MARKOV BASES OF LATTICE IDEALS

arXiv:1303.2303v2 [math.AC] 4 Apr 2013

MARKOV BASES OF LATTICE IDEALS HARA CHARALAMBOUS, APOSTOLOS THOMA, AND MARIUS VLADOIU Abstract. Let L ⊂ Zn be a lattice, R = k[x1 , . . . , xn ] where k is a field and IL = hxu − xv : u − v ∈ Li the corresponding lattice ideal. Most results in the literature on generating sets of lattice ideals concern the case L ∩ Nn = {0}. In this paper, using appropriate graphs for each fiber, we characterize minimal generating sets of IL of minimal cardinality for all lattices and give invariants for these generating sets. As an application we characterize all binomial complete intersection lattice ideals.

1. Introduction Let R = k[x1 , . . . , xn ] where k is a field, and let L be a lattice in Zn . The lattice ideal IL is defined to be the ideal generated by the following binomials: IL := hxu − xv : u − v ∈ Li . Let µ(IL ) be the least cardinality of any minimal generating set of IL consisting of binomials. We call Markov basis of IL a minimal system of binomial generators of IL of cardinality µ(IL ). The study of lattice ideals is a rich subject on its own, see [22, 33] for the general theory and [21] for recent developments. Moreover lattice ideals have applications in diverse areas in mathematics, such as algebraic statistics [8, 26], integer programming [10], hypergeometric differential equations [9], graph theory [25], etc. We note that such ideals were first systematically studied in [11] and that toric ideals are lattice ideals IL for which the lattice L is the kernel of an integer matrix. We note that almost all results in the literature are about lattices L such that L ∩ Nn = {0}, with very few exceptions like in [11, 19, 13, 15, 20]. If L is such that L ∩ Nn = {0} we say that L is positively graded. Let A be the subsemigroup of Zn /L generated by the elements {ai = ei + L : 1 ≤ i ≤ n}, where {ei : 1 ≤ i ≤ n} is the canonical basis of Zn and set degA (xv ) := v1 a1 + · · · + vn an ∈ A where xv = xv11 · · · xvnn . It follows that IL = hxu − xv : degA (xu ) = degA (xv ) i and that IL is A-graded. When L is positively graded, the semigroup A is partially ordered: c ≥ d ⇐⇒ there is e ∈ A such that c = d + e . Then the grading of A forces the IL -fiber of xu , i.e. the set {xv : xv − xu ∈ IL } = {xv : degA (xv ) = degA (xu )}, to be finite. The homogeneous Nakayama Lemma applies and guarantees that all minimal binomial generating sets of IL are Markov bases of IL , since they have the same cardinality. Let S be a Markov basis of IL 1991 Mathematics Subject Classification. 13C40, 14M25, 60J10. Key words and phrases. Binomial ideals, Markov basis, lattices. 1

2

H. CHARALAMBOUS, A. THOMA, AND M. VLADOIU

and form the multiset of all IL -fibers corresponding to the elements of S. This multiset is an invariant of IL and does not depend on the choice of S. Moreover the binomials of S are primitive, see [33, 22], and thus S is a subset of the Graver basis of IL . Since the Graver basis of IL is a finite set, see [22], it follows that the Universal Markov basis of IL , (see [17]), i.e. the union of all Markov bases, is finite. The situation for a general lattice ideal is completely different. Take for example the lattice L generated by {(1, 1), (5, 0)}. It can be shown that the following are minimal generating sets of IL in k[x, y]: {1 − xy, 1 − x5 }, {1 − xy, x3 − y 2 }, {1 − x2 y 2 , 1 − x3 y 3 , 1 − x5 }. It is clear that IL is not a principal ideal and that µ(IL ) = 2. It is not hard to produce minimal generating sets of IL of any desired cardinality greater than 2. For example let p1 , . . . , ps be s distinct primes and let ai = p1 · · · ps /pi . The elements a1 , . . . , as are relatively prime and the greatest common divisor of (1 − z a1 , . . . , 1 − z as ) is 1 − z, while the greatest common divisor of {1 − z aj : j 6= i} is 1 − z pi . It follows that h1 − (xy)a1 , . . . , 1 − (xy)as i = h1 − xyi and that {1 − x5 , 1 − (xy)a1 , . . . , 1 − (xy)an } is a minimal generating set of IL . Even if we restrict our attention to Markov bases of IL , we get some very interesting behaviour. The set {1 − x2012 y 2017 , y 4 − x2013 y 2022 } is a Markov basis of IL , but it is easily seen that its elements are not primitive binomials. It is easy to produce an infinite set of Markov bases of IL : in Section 4 we discuss how to obtain Markov bases of lattice ideals in general. It follows that in this example, the Universal Markov basis of IL is infinite. Moreover, there is no unique multiset of IL -fibers corresponding to the Markov bases of IL . Indeed the monomials of k[x, y] are partitioned into exactly five infinite IL -fibers: Fk = {xi y j : i − j ≡ k

mod 5}, 5

0≤k≤4.

