The minimal components of the Mayr-Meyer ideals

The minimal components of the Mayr-Meyer ideals Irena Swanson 24 April 2003 Grete Hermann proved in [H] that for any ideal I in an n-dimensional polynomial ring over the field of rational numbers, if I is generated by polynomials f 1,..., f k of degree at most d, then it is possible to write f = r i f i such that each r i has degree at most deg f+kd) 2n). Mayr and Meyer in [MM] found generators) of a family of ideals for which a doubly exponential bound in n is indeed achieved. Bayer and Stillman [BS] showed that for these Mayr-Meyer ideals any minimal generating set of syzygies has elements of doubly exponential degree in n. Koh [K] modified the original ideals to obtain homogeneous quadric ideals with doubly exponential syzygies and ideal membership equations. Bayer, Huneke and Stillman asked whether the doubly exponential behavior is due to the number of minimal and/or associated primes, or to the nature of one of them? This paper examines the minimal components and minimal primes of the Mayr-Meyer ideals. In particular, in Section 2 it is proved that the intersection of the minimal components of the Mayr-Meyer ideals does not satisfy the doubly exponential property, so that the doubly exponential behavior of the Mayr-Meyer ideals must be due to the embedded primes. The structure of the embedded primes of the Mayr-Meyer ideals is examined in [S2]. There exist algorithms for computing primary decompositions of ideals see Gianni- Trager-Zacharias [GTZ], Eisenbud-Huneke-Vasconcelos [EHV], or Shimoyama-Yokoyama [SY]), and they have been partially implemented on the symbolic computer algebra programs Singular and Macaulay2. However, the Mayr-Meyer ideals have variable degree and a variable number of variables over an arbitrary field, and there are no algorithms to deal with this generality. Thus any primary decomposition of the Mayr-Meyer ideals has to be accomplished with traditional proof methods. Small cases of the primary decomposition analysis were partially verified on Macaulay2 and Singular, and the emphasis here is on partially : the computers quickly run out of memory. The author thanks the NSF for partial support on grants DMS-0073140 and DMS- 9970566. 1991 Mathematics Subject Classification. 13C13, 13P05 ideals. Key words and phrases. Primary decomposition, Mayr-Meyer, membership problem, complexity of 1

The Mayr-Meyer ideals are binomial, so by the results of Eisenbud-Sturmfels in [ES] all the associated primes themselves are also binomial ideals. It turns out that many minimal primes are even monomial, which simplifies many of the calculations. The Mayr-Meyer ideals depend on two parameters, n and d, where the number of variables in the ring is On) and the degree of the given generators of the ideal is Od). Both n and d are positive integers. Here is the definition of the Mayr-Meyer ideals: let n, d 2 be integers and k a field. Let s, f, s r+1, f r+1, b r1, b r2, b r3, b r4, c r1, c r2, c r3, c r4 be variables over k, with r = 0, 1,..., n 1. The notation here closely follows that of [K]. Set S = k[s = s 0, f = f 0, s r+1, f r+1, b ri, c ri r = 0,..., n 1; i = 1,..., 4]. Thus S is a polynomial ring of dimension 10n + 2. The following generators define the Mayr-Meyer ideal J l n, d) subscript l for long, there will be a shortened version later on): first the four level 0 generators: ) H 0i = c 0i s fb d 0i, i = 1, 2, 3, 4; then the first six level r generators, r = 1,..., n: H r1 = s r s r 1 c r 1,1, H r2 = f r s r 1 c r 1,4, H r3 = f r 1 c r 1,1 s r 1 c r 1,2, H r4 = f r 1 c r 1,4 s r 1 c r 1,3, H r5 = s r 1 c r 1,3 c r 1,2 ), H r6 = f r 1 c r 1,2 b r 1,1 c r 1,3 b r 1,4 ), the last four level r generators, r = 2,..., n 1: H r,6+i = f r 1 c r 1,2 c ri b r 1,2 b ri b r 1,3 ),i = 1,..., 4, and the last level n generator: H n7 = f n 1 c n 1,2 b n 1,2 b n 1,3 ). The maximum degree of a given generator of J l n, d) is d+2. The degree 1 element s n f n of S is in J l n, d), and when written as an S-linear combination of the given generators, the S-coefficient of H 04 has degree which is doubly exponential in n see [MM], [BS], [K]). 2

