Powers of sums and their associated primes
aa r X i v : . [ m a t h . A C ] J a n POWERS OF SUMS AND THEIR ASSOCIATED PRIMES
HOP D. NGUYEN AND QUANG HOA TRAN
Abstract.
Let
A, B be polynomial rings over a field k , and I ⊆ A, J ⊆ B proper homogeneous ideals. We analyze the the associated primes of powersof I + J ⊆ A ⊗ k B given the data on the summands. The associated primesof large enough powers of I + J are determined. We then answer positively aquestion about the persistence property of I + J in many new cases. Introduction
Inspired by work of Ratliff [19], Brodmann [2] proved that in any noetherian ring,the set of associated primes of powers of an ideal is eventually constant for largeenough power. Subsequent work by many researchers have shown that importantinvariants of powers of ideals, for example, the depth, the Castelnuovo–Mumfordregularity, also eventually stabilize in the same manner. For a recent survey onassociated primes of powers and related questions, we refer to Hoa’s paper [12].Our work is inspired by the afore-mentioned result of Brodmann, and recentstudies about powers of sums of ideals [8, 9, 18]. Let
A, B be standard gradedpolynomial rings over a field k , and I ⊆ A, J ⊆ B proper homogeneous ideals.Denote R = A ⊗ k B and I + J the ideal IR + JR . In [8, 9, 18], certain homologicalinvariants of powers of I + J , notably the depth and regularity, are computed interms of the corresponding invariants of powers of I and J . In particular, we haveexact formulas for depth R/ ( I + J ) n and reg R/ ( I + J ) n if either char k = 0, or I and J are both monomial ideals. It is therefore natural to ask: Is there an exact formulafor Ass( R/ ( I + J ) n ) in terms of the associated primes of powers of I and J ? The case n = 1 is simple and well-known: Using the fact that R/ ( I + J ) ∼ = ( A/I ) ⊗ k ( B/J ),we deduce ([8, Theorem 2.5]):Ass R/ ( I + J ) = [ p ∈ Ass A ( A/I ) q ∈ Ass B ( B/J ) Min R ( R/ p + q ) . One of our main results is the following partial answer to the last question.
Theorem 1.1 (Theorem 4.1) . Let I be a proper homogeneous ideal of A such that Ass(
A/I n ) = Ass( I n − /I n ) for all n ≥ . Let J be any proper homogeneous idealof B . Then for all n ≥ , there is an equality Ass R R ( I + J ) n = n [ i =1 [ p ∈ Ass A ( A/I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q ) . Mathematics Subject Classification.
Key words and phrases.
Associated prime; powers of ideals; persistence property.
If furthermore
Ass(
B/J n ) = Ass( J n − /J n ) for all n ≥ , then for all such n , thereis an equality Ass R R ( I + J ) n = n [ i =1 [ p ∈ Ass A ( A/I i ) q ∈ Ass B ( B/J n − i +1 ) Min R ( R/ p + q ) . The proof proceeds by filtering R/ ( I + J ) n using exact sequences with terms ofthe form M ⊗ k N , where M, N are modules over
A, B , respectively, and applyingthe formula for Ass R ( M ⊗ k N ).Concerning Theorem 1.1, the hypothesis Ass( A/I n ) = Ass( I n − /I n ) for all n ≥ I is a monomial ideal of A , or if char k = 0 anddim( A/I ) ≤ A/I n ) = Ass( I n − /I n ) holds for all I and all n ifdim A ≤
3. If char k = 2 and A has Krull dimension four, using the Buchsbaum–Eisenbud structure theory of perfect Gorenstein ideal of height three and work byKustin and Ulrich [15], we establish the equality Ass( A/I ) = Ass( I/I ) for all I ⊆ A (Theorem 3.4).Another motivation for this work is the so-called persistence property for asso-ciated primes. The ideal I is said to have the persistence property if Ass( A/I n ) ⊆ Ass(
A/I n +1 ). Ideals with this property abound, including for example completeintersections. The persistence property was considered by many people; see, e.g.,[5, 11, 14, 22]. As an application of Theorem 1.1, we prove that if I is a monomialideal satisfying the persistence property, and J is any ideal, then I + J also hasthe persistence property (Corollary 5.1). Moreover, we generalize previous workdue to I. Swanson and R. Walker [22] on this question: If I is an ideal such that I n +1 : I = I n for all n ≥
1, then for any ideal J of B , I + J has the persistenceproperty (see Corollary 5.1(ii)). In [22, Corollary 1.7], Swanson and Walker provethe same result under the stronger condition that I is normal. It remains an openquestion whether for any ideal I of A with the persistence property and any ideal J of B , the sum I + J has same property.The paper is structured as follows. In Section 3, we provide large classes ofideals I such that the equality Ass( A/I n ) = Ass( I n − /I n ) holds true for all n ≥
1. An unexpected outcome of this study is a counterexample to [1, Question 3.6], onthe vanishing of the map Tor Ai ( k, I n ) → Tor Ai ( k, I n − ). Namely in characteristic2, we construct a quadratic ideal I in A such that the natural map Tor A ∗ ( k, I ) → Tor A ∗ ( k, I ) is not zero (even though A/I is a Gorenstein artinian ring, see Example3.8). This example might be of independent interest. Using the results in Section3, we give a set-theoretic upper bound and a lower bound for Ass( R/ ( I + J ) n )(Theorem 4.1). Theorem 4.1 also gives an exact formula for the asymptotic primesof I + J without any condition on I and J . In the last section, we apply our resultsto the question on the persistence property raised by Swanson and Walker.2. Preliminaries
For standard notions and results in commutative algebra, we refer to the books[3, 4].Throughout the section, let A and B be two commutative Noetherian algebrasover a field k such that R = A ⊗ k B is also Noetherian. Let M and N be two SSOCIATED PRIMES OF POWERS 3 nonzero (not necessarily finitely generated) modules over A and B , resp. Denoteby Ass A M and Min A M the set of associated primes and minimal primes of M asan A -module, resp.By a filtration of ideals ( I n ) n ≥ in A , we mean the ideals I n , n ≥ I = A and I n +1 ⊆ I n for all n ≥
0. Let ( I n ) n ≥ and ( J n ) n ≥ be filtrations of ideals in A and B respectively. Consider the filtration ( W n ) n ≥ of A ⊗ k B given by W n = P ni =0 I i J n − i . The following result is useful for the discussionin Section 4. Proposition 2.1 ([8, Lemma 3.1, Proposition 3.3]) . For arbitrary ideals I ⊆ A and J ⊆ B , we have I ∩ J = IJ . Moreover with the above notation for filtrations,for any integer n ≥ , there is an isomorphism W n /W n +1 ∼ = n M i =0 (cid:0) I i /I i +1 ⊗ k J n − i /J n − i +1 (cid:1) . We recall the following description of the associated primes of tensor products;see also [21, Corollary 3.7].
