IDEAL AS AN INTERSECTION

OF ZERO–DIMENSIONAL IDEALS

AND THE NOETHER EXPONENT

Witold Jarnicki, Liam O’Carroll and Tadeusz Winiarski^{1}

IMUJ PREPRINT 2000/25

Abstract: The main goal of this paper is to present a method of expressing a given ideal I in the polynomial ring 𝕂[X_{1}, …, X_{n}] as an intersection of zero–dimensional ideals. As an application, we get an elementary proof for some cases of the Kollár estimation of the Noether exponent of a polynomial ideal presented in [6], [7]. Moreover, an outline of the effective algorithm is given.

1. Introduction

Let 𝕂 be an algebraically closed field. It is well known that any radical ideal I can be expressed as I = ⋂_{P ∈V (I)} 𝔪_{P}, where 𝔪_{P} is the maximal ideal corresponding to P. A natural question arises whether such an intersection is possible for an arbitrary ideal I, i.e. if we can attach an 𝔪_{P} –primary ideal 𝔄_{P} to each P ∈ V (I) such that I = ⋂_{P ∈V (I)} 𝔄_{P}.

In this paper a positive answer to this question is given. Using primary decomposition, we reduce the problem to the case where I is primary. This case can be done easily and effectively, so finding the family {𝔄_{P} } is as difficult as is a primary decomposition. The proof of the primary case is based on the theory of Gröbner bases.

As an application we present a simple proof of Kollár’s Noether exponent estimate for ideals in the polynomial ring of one and two variables and for ideals without embedded primary components.

1Research partially supported by KBN grant 2 P03A 001 15

2. Notation

Let 𝕂 be an algebraically closed field. For a given ideal I, V (I) denotes its zero set in 𝕂^{n}.

3. Zero–dimensional case

Let I = Id(f_{1}, ⋯, f_{m}) be an ideal in the polynomial ring 𝕂[X] = 𝕂[X_{1}, ⋯, X_{n}] generated by {f_{1}, …, f_{m}} ⊂ 𝕂[X]. For a given point P ∈ 𝕂^{n} we define

(1)  M_{P} := {α ∈ ℕ^{n} : (X − P)^{α} ∈ IO_{P} },
(2)  D_{P} := ℕ^{n} \ M_{P} .

Observe that

α ∈ M_{P} ⟹ α + ℕ^{n} ⊂ M_{P} ⟹ M_{P} = ⋃_{α∈M_{P}}(α + ℕ^{n}).

One can prove that there exists a unique finite set α^{(1)}, ⋯, α^{(s)} ∈ M_{P} such that

(3)  α^{(i)} ∉ (α^{(j)} + ℕ^{n}), for i ≠ j,
(4)  M_{P} = ⋃_{j=1}^{s}(α^{(j)} + ℕ^{n}).

Lemma 1. I is zero-dimensional if and only if #D_{P} (I) < +∞ for every P ∈ V (I).

Proof. It suffices to prove that P is an isolated point of V (I) if and only if #D_{P} (I) < +∞. Take a P ∈ V (I). We may assume that P = 0. Observe that any of the conditions implies that for every j = 1, …, n, there exists a nonzero polynomial f_{j} ∈ I of the form f_{j} = x_{j}^{k_{j}}g_{j}(x) with g_{j}(0) ≠ 0 and k_{j} ≥ 1. On the other hand, this fact implies both that P is an isolated point of V (I) and that (0, ⋯, k_{j}, ⋯, 0) ∈ M_{0}(I) for j = 1, ⋯, n, which proves that D_{0}(I) is finite. □

Definition 2. For a given isolated point P ∈ V (I), d_{P} = d_{P} (I) := 1 + max{|α| : α ∈ D_{P} (I)} is defined to be the d–multiplicity of I at P.

Remark 3. d_{P} (I) = min{k ∈ ℕ : 𝔪^{k} ⊂ I}, where 𝔪 = Id(X − P) is the ideal corresponding to the point P.

Proposition 4. The following hold

(1) P ∉ V (I) ⟺ M_{P} = ℕ^{n}.

(2) If P is an isolated point of V (I) and Q_{P} is the primary component of I with associated prime I(P), then

M_{P} (I) = M_{P} (Q_{P}) and D_{P} (I) = D_{P} (Q_{P}).