The following summarizes the elementary facts used in the paper: Facts: 0.1: For any ideals I, I and I with I I, I + I ) I = I + I I. 0.2: For any ideal I and element x, x) I = xi : x). 0.3: Let x 1,..., x n be variables over a ring R. Let S = R[x 1,..., x n ]. For any f 1 R, f 2 R[x 1 ],..., f n R[x 1,..., x n 1 ], let L be the ideal x 1 f 1,..., x n f n )S in S. Then an ideal I in R is primary respectively, prime) if and only if IS+L is a primary respectively, prime) in S. Furthermore, i q i = I is a primary decomposition of I if and only if i q i S + L) is a primary decomposition of IS + L. 0.4: Let x be an element of a ring R and I an ideal. Suppose that there is an integer k such that for all m, I : x m I : x k. Then I = I : x k) I + x k ) ). Thus to find a possibly redundant) primary decomposition of I it suffices to find primary decompositions of possibly larger) I : x k and of I + x k ). We immediately apply the last fact: in order to find a primary decomposition of the Mayr-Meyer ideals J l n, d), by the structure of the H r1, H r2 and by Fact 0.3, it suffices to find a primary decomposition of the ideals Jn, d) obtained from J l n, d) by rewriting the variables s r, f r in terms of other variables, and then omitting the generators H r1, H r2, r 1. An ideal q is a component resp. associated prime) of Jn, d) if and only if q + H r1, H r2 r))s is a component resp. associated prime) of J l n, d). Thus to simplify the notation, throughout we will be searching for the primary components and associated primes of the shortened Mayr-Meyer ideals Jn, d) in a smaller polynomial ring R obtained as above. When we list the new generators explicitly, the case n = 1 is rather special. In fact, the primary decomposition in the case n = 1 is very different from the case n 2, and is given in [S1]. In this paper it is always assumed that n 2. Thus explicitly, we will be working with the following shortened Mayr-Meyer ideals: for any fixed integers n, d 2, R = k[s, f, b ri, c ri r = 0,..., n 1; i = 1,..., 4], a polynomial ring in 8n+2 variables, and Jn, d) is the ideal in R generated by the following polynomials h ri : first the four level 0 generators: ) h 0i = c 0i s fb d 0i, i = 1, 2, 3, 4; then the eight level 1 generators: h 13 = fc 01 sc 02, h 14 = fc 04 sc 03, h 15 = s c 03 c 02 ), 3

h 16 = f c 02 b 01 c 03 b 04 ), h 1,6+i = fc 02 c 1i b 02 b 1i b 03 ),i = 1,..., 4, the first four level r generators, r = 2,..., n: h r3 = sc 01 c 11 c r 3,1 c r 2,4 c r 1,1 c r 2,1 c r 1,2 ), h r4 = sc 01 c 11 c r 3,1 c r 2,4 c r 1,4 c r 2,1 c r 1,3 ), h r5 = sc 01 c 11 c r 2,1 c r 1,3 c r 1,2 ), h r6 = sc 01 c 11 c r 3,1 c r 2,4 c r 1,2 b r 1,1 c r 1,3 b r 1,4 ), the last four level r generators, r = 2,..., n 1: h r,6+i = sc 01 c 11 c r 3,1 c r 2,4 c r 1,2 c ri b r 1,2 b ri b r 1,3 ),i = 1,..., 4, and the last level n generator: h n7 = sc 01 c 11 c n 3,1 c n 2,4 c n 1,2 b n 1,2 b n 1,3 ). For simpler notation, Jn, d) will often be abbreviated to J. Observe that the maximum degree of the given generators of Jn, d) is max{n+2, d+ 2}. The image sc 01 c 11 c n 2,1 c n 1,1 c n 1,4 ) of s n f n by construction lies in Jn, d) and has degree n + 1. When this element is written as an R-linear combination of the h ri, the coefficient of h 04 is doubly exponential in n. Note that the contrast between a number doubly exponential in n and the degree n+1 of the input polynomial arising from this instance of the ideal membership problem for Jn, d) is not as striking as the contrast between a number doubly exponential in n and the degree 1 of the input polynomial arising from the ideal membership example s n f n for J l n, d). Thus while Jn, d) is a useful simplification of J l n, d) as far as the primary decomposition and associated primes are concerned, its doubly exponential nature is partially concealed. This paper consists of two sections. Section 1 is about all the minimal primes, their components, and their heights. For simplicity we assume that the underlying field k is algebraically closed. Then the number of minimal primes over Jn, d) is nd ) 2 + 20 Proposition 1.5), where d = d if the characteristic is 0 and is otherwise the largest divisor of d which is relatively prime to the characteristic of the field. All except 18 of the minimal components are simply the primes Proposition 1.6). Section 2 shows that the doubly exponential behavior of the Mayr-Meyer ideals is due to the existence of embedded primes. 4

The computation of embedded primes is tackled in [S2]. [S2] also constructs a new family of ideals with the doubly exponential ideal membership problem. Recursion can be applied to this new family in the construction of the embedded prime ideals, see [S3]. Acknowledgement. I thank Craig Huneke for suggesting this problem all for all the conversations and enthusiasm for this research. 1. Minimal primes and their components The minimal primes over Jn, d) and their components are quite easy to compute. Let d denote the largest divisor of d which is relatively prime to the characteristic of the field if that is positive, and otherwise d = d. Then there are are nd ) 2 + 20 minimal primes, all but 18 of which are their own primary components of Jn, d). The minimal primes are analyzed in two groups: those on which s and f are nonzerodivisors, and the rest of them. The first group consists of nd ) 2 + 1 prime ideals. The minimal primes not containing sf are denoted P r, where r varies from 0 to n, and the other part of the subscript depends on r. For the rest of the minimal primes the front part of the subscript varies from 1 to 4. Lemma 1.1: Let P be an ideal of R containing J such that s and f are non-zerodivisors modulo P in particular sf P). Let r {0,..., n 1}. Suppose that for all j < r and all i = 1, 2, 3, 4, c ji is not a zero-divisor modulo P. Then 1) For all j {0,..., r}, c j3 c j2, c j4 c j1, c 01 c 02 b d 01 P, and if j > 0, c j2 c j1 P. 2) If r > 0, c ri P for some i {1, 2, 3, 4} if and only if c ri P for all i {1, 2, 3, 4}. 3) For all j {0,..., r 1}, b j4 b j1 P. Also, for all j {0,..., r 2}, b j2 b j+1,i b j3 P, i = 1, 2, 3, 4. 4) Assume that r > 0. Then for all i, j {1, 2, 3, 4}, b d 0i bd 0j P. 5

5) Assume that r > 0 and that P is a primary ideal such that no b 0i lies in P. Then s fb d 01 P, and whenever 1 i < j 4, there exists a d )th root of unity α ij k such that b 0i α ij b 0j P and α 14 = 1, α 24 = α 1 12, α 34 = α 1 13. Proof: By the assumption that sf is a non-zerodivisor modulo J, if j = 0, h 15 = s c 03 c 02 ) being in P implies that c 03 c 02 is in P. Also, h 14 h 13 equals f c 04 c 01 ), so that c 04 c 01 P. Note that h 01 b d 01h 13 = sc 01 c 02 b d 01), so that c 01 c 02 b d 01 P. This proves 1) for j = 0. Now assume that j > 0. If j r < n, h j+1,5 = sc 01 c 11 c j 1,1 c j3 c j2 ) being in P implies that c j3 c j2 is in P. Furthermore, h j+1,4 h j+1,3 = sc 01 c 11 c j 2,1 c j 1,4 c j4 c j1 ) so that c j4 c j1 is in P. Then h j+1,3 equals sc 01 c 11 c j 1,1 c j1 c j2 ) modulo c j 1,4 c j 1,1 ), so that c j1 c j2 lies in P. This proves 1). With 1) established, 2) is an easy consequence. To prove 3), observe that modulo c 03 c 02 ) P, h 16 equals fc 02 b 01 b 04 ). Hence if r > 0, b 01 b 04 is in P. If 0 j < r, h j+1,6 sc 01 c 11 c j 2,1 c j 1,4 c j2 b j1 b j4 ) modulo c j3 c j2 ), hence b j1 b j4 is in P. Furthermore, for all i = 1,..., 4, h 1,6+i = fc 02 c 14 b 02 b 1i b 03 ) P, h j,6+i = sc 01 c 11 c j 3,1 c j 2,4 c j 1,2 c ji b j 1,2 b ji b j 1,3 ) P for j > 1, so that b j 1,2 b ji b j 1,3 is in P for all j = 1,..., r 1 and all i = 1,..., 4. This proves 3). If r > 0, h 0i = c 0i s fb d 0i ) P implies that s fbd 0i P. Hence whenever 1 i < j 4, f b d 0i bd 0j) is in P so that b d 0i b d 0j follows easily. For notational purposes define the following ideals in R: E = s fb d 01) + b 01 b 04, b d 02 b d 03, b d 01 b d 02), F = b 02 b 11 b 03, b 14 b 11, b 13 b 11, b 12 b 11, b d 12 1) C r = c r1, c r2, c r3, c r4 ), r = 0,..., n 1 C n = 0), D 0 = c 04 c 01, c 03 c 02, c 01 c 02 b d 01), 6 is in P. This proves 4), and then 5)

D r = c r4 c r1, c r3 c r2, c r2 c r1 ),r = 1,..., n 1, D n = 0), B 0 = B 1 = 0), B r = 1 b 2i, 1 b 3i,..., 1 b ri i = 1,..., 4),r = 2,..., n 1. With the previous lemma and this notation then: Proposition 1.2: Let P be a minimal prime ideal containing J and not containing sf. 1) If P contains one of the c 0i, then P equals the height four prime ideal P 0 = c 01, c 02, c 03, c 04 ) = C 0. 2) If P contains no c ji, set r = n, otherwise set r to be the smallest integer such that P contains some c ri. If r = 1, P contains and if r > 1, P contains 3) For all r = 1,..., n, J. p 1 = C 1 + E + D 0, = C r + E + F + B r 1 + D 0 + D 1 + + D r 1. Proof: Suppose that P contains c 01 or c 04. Then by Lemma 1.1, P contains c 02 b d 0i. If c 02 is not in P, then b 02 P, hence as h 02 = c 02 s fb d 02 ) P, necessarily c 02s P, contradicting the choice. Thus necessarily P contains c 02. Then by Lemma 1.1, P contains all the c 0i. As P 0 contains J, this verifies 1). If r 1, obviously contains J, thus verifying 3). If b 02 is in P, then as above also c 02 s is in P, contradicting the assumptions. Thus b 02 is not in P and 2) follows for the case r = 1 by Lemma 1.1. Now let r > 1. By Lemma 1.1, it remains to prove that F + B r 1 P when r > 1. As b j2 b j+1,i b j3 P for all j = 0,..., r 2, i = 1,..., 4, it follows that b j+1,i b j+1,i )b j3 is in P for any i, i {1, 2, 3, 4}. If b j3 P, by an application of Lemma 1.1 3), b j 1,2 P, whence b j 2,2 P,..., b 02 is in P, which is a contradiction. Thus necessarily b j+1,i b j+1,i is in P for all j = 0,..., r 2, or that b j 1,i b j 1,i is in P for all j = 2,..., r. Once this is established, then h j,6+i equals sc 01 c 11 c j 3,1 c j 2,4 c j 1,2 c ji b j 1,3 1 b ji ) modulo P so that 1 b ji is in P for all i = 1,..., 4 and all j = 2,..., r 1. A similar argument shows that b d 11 1 is in P. The remaining case r = n has essentially the same proof. From this one can read off the minimal primes and components: 7