Theorem 2.2 ([8, Theorem 2.5]) . Let M and N be non-zero modules over A and B , resp. Then there is an equality Ass R ( M ⊗ k N ) = [ p ∈ Ass A ( M ) q ∈ Ass B ( N ) Min R ( R/ p + q ) . The following simple lemma turns out to be useful in the sequel.
Lemma 2.3.
Assume that char k = 0 . Let A = k [ x , . . . , x r ] be a standard gradedpolynomial ring over k , and m its graded maximal ideal. Let I be proper homo-geneous ideal of A . Denote by ∂ ( I ) the ideal generated by partial derivatives ofelements in I . Then there is a containment I : m ⊆ ∂ ( I ) . In particular, I n : m ⊆ I n − for all n ≥ . If for some n ≥ , m ∈ Ass(
A/I n ) then m ∈ Ass( I n − /I n ) .Proof. Take f ∈ I : m . Then x i f ∈ I for every i = 1 , . . . , r . Taking partialderivatives, we get f + x i ( ∂f /∂x i ) ∈ ∂ ( I ). Summing up and using Euler’s formula,( r + deg f ) f ∈ ∂ ( I ). As char k = 0, this yields f ∈ ∂ ( I ), as claimed.The second assertion holds since by the product rule, ∂ ( I n ) ⊆ I n − .If m ∈ Ass(
A/I n ) then there exists an element a ∈ ( I n : m ) \ I n . Thus a ∈ I n − \ I n , so m ∈ Ass( I n − /I n ). (cid:3) The condition on the characteristic is indispensable: The inclusion I : m ⊆ I may fail in positive characteristic; see Example 3.8.The following lemma will be employed several times in the sequel. Denote bygr I ( A ) the associated graded ring of A with respect to the I -adic filtration. Lemma 2.4.
Let A be a noetherian ring, and I an ideal. Then the following areequivalent: (i) I n +1 : I = I n for all n ≥ , (ii) ( I n +1 : I ) ∩ I n − = I n for all n ≥ , (iii) depth gr I ( A ) > , (iv) I n = f I n for all n ≥ , where e I = S i ≥ ( I i +1 : I i ) denotes the Ratliff-Rushclosure of I . HOP D. NGUYEN AND QUANG HOA TRAN
If one of these equivalent conditions holds, then
Ass(
A/I n ) ⊆ Ass(
A/I n +1 ) for all n ≥ , namely I has the persistence property .Proof. Clearly (i) = ⇒ (ii). We prove that (ii) = ⇒ (i).Assume that ( I n +1 : I ) ∩ I n − = I n for all n ≥
1. We prove by induction on n ≥ I n : I = I n − .If n = 1, there is nothing to do. Assume that n ≥
2. By the induction hypothesis, I n : I ⊆ I n − : I = I n − . Hence I n : I = ( I n : I ) ∩ I n − = I n − , as desired.That (i) ⇐⇒ (iii) ⇐⇒ (iv) follows from [10, (1.2)] and [20, Remark 1.6]. (cid:3) Associated primes of quotients of consecutive powers
The following question is quite relevant to the task of finding the associatedprimes of powers of sums.
Question 3.1.
Let A be a standard graded polynomial ring over a field k (ofcharacteristic zero), and I a proper homogeneous ideal. Is it true thatAss( A/I n ) = Ass( I n − /I n ) for all n ≥ I ) the Rees algebraof I . The ideal I is said to be unmixed if it has no embedded primes. It is called normal if all of its powers are integrally closed ideals. Theorem 3.2.
