Ordinary and symbolic powers are Golod

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Advances in Mathematics 246 (2013) 89–99 www.elsevier.com/locate/aim

Ordinary and symbolic powers are Golod✩ J¨urgen Herzog a,∗ , Craig Huneke b a Fachbereich Mathematik, Universit¨at Duisburg–Essen, Campus Essen, 45117 Essen, Germany b Department of Mathematics, University of Virginia, 1404 University Ave, Charlottesville, VA 22903-2600,

United States Received 7 January 2013; accepted 1 July 2013

Communicated by Henning Krause

Abstract Let S be a positively graded polynomial ring over a field of characteristic 0, and I ⊂ S a proper graded ideal. In this note it is shown that S/I is Golod if ∂(I )2 ⊂ I . Here ∂(I ) denotes the ideal generated by all the partial derivatives of elements of I . We apply this result to find large classes of Golod ideals, including powers, symbolic powers, and saturations of ideals. c 2013 Elsevier Inc. All rights reserved. ⃝ MSC: 13A02; 13D40 Keywords: Powers of ideals; Golod rings; Koszul cycles

0. Introduction Let (R, m) be a Noetherian local ring with residue class field K , or a standard graded K -algebra with graded maximal ideal m. The formal power series  PR (t) = dim K ToriR (R/m, R/m)t i i≥0

✩ Part of the paper was written while the authors were visiting MSRI at Berkeley. They wish to acknowledge the support, the hospitality and the inspiring atmosphere of this institution. The second author was partially supported by NSF grant 1259142. ∗ Corresponding author. E-mail addresses: [email protected], [email protected] (J. Herzog), [email protected] (C. Huneke).

c 2013 Elsevier Inc. All rights reserved. 0001-8708/$ - see front matter ⃝ http://dx.doi.org/10.1016/j.aim.2013.07.002

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is called the Poincar´e series of R. Though the ring is Noetherian, in general the Poincar´e series of R is not a rational function. The first example that showed that PR (t) is not necessarily rational was given by Anick [1]. In the meantime more such examples are known; see [19] and its references. On the other hand, Serre showed that PR (t) is coefficientwise bounded above by the rational series 1−t



(1 + t)n , dim K Hi (x; R)t i

i≥1

where x = x1 , . . . , xn is a minimal system of generators of m and Hi (x; R) denotes the ith Koszul homology of the sequence x. The ring R is called Golod, if PR (t) coincides with this upper bound given by Serre. Obviously the residue field of a Golod ring has a rational Poincar´e series. Remarkably, the Poincar´e series of all finitely generated modules over a Golod ring are also rational, sharing a common denominator. There is also a relative version of Golodness which is defined for local homomorphisms as an obvious extension of the above concept of Golod rings. We refer the reader for details regarding Golod rings and Golod homomorphism to the survey article [2] by Avramov. Here we just want to quote the following result of Levin [15] which says that for any Noetherian local ring (R, m), the canonical map R → R/mk is a Golod homomorphism for all k ≫ 0. It is natural to ask whether in this statement m could be replaced by any other proper ideal of R. Some very recent results indicate that this question may have a positive answer. In fact, in [12] it is shown that if R is regular, then for any proper ideal I ⊂ R the residue class ring R/I k is Golod for k ≫ 0, which, since R is regular, is equivalent to saying that the residue class map R → R/I k is a Golod homomorphism for k ≫ 0. But how big must k 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 is that R/I k is Golod for all k ≥ 2, as has been shown by Fakhari and Welker in [20]. The authors show even more: if I and J are proper monomial ideals, then R/I J is Golod. Computational evidence suggests that R/I J is Golod for any two proper ideals I, J in a local ring (or graded ideals in a graded ring). This is consistent with a result of Huneke [14] which says that for an unramified regular local ring R, the residue class ring R/I J is never Gorenstein, unless I and J are principal ideals: indeed, being Golod implies in particular that the Koszul homology H (x; R) admits trivial multiplication, while for a Gorenstein ring, by a result of Avramov and Golod [3], the multiplication map induces for all i a non-degenerate pairing Hi (x; R) × H p−i (x; R) → H p (x : R) where p is the place of 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/I J is Golod if I J = I ∩ J ; see [11]. In the present note we consider graded ideals in the graded polynomial ring S = K [x1 , . . . , xn ] over a field K of characteristic 0 with deg xi = ai > 0 for i = 1, . . . , n. The main result of Section 1 is given in Theorem 1.1 which says that S/I is Golod if ∂(I )2 ⊂ I . Here ∂(I ) denotes the ideal which is generated by all elements f ∈ I . We call an ideal strongly Golod if ∂(I )2 ⊂ I . In Section 2 it is shown that the class of strongly Golod ideals is closed under several important ideal operations like products, intersections and certain colon ideals. In particular it is shown that for any k ≥ 2, the kth power of a graded ideal, as well as its kth symbolic power and its kth saturated power, is strongly Golod. We also prove the surprising fact that all the primary components of a graded ideal I which belong to the minimal prime ideals of I are strongly Golod, if I is so. Even more, we prove that every strongly Golod ideal has a primary decomposition in which each primary component is strongly Golod. We are also able to

