# Degree of a variety via Hilbert polynomial

$$\newcommand{\dprime}{^{\prime\prime}} \newcommand{\kk}{\mathbb{K}} \newcommand{\pp}{\mathbb{P}} \newcommand{\qq}{\mathbb{Q}} \newcommand{\rr}{\mathbb{R}} \newcommand{\zz}{\mathbb{Z}}$$
As presented in the preceding post, the degree of a subvariety $$X$$ of a projective space $$\pp^n$$ is the number of points of intersection of $X$ and a “generic” $n – m$ dimensional linear subspace of $\pp^n$, where $m:= \dim(X)$. In this post we show that the degree of a variety can be computed in terms of the Hilbert Polynomial of its homogeneous coordinate ring. The presentation follows Mumford’s Algebraic Geometry I: Complex Projective Varieties, Section 6C.

The basis of this theory, as in many parts of algebraic geometry, is a beautiful result of Hilbert.

Theorem 1. Let $$M$$ be a finitely generated graded module over $$R_n := \kk[x_0, \ldots, x_n]$$, i.e. $$M$$ is an $$R_n$$-module and there is a direct sum decomposition $$M = \bigoplus_{k \geq k_0} M_k$$ such that for all homogeneous polynomial $$f$$ in $$R_n$$, $fM_k \subseteq M_{k + \deg(f)}$
Then there is a polynomial of $$P_M(t) \in \qq[t]$$ of degree at most $$n$$ such that $$\dim(M_k) = P_M(k)$$ for all sufficiently large $$k$$.

$P_M$ as in Theorem 1 is called the Hilbert polynomial of $M$. If $M = \kk[x_0, \ldots, x_n]/I(X)$, where $X$ is a subvariety of $\pp^n$ and $I(X)$ is the homogeneous coordinate ring of $X$, then $P_M$ is called the Hilbert polynomial of $X$.

Proof of Theorem 1. Proceed by induction on $$n$$. If $$n = -1$$, then $$R_n = \kk$$, and $$M$$, being a finitely generated module over $$\kk$$, must be a finite dimensional vector space over $$\kk$$. Consequently, $$M_k = 0$$ for all sufficiently large $$k$$, and the theorem holds with $$P_M \equiv 0$$. In the general case, consider the map $$M \to M$$ given by $$m \to x_nM$$. The kernel of the map is
$M’ := \{m \in M: x_nm = 0\}$
and the cokernel is
$M\dprime := M/x_nM$
Both $$M’$$ and $$M\dprime$$ are finitely generated graded $$R_n$$-modules in which multiplication by $$x_n$$ is $$0$$. Consequently, these are finitely generated graded $$R_{n-1}$$-modules and by the induction hypothesis, there are polynomials $$P’$$ and $$P\dprime$$ of degree at most $$n-1$$ such that $$\dim(M’_k) = P'(k)$$ and $$\dim(M\dprime_k) = P\dprime(k)$$ for $$k \gg 0$$. Now for all $$k$$ the multiplication by $$x_n$$ maps $$M_k$$ to $$M_{k+1}$$ and induces an exact sequence of vector spaces over $$\kk$$:
$0 \to M’_k \to M_k \to M_{k+1} \to M\dprime_{k+1} \to 0$
Consequently,
\begin{align*} \dim(M_{k+1}) – \dim(M_k) &= \dim(M\dprime_{k+1}) – \dim(M’_k) \\ &= P\dprime(k+1) – P'(k) \end{align*}
where the first equality holds for all $$k$$ and the second equality holds for $$k \gg 0$$. Now, given any $$f(t) \in \qq(t)$$ of degree $$d$$, there is a polynomial $$g(t)$$ of degree $$d+1$$ such that
$g(t+1) – g(t) \equiv f(t)$
(note that $$(t+1)^{d+1} – t^{d+1} = (d+1)t^d +$$ a polynomial with a degree smaller than $$d$$, and use induction on $$d$$). Consequently there is a polynomial $$Q(t) \in \qq(t)$$ of degree at most $$n$$ such that
$Q(k+1) – Q(k) \equiv P\dprime(k+1) – P'(k)$
Then for $$k \gg 0$$,
$\dim(M_{k+1}) – Q(k+1) = \dim(M_k) – Q(k)$
In other words,
$\dim(M_k) = Q(k) + \text{a constant}$
for $$k \gg 0$$, as required to complete the proof of Theorem 1.

