Course Notes for Math 737 (Algebraic Geometry)

3. Hilbert Bases and Polynomials. Two important theorems of Hilbert are proved, and some consequences are investigated.

Recall that in §2, we observed:

V ({L_{s}}_{s∈S}) = V (L_{s_{1}}, ..., L_{s_{r}}) ⊂ P(V)

for linear forms L_{s} and suitable (finite!) choice of s_{1}, ..., s_{r}. Our first task is to generalize this finiteness to all algebraic sets.

Lemma 3.1: Suppose that f_{s}, g_{t} ∈ k[x_{1}, ..., x_{n}] for each s ∈ S and t ∈ T. Let I be the ideal generated by {f_{s}}_{s∈S} and J be the ideal generated by {g_{t}}_{t∈T}. Then:

J ⊆ I ⇒ V ({f_{s}}_{s∈S}) ⊆ V ({g_{t}}_{t∈T}) ⊆ A_{k}^{n}.

Proof: Any g_{t} ∈ J ⊂ I, so by definition, there are polynomials f_{s_{1}}, ..., f_{s_{r}} and h_{1}, ..., h_{r} ∈ k[x_{1}, ..., x_{n}] such that

g_{t} = h_{1}f_{s_{1}} + ... + h_{r}f_{s_{r}}

If p ∈ V ({f_{s}}_{s∈S}) then f_{s}(p) = 0 for all s ∈ S, so by the sum above, g_{t}(p) = 0 for all t ∈ T. Thus V ({f_{s}}_{s∈S}) ⊆ V ({g_{t}}_{t∈T}).

Note: In particular, applying the Lemma twice gives us:

I = J ⇒ V ({f_{s}}_{s∈S}) = V ({g_{t}}_{t∈T})

so that we can really think of V as a function from the ideals in k[x_{1}, ..., x_{n}] to algebraic subsets of A_{k}^{n} (and similarly in the projective case). Note that the reverse implication is often false(!)

Definition: A commutative ring A with 1 is Noetherian if every ascending chain of ideals in A:

I_{1} ⊆ I_{2} ⊆ I_{3} ⊆ ...

stabilizes (i.e. if there is an r such that I_{r} = I_{r+1} = ...).

Examples: (a) Any ring with finitely many ideals, e.g. a field, is Noetherian.

(b) Z is Noetherian. Every ideal is of the form nZ for some n ∈ Z, and nZ ⊆ mZ iff m|n, so in particular, nZ ⊆ mZ implies |m| ≤ |n|.

(c) A polynomial ring k[x_{1}, x_{2}, ...,] in infinitely many variables is not Noetherian. The chain of ideals: ⟨x_{1}⟩ ⊂ ⟨x_{1}, x_{2}⟩ ⊂ ... does not stabilize.

Lemma 3.2: A is Noetherian iff every ideal I ⊂ A is finitely generated.

Proof: Suppose that every ideal is finitely generated. If I_{1} ⊆ I_{2} ⊆ ... is an ascending chain of ideals in A, let I = ∪_{j=1}^{∞}I_{j}. Then I is an ideal, which must be finitely generated by assumption. But the generators must all appear in some I_{r} (why?) which implies I = I_{r} = I_{r+1} = ....

On the other hand, suppose that I ⊂ A is not finitely generated. Then we can inductively choose elements a_{1}, a_{2}, ... ∈ I so that a_{r+1} ∉ ⟨a_{1}, ..., a_{r}⟩ for all r. These ideals generated by a_{1}, ..., a_{r} form an ascending chain which does not stabilize.

Hilbert Basis Theorem: If A is Noetherian, then A[x] is Noetherian.

Proof: Suppose I ⊂ A[x] is an ideal. Consider the following ideals in A:

J_{d} := {a ∈ A| ∃f(x) ∈ I of the form f(x) = ax^{d} + {lower order in x}}.

Then the ideals J_{d} form an ascending chain, so since A is Noetherian, there must be an r such that J_{r} = J_{r+1} = ....