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prove that the integral closure of a strongly Golod monomial ideal is again strongly Golod; in particular, the integral closure of I k is always strongly Golod if I is monomial and k ≥ 2. We do not know if this last assertion is true for general ideals; however, we are able to prove that for all k ≥ n + 1, the integral closure of I k is strongly Golod for an arbitrary homogeneous ideal I . It should be noted that our results, though quite general, do not imply the result of Fakhari and Welker concerning products of monomial ideals. One can easily find products of monomial ideals which are not strongly Golod. Moreover we have to require that the base field is of characteristic 0. Actually as the proof will show, it is enough to require in Theorem 1.1 that the characteristic of K is big enough compared with the shifts in the graded free resolution of the ideal. A preliminary version of this paper by the first author was posted on ArXiv, proving that powers are Golod. After some discussions with the second author, this final version emerged. 1. A differential condition for Golodness Let K be a field of characteristic 0, and S = K [x1 , . . . , xn ] the graded polynomial ring over K with deg xi = ai > 0 for i = 1, . . . , n, and let I ⊂ S be a graded ideal different from S. We denote by ∂(I ) the ideal which is generated by the partial derivatives ∂ f /∂ xi with f ∈ I and i = 1, . . . , n. Theorem 1.1. Suppose that ∂(I )2 ⊂ I . Then S/I is Golod. Proof. We set R = S/I , and denote by K (R) the Koszul complex of R with respect to the sequence x = x1 , . . . , xn . Furthermore we denote by Z (R), B(R) and H (R) the module of cycles, boundaries and the homology of K (R). Golod [8] showed that Serre’s upper bound for the Poincar´e series is reached if and only if all Massey operations of R vanish. By definition, this is the case (see [4, Def. 5.5 and 5.6]), if n for each subset S of homogeneous elements of i=1 Hi (R) there exists a function n γ , which is defined on the set of finite sequences of elements from S with values in m ⊕ i=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 1 , . . . , h m is a sequence in S with m > 1, then dγ (h 1 , . . . , h m ) =

m−1 

γ (h 1 , . . . , h ℓ )γ (h ℓ+1 , . . . , h m ),

ℓ=1

where a¯ = (−1)i+1 a for a ∈ K i (R) and where d denotes the differential on the Koszul complex. Note that (G2) implies, that γ (h 1 )γ (h 2 ) is a boundary for all h 1 , h 2 ∈ S (which in particular implies that the Koszul homology of a Golod ring has trivial multiplication). Suppose now that for each S we can choose a functions γ such that γ (h 1 )γ (h 2 ) is not only a boundary but that γ (h 1 )γ (h 2 ) = 0 for all h 1 , h 2 ∈ S. Then obviously we may set γ (h 1 , . . . , h r ) = 0 for all r ≥ 2, so that in this case (G2) is satisfied and R is Golod. The proof of Theorem 1.1 follows, once we have shown that γ can be chosen that γ (h 1 )γ (h 2 ) = 0 for all h 1 , h 2 ∈ S. For the proof of this fact we use the following result from [9]: let 0 → F p → F p−1 → · · · → F1 → F0 → S/I → 0 be the graded minimal free S-resolution of S/I , and for each i let f i1 , . . . , f ibi be a homogeneous basis of Fi . Let ϕi : Fi → Fi−1 denote the chain maps in the resolution, and let

