The Strong Lefschetz Property and Simple Extensions

Stanley [4] showed that monomial complete intersections have the strong Lefschetz property. Extending this result we show that a simple extension of an Artinian Gorenstein algebra with the strong Lefschetz property has again the strong Lefschetz property. Introduction Let K be a field, A be a standard graded Artinian K-algebra and a ∈ A a homogeneous form of degree k. The element a is called a Lefschetz element if for all integers i the K-linear map a : Ai → Ai+k (induced by multiplication with a) has maximal rank. One says that A has the weak Lefschetz property if there exists a Lefschetz element a ∈ A of degree 1. An element a ∈ A1 for which all powers a r are Lefschetz is called a strong Lefschetz element, and A is said to have the strong Lefschetz property if A admits a strong Lefschetz element. Note that the set of Lefschetz elements a ∈ A1 form a Zariski open subset of A1. The same holds true for the set of strong Lefschetz elements. Assuming that the characteristic of K is zero and the defining ideal of A is generated by generic forms, it is conjectured that A has the strong Lefschetz property. Thus in particular, A = K[x1, . . . , xn]/(f1, . . . , fn) should have the strong Lefschetz property for generic forms f1, . . . , fn. Note that such an algebra is an Artinian complete intersection. It is expected that any standard graded Artinian complete intersection over a base field of characteristic 0 has the strong Lefschetz property. Stanley [4] and later J. Watanabe [5] proved this in case A is a monomial complete intersection. Stanley used the Hard Lefschetz Theorem to prove this result, while Watanabe used the representation theory of the Lie algebra sl(2). As a main result of this paper we prove the following Theorem: Let K be a field of characteristic 0, A be a standard graded Artinian Gorenstein K-algebra having the strong Lefschetz property, and let f ∈ A[x] be a monic homogeneous polynomial. Then the algebra B = A[x]/(f) has the strong Lefschetz property. The proof only uses techniques from linear algebra. The result implies in particular Stanley’s theorem. More generally it implies that a complete intersection 1991 Mathematics Subject Classification. 13D40, 13P10, 13C40, 13D07. The second author was mainly supported by Marie Curie Intra-European Fellowships MEIFCT-2003-501046 and partially supported by CNCSIS and the Ceres program 4-131/2004 of the Romanian Ministery of Education and Research. 1 K[x1, . . . , xn]/(f1, . . . , fn) with fi ∈ K[x1, . . . , xi] for i = 1, . . . , n has the strong Lefschetz property. 1. The proof of the main theorem Let A be a standard graded K-algebra and I ⊂ A a graded ideal. For convenience we will say that a ∈ A is Lefschetz for A/I if the residue class a + I is a Lefschetz element of A/I. In the proof of the main theorem we shall use the following two lemmata. Lemma 1.1. Let A be a standard graded K-algebra, f, g ∈ A homogeneous elements which are nonzero divisors on A. Then f is Lefschetz for A/(g) if and only if g is Lefschetz for A/(f). Proof. Consider the long exact sequence for Koszul homology (see [1, Corollary 1.6.13]) · · · → H1(g;A) → H1(f, g;A) → H0(g;A) f −−−→ H0(g;A) → H0(f, g;A) → 0. Since g is a non-zerodivisor on A this yields the exact sequence 0 −→ H1(f, g;A) −→ A/(g) f −−−→ A/(g) → H0(f, g;A) → 0. Similarly we obtain an exact sequence 0 −→ H1(f, g;A) −→ A/(f) g −−−→ A/(f) → H0(f, g;A) → 0. Comparing this two exact sequences, the assertion follows. Lemma 1.2. Let K be field of characteristic 0, A a standard graded Artinian Kalgebra with strong Lefschetz property and f ∈ A[y] a monic homogeneous polynomial. Then for any strong Lefschetz element a ∈ A1 there exists a non-zero element c ∈ K such that f(a/c) is a Lefschetz element of A. Proof. Let f = y + a1y d−1 + · · · + ad, and s = max{i : Ai 6= 0}. We may assume that d ≤ s because otherwise the statement is trivial. For c ∈ K we set fc = y+ ∑d i=1 c aiy . Let a ∈ A1 be a strong Lefschetz element. Then a d is a Lefschetz element, that is, for all i the multiplication map f0(a) : Ai → Ai+d has maximal rank. Fix i ≤ s−d and K-bases of the nonzero K-vector spaces Ai and Ai+d, and let Dc be the matrix describing the K-linear map fc(a) : Ai → Ai+d. Note that the entries of Dc are polynomial expressions in c with coefficients in K. Now Pc(a) has maximal rank if and only if one maximal minor Mj(Dc) of Dc does not vanish. In particular, Mj0(D0) 6= 0 for some j0. Since Mj0(Dc) is a (non-zero) polynomial expression in c with coefficients in K, there exist only finitely many c ∈ K such that Mj0(Dc) = 0. Thus, since K is infinite, we have Mj0(Dc) 6= 0 for infinitely many c ∈ K, and so fc(a) : Ai → Ai+d has maximal rank for infinitely many c ∈ K. Since A has only finitely many non-zero components, we can therefore find c ∈ K, c 6= 0 such that fc(a) has maximal rank for all i. Then a/c ∈ A1 has the desired property, since f(a/c) = fc(a)/c . 2 Now we are ready to begin with the proof of the main theorem. Let A be a standard graded Artinian Gorenstein K-algebra having the strong Lefschetz property. In a first step we will prove: suppose C = A[x]/(x) has the strong Lefschetz property for all r > 1, then B = A[x]/(f) has the strong Lefschetz property for any monic homogeneous polynomial f ∈ A[x]. Let Ur ⊂ B1 be the Zariski open set of elements b ∈ B1 for which b r is a Lefschetz element. If Ur 6= ∅ for all r ≥ 1, then the finite intersection U = ⋂ r Ur is non-empty, as well, and any b ∈ U is then a strong Lefschetz element. Thus it suffices to show that for each r ≥ 1 there exists an element br ∈ B1 such that b r r is a Lefschetz element. By Lemma 1.2 we may choose an element a ∈ A1 such that f(a) is a Lefschetz element of A. It follows that f(x) is Lefschetz for A[x]/(a − x). Thus by Lemma 1.1, the element b1 = a− x is Lefschetz for B. In case r > 1, we may view f(y) as a polynomial in C[y] where C = A[x]/(x). By our assumption C has a strong Lefschetz element. Now Lemma 1.2 implies that we can find a strong Lefschetz element c ∈ C1 such that f(c) is a Lefschetz element of C. Since the strong Lefschetz elements form a nonempty Zariski open set in C1, we may assume that c = a + λx with a ∈ A1 and λ ∈ K, λ 6= 0. Applying the substitution x 7→ br = λ (x− a) it follows that f(x) is Lefschetz for A[x]/brr. Thus by Lemma 1.2, the element brr is Lefschetz for A[x]/(f). In order to complete the proof of the theorem it remains to be shown that if A is a standard graded Artinian Gorenstein K-algebra having the strong Lefschetz property, then A[x]/(x) has the strong Lefschetz property. We use Lemma 1.1 and show instead that if a ∈ A1 is a strong Lefschetz element, then for all k the element x is Lefschetz for B = A[x]/(a + x). In B we have x = − k−1