1 Hilbert function

1.1 Graded rings

Let G be a commutative semigroup. A commutative ring R is called G-graded when it has a (weak) direct sum decomposition R = ∑_{i∈G} R_{i} (that is, the R_{i} are additive subgroups, and every element r of R can be written in a unique way as finite sum r = r_{1} + ... + r_{m} where the r_{j} are nonzero and belong to distinct R_{i}) and moreover R_{i}R_{j} ⊆ R_{i+j}.

Elements that belong to one of the R_{i} are called homogeneous. The r_{j} that occur in the unique representation of r are called the homogeneous components of r.

Exercise Give an example of a graded commutative ring R that has an identity element 1 that is not homogeneous. Show that this cannot happen when R is N-graded.

Now let G be a commutative monoid. A commutative ring R with identity 1 is called G-graded when it is G-graded as commutative ring, and moreover 1 ∈ R_{0}, where 0 is the zero element of G.

Clearly, if H is a monoid containing G, and R is G-graded, then R is also H-graded.

Example A polynomial ring R = k[x_{1}, ..., x_{m}] is N-graded (and therefore also Z-graded): Take for R_{i} the set of all polynomials that are homogeneous of total degree i. It is also N^{m}-graded: Take for R_{i} the set of all polynomials that are homogeneous of multidegree i.

1.2 Graded modules

Let R be G-graded, and let H be a monoid containing G. An R-module M is called H-graded when it has a (weak) direct sum decomposition M = ∑_{i∈H} M_{i} such that R_{i}M_{j} ⊆ M_{i+j}.

For example, it is natural to work with N-graded rings and Z-graded modules. Or perhaps N^{n}-graded rings and Z^{n}-graded modules.

An important special case of a graded R-module, is R itself, but with shifted grading: put M = R and M_{j} = R_{h+j}. Let us call this module R^{(h)}.

A submodule N of M is called a graded submodule when it is generated by the intersections N_{i} = N ∩ M_{i}. (Then it is called homogeneous.) If this is the case, then the quotient module M/N is graded, with grading (M/N)_{i} = M_{i}/N_{i}.

Example In k[x, y, z] the ideal (x^{2} + y^{3} + z^{5}) will be homogeneous if we choose the grading that assigns degrees 15, 10, 6 to x, y, z, respectively.

1.3 Hilbert function

Consider the situation of a G-graded k-algebra R: all R_{i} are vector spaces over k. Given an H-graded R-module M, put

F (M, λ) = ∑_{i∈H} H(M, i)λ^{i}

where H(M, i) = dim_{k}M_{i}.

The function H(M, .) from H to N is called the Hilbert function of M.

Theorem 1.1 Suppose R is finitely generated by homogeneous elements r_{1}, ..., r_{s} of degrees g_{1}, ..., g_{s}, respectively. If M is finitely generated H-graded Rmodule, where H is an additive group, then

F (M, λ) = \frac{P (M, λ)}{∏_{j=1}^{s}(1 − λ^{g_{j}})}

for some finite sum P (M, λ) = ∑_{h} a_{h}λ^{h} with integral coefficients a_{h}.

Proof: Induction on the number s of generators of the k-algebra R. If s = 0, then R = R_{0} = k, and F (M, λ) has only finitely many terms.

If s > 0, then let r be one of the generators of R (of degree g, say). The module M/rM is a finitely generated S-module, where S is the subring of R generated by the generators different from r. By induction F (M/rM, λ) has a representation of the required form.

Next, consider K_{r} = {m ∈ M | rm = 0}. Again, this is a finitely generated S-module, so F (K_{r}, λ) has the required form.

Finally, all will be proved if we show that

F (M, λ) = \frac{F (M/rM, λ) − λ^{g}F (K_{r}, λ)}{1 − λ^{g}}

But that is the same as saying that

F (M, λ) − F (M/rM, λ) = λ^{g}(F (M, λ) − F (K_{r}, λ))

that is,

dim(rM)_{j+g} = dim(M)_{j} − dim(K_{r})_{j}

and that is clear (since i + g = j + g implies i = j). □

1.4 Examples

Below, R will be N-graded.

Consider R = k. We have F (R, λ) = 1.

Consider R = k[x]. We have F (R, λ) = 1 + λ + λ^{2} + ... = 1/(1 − λ).