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ϕi ( f i j ) =

b i−1

(i)

α jk f i−1,k ,

k=1 (i) α jk

where the are homogeneous polynomials. In [9, Corollary 2] it is shown that for all l = 1, . . . , p the elements 

ai1 ai2 · · · ail

1≤i 1
bl−1  j2 =1

···

b1  jl =1

(l)

c j1 ,..., jl

(l−1)

(1)

∂(α j1 , j2 , α j2 , j3 , . . . , α jl ,1 ) ∂(xi1 , . . . , xil )

ei1 ∧ · · · ∧ eil ,

j1 = 1, . . . , bl are cycles of K (R) whose homology classes form a K -basis of Hl (R). Here the ∂(g1 ,...,gl ) of the polynomials g1 , . . . , gl with respect to the variables xi1 , . . . , xil is Jacobian ∂(x i 1 ,...,xil ) defined to be the determinant of the matrix (∂gi /∂ xi j ) i=1,...,l . j=1,...,l

Thus we see that a K -basis of Hl (R) is given by cycles whose coefficients are linear (i) combinations of Jacobians determined by the entries α jk of the matrices describing the resolution of S/I . The coefficients c j1 ,..., jl which appear in these formulas are rational numbers determined (i) by the degrees of the α jk , and the elements ei1 ∧ · · · ∧ eil form the natural R-basis of the free l n module K l (R) = ( i=1 Rei ). From this result it follows that any homology class of Hl (R) can be represented by a cycle whose coefficients are linear combinations of Jacobians of the form (l)

(l−1)

(1)

∂(α j1 , j2 α j2 , j3 , . . . , α jl ,1 ) ∂(xi1 , . . . , xil )

.

(1)

We choose such representatives for the elements of the set S. Thus we may choose the map γ in such a way that it assigns to each element of S a cycle whose coefficients are linear combinations of Jacobians as in (1). (1) The elements α jl ,1 , generate I . Thus it follows that γ (h) ∈ Z (R) ∩ (∂ I )K (R) for all h ∈ S, and hence our assumption, ∂(I )2 ⊂ I , implies that γ (h 1 )γ (h 2 ) = 0 for any two elements h 1 , h 2 ∈ S.  2. Classes of graded Golod ideals We keep the notation and the assumptions of Section 1 and apply Theorem 1.1 to exhibit new classes of Golod rings. It is customary to call a graded ideal I ⊂ S a Golod ideal, if S/I is Golod. For convenience, we call a graded ideal I ⊂ S as well as the factor ring R = S/I strongly Golod, if ∂(I )2 ⊂ I . As we have shown in Theorem 1.1, any strongly Golod ideal is Golod. Remark 2.1. In the case that S is standard graded, the property of an ideal I ⊂ S to be strongly Golod, does not depend on the chosen coordinates. Indeed let y1 , . . . , yn be linear forms in S such that S = K [y1 , . . . , yn ]. We denote by ∂ ′ (I ) the ideal which is generated by the partial ′ (I )2 ⊂ I . derivatives ∂ f /∂ yi with f ∈ I and i = 1, . . . , n. Then ∂(I )2 ⊂ I if and only if ∂ n ′ 2 2 By symmetry it suffices to show that ∂ (I ) ⊂ I if ∂(I ) ⊂ I . Let xi = j=1 ai j y j for i = 1, . . . , n and ϕ: S1 → S1 be the linear automorphism defined as   n n   ϕ(y1 , . . . , yn ) = a1i yi , . . . , ani y j . i=1