(3) P is an isolated point of V (I) if and only if #D_{P} (I) < +∞.

Proof. To prove (1) it is enough to observe that if P ∉ V (I) then 1 ∈ IO_{P}.

Without loss of generality we may assume that P = 0. Since 0 is an isolated point of V (I), Q_{0} does not depend on a primary decomposition of I (see e.g. [1], Thm. 8.56). Thus I = Q_{0} ∩ J_{0}, where J_{0} contains the rest of primary components. Since Q_{0}O_{0} = IO_{0}, M_{0}(Q_{0}) = M_{0}(I). To prove the opposite inclusion, take β ∈ M_{0}(I). Then there exists f = x^{β}(1 + ∑_{α>0} a_{α}X^{α}) ∈ I. Therefore f ∈ Q_{0}, and finally x^{β} ∈ Q_{0}O_{0}.

Because of (2) we may assume that I is zero–dimensional. Applying Lemma 1 finishes the proof. □

Effective construction of D_{P}. Again, it is enough to consider the case P = 0. Let J_{α} = I : Id(X^{α}). Observe that

(5) M_{0}(I) = {α ∈ ℕ^{n} : ∃g ∈ 𝕂[X] : g(0) ≠ 0, X^{α}g ∈ I} = {α ∈ ℕ^{n} : ∃g ∈ J_{α}, g(0) ≠ 0}.

Now it suffices to compute the Gröbner basis G_{α} = {g_{α}^{(1)}, ⋯, g_{α}^{(s_{α})}} of J_{α} (see e.g.

[3] or [1]) for “all” α (because of (1) and Lemma 1 the computation ends after a finite number of steps) and check whether there exists j ∈ {1, …, s_{α}} such that g_{α}^{(j)}(0) ≠ 0.

Theorem 5. If I is a zero–dimensional ideal, then d(I) := max_{P ∈V (I)} d_{P} (I) is the Noether exponent N(I) = min{k ∈ ℕ : (rad(I))^{k} ⊂ I}.

Proof. The proof that d(I) ≥ N(I) follows directly from the fact that for every P ∈ V (I) we have 𝔪^{d_{P} (I)} ⊂ IO_{P}. To prove the opposite, one can assume that 0 ∈ V (I), d_{0}(I) = d(I). Take an α ∈ D_{0} such that |α| = d_{0}(I)−1, and g ∈ 𝕂[X] such that g(0) ≠ 0 and g(P) = 0 for all P ∈ V (I) \ {0}. Observe that X_{j}g(X) ∈ rad(I), j = 1, …, n, but (rad(I))^{d_{0}(I)−1} ∋ X^{α}(g(X))^{|α|} ∉ I. □

Lemma 6. Let I be a zero–dimensional ideal. Then d(I) ≤ dim 𝕂[X]/I.

Proof. Let V (I) = {P_{1}, …, P_{s}} and let Q_{1}, …, Q_{s} be the primary decomposition of I such that V (Q_{i}) = P_{i}. Since I ⊂ Q_{i}, d_{P_{i}}(I) = d_{P_{i}}(Q_{i}) (by Proposition 4) and dim 𝕂[X]/Q_{i} ≤ dim 𝕂[X]/I, it suffices to prove the case s = 1 and P := P_{1} = 0.

Let l = dim 𝕂[X]/I. To end the proof, it is enough to show that for any a = (a_{1}, …, a_{n}) ∈ 𝕂^{n}, (a_{1}X_{1} + ⋯ + a_{n}X_{n})^{l} ∈ IO_{0}. Fix an a and let T be a linear isomorphism such that T (X_{1}) = a_{1}X_{1} + ⋯ + a_{n}X_{n}. Let T^{∗}I ∩ 𝕂[X_{1}] = Id(f). Since I is primary we may take f = X_{1}^{k}. Observe that

k = dim 𝕂[X_{1}]/(X_{1}^{k}) ≤ dim 𝕂[X]/T^{∗}I = dim 𝕂[X]/I = l

— this implies that X_{1}^{l} ∈ T^{∗}I. □

Lemma 7. Let I = Id(f_{1}, …, f_{m}) be an ideal, where the f_{i} are non–zero polynomials, and let 1 ≤ k ≤ n. Then there exist linear forms L_{j} ∈ L(𝕂^{m}, 𝕂), j = 2, …, k such that for I_{k} = Id(f_{1}, L_{2} ∘ f, …, L_{k} ∘ f) where f denotes the m–tuple f_{1}, …, f_{m}, the components of V (I_{k}) not contained in V (I) are at most n − k–dimensional.

