z has $degree = 1.$

This result does not contradict intuition: The general line in $\mathbb P^2$ does not pass through the distinguished point of $Z$. All lines in $\mathbb P^2$ form the dual $\mathbb P^2$, those passing through the distinguished point form a 1-dimensional subset, isomorphic to $\mathbb P^1.$

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