i=1

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Then by the chain rule for composite functions we have  n  n  ∂ϕ j ∂( f ◦ ϕ)  ∂f ∂f = ◦ϕ = a ji ◦ ϕ. ∂ yi ∂x j ∂ yi ∂x j j=1 j=1 Note that f ∈ K [x1 , . . . , xn ] as a polynomial in the variables y1 , . . . , yn is just f ◦ ϕ, and ∂ f /∂ yi is understood to be ∂( f ◦ ϕ)/∂ yi . Thus the desired conclusion follows from the above chain rule identity. Corollary 2.2. Let R = S/I be a standard graded strongly Golod K -algebra, and let x ∈ R be a linear form. Then R/x R is again strongly Golod. Proof. By Remark 2.1 we may change the coordinates, and hence may assume that x is the ¯ I¯ where S¯ = K [x1 , . . . , xn−1 ] and I¯ is generated residue class of xn modulo I . Then R/x R ∼ = S/ ¯ by the polynomials f (x1 , . . . , xn−1 , 0) ∈ S with f ∈ I . Let f 0 , g0 ∈ I¯. Then there exist f 1 , g1 ∈ S¯ such that f = f 0 + f 1 xn and g = g0 + g1 xn belong to I . By assumption, (∂ f /∂ xi )(∂g/∂ x j ) ∈ I for all i and j. In particular,    ∂ f0 ∂g0 ∂ f1 ∂g1 + xn + xn ∈ I ∂ xi ∂ xi ∂x j ∂x j for all i, j < n. It follows that (∂ f 0 /∂ xi )(∂g0 /∂ x j ) ∈ I¯ for all i, j < n, as desired.



Suppose the linear form x is a non-zerodivisor on R. It is a known fact, and easy to prove, that R is Golod if and only if R/x R is Golod. As we just have seen, R/x R is strongly Golod for any linear form x ∈ R provided R is strongly Golod. On the other hand, even if x is a non-zerodivisor on R and R/x R is strongly Golod, then R need not to be strongly Golod (but of course Golod), as the following simple example shows: the hypersurface ring R = K [x, y]/(x y) is not strongly Golod, x − y is a non-zerodivisor on R and R/(x − y)R ∼ = K [x]/(x 2 ) is strongly Golod. (k) As usual we denote by I the symbolic powers and by Ik the saturated powers of I . Recall (k) that I = t≥1 I k : L t , where L is the intersection of all associated, non-minimal prime ideals  of I k , while Ik = I k : mt where m is the graded maximal ideal of S. t≥1

Since S is Noetherian, there exists an integer t0 such that I (k) = I k : L t for all t ≥ t0 . In particular, if we let J = L t for some t ≥ t0 , then I (k) = I : J = I : J 2 . The next result shows that strongly Golod ideals behave well with respect to several important ideal operations; in particular combining the various parts of the theorem yields a quite large class of Golod ideals. Theorem 2.3. Let I, J ⊂ S be graded ideals. Then the following hold: (a) (b) (c) (d) (e)

if I and J are strongly Golod, then I ∩ J and I J are strongly Golod; if I and J are strongly Golod and ∂(I )∂(J ) ⊂ I + J , then I + J is strongly Golod; if I is strongly Golod, J is arbitrary, and I : J = I : J 2 , then I : J is strongly Golod; I k , I (k) and Ik are strongly Golod for all k ≥ 2; if I is strongly Golod and I ⊂ J , then I J is strongly Golod.

Proof. To simplify notation we write ∂ f to mean anyone of the partials ∂ f /∂ xi . (a) Let f, g ∈ I ∩ J . Since I and J are strongly Golod, it follows that (∂ f )(∂g) ∈ I and (∂ f )(∂g) ∈ J , and hence (∂ f )(∂g) ∈ I ∩ J . This shows that I ∩ J is strongly Golod.

