aa r X i v : . [ m a t h . A C ] D ec POWERS ARE GOLOD
J ¨URGEN HERZOG
Abstract.
Let I be a proper graded ideal in a positively graded polynomial ring S over a field of characteristic 0. In this note it is shown that S/I k is Golod forall k ≥ Introduction
Let ( R, m ) be a Noetherian local ring with residue class field K , or a standardgraded K -algebra with graded maximal ideal m . The formal power series P R ( t ) = P i ≥ dim K Tor Ri ( R/ m , R/ m ) t i is called the Poincar´e series of R . Though the ringis Noetherian, the Poincar´e series of R is in general not a rational function. Thefirst example that showed that P R ( t ) may not rational was given by Anik [1]. Inthe meantime more such examples are known, see [13] and its references. On theother hand, Serre showed that P R ( t ) is coefficientwise bounded above by the rationalseries (1 + t ) n − t P i ≥ dim K H i ( x ; R ) t i , where x = x , . . . , x n is a minimal system of generators of m and where H i ( x ; R )denotes the i th Koszul homology of the sequence x .The ring R is called Golod , if P R ( t ) coincides with this upper bound given bySerre. There is also a relative version of Golodness which is defined for local ho-momorphisms as an obvious extension of the above concept of Golod rings. Werefer the reader for details regarding Golod rings and Golod homomorphism to thesurvey article [3] by Avramov. Here we just want to quote the following result ofLevin [12] which says that for any Noetherian local ring ( R, m ), the canonical map R → R/ m k is a Golod homomorphism for all k ≫
0. It is natural to ask whetherin this statement m could be replaced by any other proper ideal of I . Some veryrecent results indicate that this question may have a positive answer. In fact, in[11] it is shown that if R is regular, then for any proper ideal I ⊂ R the residueclass ring R/I k is Golod for k ≫
0, which, since R is regular, is equivalent to sayingthat the residue class map R → R/I k is a Golod homomorphism for k ≫
0. Buthow big k has to be chosen to make sure that R/I k is Golod? In the case that R is the polynomial ring and I is a proper monomial ideal, the surprising answer isthat R/I k is Golod for all k ≥
2, as has been shown by Fakhari and Welker in [5].
Mathematics Subject Classification.
Key words and phrases.
Powers of ideals, Golod rings, Koszul cycles.The paper was written while the author was visiting MSRI at Berkeley. He thanks for thesupport, the hospitality and the inspiring atmosphere of this institution. he authors show even more: if I and J a proper monomial ideals, then R/IJ isGolod. Computational evidence as well as a result of Huneke [10] which says thatfor a regular local ring R , the residue class ring R/IJ is never Gorenstein, unless I and J are principal ideals, suggest that R/IJ is Golod for any two proper ideals
I, J in a local ring (or graded ideals in graded ring). Indeed, being Golod impliesin particular that the Koszul homology H ( x ; R ) admits trivial multiplication, whilefor a Gorenstein ring, by a result of Avramov and Golod [2], the multiplication mapinduces for all i a non-degenerate pairing H i ( x ; R ) × H p − i ( x ; R ) → H p ( x : R ) where p is the top non-vanishing homology of the Koszul homology. In the case that I and J are not necessarily monomial ideals, it is only known that R/IJ is Golod if IJ = I ∩ J , see [9].In the present note we prove Theorem 1.
Let K be a field of characteristic and S = K [ x , . . . , x n ] the gradedpolynomial ring over K with deg x i = a i > for i = 1 , . . . , n . Let I ⊂ S be a gradedideal different from S . Then S/I k is Golod for all k ≥ . It should be noted that this theorem does not imply the result of Fakhari andWelker, since here we consider only powers and not products of ideals, and moreoverwe have to require that the base field is of characteristic 0. Actually as the proofwill show, it is enough to require in Theorem 1 that the characteristic of K is bigenough compared with the shifts in the graded free resolution of the ideal.1. Proof of Theorem 1
Let J ⊂ S K [ x , . . . , x n ] be a graded ideal different from S and set R = S/J .We denote by ( K ( R ) , ∂ ) the Koszul complex of R with respect to the sequence x = x , . . . , x n . Let Z ( R ), B ( R ) and H ( R ) denote the module of cycles, boundariesand the homology of K ( R ).Golod [6] showed that Serre’s upper bound for the Poincar´e series is reached ifand only if all Massey operations of R vanish. By definition, this is the case (see[4, Def. 5.5 and 5.6]), if for each subset S of homogeneous elements of L ni =1 H i ( R )there exists a function γ , which is defined on the set of finite sequences of elementsfrom S with values in m ⊕ L ni =1 K i ( R ), subject to the following conditions:(G1) if h ∈ S , then γ ( h ) ∈ Z ( R ) and h = [ γ ( h )];(G2) if h , . . . , h m is a sequence in S with m >
1, then ∂γ ( h , . . . , h m ) = m − X ℓ =1 γ ( h , . . . , h ℓ ) γ ( h ℓ +1 , . . . , h m ) , where ¯ a = ( − i +1 a for a ∈ K i ( R ).Note that (G2) implies, that γ ( h ) γ ( h ) is a boundary for all h , h ∈ S (whichin particular implies the Koszul homology of Golod ring has trivial multiplication).Suppose now that for each S we can choose a functions γ such that γ ( h ) γ ( h ) isnot only a boundary but that γ ( h ) γ ( h ) = 0 for all h , h ∈ S . Then obviously we ay set γ ( h , . . . h r ) = 0 for all r ≥
2, so that in this case (G2) is satisfied and R isGolod.The proof of Theorem 1 follows, once we have shown that γ can be chosen that γ ( h ) γ ( h ) = 0 for all h , h ∈ S , in the case that J = I k where I is a graded idealand k ≥
2. For the proof of this fact we use the following result from [8]: Let0 → F p → F n − → · · · → F → F → S/J be the graded minimal free S -resolution of S/J , and for each i let f , . . . , f ib i ahomogeneous basis of F i . Let ϕ i : F i → F i − denote the chain maps in the resolution,and let ϕ i ( f ij ) = b i − X k =1 α ( i ) jk f i − ,k , where the α ( i ) jk are homogeneous polynomials.In [8, Corollary 2] it is shown that for all l = 1 , . . . , p the elements X ≤ i
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