Certain monomial ideals whose numbers of generators of powers descend
aa r X i v : . [ m a t h . A C ] M a y CERTAIN MONOMIAL IDEALS WHOSE NUMBERS OFGENERATORS OF POWERS DESCEND
REZA ABDOLMALEKI AND SHINYA KUMASHIRO
Abstract.
This paper studies the numbers of minimal generators of powers ofmonomial ideals in polynomial rings. For a monomial ideal I in two variables,Eliahou, Herzog, and Saem gave a sharp lower bound µ ( I ) ≥ I . Recently, Gasanova constructed monomial ideals suchthat µ ( I ) > µ ( I n ) for any positive integer n . In reference to them, we construct acertain class of monomial ideals such that µ ( I ) > µ ( I ) > · · · > µ ( I n ) = ( n + 1) for any positive integer n , which provides one of the most unexpected behaviorsof the function µ ( I k ). Introduction
The number of minimal generators of ideals plays an important role to investigatecommutative rings. As a classical fact, the notion of regular local rings is defined bythe number of minimal generators of the maximal ideal, and the numbers of minimalgenerators of powers of the maximal ideal play as a polynomial function. Moregenerally, for each ideal, the number of generators of powers eventually agrees withunique polynomial function. In particular, the function diverges infinitely. However,before being a polynomial, the numbers of generators of powers sometimes behavein an unexpected way even in polynomial rings (see for example [2] and [5]).The study of this direction may originate from the question of Judith Sally in 1974.She asked whether there exists a 1-dimensional local domain for which the squareof the maximal ideal has less generators than the maximal ideal itself. Such anexample is given in [6], then many researcher study the numbers of generators of anideal I and I . Especially, there are some interesting results for monomial ideals in 2variables. Let µ ( I ) denote the minimal number of generators of a monomial ideal I .We should note that it is not too difficult to construct examples of monomial ideals I in a polynomial ring S with at least 4 variables such that µ ( I ) > µ ( I ) since I admits height( I ) < dim S . Hence the assumption of 2 variables makes the problemmore interesting, and it provides ideals in arbitrary variables by the polynomialextension of ideals.Let K [ x, y ] be the polynomial ring over a field K and I a monomial ideal of K [ x, y ]. Then Eliahou, Herzog, and Saem [4, Theorem 1.2] showed a sharp lowerbound µ ( I ) ≥ µ ( I ) ≥
6, and Gasanova [5] construct monomial ideals such
Mathematics Subject Classification.
Key words and phrases. number of generators, polynomial ring, monomial ideal.The second author was supported by JSPS KAKENHI Grant Number JP19J10579 and JSPSOverseas Challenge Program for Young Researchers. hat µ ( I ) > µ ( I n ) for any positive integer n . With this background, our result isstated as follows. Theorem 1.1.
For any positive integer n , there exists a monomial ideal I satisfyingthe following conditions. µ ( I ) > µ ( I ) > · · · > µ ( I n ) = ( n + 1) and µ ( I k ) = ( n + 2) k + 1 for all k ≥ n . Let us explain about this result. The inequalities µ ( I ) > µ ( I ) > · · · > µ ( I n ) aremore unexpected than the behavior obtained in [5], and the equality µ ( I n ) = ( n +1) is related to the lower bound of [4, Theorem 1.2] regarding ( n + 1) as 9 by n = 2.The latter assertion claims µ ( I k ) agrees with the polynomial for k ≥ n . Thus ourconstruction provides one of the most unexpected numbers of generators of monomialideals before being a polynomial function. We also note that our ideal to describeTheorem 1.1 are constructed by n equigenerated ideals in contrast that [4, 5] and[2] basically explore ideals constructed by two equigenerated ideals.In Section 2 we construct certain monomial ideals which yield Theorem 1.1. Inwhat follows, let S = K [ x, y ] be the polynomial ring over a field K with deg( x ) =deg( y ) = 1. Set m = ( x, y ). For a graded ideal I of S , deg( I ) denotes the smallestdegree of homogeneous elements of I . We denote by [ I ] k the K -vector space spannedby the elements in I of degree k . For two integers a and b , [ a, b ] denotes the set ofall integers i such that a ≤ i ≤ b .2. Proof of Theorem 1.1
Our main result is precisely stated as follows.
Theorem 2.1.