Question 3.1 has a positive answer if either of the following condi-tions holds: (1) I is a monomial ideal. (2) depth gr I ( A ) ≥ . (3) depth Rees( I ) ≥ . (4) I is normal. (5) I n is unmixed for all n ≥ , e.g. I is generated by a regular sequence. (6) All the powers of I are primary, e.g. dim( A/I ) = 0 . (7) char k = 0 and dim( A/I ) ≤ . (8) char k = 0 and dim A ≤ .Proof. (1): See [17, Lemma 4.4].(2): By Lemma 2.4, since depth gr I ( A ) ≥ I n : I = I n − for all n ≥
1. Inducton n ≥ A/I n ) = Ass( I n − /I n ).Let I = ( f , . . . , f m ). For n ≥
2, as I n : I = I n − , the map I n − → I n − ⊕ · · · ⊕ I n − | {z } m times , a ( af , . . . , af m ) , induces an injection I n − I n − ֒ → (cid:18) I n − I n (cid:19) ⊕ m . Hence Ass(
A/I n − ) = Ass( I n − /I n − ) ⊆ Ass( I n − /I n ). The exact sequence0 → I n − /I n → A/I n → A/I n − → A/I n ) ⊆ Ass( I n − /I n ), which in turn implies the desired equality.Next we claim that (3) and (4) all imply (2). SSOCIATED PRIMES OF POWERS 5 (3) = ⇒ (2): This follows from a result of Huckaba and Marley [13, Corollary 3.12]which says that either gr I ( A ) is Cohen-Macaulay (and hence has depth A = dim A ),or depth gr I ( A ) = depth Rees( I ) − ⇒ (2): If I is normal, then I n : I = I n − for all n ≥
1. Hence we are doneby Lemma 2.4.(5): Take P ∈ Ass(
A/I n ), we show that P ∈ Ass( I n − /I n ). Since A/I n isunmixed, P ∈ Min(
A/I n ) = Min( I n − /I n ).Observe that (6) = ⇒ (5).(7): Because of (6), we can assume that dim( A/I ) = 1. Take P ∈ Ass(
A/I n ),we need to show that P ∈ Ass( I n − /I n ).If dim( A/P ) = 1, then as dim(
A/I ) = 1, P ∈ Min(
A/I n ). Arguing as for case(5), we get P ∈ Ass( I n − /I n ).If dim( A/P ) = 0, then P = m , the graded maximal ideal of A . Since m ∈ Ass(
A/I n ), by Lemma 2.3, m ∈ Ass( I n − /I n ).(8) It is harmless to assume that I = 0. If dim( A/I ) ≤ A/I ) ≥
2, then the hypothesis forces dim A = 3 and ht I = 1.Thus we can write I = xL where x is a form of degree at least 1, and L = R orht L ≥
2. The result is clear when L = R , so it remains to assume that L is properof height ≥
2. In particular dim(
A/L ) ≤
1, and by (7), for all n ≥ A/L n ) = Ass( L n − /L n ) . Take p ∈ Ass(
A/I n ). Since A/I n and I n − /I n have the same minimal primes, wecan assume ht p ≥
2. From the exact sequence0 → A/L n · x n −−→ A/I n → A/ ( x n ) → p ∈ Ass(
A/L n ). Thus p ∈ Ass( L n − /L n ). There is an exact sequence0 → L n − /L n · x n −−→ I n − /I n so p ∈ Ass( I n − /I n ), as claimed. This concludes the proof. (cid:3) Example . Here is an example of a ring A and an ideal I not satisfying any ofthe conditions (1)–(8) in Theorem 3.2. Let I = ( x + y z, x y, x t , y , y z ) ⊆ A = k [ x, y, z, t ]. Then depth gr I ( A ) = 0 as x y z ∈ ( I : I ) \ I . So I satisfies neither (1)nor (2).Note that √ I = ( x, y ), so dim( A/I ) = 2. Let m = ( x, y, z, t ). Since x y zt ∈ ( I : m ) \ I , depth( A/I ) = 0, hence
A/I is not unmixed. Thus I satisfies neither (5) nor(7). By the proof of Theorem 3.2, I satisfies none of the conditions (3), (4), (6).Unfortunately, experiments with Macaulay2 [6] suggest that I satisfies the con-clusion of Question 3.1, namely for all n ≥ A/I n ) = Ass( I n − /I n ) = { ( x, y ) , ( x, y, z ) , ( x, y, t ) , ( x, y, z, t ) } . Partial answer to Question 3.1 in dimension four.
We prove that ifchar k = 2 and dim A = 4, the equality Ass( A/I ) = Ass( I/I ) always holds, insupport of a positive answer to Question 3.1. The proof requires the structuretheory of perfect Gorenstein ideals of height three and their second powers. Theorem 3.4.
Assume char k = 2 . Let ( A, m ) be a four dimensional standardgraded polynomial ring over k . Then for any proper homogeneous ideal I of A ,there is an equality Ass(
A/I ) = Ass( I/I ) . HOP D. NGUYEN AND QUANG HOA TRAN
Proof.
It is harmless to assume I is a proper ideal. If ht I ≥ A/I ) ≤ I = 1, then I = f L , where f ∈ A is a form of positive degree and ht L ≥ → AL · f −→ AI → A ( f ) → , yields Ass( A/I ) = Ass(
A/L ) S Ass( A/ ( f )), as Min( I ) ⊇ Ass( A/ ( f )). An analogousformula holds for Ass( A/I ), as I = f L . If we can show that Ass( A/L ) ⊆ Ass(
L/L ), then from the injection L/L · f −−→ I/I we haveAss( A/I ) = Ass( A/L ) [ Ass( A/ ( f ))= Ass( L/L ) [ Ass( A/ ( f )) ⊆ Ass(
I/I ) . Hence it suffices to consider the case ht I = 2. Assume that there exists p ∈ Ass(
A/I ) \ Ass(
I/I ). The exact sequence0 → I/I → A/I → A/I → p ∈ Ass(
A/I ).By Lemma 2.3, p = m . Since Min( I ) = Min( I/I ), p / ∈ Min( I ), we get ht p = 3.Localizing yields p A p ∈ Ass( A p /I p ) \ Ass( I p /I p ). Then there exists a ∈ ( I p : p A p ) \ I p . On the other hand, since A p is a regular local ring of dimension 3containing one half, Lemma 3.5 below implies I p : p A p ⊆ I p , so a ∈ I p \ I p . Hence p A p ∈ Ass( I p /I p ). This contradiction finishes the proof. (cid:3) To finish the proof of Theorem 3.4, we have to show the following
Lemma 3.5.