Proposition 1.3: All the minimal prime ideals over J which do not contain sf are P 0, P 1αβ = p 1 + b 01 αb 02, b 02 βb 03 ), P rαβ = + b 01 αb 02, b 02 βb 03, β b 1i i = 1,..., 4), where α and β vary over the d )th roots of unity. The heights of these ideals are as follows: ht P 0 ) = 4, for r {1,..., n 1}, ht P rαβ ) = 7r + 4, and ht P nαβ ) = 7n. The components of Jn, d) corresponding to these prime ideals are the primes themselves.. Furthermore, with notation as in the previous proposition, for all r 1, α,β P rαβ = Proof: The case of P 0 is trivial. It is easy to see that for r > 0, the listed primes P rαβ are minimal over and that the intersection of the d ) 2 P rαβ equals. It is trivial to calculate the heights, and it is straightforward to prove the last statement. This completes the list of all the minimal primes over Jn, d) which do not contain s and f. The next group of minimal primes all contain s: Proposition 1.4: Let P be a prime ideal minimal over J. If P contains s, then P is one of the following 19 prime ideals: P 1 = s, f), P 2 = s, c 01, c 02, c 04, b 03, b 04 ) P 3 = s, c 01, c 04, b 02, b 03, c 02 b 01 c 03 b 04 ), P 4Λ = c 1i i Λ) + b 1i i Λ) + s, c 01, c 03, c 04, b 01, b 02 ), as Λ varies over the subsets of {1, 2, 3, 4}. The heights of these primes are 2, 6, 6 and 10, respectively. Proof: Note that J + s) = c 0i fb d 0i, fc 02c 1i b 02 b 1i b 03 ) i = 1, 2, 3, 4 ) + s, fc 01, fc 04, f c 02 b 01 c 03 b 04 )). If P contains f, it certainly equals P 1. Now assume that P does not contain f. Then P is minimal over c0i b d 0i, c 02c 1i b 02 b 1i b 03 ) i = 1, 2, 3, 4 ) + s, c 01, c 04, c 02 b 01 c 03 b 04 ). If c 02 P, then P is minimal over c03 b d 03, s, c 01, c 02, c 04, c 03 b 04 ), 8

so it is either s, c 01, c 02, c 03, c 04 ) or s, c 01, c 02, c 04, b 03, b 04 ) = P 2. However, the first option is not minimal over J as it strictly contains P 0 from Proposition 1.2. Now assume that P does not contain fc 02. Then P is minimal over b 02, c 03 b d 03 ) + c 1ib 1i b 03 i = 1, 2, 3, 4) + s, c 01, c 04, c 02 b 01 c 03 b 04 ). If P contains b 03, then P = s, c 01, c 04, b 02, b 03, c 02 b 01 c 03 b 04 ), which is P 3. Finally, assume that P does not contain fc 02 b 03. Then P is minimal over b 02, c 03 ) + c 1i b 1i i = 1, 2, 3, 4) + s, c 01, c 04, b 01 ), whence P is one of the P 4Λ. It turns out that there are no other minimal primes over Jn, d): Proposition 1.5: The prime ideals from the previous two propositions are the only prime ideals minimal over J. Thus there are 1 + nd ) 2 + 3 + 2 4 = nd ) 2 + 20 minimal primes. Proof: Proposition 1.3 determined all the minimal primes over J not containing sf, and Proposition 1.4 determined all those minimal primes which contain s. It remains to find all the prime ideals containing f and J but not s. As J+f) contains c 0i s i = 1, 2, 3, 4), a minimal prime ideal containing J +f) but not s contains, and even equals f, c 01, c 02, c 03, c 04 ). However, this prime ideal properly contains P 0, and hence is not minimal over J. The proposition follows as there are no containment relations among the given primes. The nd ) 2 + 20 minimal primary components can be easily computed: Proposition 1.6: For all possible subscripts, let p be the P -primary component of J. Then p 2 = s, c 01, c 02, c 04, b d 03, b 04 ), p 4Λ = c 1i i Λ) + b d 1i, b 02 b 1i b 03, b 1i b 1j i, j Λ ) + s, c 01, c 03, c 04, b 01, b d 02), and all the other p equal P. Proof: By Proposition 1.3, it remains to calculate p 1, p 2, p 3 and p 4Λ. As c 03 c 02 is not an element of P 1, P 2, P 3 and P 4Λ, and since h 15 = s c 03 c 02 ) is in J, it follows that s p 1, p 2, p 3 and p 4Λ. Then c 01 fb d 01 p 1, so that f p 1, and so p 1 = P 1. As h 13 = fc 01 sc 02, h 14 = fc 04 sc 03 are in J, then fc 01, fc 04 p 2, p 3 and p 4Λ, whence c 01, c 04 p 2, p 3, p 4Λ. For all i = 1,..., 4, as h 1,6+i = fc 02 c 1i b 02 b 1i b 03 ) J, 9

