THE FIRST MAYR-MEYER IDEAL

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1 THE FIRST MAYR-MEYER IDEAL IRENA SWANSON New Mexico State University - Department of Mathematical Sciences, Las Cruces, New Mexico , USA iswanson@nmsueu Summary This paper gives a complete primary ecomposition of the first, that is, the smallest, Mayr-Meyer ieal, its raical, an the intersection of its minimal components The particular memership prolem which makes the Mayr-Meyer ieals complexity ouly exponential in the numer of variales is here examine also for the raical an the intersection of the minimal components It is prove that for the first Mayr-Meyer ieal the complexity of this memership prolem is the same as for its raical This prolem was motivate y a question of Bayer, Huneke an Stillman Grete Hermann prove in [H] that for any ieal I in an n-imensional polynomial ring over the fiel of rational numers, if I is generate y polynomials f 1,, f k of egree at most, then it is possile to write f = r i f i, where each r i has egree at most eg f + k n Mayr an Meyer in [MM] foun ieals Jn, for which a ouly exponential oun in n is inee achieve Bayer an Stillman [BS] showe that for these same ieals also any minimal generating set of syzygies has elements of egree which is ouly exponential in n Koh [K] moifie the original ieal to otain homogeneous quaric ieals with ouly exponential egrees of syzygies an ieal memership coefficients Bayer, Huneke an Stillman have raise questions aout the structure of these Mayr- Meyer ieals: is the ouly exponential ehavior ue to the numer of minimal primes, to the numer of associate primes, or to the structure of one of them? This paper, together with [S], is an attempt at answering these questions More precisely, the Mayr-Meyer ieal Jn, is an ieal in a polynomial ring in 10n + variales whose generators have egree at most + This paper analyzes the case n = 1 an shows that in this ase case the emee components o not play a role Theorem 1 of this paper gives a complete primary ecomposition of J1,, after which the intersection of the minimal components an the raical come as easy corollaries The last proposition shows that the complexity of the particular memership prolem from 1

2 [MM, BS, K] for the raical of J1, is the same as the complexity of the memership prolem for J1, Thus at least for the case n = 1, neither the emee components nor the non-reuceness play a role in the complexity In a eveloping paper Primary ecomposition of the Mayr-Meyer ieal [S], partial primary ecompositions are etermine for the Mayr-Meyer ieals Jn, for all n, 1 Uner the assumption that the characteristic of the fiel oes not ivie, for n, the numer of minimal primes is exactly n + 0, an the numer of emee primes likewise epens on n an However, a precise numer of emee components is not known The case n = 1 is very ifferent from the case n For example, uner the same assumption on the characteristic of the fiel, the numer of minimal primes of the first Mayr-Meyer ieal is + 4, an there is exactly one emee prime For unerstaning the asymptotic ehavior of the Mayr-Meyer ieals Jn,, the case n = 1 may not seem interesting, however, it is a asis of the inuction arguments for the ehavior of the other Jn, Furthermore, the case n = 1 is computationally an notationally more accessile All results of this paper were verifie for specific low values of on Macaulay Acknowlegement This research was one uring 000/01 at University of Kansas, supporte y the NSF-POWRE grant I thank Craig Huneke for introucing me to the Mayr-Meyer ieals, for all the conversations aout them, an for his enthusiasm for this project I also thank the NSF for partial support on DMS The first Mayr-Meyer ieal J1, is efine as follows Let K e a fiel, an a positive integer In case the characteristic of K is a positive prime p, write = i, where i is a power of p, an an p are relatively prime integers In case the characteristic of K is zero, let =, i = 1 For notational simplicity we assume that K contains all the i th roots of unity Let s, f, s 1, f 1, c 1,, c 4, 1,, 4 e ineterminates over K, an R = K[s, f, s 1, f 1, c 1,, c 4, 1,, 4 ] Note that R has imension 1 The Mayr-Meyer ieal for n = 1 is the ieal in R with the generators as follows: J = J1, = s 1 sc 1, f 1 sc 4 + c i s f i i = 1,, 3, 4 + fc 1 sc, fc 4 sc 3, s c 3 c, f c 1 c 3 4, fc 3 Theorem 1: A minimal primary ecomposition of J = J1, is as follows: J = s 1 sc 1, f 1 sc 4, c 1, c, c 3, c 4