Each J_{d} is finitely generated, since A is Noetherian, say by elements a_{d,i}, i = 1, ..., n_{i}. For all d from 1 to r and generators a_{d,i}, i = 1, ..., n_{i}, let f_{d,i} ∈ I be polynomials of degree d with leading term a_{d,i}. Then because of the fact that the chain of ideals stabilizes at J_{r}, given any polynomial f ∈ I, we can match its leading coefficient and degree with a polynomial in the ideal spanned by the f_{d,i}, so by induction on the degree of f, we find that the finite collection of polynomials f_{d,i} generate I, which proves A[x] is Noetherian.

Now we are ready for the:

Corollary 3.3: If V ({f_{s}}_{s∈S}) ⊆ A_{k}^{n} is an algebraic subset, then there are finitely many elements s_{1}, ..., s_{r} ∈ S so that:

V ({f_{s}}_{s∈S}) = V (f_{s_{1}}, ..., f_{s_{r}})

Proof: By the note following Lemma 3.1, it suffices to show that the ideal generated by the f_{s} can be generated by finitely many of them. By the Basis Theorem and induction, the ring k[x_{1}, ..., x_{n}] is Noetherian.

Thus at least we know that the ideal generated by the f_{s} is finitely generated. But in addition, if the inclusions of ideals ⟨f_{s_{1}}, ..., f_{s_{r}}⟩ ⊂ ⟨{f_{s}}_{s∈S}⟩ were proper for all r, then we could find elements s_{1}, s_{2}, ... ∈ S such that f_{s_{r+1}} ∉ ⟨f_{s_{1}}, ..., f_{s_{r}}⟩ for all r. As in the proof of Lemma 3.2, the chain of ideals generated by f_{s_{1}}, ..., f_{s_{r}} would not stabilize, violating the Noetherian property. The upshot of this is that not only is the ideal generated by the f_{s} finitely generated, but a finite set of generators may be chosen from among the f_{s} themselves.

Remark: The same proof gives the same result for algebraic subsets of projective space, too.

The next goal of this section is to find a polynomial attached to each projective variety. This will follow from some general results on graded modules over a polynomial ring which we list below.

Let A = k[x_{0}, ..., x_{n}] be the polynomial ring graded by degree.

Definitions: (a) A graded A-module M is a module over A together with a decomposition (as a vector space over k) as:

M = ⨁_{d∈Z} M_{d}

with the property that if f ∈ A_{d} and m ∈ M_{e} then fm ∈ M_{d+e}. Notice that in contrast to graded rings, we allow summands indexed by negative integers.

(b) An A-module homomorphism α : M → N is a homomorphism of graded modules if M and N are graded and α(M_{d}) ⊂ N_{d} for all d ∈ Z.

(c) If M is a graded A-module and p ∈ Z, define the twist M(p) to be the graded A-module defined by M(p)_{d} = M_{d+p}. (This operation is invertible!) For instance, multiplication by a homogeneous polynomial F ∈ A_{e} determines graded homomorphisms m_{F} : M(p) → M(p + e) for all twists p.

(d) If M is a graded A-module and all the M_{d} are finite-dimensional vector spaces, the associated Hilbert function is:

h_{M} : Z → Z_{≥0}; d ↦ dim_{k}M_{d}

Examples: (a) A itself with its grading is a graded A-module. Its Hilbert function is:

h_{A}(d) = \{ \begin{matrix}\binom{n+d}{n} = \frac{(n+d)(n+d−1)...(d+1)}{n!} if d ≥ 0 and \\ 0 if d < 0 \end{matrix}

Notice that \binom{n+d}{n} is polynomial of degree n in d with leading term \frac{1}{n!}d^{n}.

(b) Let I ⊂ A be a homogeneous ideal. We will be particularly interested in the Hilbert function of A/I (thought of as a graded A- module). We can compute some special cases:

(b1) Suppose F ∈ A is homogeneous of degree e and I = ⟨F ⟩. Then I ≅ A(−e) as graded A-modules(!) so that:

h_{A/I}(d) = h_{A}(d) − h_{I}(d) = h_{A}(d) − h_{A}(d − e)

and we get the following Hilbert function for A/I:

h_{A/I}(d) = \{ \begin{matrix}0 if d < 0 \\ \binom{n+d}{n} & if 0 ≤ d < e and \\ \binom{n+d}{n} − \binom{n+d−e}{n} if d ≥ e \end{matrix}

Notice that h_{A/I}(d) is a polynomial in d with leading term \frac{e}{(n−1)!}d^{n−1} as soon as d ≥ e.