it follows that b 02 b 1i b 03 p 3. Thus as b 11 b 12 P 3, it follows that b 03 and hence also b 02 are in p 3. Now it is clear that p 3 is the P 3 -primary component of J. Further for i = 2, 3, c 0i fb d 0i p j implies that b d 02 p 4Λ, b d 03 p 2, c 02 p 2, and c 03 p 4Λ. As h 16 = f c 02 b 01 c 03 b 04 ) is in J, then fc 03 b 04 is in p 2 so that b 04 is in p 2. Also fc 02 b 01 is in p 4Λ so that b 01 is in p 4Λ. Thus the P 2 -primary component contains p 2. But p 2 contains J, so p 2 is the P 2 -primary component of J. Lastly, as J contains h 1,6+i, i = 1,..., 4, each p 4Λ contains each c 1i b 02 b 1i b 03 ). If i Λ, then b 02 b 1i b 03 is not in P 4Λ, so that c 1i p 4Λ. If instead i Λ, then c 1i P 4Λ, so that b 02 b 1i b 03 is in p 4Λ. Hence b d 02 bd 1i bd 03 is in p 4Λ, so that as b d 02 p 4Λ, so is b d 1i bd 03. Hence b d 1i is in p 4Λ. Furthermore, for i, j Λ, b 03 b 1j b 1i ) = b 02 b 1i b 03 ) b 02 b 1j b 03 ) is in p 4Λ, so that b 1j b 1i is in p 4Λ. Thus p 4Λ c 1i i Λ) + b d 1i, b 02 b 1i b 03, b 1i b 1j i, j Λ ) + s, c 01, c 03, c 04, b 01, b d 02), but the latter ideal is primary and contains J, so equality holds. The structure of p 2 says that Proposition 1.7: For n, d 2, Jn, d) is not a radical ideal. Here is the table of all the minimal primes over Jn, d), where α and β are d )th roots of unity: minimal prime height component of Jn, d) P 0 = c 01, c 02, c 03, c 04 ) 4 p 0 = P 0 P 1αβ = p 1 + b 01 αb 02, b 02 βb 03 ) 11 p 1αβ = P 1αβ P rαβ = 7r + 4 αβ = P rαβ, 2 r < n +b 01 αb 02, b 02 βb 03, β b 1i ) 7n αβ = P rαβ, r = n P 1 = s, f) 2 p 1 = P 1 P 2 = s, c 01, c 02, c 04, b 03, b 04 ) 6 p 2 = s, c 01, c 02, c 04, b d 03, b 04) P 3 = s, c 01, c 04, b 02, b 03, c 02 b 01 c 03 b 04 ) 6 p 3 = P 3 P 4Λ = s, c 01, c 03, c 04, b 01, b 02 ) 10 p 4Λ = s, c 01, c 03, c 04, b 01, b d 02) +c 1i, b 1j i Λ, j Λ) +c 1i i Λ) +b d 1j, b 02 b 1j b 03, b 1j b 1j j, j Λ) 10