3 s1 sc 1, f 1 sc 4, c 4 c 1, c 3 c, c 1 c 1, s f 1, 1 4, 3, i 1 α i α s 1 sc 1, f 1 sc 4, s, f s 1 sc 1, f 1 sc 4, s, c 1, c, c 4, 3, 4 s 1 sc 1, f 1 sc 4, s, c 1, c 4, 3, 3, c 1 c 3 4 s 1 sc 1, f 1 sc 4, s, f, c 4 s f 4, c 3 s f 3, sc 3 fc 4, c 3, c 4, c 1 c 4, c c 3, 3, 1 4, where the α vary over the i th roots of unity in K It is easy to verify that J = J1, is containe in the intersection, an that all ut the last ieal on the right-han sie of the equality are primary The following lemma proves that the last ieal is primary as well: Lemma : The last ieal in the intersection in Theorem 1 is primary Proof: Here is a simple fact: let x 1,, x n e variales over a ring A, S = A[x 1,, x n ], an I an ieal in A Then I is primary respectively, prime if an only if for any f 1,, f n A, IS + x 1 f 1,, x n f n S is a primary respectively, prime ieal in S By this fact it suffices to prove that the ieal L = s, f, c 4 s f 4, c 3 s f 3, sc 3 fc 4, c 3, c 4 is primary Note that L = s, f, c 3, c 4 is a prime ieal It suffices to prove that the set of associate primes of L is { L} It is an easy fact that for any x R, Ass R R R Ass Ass L L : x L + x In particular, when x = f, L + f = s, f, sc 4, sc 3, c 3, c 4 is clearly primary to L Thus it suffices to prove that L : f is primary to L We fix the monomial lexicographic orering s > f > c 4 > c 3 > 4 > 3 Clearly L : f contains s, f, c 4 c 3 3, sc 3, c 3 If r L : f, then the leaing term of r times f is containe in the ieal of leaing terms of L, namely in s, f, sc 4, sc 3, c 3, c 4, fc 4, so that the leaing term of r lies in s, f, c 4, sc 3, c 3 This proves that L : f = s, f, c 4 c 3 3, sc 3, c 3 This ieal is clearly primary to L, which proves the lemma 3

4 We next prove that the intersection of ieals in Theorem 1 equals J = J1, Note that it suffices to prove the shortene equality: c 1, c, c 3, c 4 c4 c 1, c 3 c, c 1 c 1, s f 1, 1 4, 3, i 1 α i α s, f s, c 1, c, c 4, 3, 4 s, c 1, c 4, 3, 3, c 1 c 3 4 s, f, c 4 s f 4, c 3 s f 3, sc 3 fc 4, c 3, c 4, c 1 c 4, c c 3, 3, 1 4 = c i s f i, fc1 sc, fc 4 sc 3, s c 3 c, f c 1 c 3 4, fc 3 The intersection of the first two rows equals: c 1, c, c 3, c 4 c 4 c 1, c 3 c, c 1 c 1, s f 1, 1 4, 3, 1 = c 4 c 1, c 3 c, c 1 c 1 + c1, c, c 3, c 4 s f 1, 1 4, 3, 1 = J + c 4 c 1, c 3 c, c 1 c 1 + c 1 4, 3, 1 This intersecte with the thir row, namely with s, f, equals Moulo J, J + s, f c 4 c 1, c 3 c, c 1 c 1 + c s, f 1 4, 3, 1 sc 1 fc 1 1 sc 1 fc 1 fc 1 3 fc sc sc fc fc fc fc sc 3 4 fc 4 4 sc 4, so that sc 1 c 4 J Also it is clear that fc 1 c 4, sc 3 c, sc 1 c 1 J, that sc J + fc, an that fc 3 J Thus the intersection of the ieals in the first three rows of Theorem 1 simplifies to J + f c 3 c, c 1 c 1 + fc 1 4, 1 Furthermore, fc 1 4 fc 3 c + J an moulo J, fc 1 c 1 c s f 1 fc 1, so that finally the intersection of the first three rows simplifies to J + fc 3 c, fc 1 4

5 We intersect this with the shortene ieal in the fourth row of Theorem 1, namely with s, c 1, c, c 4, 3, 4, to get J + fc 1 + fc 3 c s, c 1, c, c 4, 3, 4 = J + fc 1 + fc 3 c s, c 1, c, c 4, 3, 4 = J + fc 1 + fc 3 c c, 3, 4 As moulo J, fc 3 c 4 fc 1 4, an fc 3 c 3 fc 3 3 fc sc 3 sc 0, the intersection of the first four rows simplifies to J + fc 1, c 3 c, 1 4 Next we intersect this with the ieal in the fifth row of Theorem 1 namely with s, c 1, c 4, 3, 3, c 1 c 3 4, to get: As moulo J, J + fc 1, c 3 c, 1 4 s, c1, c 4, 3, 3, c 1 c 3 4 = J + fc 1, c 3 c, 1 4 s, c1, c 4, 3, 3, c 1 c 3 4 = J + fc c 1 c , c 3 c, 1 4 s, c1, c 4, 3, 3 = J + fc c 1 c , c 3 c, 1 4 s, c1, c 4, 3, 3 = J + fc 1, c 3 c, 1 4 s, c1, c 4, 3 sc 1 4 fc f 3c 3 c 1 sc 3 f c 1 sc 3 c 1 0, sfc 1 f c 1 1 s c sfc 1 sfc 1 = 0, fc 1 fc 4, the intersection of the ieals in the first five rows simplifies to J + fc 1, c 3 c, 1 4 c1, 3 = J + sc 1, c 3 c, 1 4 = J + sc 1 Finally we intersect this intersection of the ieals in the first five rows in the statement of Theorem 1 with the shortene last ieal there, namely with L = s, f, c 4 s f 4, c 3 s f 3, sc 3 fc 4, c 3, c 4, c 1 c 4, c c 3, 3, 1 4, to get: J + sc 1 L 5