Proof. The case k = 1 is trivial. Fix a k ≥ 2 and suppose than the forms L_{2}, …, L_{k−1} are constructed. Let V_{1}, …, V_{s} be components of V (I_{k}) not contained in V (I). For each j = 1, …, s, there exist P_{j} ∈ V_{j} such that the sequence f_{1}(P_{j}), …, f_{m}(P_{j}) has at least one non–zero element. Let H_{j} = {α = (α_{1}, …, α_{m}) ∈ 𝕂^{m} : α_{1}f_{1}(P_{j})+…+ α_{m}f_{m}(P_{j}) ≠ 0}, j = 1, …, s. Sine the H_{j} are Zariski–open, the set H := ∩_{j=1}^{s}H_{j} ≠ ∅. Take α ∈ H and let L_{k}(Y_{1}, …, Y_{m}) = α_{1}Y_{1} + … + α_{m}Y_{m}. Since V (L_{k} ∘ f) intersects each V_{j} properly, the dimension decreases. □

Corollary 8. Let I = Id(f_{1}, …, f_{m}) be an ideal and suppose the d_{i} := deg f_{i} form a non–increasing sequence. Then there exists an ideal J ⊂ I such that all components of V (J) not contained in V (I) are zero–dimensional, J = Id(g_{1}, …, g_{n}), deg g_{n} = deg f_{m} and deg g_{i} ≤ deg f_{i} for i = 1, …, n − 1, where if n > m, set f_{(}1) = ⋯ = f_{n−m−1} = 0.

Proof. If is enough to renumber the generators and apply Lemma 7 followed by Gaussian elimination of the forms L_{i}. □

Theorem 9. Let I be as in Corollary 8. Then N(I) ≤ d_{1}d_{2} · · · · · d_{n−1}d_{m}.

Proof. Let J be as in Corollary 8. Applying Lemma 5, Lemma 6 and Bezout’s theorem, we get

N(I) = d(I) = dim 𝕂[X]/I ≤ dim 𝕂[X]/J ≤ g_{1}g_{2} · · · · · g_{n} ≤ f_{1}f_{2} ⋯ f_{n−1}f_{m}.

□

4. The case of one and two variables

For the ring of polynomials of one variable all ideals are zero–dimensional.

Theorem 10. The estimate is true for ideals in the ring of polynomials of two variables.

Proof. Take an ideal I = Id(f_{1}, …, f_{m}) as in Corollary 8 and assume that I is one–dimensional. Let g_{1}, …, g_{s} be irreducible polynomials corresponding to the hypersurfaces contained in V (I). Let r_{1}, …, r_{s} be such that g := g_{1}^{r_{1}} · · · · · g_{s}^{r_{s}} = GCD(f_{1}, …, f_{m}). Put f̃_{i} := f_{i}/g, i = 1, …, m. Observe that J := Id(f̃_{1}, …, f̃_{m}) is zero–dimensional.

Put d = (d_{1} − deg g)(d_{m} − deg g) + max{r_{1}, r_{2}, …, r_{s}} ≤ d_{1}d_{m} and let p_{1}, …, p_{d} ∈ rad(I). Obviously p_{1}, …, p_{d} ∈ rad(J) and, consequently, p_{1} · · · · · p_{d} ∈ I. □

Remark 11. The above technique can be used to “remove” components of codimension one during the effective calculation of the Noether exponent.

5. Higher–dimensional case

The main goal of this part is to present a given ideal I as an intersection of primary ideals and then to use induction. The proof does not work for ideals with embedded primary components.