Let ( R, m ) be a three dimensional regular local ring such that / ∈ R .Then for any ideal I of R , there is a containment I : m ⊆ I . We will deduce it from the following result.
Proposition 3.6.
Let ( R, m ) be a regular local ring such that / ∈ R . Let J be aperfect Gorenstein ideal of height . Then for all i ≥ , the maps Tor Ri ( J , k ) → Tor Ri ( J, k ) is zero. In particular, there is a containment J : m ⊆ J .Proof. We note that the second assertion follows from the first. Indeed, the hy-potheses implies that dim( R ) = d ≥
3. Using the Koszul complex of R , we seethat Tor Rd − ( J, k ) ∼ = Tor Rd ( R/J, k ) ∼ = J : m J .
Since the map Tor Ri ( J , k ) → Tor Ri ( J, k ) is zero for i = d −
1, the conclusion is J : m ⊆ J . Hence it remains to prove the first assertion. We do this by exploitingthe structure of the minimal free resolution of J and J , and constructing a mapbetween these complexes.Since J is Gorenstein of height three, it has a minimal free resolution P : 0 → R δ −→ F ∗ ρ −→ F → . Here F = Re ⊕ · · · ⊕ Re g is a free R -module of rank g – an odd integer. Themap τ : F → J maps e i to f i , where J = ( f , . . . , f g ). The free R -module F ∗ has SSOCIATED PRIMES OF POWERS 7 basis e ∗ , . . . , e ∗ g . The map ρ : F ∗ → F is alternating with matrix ( a i,j ) g × g , namely a i,i = 0 for 1 ≤ i ≤ g and a i,j = − a j,i for 1 ≤ i < j ≤ g , and ρ ( e ∗ i ) = g X j =1 a j,i e j for all i. The map δ : R → F ∗ has the matrix ( f . . . f g ) T , i.e. it is given by δ (1) = f e ∗ + · · · + f g e ∗ g .It is known that if J is Gorenstein of height three, then J ⊗ R J ∼ = J , andby constructions due to Kustin and Ulrich [15, Definition 5.9, Theorems 6.2 and6.17], J has a minimal free resolution Q as below. Note that in the terminologyof [15] and thanks to the discussion after Theorem 6.22 in that work, J satisfiesSPC g − , hence Theorem 6.17, parts (c)(i) and (d)(ii) in ibid. are applicable. Theresolution Q given in the following is taken from (2.7) and Definition 2.15 in Kustinand Ulrich’s paper. Q : 0 → ∧ F ∗ d −→ ( F ⊗ F ∗ ) /η d −→ S ( F ) d −→ J → . Here S ( F ) = L ≤ i ≤ j ≤ g R ( e i ⊗ e j ) is the second symmetric power of F , η = R ( e ⊗ e ∗ + · · · + e g ⊗ e ∗ g ) ⊆ F ⊗ F ∗ , and ∧ F ∗ is the second exterior power of F ∗ .The maps d : S ( F ) → J , d : ( F ⊗ F ∗ ) /η → S ( F ), d : ∧ F ∗ → ( F ⊗ F ∗ ) /η are given by: d ( e i ⊗ e j ) = f i f j for 1 ≤ i, j ≤ g,d ( e i ⊗ e ∗ j + η ) = g X l =1 a l,j ( e i ⊗ e l ) for 1 ≤ i, j ≤ g,d ( e ∗ i ∧ e ∗ j ) = g X l =1 a l,i ( e l ⊗ e ∗ j ) − g X v =1 a v,j ( e v ⊗ e ∗ i ) + η for 1 ≤ i < j ≤ g. We construct a lifting α : Q → P of the natural inclusion map J → J such that α ( Q ) ⊆ m P . Q : 0 / / ∧ F ∗ α (cid:15) (cid:15) d / / ( F ⊗ F ∗ ) /η d / / α (cid:15) (cid:15) S ( F ) d / / α (cid:15) (cid:15) J / / (cid:127) _ ι (cid:15) (cid:15) P : 0 / / R δ / / F ∗ ρ / / F τ / / J / / . In detail, this lifting is • α ( e i ⊗ e j ) = f i e j + f j e i ≤ i, j ≤ g, • α ( e i ⊗ e ∗ j + η ) = f i e ∗ j , if ( i, j ) = ( g, g ) , − P g − v =1 f v e ∗ v , if ( i, j ) = ( g, g ) , • α ( e ∗ i ∧ e ∗ j ) = , if 1 ≤ i < j ≤ g − , − a g,i , if 1 ≤ i ≤ g − , j = g. Note that α is well-defined since α ( e ⊗ e ∗ + · · · + e g ⊗ e ∗ g + η ) = P g − v =1 f v e ∗ v − P g − v =1 f v e ∗ v . HOP D. NGUYEN AND QUANG HOA TRAN
Observe that α ( Q ) ⊆ m P since f i , a i,j ∈ m for all i, j . It remains to check thatthe map α : Q → P is a lifting for J ֒ → J . For this, we have: • τ ( α ( e i ⊗ e j )) = τ (cid:18) f i e j + f j e i (cid:19) = f i f j = ι ( d ( e i ⊗ e j )).Next we compute α ( d ( e i ⊗ e ∗ j + η )) = α g X l =1 a l,j ( e i ⊗ e l ) ! = g X l =1 a l,j f i e l + f l e i f i ( P gl =1 a l,j e l )2 (since g X l =1 a l,j f l = 0) . • If ( i, j ) = ( g, g ) then ρ ( α ( e i ⊗ e ∗ j + η )) = ρ ( f i e ∗ j /