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Due to the product rule for partial derivatives it follows that ∂(I J ) ⊂ ∂(I )J + I ∂(J ). This implies that ∂(I J )2 ⊂ ∂(I )2 J 2 + ∂(I )∂(J )I J + ∂(J )2 I 2 . Obviously, the middle term is contained in I J , while ∂(I )2 J 2 ⊂ I J 2 ⊂ I J , since I is strongly Golod. Similarly, ∂(J )2 I 2 ⊂ I J . This shows that ∂(I J )2 ⊂ I J , and proves that I J is strongly Golod. (b) is proved in the same manner as (a). (c) Let f, g ∈ I : J . Then for all h 1 , h 2 ∈ J one has f h 1 ∈ I and gh 2 ∈ I . This implies that (∂ f )h 1 + f ∂h 1 ∈ ∂(I ) and (∂g)h 2 + g∂h 2 ∈ ∂(I ). Thus ((∂ f )h 1 + f ∂h 1 )((∂g)h 2 + g∂h 2 ) = (∂ f )(∂g)h 1 h 2 + (∂ f )(∂h 2 )gh 1 + (∂g)(∂h 1 ) f h 2 + f g(∂h 1 )(∂h 2 ) ∈ (∂ I )2 ⊂ I. Since gh 1 , f h 2 ∈ I , it follows that (∂ f )(∂h 2 )gh 1 + (∂g)(∂h 1 ) f h 2 ∈ I . Moreover, J f g(∂h 1 )(∂h 2 ) ⊂ I . Hence J (∂ f )(∂g)h 1 h 2 ⊂ I . Since h 1 , h 2 were arbitrary in J , it then follows that (∂(I : J ))2 ⊂ I : J 3 = I : J , as desired. (d) Let k ≥ 2. Then ∂(I k ) ⊂ I k−1 ∂(I ). It follows that ∂(I k )2 ⊂ I 2k−2 ∂(I )2 ⊂ I k . Thus I k is strongly Golod. As explained above, for appropriately chosen J , I (k) = I k : J = I k : J 2 . Now it follows from (c) that I (k) is strongly Golod. The same arguments show that Ik is strongly Golod. (e) As in part (a), we have that ∂(I J ) ⊂ ∂(I )J + I ∂(J ). This implies that ∂(I J )2 ⊂ ∂(I )2 J 2 + ∂(I )∂(J )I J + ∂(J )2 I 2 . Obviously, the middle term is contained in I J , while ∂(I )2 J 2 ⊂ I J 2 ⊂ I J , since I is strongly Golod. Finally, ∂(J )2 I 2 ⊂ I J since by assumption I ⊂ J . This shows that ∂(I J )2 ⊂ I J , and proves that I J is strongly Golod.  Corollary 2.4. Let P be a homogeneous prime ideal of S containing I , a strongly Golod ideal. Then I + P k is strongly Golod for all k ≥ 2. Proof. First note that ∂(I ) ⊂ P. If not, then ∂(I )2 will also not be in P, and therefore not in I , a contradiction. We also observe that ∂(P k ) ⊂ P k−1 . It follows that (∂ I )(∂ P k ) ⊂ P k ⊂ I + P k . Thus the assertion follows from Theorem 2.3(b) and (d) (we apply (d) to ensure that P k is strongly Golod).  Let R = S/I and n the graded maximal ideal of R, and suppose that I is strongly Golod. Then Corollary 2.4 implies that R/nk is Golod for all k ≥ 2. Also note that by the theorem of Zariski–Nagata (see, e.g., [18, p.143]), the fact that ∂(I ) is contained in every homogeneous prime containing I implies that I is in the second symbolic power of every such prime. In particular, self-radical ideals are never strongly Golod, though they may be Golod. Corollary 2.5. The (uniquely determined) primary components belonging to the minimal prime ideals of a strongly Golod ideal are strongly Golod. Proof. Let I be strongly Golod and P1 , . . . , Ps its minimal  prime ideals. Let Q i be the primary component of I with Ass(S/Q i ) = {Pi }, and set L i = j̸=i P j . Then there exists an integer r > 1 such that Q i = I : L ri = I : L i2r . It follows from Theorem 2.3(c) that Q i is strongly Golod. 