Example 2. If $$M = \kk[x_0, \ldots, x_n]$$ itself, then $$\dim(M_k)$$ is the dimension of the vector space of homogeneous polynomials of degree $$k$$, which is equal to the number of distinct monomials of degree $$k$$ in $$n+1$$ variables. The latter number is the same as the number of ways to place $$n$$ identical dividers in $$k+n$$ positions, which is
\begin{align*} \binom{k+n}{n} &= \frac{(k+n)(k+n-1) \cdots (k+1)}{n!} \\ &= \frac{k^n}{n!} + \text{a polynomial of degree lower than}\ n \end{align*}
In particular, the Hilbert polynomial of $$M$$ is
$P_M(t) = \frac{(t+n)(t+n-1) \cdots (t+1)}{n!}$

Example 3. If $$M = \kk[x_0, \ldots, x_n]/\langle f \rangle$$, where $$f$$ is homogeneous of degree $$d$$, then multiplication by $$f$$ induces an exact sequence
$0 \to \kk[x]_{k-d} \to \kk[x]_k \to M_k \to 0$
where we write $$\kk[x]$$ for $$\kk[x_0, \ldots, x_n]$$. Then for all $$k$$
\begin{align*} \dim(M_k) &= \dim(\kk[x]_k) – \dim(\kk[x]_{k-d}) \\ &= \binom{k+n}{n} – \binom{k+n-d}{n} \end{align*}
Write $$\binom{k+n}{n} = \sum_{j=0}^n a_jk^j$$. In Example 2 above we have seen that $$a_n = 1/n!$$. Consequently,
\begin{align*} P_M(t) &= \sum_{j=0}^n a_jt^j – \sum_{j=0}^n a_j(t-d)^j \\ &= \frac{t^n – (t-d)^n}{n!} + \sum_{j=0}^{n-1} a_j(t^j – (t-d)^j) \\ &= \frac{dt^{n-1}}{(n-1)!} + \text{a polynomial of degree lower than}\ n – 1 \end{align*}

Note that in both the above examples, the Hilbert polynomial equals $$\dim(M_k)$$ for all $$k$$, not only when $$k$$ is large. The following is an example where the equality holds only when $$k$$ is sufficiently large.