Proposition 12. Let I = Id(f_{1}, …, f_{m}) be a primary ideal and let k ∈ ℕ such that for every y = (y_{1}, …, y_{k}) ∈ 𝕂^{k} the ideal I_{y} = Id(f_{1}(y, Z), …, f_{m}(y, Z)) in the ring 𝕂[Z] = 𝕂[Z_{1}, …, Z_{n−k}] is proper and zero–dimensional. Let f ∈ 𝕂[X]. Then the following conditions are equivalent, writing X = Y ∪ Z in an obvious notation:

(1) f ∈ I,

(2) f ∈ ⋂_{y∈𝕂^{k}}(I + Id(Y − y)),

(3) ∀y ∈ 𝕂^{k} : f_{y} = f(y, Z) ∈ I_{y},

(4) there exists a nonempty Zariski–open set U ⊂ 𝕂^{k} such that ∀y ∈ U : f_{y} ∈ I_{y}.

Observe that the theorem is not true without the assumption that I is primary. For example, take I = Id(Y Z, Z^{2}) = Id(Z) ∩ Id(Y, Z^{2}), k = 1 and f = Z. Then f satisfies condition (4) with U = 𝕂 \ {0}, but f ∉ I.

Proof. The implications 1⟹2⟹3⟹4 are trivial. To prove 4⟹1 suppose that G = (g_{1}, …, g_{s}), g_{i} ∈ 𝕂[Y][Z] is the comprehensive Gröbner basis ([10], see also [1]) of I for parameters y ∈ U. Observe that for y from a Zariski–open set U′ ⊂ U ⊂ 𝕂^{k} the division of f(y, Z) by (g_{1}(y, Z), …, g_{s}(y, Z)) is conducted the same way (i.e. before each step the multidegree of all the polynomials involved do not depend on y). Since ∀y ∈ 𝕂^{k}, f_{y} ∈ I_{y}, the remainders of the divisions are 0. Let q_{1}, …, q_{s} ∈ 𝕂(Y)[Z] be such that f(y, Z) = ∑_{i=1}^{s} q_{i}(y, Z)g_{i}(y, Z) for y ∈ U′. Multiplying the equation by the common denominator s(Y) of coefficients of all q_{i} we get s(Y)f(Y, Z) = ∑_{i=1}^{s} r_{i}(Y, Z)g_{i}(Y, Z), where r_{i} ∈ 𝕂[Y][Z]. This implies that s(Y)f(Y, Z) ∈ I. Since I is primary and I ∩ 𝕂[Y] = {0}, we get f ∈ I. □

Theorem 13. The estimate is true for ideals without embedded primary components.

Proof. We apply induction on the number of variables. The cases n = 1 and n = 2 are already solved.

Take n ≥ 3 and an I = Id(f_{1}, …, f_{m}) as in Corollary 8 and let d := d_{1}d_{2}·· · ··d_{n−1}d_{m}. Let Q_{1} ∩ ⋯ ∩ Q_{s} be a primary decomposition of I. Observe that for a generic linear isomorphism T, each of the components T^{∗}Q_{1}, …, T^{∗}Q_{s} of T^{∗}I satisfy the hypotheses of Proposition 12. It suffices to prove that for any p_{1}, …, p_{d} ∈ rad(T^{∗}I), p_{1} · · · · · p_{d} ∈ T^{∗}Q_{i} for any i = 1, …, s.

Let Q be a primary component of T^{∗}I. If Q is zero–dimensional then it is an isolated component and (rad Q)^{d} ⊂ Q since the multiplicity of Q does not exceed d.

Assume now that k := dim Q > 0. Let U ⊂ 𝕂^{k} be such that, for y ∈ U, T^{∗}I_{y} has no embedded primary components. Fix y ∈ U. Since (p_{1})_{y}, …, (p_{d})_{y} ∈ rad(T^{∗}I_{y}) and T^{∗}I_{y} has no embedded primary components, we get (p_{1} · · · · · p_{d})_{y} ∈ T^{∗}I_{y} ⊂ Q_{y} by the inductive hypothesis. Applying Proposition 12 ends the proof.

□

6. Reducing the number of generators

Lemma 14. Let I, J and Q be ideals such that V (I) ∪ V (Q) = V (J) and V (I) ∩ V (Q) = ∅. Fix d ∈ ℕ. If rad(J)^{d} ⊂ J then rad(I)^{d} ⊂ I.

Proof. Let I = Q_{1} ∩ … ∩ Q_{k} be a primary decomposition. For j = 1, …, k there exists a polynomial h_{j} ∈ Q \ rad(Q_{j}). Let P_{j} ∈ V (Q_{j}) be such that h_{j}(P_{j}) ≠ 0. Using the construction in 7 we get a polynomial h ∈ Q such that h(P_{j}) ≠ 0 for j = 1, …, k.