Let m, p , . . . , p m , a , . . . , a m ≥ be positive integers. Let I =( x p , y p )( x ( m +1) p , y ( m +1) p ) and I i = x ip + p i · y ( m +2 − i ) p + p + ··· + p i · ( x p i , y p i ) a i − be ideals of S for ≤ i ≤ m . Set I = I + · · · + I m . Suppose that p = ( a i + 1) p i for ≤ i ≤ m and p + · · · + p m − < p .Then (a) I k = ( I k − ( I + · · · + I m +1 − k ) if ≤ k ≤ m − I k if m ≤ k. (b) µ ( I k ) = ( ( k + 1) + k ( a + · · · + a m +1 − k ) if ≤ k ≤ m − m + 2) k + 1 if m ≤ k. We prove this theorem in several steps. Let us introduce our notations to statesimply.
Notation 2.2.
Let 0 ≤ a < · · · < a ℓ ≤ d be integers. We denote by ( a , . . . , a ℓ ) d the equigenerated monomial ideal generated by x a y d − a , x a y d − a , . . . , x a ℓ y d − a ℓ . imilarly, K ( a , . . . , a ℓ ) d denotes the K -vector space Kx a y d − a + Kx a y d − a + · · · + Kx a ℓ y d − a ℓ . We write ( a , . . . , a ℓ ) = ( a , . . . , a ℓ ) d and K ( a , . . . , a ℓ ) = K ( a , . . . , a ℓ ) d if a ℓ = d .The followings easily follows from Notation 2.2. Lemma 2.3.
For ≤ i ≤ m , let I i be a monomial ideal of S generated in singledegree d i . Set I i = ( a i , . . . , a in i ) d i for ≤ i ≤ m . Then we have the following. (a) Q mi =1 I i = ( P mi =1 a ik i | ≤ k i ≤ n i ) d + ··· + d m . (b) Suppose that d < · · · < d m ≤ ℓ . Then [ I + · · · + I m ] ℓ = K ( a ik i + t | ≤ i ≤ m, ≤ k i ≤ n i , ≤ t ≤ ℓ − d i ) ℓ . In what follows, unless otherwise stated, we suppose that the assumption of The-orem 2.1. Set d u = deg( I u ) for 1 ≤ u ≤ m . Lemma 2.4. (a) d = ( m + 2) p and d u = d + p + p + · · · + p u − for ≤ u ≤ m . (b) deg( I ) < deg( I I ) < · · · < deg( I I m ) and deg( I I m ) < deg( I i I j ) for all i, j ∈ [2 , m ] . (c) I k = ( s ( m + 1) p + tp | s, t ∈ [0 , k ]) kd for k > and I k − I u = ( { s ( m + 1) + t + u } p + vp u | s, t ∈ [0 , k − , v ∈ [1 , a u ]) ( k − d + d u for ≤ u ≤ m .Proof. (a) follows from the definition of I i , and (b) follows from (a) and p + · · · + p m − < p .(c) follows from the following computations: I k =(0 , p ) k (0 , ( m + 1) p ) k =(0 , p , p , . . . , kp )(0 , ( m + 1) p , m + 1) p , . . . , k ( m + 1) p )=( s ( m + 1) p + tp | s, t ∈ [0 , k ]) kd and I k − I u =(0 , p ) k − (0 , ( m + 1) p ) k − x up + p u y ( m +2 − u ) p + p + ··· + p u (0 , p u ) a u − =( { s ( m + 1) + t + u } p + vp u | s, t ∈ [0 , k − , v ∈ [1 , a u ]) ( k − d + d u . (cid:3) The following observation is the heart of the proof of Theorem 2.1.
Proposition 2.5.