6 It is easy to see that L : sc contains L As sc is not in L, then L : sc = L, so that the intersection of all the ieals in Theorem 1 equals J+sc 1 L : sc 1 = J + sc 1 L : 1 = J + sc 1 L = J + sc 1 s, f, c1, c, c 3, c 4, 3, 1 4 = J + sc 1 f, c It has een prove that sfc 1 J, an similarly sc 1 J This proves that the intersection of all the liste ieals in Theorem 1 oes equal J In orer to finish the proof of Theorem 1, it remains to prove that none of the liste components is reunant The last component is primary to a non-minimal prime, whereas there are no inclusion relations among the rest of the primes Thus the first + 4 liste components elong to minimal primes an are not reunant With this it suffices to prove that J has an emee prime: Lemma 3: When n = 1, c 4 s f 3 is in every minimal component ut not in J Thus there exists an emee component Proof: It has een estalishe that c 4 s f 3 is in every minimal component Suppose that c 4 s f 3 is in J Then c 4 s f 3 = 4 r i c i s f i + r5 fc 1 sc + r 6 fc 4 sc 3 + r 7 s c 3 c i=1 + r 8 f c 1 c r 9 fc 3, for some elements r i in the ring By the homogeneity of all elements in the two sets of variales {s, f} an {c 1, c, c 3, c 4 }, without loss of generality each r i is an element of K[ i i = 1,, 3, 4] Therefore the coefficients of the fc i, sc i yiel the following equations: sc 4 : 1 = r 4, fc 4 : 3 = r r 6, so r 6 = 4 3, fc 3 : 0 = r r 8 4, so r 3 = r 4, r 8 = r 3 for some r R, sc 3 : 0 = 4 3 r 3 r 7, so r 7 = 4 3 r 4, 6

7 sc 1 : 0 = r 1, fc 1 : 0 = r r 5, so r 5 = 0, sc : 0 = r r 4, so r = 4 3 r 4, fc : 0 = r r r 9 3 After expaning r in the last equation, 0 = 4 3 r 4 r r 9 3, so that 3 1, 4, 3, which is a contraiction As one emee component has een estalishe, this proves the Theorem Thus in the case n = 1, the Mayr-Meyer ieal J1, has + 4 minimal primes an one emee one, an these associate prime ieals are as follows α varies over the th roots of unity: associate prime ieal height s 1 sc 1, f 1 sc 4, c 1, c, c 3, c 4 6 s1 sc 1, f 1 sc 4, c 4 c 1, c 3 c, c 1 c 1, s f 1, 1 4, 3, 1 α 9 s 1 sc 1, f 1 sc 4, s, f 4 s 1 sc 1, f 1 sc 4, s, c 1, c, c 4, 3, 4 8 s 1 sc 1, f 1 sc 4, s, c 1, c 4,, 3, c 1 c s 1 sc 1, f 1 sc 4, s, f, c 1, c, c 3, c 4, 3, The proof of the Theorem also explicitly computes the intersection of the first five rows of the primary ecomposition, so that: Proposition 4: J + sc 1 The intersection of all the minimal components of J1, equals Furthermore, it is straightforwar to compute the raical of J1, : Proposition 5: The raical of J1, equals J1, + f 3 c 3 c, c 1 Proof: It is straightforwar to compute the raical of each component Note that as in the previous proof it suffices to compute the shortene intersection: c 1, c, c 3, c 4 7