Take p_{1}, …, p_{d} ∈ rad(I). Observe that hp_{1}, …, hp_{d} ∈ rad(J). It follows that h^{d}p_{1} · · · · · p_{d} ∈ J. Assume that p_{1} · · · · · p_{d} ∉ Q_{j} for some j ∈ {1, …, k}. Since Q_{j} is primary, h^{dn} ∈ Q_{j} for some n ∈ ℕ. It follows that h ∈ rad(Q_{j}) — contradiction. This proves that rad(I)^{d} ⊂ I. □

Corollary 15. To prove the Kollár estimate it is enough to consider the case of n generators.

Proof. It suffices to apply Corollary 8, take Q corresponding to the components of V (J) not contained in V (I) and then apply Lemma 14. □

7. Closing remarks

(1) Let A be a Noetherian ring. Then we note that:

0_{A} is an intersection of ideals which are powers of maximal ideals (and so are zero-dimensional).

For consider a ∈ A\{0_{A}} and let I = 0 : a. Then I ⊂ M, for some maximal ideal M. By Krull’s Intersection Theorem, there exists n ∈ ℕ such that a/1 ∉ M_{M}^{n}. Hence a ∉ M^{n} and the result follows.

(2) Let A be an excellent (or indeed J-2) ring. Then by a general version of Zariski’s Main Lemma on holomorphic functions (see [5], [4]), 0_{A} is an intersection of ideals of the form 𝔪^{e}, where 𝔪 is a maximal ideal and e is the maximum of the Noether exponents of the primary components in a primary decomposition of 0_{A}.

(3) We remark that Proposition 12 supplies a proof using comprehensive Gröbner bases of the following striking result, which can be regarded as a ‘Nullstellensatz with Normalization’:

Let A be an affine ring over the field 𝕂 with 0_{A} a primary ideal. Via Noether Normalization, write A = 𝕂[Y_{1}, ..., Y_{k}, z_{1}, ..., z_{n−k}] where Y_{1}, ..., Y_{k} are algebraically independent and A is integral over 𝕂[Y_{1}, ..., Y_{k}]. Then if U ⊂ 𝕂^{k} is a non-empty Zariski-open set, ∩_{y∈U}Id( Y − y).A = 0_{A}.

In this connection, we note the following proof of Theorem 13 in the unmixed case that avoids this result (for which it would be of interest to have a ‘classical’ proof).

Alternative proof of Theorem 13 in the unmixed case:

Let I be an ideal in 𝕂[X_{1}, ..., X_{n}] with primary decomposition I = Q_{1}∩...∩Q_{s}, with ht I = ht Q_{i}, i = 1, ..., s. Set P_{i} = rad(Q_{i}), i = 1, ..., s. Let A denote 𝕂[X_{1}, ..., X_{n}]/I, and for each i let 𝔭_{i} denote P_{i}/I. Via Noether Normalization, we have an integral extension B ⊂ A, with B a polynomial ring. By the basic properties of integral extensions, 𝔭_{i} ∩ B = 0, i = 1, ..., s. Hence, setting S = B\{0}, S consists of nonzerodivisors in A. Then S^{−1}A is a zero-dimensional affine ring integral over 𝕃, the quotient field of B, and the Noether exponent of 0_{A} is the same as the Noether exponent of 0_{S^{−1}A}. Moreover, the defining ideal of S^{−1}A arises from I by a linear (even triangular) transformation of variables, so the degrees of the generators get no worse. Hence we reduce to the zero-dimensional case.

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Witold Jarnicki Instytut Matematyki Uniwersytetu Jagiellońskiego Reymonta 4 30–059 Kraków Poland e–mail: wmj@im.uj.edu.pl

Liam O’Carroll Department of Mathematics University of Edinburgh King’s Buildings Edinburgh EH9 3JZ Scotland e–mail: loc@maths.ed.ac.uk

Tadeusz Winiarski Instytut Matematyki Uniwersytetu Jagiellońskiego Reymonta 4 30–059 Kraków Poland e–mail: winiarsk@im.uj.edu.pl