Let k > and ≤ u ≤ m . (a) For m + 1 − k < u ≤ m , I k − ( I + · · · + I u − ) ⊇ m ( k − d + d u . (b) For ≤ u ≤ m + 1 − k , (cid:2) I k − ( I + · · · + I u ) /I k − ( I + · · · + I u − ) (cid:3) ( k − d + d u ∼ = K ( { s ( m + 1) + k + u − } p + tp u | s ∈ [0 , k − , t ∈ [1 , a u ]) ( k − d + d u . roof. We prove by induction on u . Assume that u = 2. By Lemma 2.3(b) andLemma 2.4(a) and (c),[ I k ] ( k − d + d = K ( s ( m + 1) p + tp + w ′ | s, t ∈ [0 , k ] , w ′ ∈ [0 , p ])= K ( s ( m + 1) p + w | s ∈ [0 , k ] , w ∈ [0 , ( k + 1) p ])[ I k − I ] ( k − d + d = K ( { s ( m + 1) + t ′ + 2 } p + vp | s, t ′ ∈ [0 , k − , v ∈ [1 , a u ]) . Hence, if m + 1 − k < u = 2( ⇔ m + 1 ≤ k + 1), then I k = m ( k − d + d . Suppose that u = 2 ≤ m + 1 − k . Then, for 0 ≤ t ′ ≤ k − ≤ v ≤ a ,( t ′ + 2) p + vp ≤ ( k + 1) p + a p < ( k + 2) p ≤ ( m + 1) p since ( a + 1) p = p . Hence (cid:2) I k − ( I + I ) /I k (cid:3) ( k − d + d ∼ = K (cid:18) { s ( m + 1) + t ′ + 2 } p + vp (cid:12)(cid:12)(cid:12)(cid:12) s, t ′ ∈ [0 , k − , v ∈ [1 , a ] , and( t ′ + 2) p + vp > ( k + 1) p (cid:19) ( k − d + d = K ( { s ( m + 1) + k + 1 } p + vp | s ∈ [0 , k − , v ∈ [1 , a ]) ( k − d + d . Assume that u > , . . . , u −
1. Then, byinduction hypothesis, (cid:2) I k − ( I + · · · + I u − ) (cid:3) ( k − d + d u =[ I k ] ( k − d + d u + u − X i =2 K (cid:18) s ( m + 1) p + ( k + i − p + t i p i + w i (cid:12)(cid:12)(cid:12)(cid:12) s ∈ [0 , k − , t i ∈ [1 , a i ] ,w i ∈ [0 , d u − d i ] (cid:19) = K ( s ( m + 1) p + w ′ | s ∈ [0 , k ] , w ′ ∈ [0 , kp + d u − d ])+ u − X i =2 K (cid:18) s ( m + 1) p + ( k + i − p + t i p i + w i (cid:12)(cid:12)(cid:12)(cid:12) s ∈ [0 , k − , t i ∈ [1 , a i ] ,w i ∈ [0 , d u − d i ] (cid:19) , (1)where the second equality follows from ( k − d + d u − kd = d u − d ≥ p by Lemma2.4 (a). Claim 1. h I k − ( I + · · · + I u − ) i ( k − d + d u = K (cid:0) s ( m + 1) p + w ′′ | s ∈ [0 , k ] , w ′′ ∈ [0 , ( k + u − p ] (cid:1) . Proof of Claim 1. ( ⊆ ): It is clear that kp + d u − d ≤ ( k + u − p . For i ∈ [2 , u − , t i ∈ [1 , a i ] , and w i ∈ [0 , d u − d i ], we have( k + i − p + t i p i + w i ≤ ( k + i − p + a i p i + p i + · · · + p u − =( k + i ) p + p i +1 + · · · + p u − ≤ ( k + u − p . Hence we have the inclusion ⊆ in Claim 1.( ⊇ ): Let 0 ≤ w ′′ ≤ ( k + u − p . If w ′′ ≤ kp + d u − d , then the monomialcorresponding to s ( m + 1) p + w ′′ is in the former terms of (1). Hence we may ssume that kp + d u − d < w ′′ ≤ ( k + u − p . Then ( k + 1) p + p ≤ w ′′ since u > ≤ i ≤ u − k + i − p + p i ≤ w ′′ ≤ ( k + i ) p + p i +1 . Here we regard p u as 0 only this moment. Then there exist integers t ∈ [1 , a i ] and w ∈ [0 , p i + p i +1 ] ⊆ [0 , p i + · · · + p u − ] = [0 , d u − d i ] such that w ′′ = ( k + i − p + tp i + w .It follows that the inclusion ⊇ holds. (cid:3) Assume that m + 2 − k ≤ u ≤ m . Then ( m + 1) p ≤ ( k + u − p . Hence, byClaim 1, I k − ( I + · · · + I u − ) ⊇ m ( k − d + d u .Suppose that 2 < u ≤ m + 1 − k . We have[ I k − I u ] ( k − d + d u = K ( { s ( m +1)+ t + u } p + vp u | s, t ∈ [0 , k − , v ∈ [1 , a u ]) ( k − d + d u by Lemma 2.4(c). Note that( t + u ) p + vp u ≤ ( k − u ) p + a u p u < ( k − m + 1 − k ) p + p =( m + 1) p . Therefore, (cid:2) I k − ( I + · · · + I u ) /I k − ( I + · · · + I u − ) (cid:3) ( k − d + d u ∼ = K (cid:18) { s ( m + 1) + t + u } p + vp u (cid:12)(cid:12)(cid:12)(cid:12) s, t ∈ [0 , k − , v ∈ [1 , a u ] , ( t + u ) p + vp u > ( k + u − p (cid:19) = K ( { s ( m + 1) + k − u } p + vp u | s ∈ [0 , k − , v ∈ [1 , a u ]) . (cid:3) Let us prove Theorem 2.1.