2) = f i ( P gl =1 a l,j e l )2 . • If ( i, j ) = ( g, g ) then ρ ( α ( e g ⊗ e ∗ g + η )) = ρ − P g − v =1 f v e ∗ v ! = − P g − v =1 f v ( P gl =1 a l,v e l )2= g P l =1 ( g − P v =1 a v,l f v ) e l a v,l = − a l,v )= − g P l =1 ( a g,l f g ) e l g X v =1 a v,l f v = 0)= f g ( P gl =1 a l,g e l )2 (since a g,l = − a l,g ) • Hence in both cases, α ( d ( e i ⊗ e ∗ j + η )) = ρ ( α ( e i ⊗ e ∗ j + η )).Next, for 1 ≤ i < j ≤ g −
1, we compute α ( d ( e ∗ i ∧ e ∗ j )) = α g X l =1 a l,i ( e l ⊗ e ∗ j ) − g X v =1 a v,j ( e v ⊗ e ∗ i ) + η ! = ( P gl =1 a l,i f l ) e ∗ j − ( P gv =1 a v,j f v ) e ∗ i l, j ) nor ( v, i ) is ( g, g ))= 0 (since g X v =1 a v,l f v = 0)= δ ( α ( e ∗ i ∧ e ∗ j )) . SSOCIATED PRIMES OF POWERS 9
Finally, for 1 ≤ i ≤ g − , j = g , we have α ( d ( e ∗ i ∧ e ∗ g )) = α g X l =1 a l,i ( e l ⊗ e ∗ g ) − g X v =1 a v,g ( e v ⊗ e ∗ i ) + η ! = (cid:16)P g − l =1 a l,i f l (cid:17) e ∗ g − P g − v =1 a g,i f v e ∗ v − ( P gv =1 a v,g f v ) e ∗ i α ( e l ⊗ e ∗ g ) depends on whether l = g or not)= − a g,i f g e ∗ g − P g − v =1 a g,i f v e ∗ v g X v =1 a v,l f v = 0)= − a g,i ( P gv =1 f v e ∗ v )2We also have δ ( α ( e ∗ i ∧ e ∗ g )) = δ ( − a g,i /
2) = − a g,i ( P gv =1 f v e ∗ v )2 . Hence α : Q → P is a lifting of the inclusion map J → J .Since α ( Q ) ⊆ m P , it follows that α ⊗ ( R/ m ) = 0. Hence Tor Ri ( J , k ) → Tor Ri ( J, k ) is the zero map for all i . The proof is concluded. (cid:3) Proof of Lemma 3.5.
It is harmless to assume that I ⊆ m . We can write I as afinite intersection I ∩ · · · ∩ I d of irreducible ideals. If we can show the lemma foreach of the components I j , then I : m ⊆ ( I : m ) ∩ · · · ∩ ( I d : m ) ⊆ d \ j =1 I j = I. Hence we can assume that I is an irreducible ideal. Being irreducible, I is a primaryideal. If √ I = m , then I : m ⊆ I : m = I . Therefore we assume that I is an m -primary irreducible ideal. Let k = R/ m . It is a folklore and simple result thatany m -primary irreducible ideal must satisfy dim k ( I : m ) /I = 1. Hence I is aGorenstein ideal of height three. It then remains to use the second assertion ofProposition 3.6. (cid:3) In view of Lemma 3.5, it seems natural to ask the following
Question 3.7.
Let ( R, m ) be a three dimensional regular local ring containing 1 / I be an ideal of R . Is it true that for all n ≥ I n : m ⊆ I n − ?For regular local rings of dimension at most two, Ahangari Maleki has provedthat Question 3.7 has a positive answer regardless of the characteristic [1, Proofof Theorem 3.7]. Nevertheless, if dim A is not fixed, Question 3.7 has a negativeanswer in positive characteristic in general. Here is a counterexample in dimension9(!). Example . Choose char k = 2, A = k [ x , x , x , . . . , z , z , z ] and M = x x x y y y z z z . Let I ( M ) be the ideal generated by the 2-minors of M , and I = I ( M ) + X i =1 ( x i , y i , z i ) + ( x , x , x ) + ( y , y , y ) + ( z , z , z ) . Denote m = A + . The Betti table of R/I , computed by Macaulay2 [6], is
Therefore I is m -primary, binomial, quadratic, Gorenstein ideal. The relation x y z + x y z + x y z ∈ ( I : m ) \ I implies I : m I .Computationally, the natural map Tor A ∗ ( k, I ) → Tor A ∗ ( k, I ) is not zero. For ifnot, we would have 100 = β , ( I/I ) = β , ( I ) + β , ( I ) = 0 + 101 = 101 . Inparticular, this gives a negative answer to [1, Question 3.6] in positive characteristic.4.
Powers of sums and associated primes
Bounds for associated primes.
The second main result of this paper is thefollowing. Its part (3) generalizes [7, Lemma 3.4], which deals only with squarefreemonomial ideals.
Theorem 4.1.