(b2) Let k be an infinite field, let Z ⊂ P_{k}^{n} be a set consisting of m points, and let I = I(Z) ⊂ A be the homogeneous geometric ideal defined by Z. Then I claim the following:

Lemma: h_{A/I}(d) = m for all d ≥ m − 1.

Proof: Recall from §2 that we can think of the inclusion of a point p ↪ P_{k}^{n} = P(V) as ι_{W} where W ⊂ V is a line through the origin. Let e_{p} be a nonzero vector in W, and y_{p} the dual vector. Then we have the corresponding ι_{W}^{∗} : k[x_{0}, ..., x_{n}]_{d} → k[y_{p}]_{d} ≅ k for each p ∈ Z, which is surjective for each p individually. Moreover, I_{d} is the kernel of the map:

ρ_{d} : k[x_{0}, ..., x_{n}]_{d} → k^{m} ≅ ⨁_{p∈Z} k[y_{p}]_{d}

Thus, we need to prove that ρ_{d} is surjective if d ≥ m − 1. But consider the set Z − {p}. This is algebraic because it is finite (§1 HW) and moreover, you discovered that Z − {p} = V (J) where J is generated by homogeneous polynomials of degree m − 1. But this implies that we can find homogeneous polynomials of degree m − 1 (and any larger degree..why?) which vanish at all the points of Z except for p. Such polynomials are mapped to a basis for k^{m} by ρ_{d}, hence surjectivity.

Exercise: (a) Find the natural grading on the cokernel and kernel of a graded homomorphism of graded A-modules.

(b) Prove that if

0 → M^{1} → ... → M^{r} → 0

is a long exact sequence of graded homomorphisms of graded A-modules with Hilbert functions h_{M^{i}}(d), then ∑_{i=1}^{r}(−1)^{i}h_{M^{i}}(d) is the zero function.

(c) If M is a finitely generated graded A-module, prove that M_{d} = 0 for d << 0 and that the vector spaces M_{d} are all finite dimensional. Thus finitely generated graded A-modules always have Hilbert functions.

(d) If α : M → N is a homomorphism (not necessarily graded!) of finitely generated A-modules, prove that the kernel and cokernel of α are also finitely generated. (Hint: A is Noetherian!)

Definition: If M is a graded A-module with Hilbert function h_{M}(d), then the “Laurent series”:

P_{M}(t) := ∑_{d∈Z} h_{M}(d)t^{d} ∈ Z[[t^{−1}, t]]

is called the Poincaré series of M.

Note that if M is finitely generated, then by Exercise (c) above, the Poincaré series is bounded from below (i.e. P_{M}(t) = a_{−k}t^{−k} + a_{−k+1}t^{−k+1} + ...), but that it is usually not bounded from above. If it were, then P_{M}(t) would be a Laurent polynomial, that is, an element of Z[t^{−1}, t].

More Observations: (1) Given an exact sequence of graded A-modules as in Exercise (b) above, then ∑_{i=1}^{r}(−1)^{i}P_{M^{i}}(t) ≡ 0.

(2) For any twist, P_{M(p)}(t) = t^{−p}P_{M}(t).

Hilbert’s Polynomial Theorem (with contributions by Serre):

If M is a finitely generated graded k[x_{0}, ..., x_{n}]-module, then:

(a) (1 − t)^{n+1}P_{M}(t) ∈ Z[t^{−1}, t].

(b) There is a (uniquely determined) polynomial H_{M}(x) ∈ Q[x] such that H_{M}(d) = h_{M}(d) for all sufficiently large d. Moreover, if H_{M}(d) ≢ 0, let d_{M} = deg(H_{M}(d)). Then

f(t) := (1 − t)^{d_{M} +1}P_{M}(t) ∈ Z[t^{−1}, t] and f(1) ≠ 0

(so that in particular, d_{M} ≤ n.)

Proof: We prove (a) by induction on the number of variables. If n = −1, then M is a finite dimensional vector space over k, hence M_{d} = 0 for d sufficiently large and sufficiently small, so P_{M}(t) ∈ Z[t^{−1}, t] which is (a).