2. Doubly exponential behavior is due to embedded primes In this section we compute the intersection of all the minimal components of Jn, d), and show that the element sc 01 c n 2,1 c n 1,1 c n 1,4 ), which gave the doubly exponential membership property for Jn, d), does not give the doubly exponential membership property for the intersection of the minimal components. This proves that the doubly exponential behavior of the Mayr-Meyer ideals is due to the existence of embedded primes. First define the ideal p 4 = s, c 01, c 03, c 04, b 01, b d 02) + c 1i b 02 b 1i b 03 ), c 1i b d 1i, c 1i c 1j b 1i b 1j ) i, j = 1,..., 4). Note that p 4 : c 11 = p 4 : c 2 11, so that by Fact 0.4, p 4 = p 4 : c 11 ) p 4 + c 11 )). Similarly, p 4 = p 4 : c 11 c 12 ) p 4 : c 11 ) + c 12 )) p 4 + c 11 )) : c 12 ) p 4 + c 11, c 12 )) = = Λ p 4 1 c 11 ) 2 c 12 ) 3 c 13 ) 4 c 14 ), where i vary over the operations colon and addition. But the resulting component ideals are just the various p 4Λ, so that p 4 = Λ p 4Λ. Next we compute the intersection of p 4 and p 2 using Fact 0.1): p 2 p 4 = s, c 01, c 04 ) + c 02, b d 03, b 04) p 4 = s, c 01, c 04 ) + c 02, b d 03, b 04 ) c 03, b 01, b d 02, c 1i b 02 b 1i b 03 ), c 1i b d 1i, c 1i c 1j b 1i b 1j )), so that p 2 p 3 p 4 = s, c 01, c 04, c 02 b 01 c 03 b 04 ) + b d 03 )c 03, b 01, b d 02, c 1ib 02 b 1i b 03 ), c 1i b d 1i, c 1ic 1j b 1i b 1j )) + c 02, b 04 ) b d 02, c 1i b 02 b 1i b 03 )) + c 02, b 04 ) c 03, b 01, c 1i b d 1i, c 1ic 1j b 1i b 1j )) b 02, b 03 ) = s, c 01, c 04, c 02 b 01 c 03 b 04 ) + b d 03)c 03, b 01, b d 02, c 1i b 02 b 1i b 03 ), c 1i b d 1i, c 1i c 1j b 1i b 1j )) + c 02, b 04 )b d 02, c 1ib 02 b 1i b 03 ), c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 02, c 1i b d 1i b 03). 11

Thus the intersection of the minimal components of Jn, d) which contain s equals: p 1 p 2 p 3 p 4 = s, fc 01, fc 04, fc 02 b 01 c 03 b 04 )) + fb d 03 )c 03, b 01, b d 02, c 1ib 02 b 1i b 03 ), c 1i b d 1i, c 1ic 1j b 1i b 1j )) + fc 02, b 04 )b d 02, c 1i b 02 b 1i b 03 ), c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1ib 02, c 1i b d 1ib 03 ) = J + s) + fb d 03 b 01, b d 02, c 1ib 02 b 1i b 03 ), c 1i b d 1i, c 1ic 1j b 1i b 1j )) + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03) + fb 04 b d 02, c 1i b 02 b 1i b 03 ), b 01 b 02, b 01 b 03, c 1i b d 1ib 02, c 1i b d 1ib 03 ). We can simplify this intersection in terms of the generators of J if we first intersect the intersection with the minimal component p 0 : p 0 p 4 = J + sc 0 + fb d 03C 0 b 01, b d 02, c 1i b 02 b 1i b 03 ), c 1i b d 1i, c 1i c 1j b 1i b 1j )) + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03) + fb 04 C 0 b d 02, c 1ib 02 b 1i b 03 ), b 01 b 02, b 01 b 03, c 1i b d 1i b 02, c 1i b d 1i b 03). As fc 0 J + fc 02, c 03 ) and sc 0 J + sd 0 + fc 02 b d 02), it follows that p 0 p 4 = J + sc 0 + fb d 03 c 02, c 03 )b 01, b d 02, c 1ib 02 b 1i b 03 ), c 1i b d 1i, c 1ic 1j b 1i b 1j )) + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03) + fb 04 c 02, c 03 )b d 02, c 1i b 02 b 1i b 03 ), b 01 b 02, b 01 b 03, c 1i b d 1ib 02, c 1i b d 1ib 03 ) = J + sc 0 + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03) = J + sd 0 + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03, b d 02 ). Next we compute the intersection of all the minimal components of Jn, d) which do not contain s: Lemma 2.1: For 2 r n, r 1 p 1 p 2 = E + D 0 + C 1 F + C 1 C 2 C i D i+1 + B i ) + C 1 C 2 C r. Proof: When r = 2, p 1 p 2 = C 1 + E + D 0 ) C 2 + E + F + D 0 + D 1 + B 1 ) = E + D 0 + C 1 C 2 + E + F + D 0 + D 1 + B 1 ) = E + D 0 + D 1 + C 1 C 2 + E + F + D 0 + B 1 ) = E + D 0 + D 1 + C 1 C 2 + E + F + D 0 + B 1 ) = E + D 0 + D 1 + C 1 F + C 1 C 2 + B 1 ), 12 i=0