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Corollary 2.6. Every strongly Golod ideal has a primary decomposition with strongly Golod primary ideals. Proof. Let P be an associated prime of I . By Corollary 2.4, I + P k is strongly Golod for all k ≥ 2, and then by Corollary 2.5 the unique P-primary minimal component of I + P k is also strongly Golod. We denote this component by Pk . The Corollary now follows from the general fact that I = ∩Pk for large k, where the intersection is taken over all associated primes of I . To prove this, fix a primary decomposition of I , say I = ∩Q P , where Q P is P-primary, and the intersection runs over all associated primes of I . It suffices to prove that Pk ⊂ Q P for large k, since then I ⊂ ∩Pk ⊂ ∩Q P = I . To check that Pk ⊂ Q P , we may localize at P, since both of these ideals are P-primary. Then the claim is clear.  Statement (d) of Theorem 2.3 can be substantially generalized as follows. Theorem 2.7. Let I ⊂ S be a homogeneous ideal and suppose that J is a homogeneous ideal such that (I (k−1) )2 ⊂ J ⊂ I (k) . Then J is strongly Golod. In particular, if (I (k−1) )2 ⊂ I k for some k ≥ 2, then all homogeneous ideals J with I k ⊂ J ⊂ I (k) are strongly Golod. Any homogeneous ideal J with I 2 ⊂ J ⊂ I (2) is strongly Golod. Proof. We use a theorem of Zariski–Nagata [18, p.143] according to which f ∈ S belongs to I (k) if and only if all partials of f of order
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Proof. We first prove the following claim: let L be an arbitrary homogeneous ideal, k ≥ 2 an integer, and J another homogeneous ideal such that L k ⊂ J , and J 2 ⊂ L k+2 . We claim that J is strongly Golod. To prove this claim, let f ∈ J . Then f 2 ∈ L k+2 by assumption. After differentiating twice, it follows that ∂( f )∂( f ) ∈ J + L k ⊂ J. Since (∂(J ))2 is generated by the squares of generators of ∂ J , it follows that J is strongly Golod. Apply this claim with J = I k where k ≥ n + 1. Obviously I k ⊂ J , so it suffices to prove that 2 J ⊂ I k+2 . However, J 2 ⊂ I 2k ⊂ I 2k−n+1 by the Brianc¸on–Skoda theorem; see [16]. Since k ≥ n + 1, we have that 2k − n + 1 ≥ k + 2, which proves that I n is strongly Golod.  It is worth remarking that if I is strongly Golod, and J another homogeneous ideal such that I ⊂ J , and J 2 ⊂ I 2 , then J is strongly Golod. The proof is similar to the proof of the first claim in the theorem above. Next we show that there is yet another closure, the Ratliff–Rush closure, which also behaves well with respect to the strongly Golod property. Proposition 2.12. Let I be a homogeneous ideal in S, a standard graded polynomial ring over a field K of characteristic 0. Assume that I is strongly Golod. Then the Ratliff–Rush closure of I , namely the ideal J := ∪k (I k+1 : I k ), is also strongly Golod. Proof. Let f ∈ J . Then there exists an integer k such that I k f ⊂ I k+1 . It follows that I k ∂ f ⊂ I k−1 f ∂(I ) + I k ∂(I ). Multiplying by I gives us that ∂ f ∈ (I k+1 ∂(I ) : I k+1 ). Hence if f, g ∈ J , then ∂( f )∂(g) ∈ ((I k+1 ∂(I ) : I k+1 ))2 . Multiplying by I 2k+2 gives that I 2k+2 ∂( f )∂(g) ⊂ (I k+1 ∂(I ))2 ⊂ I 2k+3 since I is strongly Golod. It follows that ∂(J )2 ⊂ J , proving that J is strongly Golod.  3. Monomial ideals It is immediate that a monomial ideal I is strongly Golod, if for all minimal monomial generators u, v ∈ I and all integers i and j such that xi |u and x j |v it follows that uv/xi x j ∈ I . Fakhari and Welker [20] showed that I J is Golod for any two proper monomial ideals. However a product of proper monomial ideals need not to be strongly Golod as the following example shows: let I = (x, y) and J = (z). Then I J = (x z, yz) and (x z)(yz)/x y = z 2 does not belong to I J . We use this characterization to show the following. Proposition 3.1. Let I be a strongly Golod monomial ideal. Then I¯ is strongly Golod. In particular, for any monomial ideal I , the ideals I k are strongly Golod for all k ≥ 2. Proof. The integral closure of the monomial ideal I is again a monomial ideal, and a monomial u ∈ S belongs to I¯ if and only if there exists an integer r > 0 such that u r ∈ I r ; see [21, Proposition 1.4.2] and the remarks preceding this proposition. Let u, v ∈ I¯ monomials and suppose that xi |u and x j |v. There exist integers s, t > 0 such that u s ∈ I s and v t ∈ I t . We may assume that both, s and t, are even. We observe that for a monomial w with xk |w and wr ∈ I r it follows that (w/xk )r ∈ I r/2 if r is even. Indeed, wr = m 1 · · · m r with m i ∈ I . Let di be the highest exponent of a power of xi that divides m i . We may assume that di = 0 for i = 1, . . . , a, di = 1 for i = a + 1, . . . , b and