8 c4 c 1, c 3 c, c 1 c 1, s f 1, 1 4, 3, 1 α α s, f s, c 1, c, c 4, 3, 4 s, c 1, c 4,, 3, c 1 c 3 4 = c i s f i, fc1 sc, fc 4 sc 3, s c 3 c, f c 1 c fc 3, f 3 c 3 c, f 3 c 1 As in the proof of the Theorem, the intersection of the first three rows equals J1, + fc 3 c, fc 1 Intersection with the ieal in the fourth row, namely with s, c 1, c, c 4, 3, 4, equals J1, + fc 1 + fc 3 c s, c 1, c, c 4, 3, 4 = J1, + fc 1 + fc 3 c s, c 1, c, c 4, 3, 4 = J1, + fc 1 + fc 3 c c, 3, 4 When this is intersecte with the ieal in the fifth row, namely with s, c 1, c 4,, 3, c 1 c 3 4, the resulting raical of J1, equals J1, + f 3 c 3 c + fc 1, fc c 3 c, f 4 c 3 c s, c 1, c 4,, 3, c 1 c 3 4 = J1, + f 3 c 3 c + fc 1, c 3 c, 1 4 s, c 1, c 4,, 3, c 1 c 3 4 = J1, + f 3 c 3 c + fc 1, c 3 c, 1 4 s, c 1, c 4,, 3, an y previous computations this simplifies to J1, + f 3 c 3 c + fc 3 1, c 3 c, 1 4 = J1, + f 3 c 3 c, c 1 Mayr an Meyer [MM] oserve that whenever the element sc 4 c 1 of J1, is expresse as an R-linear comination of the given generators of J1,, at least one of the coefficients has egree at least In fact, as the proposition elow proves, the egree of 8

9 at least one of the coefficients is at least 1, an this lower oun is achieve See also the proof of Theorem showing that sc 4 sc 1 moulo J1, Mayr an Meyer also showe the analogues for n 1, with egrees of the coefficients epening on n 1 ouly exponentially Bayer, Huneke an Stillman questione how much this ouly exponential growth epens on the existence of emee primes of Jn,, or on the structure of the components The proposition elow shows that at least for n = 1, the facts that J1, has an emee prime an that the minimal components are not raical, o not seem to e crucial for this property: Proposition 6: Let I e any ieal etween J1, an its raical Then whenever the element sc 4 c 1 is expresse as an R-linear comination of the minimal generators of I which inclue all the given generators of J1,, at least one of the coefficients has egree at least 1 Proof: All the cases can e euce from the case of I eing the raical of J1, To simplify the notation it suffices to replace y, so that I = J1, + f 3 c 3 c, c 1 Write 4 sc 4 c 1 = r i c i s f i + r5 fc 1 sc + r 6 fc 4 sc 3 + r 7 s c 3 c i=1 + r 8 f c 1 c r 9 fc 3 + r 10 f 3 c 3 c + r 11 f 3 c 1 for some elements r i in the ring Note that each of the explicit elements of I is homogeneous in the two sets of variales {s, f} an {c 1, c, c 3, c 4 } Thus it suffices to prove that in egrees 1 in each of the two sets of variales, one of the coefficients has egree at least 1 So without loss of generality each r i is an element of K[ i i = 1,, 3, 4] Therefore the coefficients of the sc i, fc i yiel sc 4 : 1 = r 4, fc 4 : 0 = r r 6, so r 6 = 4, sc 1 : 1 = r 1, fc 1 : 0 = r r 5, so r 5 = 1, sc : 0 = r + 1 r 7, so r 7 = r + 1, 9

10 fc : 0 = r + r r 9 3 r r , sc 3 : 0 = r r 7, so r 3 = 4 1 r, fc 3 : 0 = r 3 3 r r 10 3 The last equation implies that r 8 = r 8 3, so that the equations for the coefficients of fc 3 an fc can e rewritten as r 10 = r r 8 4 = r r 8 4, 0 = r + r r r 3 r r , = r 3 + r r r Thus r , 3 1 4, 1, so that r 8 3, , 1 If r 8 3, 1 Then , 1, which is a contraiction Thus r 8 has a multiple of as a summan, so r 8 has egree at least, so that r 8 has egree at least 1 In fact, y setting all the free variales r, r 11 to zero, the maximum egree of the coefficients r i is 1 Note that in the proof aove it is possile to have oth r 10 = r 11 = 0 an the egrees of the r i still at most 1, with 1 attaine on some r i Lemma 3 of [BS] erroneously claims that the egree of some r i is at least References [BS] Bayer, D; Stillman, M, On the complexity of computing syzygies, J Symolic Comput 6, 1988, [GS] Grayson, D; Stillman, M, Macaulay 1996 A system for computation in algeraic geometry an commutative algera, availale via anonymous ftp from mathuiuceu [H] Herrmann, G, Die Frage er enlich vielen Schritte in er Theorie er Polynomieale, Math Ann 95, 196, [K] Koh, J, Ieals generate y quarics exhiiting oule exponential egrees, J Algera 00, 1998, 5-45 [MM] Mayr, E; Meyer, A, The complexity of the wor prolems for commutative semigroups an polynomial ieals, Av Math 46, 198, [S] Swanson, I, Primary ecomposition of the Mayr-Meyer ieals, in preparation,