Example 4. Let $$M = \kk[x_0, \ldots, x_n]/I(S)$$ where $$S = \{P_1, \ldots, P_s\}$$ is a finite set of points in $$\pp^n$$ and $$I(S)$$ is the ideal of all homogeneous polynomials vanishing on all points of $$S$$. After an appropriate (linear) change of coordinates if necessary, we may assume that $$x_0|_{P_i} \neq 0$$ for any $$i$$. Then for all $$k$$ there is an exact sequence:
$0 \to I(S)_k \to \kk[x]_k \xrightarrow{\phi_k} \kk^s$
where we write $$\kk[x]$$ for $$\kk[x_0, \ldots, x_n]$$ and the map $$\phi_k$$ is defined as follows:
$\phi_k(f) := (\frac{f}{x_0^k}(P_1), \ldots, \frac{f}{x_0^k}(P_s))$
Consequently, $$\dim(M_k)$$ is the dimension of the image of $$\phi_k$$. Now, assume
$|\kk| \gg s$
Under this condition we are going to show that for $$k \geq s-1$$, there are homogeneous polynomials $$f_1, \ldots, f_s$$ of degree $$k$$ such that
\begin{align*} \frac{f_i}{x_0^k}(P_j) &= 0,\ i \neq j \\ \frac{f_i}{x_0^k}(P_j) &\neq 0, i = j \end{align*}
Indeed, consider the set $E_{i,j}$ of all linear polynomials $h$ such that $(h/x_0)(P_i) = (h/x_0)(P_j)$. Then $E_{i,j}$ is a proper codimension one linear subspace of the $\kk$-vector space $\kk[x]_1$ of linear homogeneous polynomials in $(x_0, \ldots, x_n)$. We claim that if $|\kk|$ is sufficiently large, then
$\kk[x]_1 \neq \bigcup_{i,j}E_{i,j}$
In fact we show more generally (following a MathOverflow answer) that an affine space $V$ over $\kk$ (recall that an affine space is a set which satisfies all axioms of a vector space except for those involving the “zero element”) can not be covered by the union of finitely many proper affine hyperplanes $V_1, \ldots, V_r$ if $|\kk| > r$. Indeed, it is clearly true for $r = 1$. For $r > 1$, consider translations $V_1 + a$ with $a \in \kk$. Since $|\kk| > r$, there is $a \in \kk\setminus\{0\}$ such that $V_1 + a$ is not equal to any $V_j$ for $j > 1$. Consequently, $V_j \cap (V_1 + a)$, $j = 2, \ldots, r$, are proper affine hyperplanes of the affine space $V_1 + a$. Now we are done by induction. Consequently, if $|\kk| > \binom{s}{2}$, we can choose
$h \in \kk[x]_1 \setminus \bigcup_{i,j}E_{i,j}$
Then $$a_i := (h/x_0)(P_i)$$ are pairwise distinct, and for $k \geq s-1$ we can construct the $$f_i$$ as follows:
$f_i := x_0^{k-s + 1}\prod_{j \neq i} (h-a_jx_0)$
In any event, this implies that for $$\dim(M_k) = s$$ for $$k$$ sufficiently large. In particular, $$P_M(t) = s$$ is a constant polynomial.

The examples above have the following properties: the leading coefficient of a Hilbert polynomial of a module is of the form $dt^n/n!$ for some integer $d$, and moreover, the degree of the Hilbert polynomial $P_X$ of a variety $X$ equals $\dim(X)$. As the following result shows, this is true in general, and moreover, the coefficient of the leading term of $P_X$ equals $\deg(X)$.

Theorem 2. Let $X$ be an irreducible variety in $\pp^n$ of dimension $m$ and degree $d$. Then its Hilbert polynomial has the form
$P_X(t) = d\frac{t^m}{m!} + \text{terms of lower degree}$
In particular, both dimension and degree of $X$ can be determined from the leading term of $P_X$.

Proof. Example 3 above and Proposition 1 in the preceding post show that the theorem holds when $X$ is a hypersurface. Now choose linear subspaces $H’, H$ of $\pp^n$ of dimension respectively $n-m-2$ and $m+1$ such that $H’ \cap X = H’ \cap H = \emptyset$ and the projection $\pi : \pp^n\setminus H’ \to H$ (which we defined in the preceding post) restricts to a birational map on $X$ (this is possible due to Lemma 2 of the preceding post). Since $\deg(X) = \deg(\pi(X))$ (Theorem 3 of the preceding post) and $\pi(X)$ is a hypersurface in $H$, it suffices to show that $P_X$ and $P_{\pi(X)}$ differ by terms of degree $< m$. Indeed, choose coordinates $[x_0: \cdots :x_n]$ on $\pp^n$ such that $H’ = V(x_0, \ldots, x_{m+1})$ and
$\pi: [x_0: \cdots :x_n] \in \pp^n \setminus H’ \mapsto [x_0: \cdots: x_{m+1}] \in \pp^{m+1}$
where we identify $H$ with $\pp^{m+1}$. Let $R, R’$ respectively be the homogeneous coordinate rings of $X$ and $X’ := \pi(X)$. Both $R$ and $R’$ are integral domains and
\begin{align*} R &= \kk[x_0, \ldots, x_n]/I(X) \\ &\supseteq \kk[x_0, \ldots, x_{m+1}]/(I(X) \cap \kk[x_0, \ldots, x_{m+1}]) \\ &= \kk[x_0, \ldots, x_{m+1}]/I(X’) \\ &= R’ \end{align*}
Claim 2.1. $R$ is integral over $R’$.