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di > 1 for i > b. Then (w/xk )r = (m 1 · · · m a )((m a+1 /xk ) · · · (m b /xk ))((m b+1 · · · m r )/xkd ),  where d = ri=b+1 di . Since I is strongly Golod, it follows from this presentation of (w/xk )r that (w/xk )r ∈ I a I ⌊b/2⌋ = I ⌊(2a+b)/2⌋ , where ⌊c⌋ denotes the lower integer part of the real number c. Note that r = d + b and that 2(r − a − b) ≤ d. This implies that 2a ≥ d, and hence 2a + b ≥ d + b = r , which implies that (w/xk )r ∈ I r/2 , as desired. Now it follows that (uv/xi x j )s+t = (u/xi )t+s (v/x j )t+s ∈ I s+t . This shows that uv/xi x j ∈ I¯, and proves that I¯ is strongly Golod.



A case of monomial ideals to consider when the hypotheses of Theorem 2.7 are satisfied is the following: let G be a finite simple graph on the vertex set [n]. A vertex cover of G is a subset C ⊂ [n] such that C ∩ {i, j} ̸= ∅  for all edges {i, j} of G. The vertex cover ideal I of G is the ideal generated by all monomials i∈C xi ⊂ S = K [x1 , . . . , xn ] where C is a vertex cover of G. It is obvious that  I = (xi , x j ), {i, j}∈E(G)

where E(G) denotes the set of edges of G. It follows that I (k) =



k {i, j}∈E(G) (x i , x j )

for all k.

Proposition 3.2. Let I be the vertex cover ideal of a finite simple graph G, and suppose that (I (2) )2 ⊂ I 3 . Then (I (k−1) )2 ⊂ I k for all k. In particular, every homogeneous ideal J between I k and I (k) is strongly Golod.  In [10, Theorem 5.1] it is shown that the graded S-algebra k≥0 I (k) t k ⊂ S[t] is generated in degrees 1 and 2. Thus the proposition follows from the more general proposition.  (k) k Proposition 3.3. Let I be an unmixed ideal of height two, and suppose that k≥0 I t is (2) 2 3 (k−1) 2 k generated in degrees 1 and 2, and (I ) ⊂ I . Then (I ) ⊂ I for all k.  Proof. The assumption that k≥0 I (k) t k ⊂ S[t] is generated in degrees 1 and 2 is equivalent to the statement that for all p ≥ 0 one has I (2 p+1) = I (I (2) ) p

and

I (2 p) = (I (2) ) p .

Since I is an ideal of height 2 it follows by a theorem of Ein, Lazarsfeld, and Smith [7] that I (2 p) ⊂ I p for all p ≥ 0. This implies that (I (2) ) p ⊂ I p for all p, since (I (2) ) p ⊂ I (2 p) . We want to prove that (I (k−1) )2 ⊂ I k . We consider the two cases where k is even or odd. We have that (I (2 p+1) )2 = I 2 ((I (2) ) p )2 ⊂ I 2 (I p )2 = I 2 p+2 . Next assume that k is even. Then (I (2 p) )2 = (I (2) )2 p . Thus (I (2 p) )2 = ((I (2) )2 ) p ⊂ (I 3 ) p ⊂ I 2 p+1 . Here we used that (I (2) )2 ⊂ I 3 .