Proof of Theorem 2.1. (a): For 2 ≤ k ≤ m −
1, by Lemma 2.4(b), I k =( I + · · · + I m ) k = I k − ( I + · · · + I m +1 − k ) + (monomials of degree > ( k − d + d m +1 − k ) . On the other hand, I k − ( I + · · · + I m +1 − k ) ⊇ m ( k − d + d m +1 − k by Proposition 2.5(a).Hence, I k = I k − ( I + · · · + I m +1 − k ).If k ≥ m , then I k = ( sp | s ∈ [0 , k ( m + 2)]) kd = ( x p , y p ) k ( m +2) by Lemma2.4(c). Hence I k ⊇ m kd + p = m ( k − d + d It follows that I k = I k .(b): Let 2 ≤ k ≤ m −
1. Then µ ( I k ) = µ ( I k − ( I + · · · + I m +1 − k ))= µ ( I k ) + m +1 − k X u =2 dim K (cid:2) I k − ( I + · · · + I u ) /I k − ( I + · · · + I u − ) (cid:3) ( k − d + d u =( k + 1) + m +1 − k X u =2 ka u , here the third equality follows from Lemma 2.4(c) and Proposition 2.5(b). For thecase where m ≤ k , one can easily check that µ ( I k ) = ( m + 2) k + 1 since I k = I k . (cid:3) Now we reach to prove Theorem 1.1.
Proof of Theorem 1.1.
Let n ≥ µ ( I n ) − µ ( I n − ) = 2 n + 1 − ( n − a µ ( I n − ) − µ ( I n − ) = 2 n − a − ( n − a , ... µ ( I k +1 ) − µ ( I k ) = 2 k + 3 + a + · · · + a m − k − ka m +1 − k , ... µ ( I ) − µ ( I ) = 7 + a + · · · + a n − − a n − , and µ ( I ) − µ ( I ) = 5 + a + · · · + a n − − a n . Therefore, by defining a , a , . . . , a n consecutively enough large, we get the assertion.We need to show that we can choose a , . . . , a n enough large with the assumptions p = ( a i + 1) p i for 2 ≤ i ≤ n and p + · · · + p n − < p .In fact, the former condition is always satisfied by choosing p , . . . , p n as follows. p i = n Y i =2 ( a i + 1) and p i = p a + 1 for 2 ≤ i ≤ n. For the latter condition, by substituting p = ( a i + 1) p i , we can rephrase by1 a + 1 + · · · + 1 a n − + 1 < . It follows that we can choose a , . . . , a n enough large as desired. (cid:3) Remark 2.6.
In the proof of Theorem 1.1, for two integers 1 ≤ i < j ≤ n , we canchoose so that a i = a j = 2. It follows that at most two of inequalities in Theorem1.1(a) can change to the converse inequalities. Example 2.7.
With the notation of Theorem 2.1, set m = 5, p = 72, p = 18, p = 12, p = 8, p = 2, a = 3, a = 5, a = 8, and a = 35. Then I = I + I + I + I + I , where I = ( x , y )( x , y ) , I = x y ( x , y ) , I = x y ( x , y ) ,I = x y ( x , y ) , and I = x y ( x , y ) . One can check by CoCoA [1] that µ ( I ) = 55, µ ( I ) = 41, µ ( I ) = 40, µ ( I ) = 37, µ ( I ) = 36 = 6 , and µ ( I ) = 43.We close this paper with the following questions. uestion 2.8. (a) For any positive integer n and ( n −
1) signs α , . . . , α n − ∈{ + , −} , does there exist a monomial ideal I satisfying the following conditions?(i) Sign( µ ( I k +1 ) − µ ( I k )) = α k for all 1 ≤ k ≤ n − µ ( I k ) = ( n + 2) k + 1 for all k ≥ n .(b) In reference to [4, Theorem 1.2], for any monomial ideal I with µ ( I ) ≫
0, does µ ( I n ) ≥ ( n + 1) hold? Acknowledgment.
Most of this work was done during the visit of the secondauthor to the University of Duisburg-Essen in 2019-2020. The second author isgrateful to J¨urgen Herzog for telling him this topic.
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E-mail address : [email protected] Shinya Kumashiro, Department of Mathematics and Informatics, Graduate Schoolof Science and Engineering, Chiba University, Yayoi-cho 1-33, Inage-ku, Chiba, 263-8522, Japan
E-mail address : [email protected]@chiba-u.jp