which starts the induction. Then by induction assumption for r 2 and r n 1, ) r 1 p 1 +1 = E + D 0 + C 1 F + C 1 C 2 C i D i+1 + B i ) + C 1 C r +1 i=0 r 1 = E + D 0 + C 1 F + C 1 C 2 C i D i+1 + B i ) + C 1 C r ) +1, i=0 and by multihomogeneity, the last intersection equals C 1 C r C r+1 + E + F + B r ) + Combining the last two displays proves the lemma. r r D j C k ). Thus the intersection of all the minimal components of Jn, d) equals: r= 4 = J + sd 0 + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03, b d 02 )) j=1 k j = J + sd 0 + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03, b d 02 )) = J + sd 0 + fc 02 c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1ib 03, b d 02) ). Let A = c 03 b 02, c 03 b 03, b 01 b 02, b 01 b 03, c 1i b d 1i b 03, b d 02 ) n. Thus the intersection of all the minimal components of Jn, d) equals J + sd 0 + fc 02 A. Finding the generators of A takes up most of the rest of this section. We will use the decomposition A = c 03, b 01, c 1i b d 1i, bd 02 ) c 03, b 01, b 03, b d 02 ) b 02, b 03 ), and start computing A via the indicated partial intersections, again using Fact 0.1: where b 02, b 03 ) = b d 02 bd 03, c 11b 02 b 11 b 03 )) + b 02, b 03 ) L = b d 02 b d 03, c 11 b 02 b 11 b 03 )) + b 03 L + b 02 L, n 1 L = L + c 11 b 1i b 1j, b d 12 1) + c 11 c i1 D i+1 + B i ) ), i=0 L = s fb d 01, b 01 b 04, b d 01 bd 03 ) + D 0 + D 1. 13

Note that L is generated by all the generators of r 1 other than b d 02 b d 03. Then the intersection of the last three components of A is c 03, b 01, b 03, b d 02) b 02, b 03 ) = b d 02 b d 03) + b 03 L + c 03, b 01, b 03, b d 02) c 11 b 02 b 11 b 03 )) + b 02 L ) = b d 02 bd 03, b 02b d 01 bd 03 )) + b 03 L + c 03, b 01, b 03, b d 02 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) = b d 02 b d 03, b 02 b d 01 b d 03)) + b 03 L + c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + b 03, b d 02 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) = b d 02 b d 03, b 02 b d 01 b d 03)) + b 03 L + c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + b 03, b d 02 ) c 11b 02 b 11 b 03 ), b 02 ) c 11 b 02 b 11 b 03 )) + s fb d 01, b 01 b 04 ) + D 0 + D 1 ) = b d 02 b d 03, b 02 b d 01 b d 03)) + b 03 L + c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + b 03 c 11 b 11, b d 02, b 02b 03 ) c 11 b 02 b 11 b 03 ), s fb d 01, b 01 b 04 ) + D 0 + D 1 ) = b d 02 b d 03, b 02 b d 01 b d 03)) + b 03 L + c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + b 03 c 11 b 11, b d 02, b 02b 03 ) s fb d 01, b 01 b 04 ) + D 0 + D 1 ) + b 03 c 11 b 11, b d 02, b 02 b 03 ) c 11 b 02 b 11 b 03 )) = b d 02 bd 03, b 02b d 01 bd 03 )) + b 03 L + c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + c 11 b 02 b 11 b 03 )b 03, b d 1 02 ) = b d 02 b d 03, b 02 b d 01 b d 03), b d 1 02 c 11b 02 b 11 b 03 )) + b 03 + c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ). Hence A, the intersection of all of its components, equals A = c 03, b 01, c 1i b d 1i, b d 02) c 03, b 01, b 03, b d 02) b 02, b 03 ) 14

= c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + c 03, b 01, c 1i b d 1i, b d 02) b d 02 b d 03, b 02 b d 01 b d 03), b d 1 02 c 11b 02 b 11 b 03 )) + b 03 = c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + c 03 b d 02 b d 03, b 02 b d 01 b d 03), b d 1 02 c 11b 02 b 11 b 03 )) + b 03 ) + b 01, c 1i b d 1i, b d 02) b d 02 b d 03, b 02 b d 01 b d 03), b d 1 02 c 11b 02 b 11 b 03 )) + b 03 ) ). Let A be the ideal in the last line. The second intersectand ideal of A decomposes as so that A = = = = = b d 02 bd 03, b 02b d 01 bd 03 ), bd 1 02 c 11b 02 b 11 b 03 )) + b 03 = b d 02, bd 03, b 02b d 01 bd 03 ), bd 1 02 c 11b 02 b 11 b 03 )) + b 03 b 01, c 1i b d 1i, b d 02) b d 02, b d 03, b 02 b d 01, b d 1 02 c 11b 11 b 03 ) + b 03 ), ) b d 02, b 02b d 01 ) + b 01, c 1i b d 1i, bd 02 ) b d 03, bd 1 02 c 11b 11 b 03 ) + b 03 b d02, b 02 b d01) + b 03 b 01, c 1i b d 1i, b d 02) b03 d 1, bd 1 02 c 11b 11 ) + b d 02, b 02b d 01, b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 ) +b 03 b 01, c 11 b d 11 ) b d 1 03, bd 1 02 c 11b 11 ) + ) )) b d 02, b 02 b d 01, b 03 c 1i b d 1i c 11 b d 11), b 03 b d 01) +b 03 b01, c 11 b d 11) b d 1 03, bd 1 02 c 11b 11 ) + L ))), ) ) ) )) where L is generated by all the given generators of r 1 other than b d 01 b d 03: n 1 L = s fb d 01, b 01 b 04, b d 02 bd 03 ) + D 0 + C 1 F + C 1 C i D i+1 + B i ). 15 i=0