Consider R = k[x, y]. We have F (R, λ) = 1+2λ+3λ^{2}+4λ^{3}+... = 1/(1−λ)^{2}. And indeed, for R = k[x_{1}, ..., x_{m}] we have F (R, λ) = 1/(1 − λ)^{m}. Consider R = k[x, y]/(xy). We have F (R, λ) = 1 + 2λ + 2λ^{2} + 2λ^{3} + ... =

(1 + λ)/(1 − λ).

Consider R = k[x, y]/(x^{2} +y^{2}). We have F (R, λ) = 1+2λ+2λ^{2} +2λ^{3} +... = (1 + λ)/(1 − λ).

1.5 Automated examples

When the rings are more complicated, it is easier to let a computer algebra pack-

age do the work. Fetch Macaulay from http://www.math.uiuc.edu/Macaulay2.

First repeat the above calculation. In the above the field k does not play any role. Let us take k = F_{101} = Z/101Z.

% ./M2
Macaulay 2, version 0.9
i1 : R=ZZ/101[x,y]
o1 = R
o1 : PolynomialRing
i2 : Q=R/ideal(x^2+y^2)
o2 = Q
o2 : QuotientRing
i3 : poincare Q
           2
o3 = 1 - $T
o3 : ZZ[ZZ^1]

The output of the function poincare is the numerator of the righthand side fraction in the theorem above. So, 1 − T^{2} really means (1 − T^{2})/(1 − T)^{2} since we have two variables, both of degree 1. And that is indeed the same as the (1 + T)/(1 − T) that we found above.

1.6 Geometric significance

Given a projective variety X, with homogeneous coordinate ring R = k[X_{0}, ..., X_{m}]/I(X), the dimension H(R, i) = dimR_{i} tells us how many independent functions of degree i there are on X.

That gives geometric information. For example, if X is a set of three points in the plane, then H(R, 1) will be 2 if the three points are collinear, and 3 otherwise. On the other hand, H(R, n) will be 3 for n > 1.

Let us confirm using Macaulay. We expect to find either 1+2λ+3λ^{2}+3λ^{3}+... or 1 + 3λ + 3λ^{2} + 3λ^{3} + ..., that is, either (1 + λ + λ^{2})/(1 − λ) or (1 + 2λ)/(1 − λ).

i1 : R=ZZ/101[x,y,z]
o1 = R
o1 : PolynomialRing
i2 : S=ideal(x,y)*ideal(x,z)*ideal(x,y+z)
             3   2     2    2              2   2      2                  2       2
o2 = ideal (x , x y + x z, x z, x*y*z + x*z , x y, x*y  + x*y*z, x*y*z, y z + y*z )
o2 : Ideal of R
i3 : T=radical S
                2       2
o3 = ideal (x, y z + y*z )
o3 : Ideal of R
i4 : Q=R/T
o4 = Q
o4 : QuotientRing
i5 : poincare Q

                3     4
o5 = 1 - $T - $T  + $T
o5 : ZZ[ZZ^1]

That was the case of three collinear points, and we find (1−T −T^{3}+T^{4})/(1− T)^{3} = (1 + T + T^{2})/(1 − T), as expected. And for three non-collinear points:

...
i2 : S=ideal(x,y)*ideal(x,z)*ideal(y,z)
...
            2      3
o5 = 1 - 3$T  + 2$T

That is, we find (1 − 3T^{2} + 2T^{3})/(1 − T)^{3} = (1 + 2T)/(1 − T), as expected.

Exercise Show that if X is a set of n points in projective space, then H(R, i) = n for sufficiently large i. How large?

1.7 The Hilbert polynomial

Consider R = k[x_{1}, ..., x_{m}]/I(X) with N-grading. Looking at the values that the Hilbert function H(R, i) takes, we see that they are a bit messy for small i, and then are described by a polynomial in i for sufficiently large i. This polynomial is called the Hilbert polynomial.

The special case of Theorem 1.1 where R = k[x_{1}, ..., x_{m}]/I and we use the Z-grading where all x_{i} have degree 1, says for any finitely generated Z-graded R-module that

F (M, λ) = \frac{P (M, λ)}{(1 − λ)^{m}}

for some finite sum P (M, λ) = ∑_{h} a_{h}λ^{h} with integral coefficients a_{h}.

