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DERIVATIONS PRESERVING a MONOMIAL IDEAL 1. Introduction

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DERIVATIONS PRESERVING a MONOMIAL IDEAL 1. Introduction PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 137, Number 9, September 2009, Pages 2935–2942 S 0002-9939(09)09922-5 Article electronically published on May 4, 2009 DERIVATIONS PRESERVING A MONOMIAL IDEAL YOHANNES TADESSE (Communicated by Bernd Ulrich) Abstract. Let I be a monomial ideal in a polynomial ring A = k[x1,...,xn] over a field k of characteristic 0, TA/k(I) be the module of I-preserving k- derivations on A and G be the n-dimensional algebraic torus on k. We compute the weight spaces of TA/k(I) considered as a representation of G.Usingthis, we show that TA/k(I) preserves the integral closure of I and the multiplier ideals of I. 1. Introduction Throughout this paper TA/k denotes the module of k-linear derivations of the polynomial ring A = k[x1,...,xn]overafieldk of characteristic 0. Let G = k∗ ×···×k∗ be the n-dimensional algebraic torus on k. There is an action of θ θ1 ··· θn · θ θ1 ··· θn · G on a monomial X = x1 xn defined by (t1,...,tn) X =(t1 tn ) Xθ, and we say that Xθ has weight θ.ThismakesA arepresentationofG and induces an action of G on TA/k which will make TA/k arepresentationsuchthat θ ∈ − ⊆ X ∂xj TA/k has weight (θ1,...,θj 1,...θn). Let I A be a monomial ideal, i.e. a G-invariant ideal. Let TA/k(I)={δ ∈ TA/k | δ(I) ⊆ I}⊂TA/k be the submodule of I-preserving derivations, which is a G-subrepresentation of TA/k. Our first result is a description of the weight spaces of TA/k(I). This implies in particular [1, Theorem 2.2.1] by Brumatti and Simis. Note that our use of the G-action significantly clarifies the structure of TA/k(I) and simplifies the proof. The action of G also gives a simple argument to the fact that the integral clo- sure I¯ is a monomial ideal; one may compare it to the more complicated proof ¯ in [7, Proposition 7.3.4]. For any ideal I,itisknownthatTA/k(I) ⊆ TA/k(I) [3, Theorem 3.2.2]. Again using the action of G and directly employing the integral equation of elements of I¯, we prove this inclusion when I is a monomial ideal. Let J (r · I) be the multiplier ideal of I for the rational number r>0; see [5]. Using Howald’s description [2] of J (r · I)whenI is a monomial ideal, we prove: Theorem 1.1. Let I be a monomial ideal and J (r · I) be any of its multiplier ideals. Then TA/k(I) ⊆ TA/k(J (r · I)). It is a useful fact that TA/k(I) preserves J (r · I), as this radically restricts its form. The proof of the inclusion for any ideal is indicated in [3]. Received by the editors November 25, 2008, and, in revised form, January 5, 2009. 2000 Mathematics Subject Classification. Primary 13A15, 13N15, 14Q99. Key words and phrases. Derivations, monomial ideals, multiplier ideals. c 2009 American Mathematical Society Reverts to public domain 28 years from publication 2935 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2936 YOHANNES TADESSE 2. The structure of TA/k(I) ∇ n ∇ ∇ The Lie algebra of the torus is G = j=1 k xj ,where xj = xj∂xj .The representation TA/k of G decomposes as (2.1) TA/k = A∇G ⊕ s, n where s = j=1 Axj ∂xj and Axj = k[x1,...,xˆj,...,xn]. Hence the representation is semi-simple. Since I is a monomial ideal, TA/k(I)isaG-subrepresentation of TA/k, hence semi-simple, having the decomposition (2.2) TA/k(I)=A∇G ⊕ s(I). Since all the weight spaces of s(I)=TA/k(I) ∩ s ⊆ s are 1-dimensional, we easily get the following lemma. ⊆ n ∈ Lemma 2.1. Let I A be a monomial ideal and δ = j=1 fj∂xj TA/k be a ∈ derivation where fj = ij mij A and mij are distinct monomial terms. Then ∈ ∈ δ TA/k(I) if and only if mij∂xj TA/k(I) for all i and j =1,...,n. { | θ ∈ }⊆Zn Put exp(I)= θ X I ≥0 and let (ej) denote the standard basis of Rn.Let{Xα1 ,...,Xαt } be the unique minimal generating set of I,where αi αi1 ··· αin X = x1 xn . Theorem 2.2. Consider the condition CI ∈ Zn − ∈ j (β): β ≥0,βj =0, and β + αi ej exp(I) for all i such that αij > 0. n We have s(I)= j=1 Ixj ∂xj ,where β |CI ⊆ Ixj =(X j (β)) Axj . β ∈ If βj =0andαij > 0 for all i =1, ..., t,thenX ∂xj / s(I). Indeed, for a − θ ∈ ≤ β θ β+θ ej ∈ monomial X I such that θj αij for all i, one has X ∂xj (X )=θjX / I. CI Hence j (β) never holds, and we then put Ixj =(0). β Proof. By Lemma 2.1 it suffices to determine when X ∂xj belongs to s(I), so in particular βj =0.SinceIxj =(0)ifαij > 0 for all i, we may assume there exists 1 ≤ k ≤ t such that αkj =0.Then β ∈ ⇔ − ∈ X Ixj β + αi ej exp(I) for all i such that αij > 0 ⇔ β αi ∈ X ∂xj (X ) I for all i =1,...,t ⇔ β ∈ X ∂xj s(I). n By [1] we have TA/k(I)= j=1[I :[I : xj]]∂xj , but the argument there does not profit on the torus action. We now show that [I :[I : xj]] = (xj)+Ixj .Note License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use DERIVATIONS PRESERVING A MONOMIAL IDEAL 2937 Xθ θ first that colon ideals of monomial ideals are monomial, and [I : xj]=( | X ∈ xj I and θj > 0). Therefore, θ β β X θ [I :[I : xj]] = (X | X · ∈ I for all X ∈ I such that θj > 0) xj β =(X | β + θ − ej ∈ exp(I) for all θ ∈ exp(I) such that θj > 0) β =(xj)+(X | βj =0andβ + θ − ej ∈ exp(I) for all θ ∈ exp(I) such that θj > 0) β =(xj)+(X | βj =0andβ + αi − ej ∈ exp(I) for all i such that αij > 0) =(xj)+Ixj . CI We assert that j (β) is equivalent to the following condition: ∈ Zn { − } ∩ Zn ⊆ (2.3) β ≥0, βj =0,and[ β ej +exp(I)] ≥0 exp(I). ⇒CI CI ⇒ It is easy to see that (2.3) j (β), so we prove only j (β) (2.3): If αij > 0for CI ≤ all i,then j (β)doesnotholdforanyβ. Now assume there exists 1 <k t such that (2.4) 0 = α1j = ···= αk−1,j <αkj ≤···≤αtj , ∈ − ∈ Zn and let θ exp(I). It is clear that θ + β ej / ≥0 if θj = 0; thus assume that θj > 0. If θj <αkj, then there exists some i =1,...,k− 1 such that θ = αi + γ with γj = θj > 0. Hence θ + β − ej = αi + β + γ − ej ∈ exp(I)sinceαi + β ∈ exp(I) − ∈ Zn ≥ ≥ and γ ej ≥0.Ifθj αkj, then there exists l k such that θ = αl + γ for ∈ Zn CI − − ∈ some γ ≥0. j (β) implies θ + β ej = γ + αl + β ej exp(I). Therefore, { − } ∩ Zn ⊆ [ β ej +exp(I)] ≥0 exp(I). It is interesting to see the structure of TA/k(I)whenA = k[x, y]. First, if I is principal, then TA/k(I) is either of the form A∇G, A∇G ⊕ Ay∂y or A∇G ⊕ Ax∂x. a1 b1 at bt Now assume I =(x y ,...,x y ) is non-principal with ai−1 <ai and bi−1 >bi for 1 <i≤ t. Then define the widths w(I)x and w(I)y of I in the direction of x and y by { − }t { − }t w(I)x =max ai ai−1 i=2 and w(I)y =max bi−1 bi i=2. This is illustrated by the following figure: y p p p p p p p p p p p p Xr p p p p p p p p p p p p (a1,b1) XX XXXz p p p p p p p p w(I)x x p p ∂yp rp p p p p p p p p p p AK p p p p p p yw(I)y ∂ A p p p p xp Arp p p p p p p p p p p (ai,bi) p rp p p p r (at,bt) x License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2938 YOHANNES TADESSE Corollary 2.3. Assume that A = k[x, y] and I =(xa1 yb1 ,...,xat ybt ) is a non- principal monomial ideal such that ai−1 <ai and bi−1 >bi for each 1 <i≤ t. Then ⎧ ⎪ (0) if a > 0 and b > 0 ⎨⎪ 1 t k[y]yw(I)y ∂ if a =0and b > 0 s(I)= x 1 t ⎪ w(I)x ⎩⎪ k[x]x ∂y if a1 > 0 and bt =0 w(I)y w(I)x k[y]y ∂x ⊕ k[x]x ∂y if a1 = bt =0. Proof. It suffices to prove the cases a1 =0andbt > 0 since the other cases are similar. l l a b a b y ∂x ∈ s(I) ⇔ y ∂x(x y ) ∈ I for all x y ∈ I ⇔ (0,l)+(a, b) − (1, 0) ∈ exp(I) for all (a, b) ∈ exp(I) such that a>0 ⇔ (ai − 1,l+ bi) ∈ exp(I) for all 1 <i≤ t ⇔ l ≥ bi−1 − bi for all 1 <i≤ t ⇔ l ≥ w(I)y.
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