With this, A = b d 02, b 02b d 01, b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 ) = + b 03 b 01 b d 1 03, bd 1 02 c 11b 11 ) + L ) +b 03 c11 b d 11 ) b d 1 b d 02, b 02 b d 01, b 03 c 1i b d 1i c 11 b d 11), b 03 b d 01) + b 03 b 01 b d 1 03, bd 1 02 c 11b 11 ) + L ) +b 03 c 11 b d 11 03, bd 1 02 c 11b 11 ) + L ))) b d 1 03, bd 1 02 ) + )) L : c 11 = b 02 b d 01 bd 02 ), b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 bd 02 ), b 01b d 03 bd 02 )) + b 01 b d 1 02 c 11b 02 b 03 b 11 )) + b 03 b 01 L + c 11 b d 11 bd 03 bd 02 ), bd 1 02 bd 1 11 c 11b 02 b 03 b 11 )) + b 03 c 11 b d 11 L : c 11 ) + = b 02 b d 01 bd 02 ), b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 bd 02 ), b 01b d 03 bd 02 )) + b 01 b d 1 02 c 11b 02 b 03 b 11 )) + b 03 b 01 L + c 11 b d 11 bd 03 bd 02 ), bd 1 02 bd 1 11 c 11b 02 b 03 b 11 )) + b 03 c 11 b d 11 L : c 11 ) + b d 02 Hence A equals A = c 03, b 01 ) c 11 b 02 b 11 b 03 )) + b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) + c 03 b d 02 b d 03, b 02 b d 01 b d 03), b d 1 02 c 11b 02 b 11 b 03 )) + b 03 ) + b 02 b d 01 bd 02 ), b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 bd 02 ), b 01b d 03 bd 02 )) + b 01 b d 1 02 c 11b 02 b 03 b 11 )) + b 03 b 01 L + c 11 b d 11b d 03 b d 02), b d 1 02 bd 1 11 c 11b 02 b 03 b 11 )) + b 03 c 11 b d 11 L : c 11 ) + b d 02 Thus finally, = J + sd 0 + fc 02 A r= 4 = J + sd 0 + fc 02 c 03, b 01 ) b 02 s fb d 01, b 01 b 04 ) + D 0 + D 1 ) ) n + fc 02 c 03 b d 02 b d 03, b 02 b d 01 b d ) 03)) + b 03 b d ) 02.. + fc 02 b 02 b d 01 bd 02 ), b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 bd 02 ), b 01b d 03 bd 02 ), c 11b d 11 bd 03 bd 02 )) + fc 02 b 03 b 01 L + fc 02 b 03 c 11 b d 11 L : c 11 ) + fc 02 b d 02, 16

or in nicer form: Theorem 2.2: The intersection of all the minimal components of Jn, d) equals = J + sd 0 + fc 02 b 02 c 03, b 01 ) ) s fb d 01, b 01 b 04 where r= 4 + fc 02 b 02 c 03, b 01 )D 0 + D 1 ) + fc 02 c 03 b d 02 bd 03, b 02b d 01 bd 03 )) n 1 + fc 02 c 03 b 03 E + D 0 + C 1 F + c 11 c i1 D i+1 + B i )) i=0 + fc 02 b 02 b d 01 bd 02 ), b 03c 1i b d 1i c 11b d 11 ), b 03b d 01 bd 02 )) + fc 02 b 01 b d 03 b d 02), c 11 b d 11b d 03 b d 02)) n 1 + fc 02 b 03 b 01 E + D 0 + C 1 F + c 11 c i1 D i+1 + B i )) n 1 + fc 02 b 03 b d 11c 11 E + D 0 + F + c 21 c i1 D i+1 + B i )) + fc 02 b d 02 E + D 0 + C 1 F + i=0 i=0 n 1 i=0 E = s fb d 01, b 01 b 04, b d 03 b d 02). c 11 c i1 D i+1 + B i )), With Macaulay2 I verified this theorem and intermediate computations in the proof above for the case n = 3, d = 2.) Set c = c 11 c 21 c n 2,1 c n 1,1 c n 1,4 ). With the listed generators, together with the generators h rj of J, sc 01 c n 2,1 c n 1,1 c n 1,4 ) = c 01 s fb d 01)c + fc 01 sc 02 )b d 01c + c 02 s fb d 02 )bd 01 c + b d 01 fc 02b d 02 c, and the degrees of the coefficients c, c, b d 01 c and b d 01 fc 02b d 02 of the generators h 01, h 13, h 02 and fc 02 b d 02c, respectively, of the intersection of the minimal components, are not doubly exponential in n. This proves: Theorem 2.3: The doubly exponential ideal membership problem of the Mayr-Meyer ideals Jn, d) and J l n, d) for the element sc 01 c n 2,1 c n 1,1 c n 1,4 ) is not due to the minimal components, but to some embedded prime ideal. 17

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