Now F (M, λ) = ∑_{i≥i_{0}} H(M, i)λ^{i} and 1/(1 − λ)^{m} = ∑_{j≥0} \binom{j+m−1}{m−1}λ^{j}, so F (M, λ) = ∑_{j≥0} ∑_{h} a_{h}\binom{j+m−1}{m−1}λ^{j+h} and

H(M, i) = ∑_{h≤i} a_{h}\binom{i − h + m − 1}{m − 1}.

This shows that for sufficiently large i (namely, for i ≥ max{h|a_{h} ≠ 0}) we have H(M, i) = p(i), where p(i) is the polynomial p(i) = ∑_{h} a_{h}\binom{i−h+m−1}{m−1}. This is the Hilbert polynomial.

We see that the leading coefficient of the Hilbert polynomial is (∑_{h} a_{h})/(m− 1)!. If ∑_{h} a_{h} = 0, then P (M, λ) has a factor (1−λ) and we can first simplify the expression for F (M, λ). This shows that if the Hilbert polynomial has degree d, then its leading coefficient (∑_{h} a_{h})/d! is an integer divided by d!.

1.8 Properties of the Hilbert polynomial

The degree of the Hilbert polynomial is the dimension of X.

Thus, for a finite set the Hilbert polynomial will be a constant, namely the size of the set.

For a curve the Hilbert polynomial is linear, say of the form ai + b, and the value 1 − b is called the arithmetic genus of the curve.

A third invariant that can be read off from the Hilbert polynomial is the degree of the variety. If a projective variety X has dimension d and lives in P^{m} then a general linear subspace of dimension m − d will hit X in finitely many points, and the number of points is called the degree of X. Now if X has degree c, the leading coefficient of the Hilbert polynomial will be c/d!.

Let us do an example. If X is a conic in the plane, then we expect to see dimension 1, genus 0, degree 2, so the Hilbert polynomial should be 2i + 1. And for X given by x^{2} + y^{2} = z^{2} we find Hilbert function (1 − T^{2})/(1 − T)^{3} = 1 + 3T + 5T^{2} + 7T^{3} + ..., indeed with Hilbert polynomial 2i + 1.

Or, with X a cubic curve in the plane we expect dimension 1, genus 1, degree 3, so the Hilbert polynomial should be 3i. And for X given by x^{3} + y^{3} + z^{3} = 0 we find Hilbert function (1−T^{3})/(1−T)^{3} = 1+3T +6T^{2}+9T^{3}+..., as expected.

1.9 Dimension

Above we said that the degree of the Hilbert polynomial equals the dimension of the variety. Let us prove this for varieties embedded in a projective space P^{n}.

As definition of dimension we use: a variety X embedded in P^{n} has dimension d when there is a linear subspace of projective dimension n − d − 1 in P^{n} disjoint from X while all linear subspaces of projective dimension n − d meet X.

So, let U be a linear subspace of projective dimension n − d − 1 disjoint from X. The quotient space P^{n}/U is a projective space P^{d}. We find a map π : X → P^{d} by sending x ∈ X to the (n − d)-space spanned by x and U. This map is onto, since no (n − d)-space is disjoint from X. Now π^{∗} is an injection of k[X_{0}, ..., X_{d}] into the coordinate ring R of X, so H(R, i) ≥ \binom{i+d}{d} for all i, and the Hilbert polynomial of X has degree not less than d.

Now conversely. This time we use as definition of dimension: the transcendence degree of its function field over k. (In the projective setting we can take as the function field the homogeneous part of degree 0 of the quotient field of R.)

Let S be the isomorphic image of k[X_{0}, ..., X_{d}] in R. Now R is finitely generated as S-module because R and S have the same transcendence degree over k.

If R has generators y_{j} of degrees d_{j} over S, so that every element of R can be written in the form ∑ s_{j}y_{j} with s_{j} ∈ S, then we find a degree-preserving surjection ⊕S^{(−d_{j})} → R given by (s_{j})_{j} ↦ ∑ s_{j}y_{j}.

It follows that H(R, i) ≤ ∑ \binom{i+d−d_{j}}{d}, and the Hilbert polynomial of X has degree not larger than d. □