Proof. Fix an arbitrary $i$, $m+1 < i \leq n$ and factor $\pi|_X$ as $\psi’_i \circ \psi_i$ where
\begin{align*} \psi_i &: [x_0: \cdots : x_n] \in X \mapsto [x_0: \cdots :x_{m+1}: x_i] \in \pp^{m+2}\\ \psi’_i &: [x_0: \cdots : x_{m+1}: x_i] \in \pp^{m+2} \setminus V(x_i) \mapsto [x_0: \cdots :x_{m+1}] \in \pp^{m+1} \end{align*}
Since $H’ \cap X = \emptyset$, $[0: \cdots : 0: 1] \not\in \psi_i(X)$ and therefore
$x_i \in \sqrt{\langle x_0, \ldots, x_{m+1} \rangle + I(\psi_i(X))}$
Since $I(\psi(X)) = I(X) \cap \kk[x_0, \ldots, x_{m+1}, x_i]$ is homogeneous, it follows that there is $k \geq 1$ such that
$(x_i)^k + \sum_{j=1}^k h_i(x_0, \ldots, x_{m+1})(x_i)^{k-j} \in I(X)$
for certain homogeneous polynomials $h_i \in \kk[x_0, \ldots, x_{m+1}]$. In particular, the image of $x_i$ in $R = \kk[x_0, \ldots, x_n]/I(X)$ is integral over $R’$. Since $i$ was arbitrary, it follows that $R$ is integral over $R’$, as claimed.

Claim 2.2. $R$ and $R’$ have the same quotient field.

Proof. In general, if $Y$ is an irreducible subvariety of $\pp^n$ such that $Y \not\subseteq V(x_0)$, then there is a map from the homogeneous coordinate ring $S$ of $Y$ to the ring of polynomials in $x_0$ over its coordinate ring $\kk[Y]$ which maps a homogeneous element $f \in S$ of degree $s$ to $(f/x_0^s)(x_0^s) \in \kk[Y][x_0]$. This map is clearly injective, and since it also maps $x_0 \mapsto x_0$, it induces an isomorphism between the quotient field of $S$ and $\kk(Y)(x_0)$. Since $X$ and $\psi(X)$ are birational and the quotient field of $R$ contains that of $R’$, this implies that these quotient fields are equal and map to $\kk(X)[x_0]$ via the above isomorphism.

Now we go back to the proof of Theorem 2. Since both $R$ and $R’$ are Noetherian, Claim 2.1 implies that $R$ is a finitely generated module over $R’$. Choose homogeneous generators $f_1, \ldots, f_r$ of $R$ as a module over $R’$. By Claim 2.2 each $f_i$ is of the form $g_i/h_i$ with $g_i, h_i \in R’$. Moreover, since both $R$ and $R’$ are graded, and the inclusion $R’ \subseteq R$ preserves the grading, it follows that $g_i, h_i$ are also homogeneous. Let $h := \prod_i h_i$. Then
$hR \subseteq R’$
If $\deg(h) = s$, then it follows that
$hR_{k – s} \subseteq R’_k \subseteq R_k$
for each $k \geq s$. Consequently,
$P_X(k-s) \leq P_{\pi(X)}(k) \leq P_X(k)$
for all $k \geq s$. But then the leading terms of $P_X$ and $P_{\pi(X)}$ must be the same. This concludes the proof of Theorem 2.