Next, suppose n ≥ 0 and consider the exact sequence:

(∗) : 0 → K → M →^{x_{n}} M(1) → N → 0

where K and N are the kernel and cokernel of the graded homomorphism given by multiplication by x_{n}. Notice that K and N are graded finitely generated modules over the ring k[x_{0}, ..., x_{n}]/(x_{n}) ≅ k[x_{0}, ..., x_{n−1}], so we are in a position to prove (a) by induction. Namely, by Observations (1) and (2), we have:

(t^{−1} − 1)P_{M}(t) = P_{M(1)}(t) − P_{M}(t) = P_{N}(t) − P_{K}(t)

so that (1 − t)^{n+1}P_{M}(t) = t(1 − t)^{n}(P_{N}(t) − P_{K}(t)) ∈ Z[t^{−1}, t].

To prove (b), we calculate. First, let P_{M}(t) = (1 − t)^{−p}g(t) such that g has neither a zero nor pole at 1. We want to show h_{M}(d) is a polynomial of degree p − 1 in d for large d. But

(1 − t)^{−p} = ∑_{r=0}^{∞} \binom{p + r − 1}{p − 1} t^{r}

and if we let g(t) = ∑_{k=−N}^{N} a_{k}t^{k}, then for large enough d, we have:

h_{M}(d) = ∑_{k=−N}^{N} a_{k} \binom{p + d − k − 1}{p − 1}

But this is a polynomial of the form:

\frac{g(1)}{(p − 1)!}d^{p−1} + lower order terms

Since g(1) ≠ 0 by assumption, this completes the proof of (b), hence of the Polynomial Theorem.

Definition: The polynomial H_{M}(d) associated to a finitely generated graded A-module M is called the Hilbert polynomial of M.

We’ll give another proof of the fact that h_{M}(d) is a polynomial for d >> 0, partly as an excuse for introducing the following important basis for the Zmodule consisting of all integer-valued polynomials:

H : Z → Z

Lemma 3.4: If H : Z → Z is a polynomial of degree p (in Q[x]), then :

H(d) = a_{0} \binom{d}{p} + a_{1} \binom{d}{p − 1} + ... + a_{p−1} \binom{d}{1} + a_{p}

where the a_{i} ∈ Z are uniquely determined. Moreover,

H(d + 1) − H(d) = a_{0} \binom{d}{p − 1} + a_{1} \binom{d}{p − 2} + ... + a_{p−1}.

Proof: The \binom{d}{p−i} are of different degrees, hence linearly independent. This proves uniqueness of the a_{i}. One computes the a_{i} by evaluating H(d) at d = 0, 1, ..., p (that’s p + 1 points!). The last statement is an identity of binomial coefficients, which is easily checked.

Alternate proof that h_{M} is a polynomial in large degrees: Using the exact sequence (∗) and induction on the number of variables again, we see that

h_{M}(d + 1) − h_{M}(d) = H_{K}(d) − H_{N}(d)

for all sufficiently large d. If we let p = deg(H_{K} − H_{N}), then from Lemma 3.4, we can find a polynomial H_{M} of degree p + 1 which has has the property that for all d,

H_{M}(d + 1) − H_{M}(d) = H_{K}(d) − H_{N}(d)

Moreover, all the coefficients of H_{M} are determined (in the canonical expansion of the Lemma) except for the constant term, which we set so that H_{M}(d) = h_{M}(d) for some (large!) d. Then h_{M}(d) = H_{M}(d) for all sufficiently large d, as desired.

Definition: If X ⊂ P_{k}^{n} is a projective variety, then the Hilbert polynomial of X is by definition the Hilbert polynomial of k[X] as a k[x_{0}, ..., x_{n}]-module.

Some Motivation: A lot of geometric information about X ⊂ P_{k}^{n} is encoded in its Hilbert polynomial. For example, we will eventually prove that if k is algebraically closed and

H_{X}(d) = \frac{a_{m}}{m!}d^{m} + lower order (where a_{m} ≠ 0), then:

(i) m is equal to the dimension of X, and

(ii) a “general” projective subspace Λ ⊂ P_{k}^{n} of codimension m intersects X in exactly a_{m} points (we will call this the degree of X).