Let
A, B be commutative Noetherian algebras over k such that R = A ⊗ k B is Noetherian. Let I, J be proper ideals of
A, B , resp. (1)
For all n ≥ , we have inclusions n [ i =1 [ p ∈ Ass A ( I i − /I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q ) ⊆ Ass R R ( I + J ) n , Ass R R ( I + J ) n ⊆ n [ i =1 [ p ∈ Ass A ( A/I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q ) . (2) If moreover
Ass(
A/I n ) = Ass( I n − /I n ) for all n ≥ , then both inclusionsin (1) are equalities. (3) In particular, if A and B are polynomial rings and I and J are monomialideals, then for all n ≥ , we have an equality Ass R R ( I + J ) n = n [ i =1 [ p ∈ Ass A ( A/I i ) q ∈ Ass B ( B/J n − i +1 ) { p + q } . Proof. (1) Denote Q = I + J . By Proposition 2.1, we have Q n − /Q n = n M i =1 ( I i − /I i ⊗ k J n − i /J n − i +1 ) . SSOCIATED PRIMES OF POWERS 11
Hence(1) n [ i =1 Ass R ( I i − /I i ⊗ k J n − i /J n − i +1 ) = Ass R ( Q n − /Q n ) ⊆ Ass R ( R/Q n ) . For each 1 ≤ i ≤ n , we have J n − i Q i ⊆ J n − i ( I i + J ) = J n − i I i + J n − i +1 . We claimthat ( J n − i I i + J n − i +1 ) /J n − i Q i ∼ = J n − i +1 /J n − i +1 Q i − , so that there is an exactsequence(2) 0 −→ J n − i +1 J n − i +1 Q i − −→ J n − i J n − i Q i −→ J n − i J n − i +1 + J n − i I i ∼ = AI i ⊗ k J n − i J n − i +1 −→ . For the claim, we have( J n − i I i + J n − i +1 ) /J n − i Q i = J n − i I i + J n − i +1 J n − i ( I i + JQ i − ) = J n − i I i + J n − i +1 J n − i I i + J n − i +1 Q i − = ( J n − i I i + J n − i +1 ) /J n − i I i ( J n − i I i + J n − i +1 Q i − ) /J n − i I i ∼ = J n − i +1 /J n − i +1 I i J n − i +1 Q i − /J n − i +1 I i ∼ = J n − i +1 J n − i +1 Q i − . In the display, the first isomorphism follows from the fact that J n − i +1 ∩ J n − i I i = J n − i +1 I i = J n − i I i ∩ J n − i +1 Q i − , which holds since by Proposition 2.1, J n − i +1 I i ⊆ J n − i I i ∩ J n − i +1 Q i − ⊆ J n − i I i ∩ J n − i +1 ⊆ I i ∩ J n − i +1 = J n − i +1 I i . Now for i = n , the exact sequence (2) yieldsAss R ( R/Q n ) ⊆ Ass R ( J/JQ n − ) ∪ Ass R ( A/I n ⊗ k B/J ) . Similarly for the cases 2 ≤ i ≤ n − i = 1. Putting everything together,(3) Ass R ( R/Q n ) ⊆ n [ i =1 Ass R ( A/I i ⊗ k J n − i /J n − i +1 ) . Combining (1), (3) and Theorem 2.2, we finish the proof of (1).(2) If A is a polynomial ring, and I is a monomial ideal, by Theorem 3.2,Ass A ( A/I i ) = Ass A ( I i − /I i ) for all i ≥
1. Hence the conclusion follows by ap-plying part (1).(3) In this situation, every associated prime of
A/I i is generated by variables.In particular, p + q is a prime ideal of R for any p ∈ Ass(
A/I i ) , q ∈ Ass B ( B/J j )and i, j ≥
1. The conclusion follows from part (2). (cid:3)
Remark . If Question 3.1 has a positive answer, then we can strengthen theconclusion of Theorem 4.1: Let
A, B be standard graded polynomial rings over k .Let I, J be proper homogeneous ideals of
A, B , respectively. Then for all n ≥ R R ( I + J ) n = n [ i =1 [ p ∈ Ass A ( A/I i ) q ∈ Ass B ( B/J n − i +1 ) Min( R/ ( p + q )) . Example . In general, for singular base rings, each of the inclusions of Theorem4.1 can be strict. First, take A = k [ a, b, c ] / ( a , ab, ac ) , I = ( b ), B = k, J = (0).Then R = A, Q = I = ( b ) and I = ( b ). Let m = ( a, b, c ). One can check that a ∈ ( I : m ) \ I and I/I ∼ = A/ ( a, b ) ∼ = k [ c ], whence depth( A/I ) = 0 < depth( I/I ).In particular, m ∈ Ass A ( A/I ) \ Ass A ( I/I ). Thus the lower bound for Ass R ( R/Q )is strict in this case.Second, take A, I as above and B = k [ x, y, z ], J = ( x , x y, xy , y , x y z ) . In this case Q = ( b, x , x y, xy , y , x y z ) ⊆ k [ a, b, c, x, y, z ] / ( a , ab, ac ). Then c + z is ( R/Q )-regular, so depth R/Q > A/I + depth B/J . Hence( a, b, c, x, y, z ) does not lie in Ass R ( R/Q ), but it belongs to the upper bound forAss R ( R/Q ) in Theorem 4.1(1).4.2. Asymptotic primes.
Let astab ∗ ( I ) denote the minimal integer m ≥ A ( A/I i ) and Ass A ( I i − /I i ) are constant sets for all i ≥ m . By a resultdue to McAdam and Eakin [16, Corollary 13], for all i ≥ astab ∗ ( I ), Ass A ( A/I i ) \ Ass A ( I i − /I i ) consists only of prime divisors of (0). Hence if grade( I, A ) ≥
1, i.e. I contains a non-zerodivisor, then Ass A ( A/I i ) = Ass A ( I i − /I i ) for all i ≥ astab ∗ ( I ).Denote Ass ∗ A ( I ) = S i ≥ Ass A ( A/I i ) = S astab ∗ ( I ) i =1 Ass A ( A/I i ) andAss ∞ A ( I ) = Ass A ( A/I i ) for any i ≥ astab ∗ ( I ).The following folklore lemma will be useful. Lemma 4.4.
