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 \(\kk\):
\[0 \to M’_k \to M_k \to M_{k+1} \to M\dprime_{k+1} \to 0\]
\dim(M_{k+1}) – \dim(M_k)
&= \dim(M\dprime_{k+1}) – \dim(M’_k) \\
&= P\dprime(k+1) – P'(k)
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
&= \frac{(k+n)(k+n-1) \cdots (k+1)}{n!} \\
&= \frac{k^n}{n!} + \text{a polynomial of degree lower than}\ n
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 \(k\)
&= \dim(\kk[x]_k) – \dim(\kk[x]_{k-d}) \\
&= \binom{k+n}{n} – \binom{k+n-d}{n}
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,
&= \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

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
\frac{f_i}{x_0^k}(P_j) &= 0,\ i \neq j \\
\frac{f_i}{x_0^k}(P_j) &\neq 0, i = j
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 $H_1, \ldots, H_r$ if $|\kk| > r$. Indeed, it is clearly true for $r = 1$. For $r > 1$, consider translations $H_1 + a$ with $a \in \kk$. Since $|\kk| > r$, there is $a \in \kk\setminus\{0\}$ such that $H_1 + a$ is not equal to any $H_j$ for $j > 1$. Consequently, $H_j \cap (H_1 + a)$, $j = 2, \ldots, r$, are proper affine hyperplanes of the affine space $H_1 + a$. 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
&= \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’
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
\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}
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.

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