Here is another observation about the Hilbert polynomial:

Lemma 3.5: Let M be a finitely generated graded A-module and suppose that F ∈ A_{e} has the property that 0 ∈ M is the only element annihilated by F. Let N = M/F M. Then:

deg(H_{N}) = deg(H_{M}) − 1

and if H_{M}(d) = \frac{a_{0}}{m!}d^{m} + lower order, then H_{N} = \frac{ea_{0}}{(m−1)!}d^{m−1} + lower order.

Proof: Since F only annihilates 0, the map M(−e) → M given by multiplication by F is injective. Thus, from the exact sequence of graded A-modules:

0 → M(−e) →^{m_{F}} M → N → 0

we see that H_{N}(d) = H_{M}(d) − H_{M}(d − e), which is a polynomial of degree exactly deg(H_{M}) − 1. The leading term is computed by applying Lemma 3.4 to the telescoping sum:

H_{M}(d) − H_{M}(d − e) = (H_{M}(d) − H_{M}(d − 1)) + (H_{M}(d − 1) − H_{M}(d − 2)) + ...

Definition: (a) An ideal I ⊂ A is called a complete intersection ideal if there are generators F_{1}, ..., F_{r} ∈ I with the property that each F_{i} is not a zero divisor in A/⟨F_{1}, ..., F_{i−1}⟩.

(b) A variety X ⊂ A_{k}^{n} is called a complete intersection if I(X) is a complete intersection ideal.

(c) A variety X ⊂ P_{k}^{n} is called a complete intersection if I(X) can be generated by homogeneous elements as in (a).

Observation: If X ⊂ P_{k}^{n} is a complete intersection, with generators F_{i} of degree d_{i} satisfying (a), then the Hilbert polynomial of X is of the form:

H_{X}(d) = \frac{d_{1} ⋯ d_{r}}{(n − r)!}d^{n−r} + lower order terms

Proof: Repeated use of Lemma 3.5.

Third Homework Assignment

1. On the function V from ideals in k[x_{1}, ..., x_{n}] to algebraic subsets of A_{k}^{n}: (a) Show that V (I) ∪ V (J) = V (IJ) (where IJ is the product of ideals).

(b) Show that ∩_{s∈S} V (I_{s}) = V (∑_{s∈S} I_{s}).

(c) Find an infinite union of algebraic subsets which is not algebraic.

2. More on V and I:

(a) If X ⊂ A_{k}^{n} is an algebraic subset, prove that V (I(X)) = X.

(b) If I ⊂ k[x_{1}, ..., x_{n}] is a geometric ideal, prove that I(V (I)) = I.

(c) Prove Craw’s converse to Lemma 1.1. Namely, if V is an algebraic set and I = I(V) is a prime (geometric) ideal, then V is irreducible.

3. Prove the exercises preceding the Polynomial Theorem.

4. Let’s compute some Hilbert functions:

(a) Suppose that m ≤ n and p_{1}, ..., p_{m} ∈ P_{k}^{n} span a projective subspace of dimension m − 1 (the maximum possible!). Let Z = {p_{1}, ..., p_{m}} and let I = I(Z). Prove that h_{A/I}(0) = 1 and h_{A/I}(d) = m for all d > 0. In other words, we match the Hilbert polynomial starting from degree 1.

(b) At the other extreme, suppose p_{1}, ..., p_{m} ∈ P_{k}^{n} are distinct (no condition on m) and only span a projective line. Prove that h_{A/I}(d) does not match the Hilbert polynomial until d = m − 1. In fact, write down the Hilbert function of A/I(!)

5. The twisted cubic and rational normal curves: let ι_{3} : P_{k}^{1} → P_{k}^{3} be defined by ι_{3}(s : t) = (s^{3} : s^{2}t : st^{2} : t^{3}) together with compatible map ι_{3}^{∗} : k[x_{0}, x_{1}, x_{2}, x_{3}] → k[s, t]. The image, which we’ll denote by C, is called the twisted cubic. Assume k is infinite, and let I = I(C).

(a) Find I_{2} explicitly. Prove that C is a variety of dimension 1.

(b) Calculate the Hilbert function of C.

(c) Prove that I(C) is not a complete intersection ideal.

(d) Do (a)-(c) for ι_{d} : P_{k}^{1} → P_{k}^{d} given by ι_{d}(s : t) = (s^{d} : s^{d−1}t : ... : t^{d}), and the image, (in case d ≥ 3) which is called the rational normal curve.