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Proposition 3.3 applied to the vertex cover ideal I of G is trivial if G does not contain an odd cycle, in other words, if G is bipartite, because in this case it is known [10, Theorem 5.1(b)] that the symbolic and ordinary powers of I coincide. The first non-trivial case is that when G is an odd cycle, say on the vertex set [n]. In that case it is known, and easy to see, that I is generated by the (n+1)/2 monomials u i = j=i xi+2 j , where for simplicity of notation xi = xi−n if i > n. It is also n (2) 2 known that I = I +(u), where u = i=1 xi ; see for example [10, Proposition 5.3]. It follows that (I (2) )2 = I 4 + I 2 (u) + (u 2 ). Since u ∈ I we see that I 2 (u) ⊂ I 3 , and since u 1 u 2 u 3 divides u 2 we also have (u 2 ) ⊂ I 3 . So that altogether, (I (2) )2 ⊂ I 3 for the vertex cover ideal of any odd cycle. We do not know whether the same inclusion holds for any graph containing an odd cycle. Example 3.4. There exist squarefree monomial ideals I for which (I (2) )2 is not contained in I 3 . Indeed, let In,d be the ideal generated by all squarefree monomials of degree d in n variables, d+1 (2) 3 , because I 3 is generated in degree and let 2 < d < n. Then u = i=1 xi ∈ In,d , but u 2 ̸∈ In,d n,d 3d, while deg u 2 = 2(d + 1) < 3d. Note that a squarefree monomial ideal is never strongly Golod. We end this paper by introducing a squarefree version of our differential condition for Golodness. Let I be a monomial ideal. We denote by I sqf the ideal generated by all squarefree monomials u ∈ I . The squarefree monomial ideal I sqf is called the squarefree part of I . The kth squarefree power of a squarefree monomial ideal I is defined to be the ideal I [k] = (I k )sqf . We say that I is squarefree strongly Golod if ∂(I )[2] ⊂ I . Theorem 3.5. Any squarefree strongly Golod ideal is Golod. Proof. We use a result of Berglund and J¨ollenbeck [5, Theorem 3] according to which a monomial ideal I is Golod if and only if the multiplication on the Koszul homology H (S/I ) is trivial. It is enough to check the triviality of the multiplication on a K -basis of H (S/I ). Recall that there is an isomorphism Tor S (K , S/I ) ∼ = H (S/I ) of Zn -graded vector spaces. By a theorem of Hochster [13] we know that S/I has a minimal Zn -graded free resolution with squarefree shifts. This implies that Tor S (K , S/I ), and hence also H (S/I ), admits a K -basis whose generators have Zn -degrees which are (0, 1)-vectors. In particular, if a and b are elements of this basis with the property that supp(deg a) ∩ supp(deg b) ̸= ∅, then necessarily ab = 0. Here, for any vector c, supp(c) = {i: ci ̸= 0}. Let us now assume that supp(deg a) ∩ supp(deg b) = ∅. We know from the proof of Theorem 1.1 that a (respectively b) can be represented by a cycle z (respectively w) whose coefficients belong to ∂(I ). Each monomial among the coefficients of z has a Zn -degree which is coefficientwise less than or equal to deg a. A similar statement holds for w. Thus if u is one of the monomials among the coefficients of z and v one of the monomials among the coefficients of w, then supp(deg u) ∩ supp(deg v) = ∅. It follows that uv ∈ ∂(I )[2] ⊂ I , and this implies that ab = 0.  Note that any principal ideal generated by a squarefree monomial is squarefree strongly Golod, but not strongly Golod. Another class of squarefree strongly Golod ideals are the ideals In,d with d ≥ 2, considered in Example 3.4. Observe that In,d is the dth squarefree power of the graded maximal ideal (x1 , . . . , xn ). More generally, and in analogy to Theorem 2.3(d), we have the following. Proposition 3.6. Let I be a squarefree monomial ideal. Then I [k] is squarefree strongly Golod for all k ≥ 2.

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This proposition is an immediate consequence of Theorem 2.3(d) and the following result. Proposition 3.7. Let I be a strongly Golod monomial ideal. Then I sq f is squarefree strongly Golod. Proof. Since ∂(I sqf ) ⊂ ∂(I ), it follows that ∂(I sqf )[2] = (∂(I sqf )2 )sq f ⊂ (∂(I )2 )sqf ⊂ I sqf , as desired.



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