For any n ≥ , we have n [ i =1 Ass A ( A/I i ) = n [ i =1 Ass A ( I i − /I i ) . In particular, if grade(
I, A ) ≥ then Ass ∗ A ( I ) = astab ∗ ( I ) [ i =1 Ass A ( I i − /I i ) = [ i ≥ Ass A ( I i − /I i ) . Proof.
For the first assertion: Clearly the left-hand side contains the right-handone. Conversely, we deduce from the inclusion Ass A ( A/I i ) ⊆ Ass A ( I i − /I i ) ∪ Ass A ( A/I i − ) for 2 ≤ i ≤ n that the other containment is valid as well.The remaining assertion is clear. (cid:3) Now we describe the asymptotic associated primes of ( I + J ) n for n ≫ ∗ ( I + J ) under certain conditions on I and J . Theorem 4.5.
Assume that grade(
I, A ) ≥ and grade( J, B ) ≥ , e.g. A and B are domains and I, J are proper ideals. Then for all n ≥ astab ∗ ( I ) + astab ∗ ( J ) − ,we have Ass R R ( I + J ) n = Ass R ( I + J ) n − ( I + J ) n = [ p ∈ Ass ∗ A ( I ) q ∈ Ass ∞ B ( J ) Min R ( R/ p + q ) [ [ p ∈ Ass ∞ A ( I ) q ∈ Ass ∗ B ( J ) Min R ( R/ p + q ) . SSOCIATED PRIMES OF POWERS 13
In particular, astab ∗ ( I + J ) ≤ astab ∗ ( I ) + astab ∗ ( J ) − and Ass ∞ R ( I + J ) = [ p ∈ Ass ∗ A ( I ) q ∈ Ass ∞ B ( J ) Min R ( R/ p + q ) [ [ p ∈ Ass ∞ A ( I ) q ∈ Ass ∗ B ( J ) Min R ( R/ p + q ) . Proof.
Denote Q = I + J . It suffices to prove that for n ≥ astab ∗ ( I )+astab ∗ ( J ) − R ( Q n − /Q n )) and upper bound forAss R ( R/Q n ) in Theorem 4.1 are equal to [ p ∈ Ass ∗ A ( I ) q ∈ Ass ∞ B ( J ) Min R ( R/ p + q ) [ [ p ∈ Ass ∞ A ( I ) q ∈ Ass ∗ B ( J ) Min R ( R/ p + q ) . First, for the lower bound, we need to show that for n ≥ astab ∗ ( I ) + astab ∗ ( J ) − n [ i =1 [ p ∈ Ass A ( I i − /I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q )(4) = [ p ∈ Ass ∗ A ( I ) q ∈ Ass ∞ B ( J ) Min R ( R/ p + q ) [ [ p ∈ Ass ∞ A ( I ) q ∈ Ass ∗ B ( J ) Min R ( R/ p + q ) . If i ≤ astab ∗ ( I ), n − i + 1 ≥ astab ∗ ( J ), hence Ass B ( J n − i /J n − i +1 ) = Ass ∞ B ( J ). Inparticular, astab ∗ ( I ) [ i =1 [ p ∈ Ass A ( I i − /I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q )= astab ∗ ( I ) [ i =1 [ p ∈ Ass A ( I i − /I i ) q ∈ Ass ∞ B ( J ) Min R ( R/ p + q ) = [ p ∈ Ass ∗ A ( I ) q ∈ Ass ∞ B ( J ) Min R ( R/ p + q ) , where the second equality follows from Lemma 4.4.If i ≥ astab ∗ ( I ) then Ass A ( A/I i ) = Ass ∞ A ( I ), 1 ≤ n + 1 − i ≤ n + 1 − astab ∗ ( I ).Hence n [ i =astab ∗ ( I ) [ p ∈ Ass A ( I i − /I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q )= n +1 − astab ∗ ( I ) [ i =1 [ p ∈ Ass ∞ A ( I ) q ∈ Ass B ( J i − /J i ) Min R ( R/ p + q ) = [ p ∈ Ass ∞ A ( I ) q ∈ Ass ∗ B ( J ) Min R ( R/ p + q ) . The second equality follows from the inequality n + 1 − astab ∗ ( I ) ≥ astab ∗ ( J ) andLemma 4.4. Putting everything together, we get (4). The argument for the equalityof the upper bound is entirely similar. The proof is concluded. (cid:3) The persistence property of sums
Recall that an ideal I in a noetherian ring A has the persistence property ifAss( A/I n ) ⊆ Ass(
A/I n +1 ) for all n ≥
1. There exist ideals which fail the persis-tence property: A well-known example is I = ( a , a b, ab , b , a b c ) ⊆ k [ a, b, c ], forwhich I n = ( a, b ) n and ( a, b, c ) ∈ Ass(
A/I ) \ Ass(
A/I n ) for all n ≥
2. However,it is still challenging to find a homogeneous prime ideal without the persistenceproperty (if it exists).Swanson and R. Walker raised the question [22, Question 1.6] whether giventwo ideals I and J living in different polynomial rings, if both of them have thepersistence property, so does I + J . The third main result answers in the positive[22, Question 1.6] in many new cases. In fact, its case (ii) subsumes [22, Corollary1.7]. Corollary 5.1.
Let A and B be standard graded polynomial rings over k , I and J are proper homogeneous ideals of A and B , resp. Assume that I has the persistenceproperty, and Ass(
A/I n ) = Ass( I n − /I n ) for all n ≥ . Then I + J has the per-sistence property. In particular, this is the case if either of the following conditionsholds: (i) I is a monomial ideal satisfying the persistence property; (ii) I n +1 : I = I n for all n ≥ . (iii) I n is unmixed for all n ≥ . (iv) char k = 0 , dim( A/I ) ≤ and I has the persistence property.Proof. From Theorem 4.1 and the hypothesis on I , we have for all n ≥ R R ( I + J ) n = n [ i =1 [ p ∈ Ass A ( A/I i ) q ∈ Ass B ( J n − i /J n − i +1 ) Min R ( R/ p + q ) . Since Ass(
A/I i ) ⊆ Ass(
A/I i +1 ) for all i ≥
1, Ass( R/ ( I + J ) n ) ⊆ Ass( R/ ( I + J ) n +1 )for all n ≥
1, as desired.The second assertion is a consequence of the first, Theorem 3.2 and Lemma2.4. (cid:3)
Acknowledgments
The first author (HDN) and the second author (QHT) are supported by theNational Foundation for Science and Technology Development (NAFOSTED, Viet-nam) under grant numbers 101.04-2019.313 and 1/2020/STS02, respectively. Nguyenis also grateful to the support of Project CT0000.03/19-21 of the Vietnam Acad-emy of Science and Technology. Part of this work was done while the second authorwas visiting the Vietnam Institute for Advance Study in Mathematics (VIASM),he would like to thank the VIASM for its hospitality and financial support. Fi-nally, the second author also acknowledge the partial support of the Core ResearchProgram of Hue University, Grant No. NCM.DHH.2020.15.
References [1] R. Ahangari Maleki,
The Golod property of powers of ideals and Koszul ideals , J. PureAppl. Algebra (2019), no. 2, 605–618. 2, 9, 10[2] M. Brodmann,
Asymptotic stability of
Ass(
M/I n M ), Proc. Amer. Math. Soc. (1979),no. 1, 16–18. 1 SSOCIATED PRIMES OF POWERS 15 [3] W. Bruns and J. Herzog,
Cohen-Macaulay rings. Rev. ed. . Cambridge Studies in AdvancedMathematics , Cambridge University Press (1998). 2[4] D. Eisenbud, Commutative algebra. With a view toward algebraic geometry . GraduateTexts in Mathematics . Springer-Verlag, New York (1995). 2[5] C. Francisco, H.T. H`a, and A. Van Tuyl,
A conjecture on critical graphs and connectionsto the persistence of associated primes . Discrete Math. (2010), no. 15–16, 2176–2182.2[6] D. Grayson and M. Stillman,
Macaulay2, a software system for research in algebraic ge-ometry.
Embedded associated primes of powers of square-free monomialideals . J. Pure Appl. Algebra (2010), 301–308. 10[8] H.T. H`a, H.D. Nguyen, N.V. Trung and T.N. Trung,
Symbolic powers of sums of ideals .Math. Z. (2020), 1499–1520. 1, 3[9] H.T. H`a, N.V. Trung and T.N. Trung,
Depth and regularity of powers of sums of ideals .Math. Z. (2016), 819–838. 1[10] W. Heinzer, D. Lantz, and K. Shah,
The Ratliff-Rush ideals in a Noetherian ring . Comm.Algebra (1992), no. 2, 591–622. 4[11] J. Herzog and A.A. Qureshi, Persistence and stability properties of powers of ideals.
J.Pure Appl. Algebra (2015), no. 3, 530–542. 2[12] L.T. Hoa,
Powers of monomial ideals and combinatorics . In
New Trends in Algebras andCombinatorics , Proceedings of the 3rd International Congress in Algebras and Combina-torics (ICAC2017), Eds: K.P. SHum, E. Zelmanov, P. Kolesnikov, and S.M. Anita Wong,World Scientific (2020), pp. 149–178. 1[13] S. Huckaba and T. Marley,
Depth formulas for certain graded rings associated to an ideal .Nagoya Math. J. (1994), 57–69. 5[14] T. Kaiser, M. Stehl´ık, and R.ˇSkrekovski,
Replication in critical graphs and the persistenceof monomial ideals . J. Comb. Theory Ser. A (1), 239 – 251 (2014) 2[15] A. Kustin, B. Ulrich,
A family of complexes associated to an almost alternating map, withapplications to residual intersection . Mem. Amer. Math. Soc. , no. 461 (1992). 2, 7[16] S. McAdam and P. Eakin, The asymptotic Ass . J. Algebra (1979), 71–81. 12[17] S. Morey and R.H. Villarreal, Edge ideals: algebraic and combinatorial properties . In:Progress in Commutative Algebra, Combinatorics and Homology, Vol. 1 (C. Francisco, L.C. Klingler, S. Sather-Wagstaff and J. C. Vassilev, Eds.), De Gruyter, Berlin, 2012, 85–126.4[18] H.D. Nguyen and T. Vu,
Powers of sums and their homological invariants . J. Pure Appl.Algebra (2019), no. 7, 3081–3111. 1[19] L. J. Ratliff, Jr.,
On prime divisors of I n , n large , Michigan Math. J. (1976), 337–352.1[20] M.E. Rossi and I. Swanson, Notes on the behavior of the Ratliff-Rush filtration.
Commu-tative algebra (Grenoble/Lyon, 2001), 313–328, Contemp. Math., , Amer. Math. Soc.,Providence, RI, 2003. 4[21] H. Sabzrou, M. Tousi, and S. Yassemi,
Simplicial join via tensor products . ManuscriptaMath. , 255–272 (2008). 3[22] I. Swanson and R.M. Walker,
Tensor-multimomial sums of ideals: Primary decompositionsand persistence of associated primes . Proc. Amer. Math. Soc. (2019), no. 12, 5071–5082. 2, 14
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