The multiset of the IL -fibers for {1 − xy, 1 − x } is {F0 , F0 } while the multiset of the IL -fibers for {1 − xy, x3 − y 2 } is {F0 , F3 }. Algorithms for computing a generating set for lattice ideals were given in [16, 2, 15]. The main problem we address in this paper is how to determine invariants of Markov bases of a lattice ideal IL and how to detect whether a set of binomials of IL is a Markov basis of IL . Instead of considering the isolated fibers of IL we consider equivalence classes of fibers and show that for all Markov bases of IL , the multiset of equivalence classes is an invariant of IL . In the last years, due to applications of Markov bases to Algebraic Statistics, there is an interest in determining the indispensable binomials of a lattice ideal, see [25, 26, 4, 1, 27]. Of particular interest is the case when all elements in the Universal Markov basis of the lattice ideal are indispensable as is the case for generic lattice ideals, see [28]. An indispensable binomial is a binomial that appears in every Markov basis of the lattice ideal up to a constant multiple. When the lattice is positively graded the problem of determining such binomials has been completely solved, see [4]. In this paper we address this problem for general lattice ideals. We show that if the lattice L is not positively graded, then there is at most one indispensable binomial. Another question we address is characterizing binomial complete intersection lattice ideals. We recall that a lattice ideal IL of height r is a complete intersection if there exist polynomials P1 , . . . , Pr such that IL = hP1 , . . . , Pr i and IL is a binomial complete intersection if there exist binomials B1 , . . . , Br such that IL = hB1 , . . . , Br i. If Nakayama’s lemma applies then complete intersection lattice ideals are automatically binomial complete intersections. The problem is completely

BINOMIAL FIBERS

3

solved when L is positively graded by a series of articles: [14, 7, 32, 18, 36, 24, 30, 29, 12, 31, 23]. The final conclusion is that IL is a complete intersection if and only if the matrix M whose rows correspond to a basis of L is “mixed dominating”: every row of M has a positive and negative entry and M contains no square submatrix with this property. When L is not positively graded the situation is less clear. In this paper we characterize all binomial complete intersection lattice ideals. The structure of this paper is as follows: in Section 2 given the lattice L ⊂ Zn we introduce and examine the properties of Lpure , a sublattice of L which will be crucial in our study. Of particular importance is the support of Lpure which we denote by σL . In Section 3 we define an equivalence relation among the IL fibers. We order the resulting equivalence classes and show that any descending chain of such classes stabilizes. We prove that the multiset of equivalence classes of fibers corresponding to a Markov basis of IL is an invariant of IL . In Section 4 we characterize all minimal binomial generating sets of pure lattices. Then we describe all Markov bases of IL for any lattice L. We explicitly compute µ(IL ) in terms of a related graph. We show that if rank Lpure ≥ 1, there is at most one indispensable binomial. In Section 5 we characterize all lattices L such that IL is a binomial complete intersection. In section 6 we work out an example in full detail. 2. Fibers, the Pure Sublattice and Bases of a Lattice Let R = k[x1 , . . . , xn ] where k is a field, L a lattice in Zn , IL = hxu −xv : u−v ∈ Li. We denote by Tn the set of monomials of R including 1 = x0 . If J is a monomial ideal of R we denote by G(J) the unique minimal set of monomial generators of J. For r ∈ N we let [r] = {1, . . . , r}. Let a = (a1 , . . . , an ), b = (b1 , . . . , bn ) ∈ Zn . We write a ≥ b if ai ≥ bi for i = 1, . . . , n. We write a ≥ 0 if a ∈ Nn . If either a ≥ 0 or −a ≥ 0 we say that a is pure. We say that a, b are incomparable if a − b is not pure. In general we let supp(a) = {i : ai 6= 0} ⊂ [n]. For any subset X of Zn we let [ supp(X) := supp(w) . w∈X

n

Definition 2.1. We say that F ⊂ T is an IL -fiber if there exists xu ∈ Tn such that F = {xv ∈ Tn : v − u ∈ L}. If xu ∈ F , and F is an IL -fiber we write Fu or Fxu for F . If B ∈ IL and B = xu − xv we write FB for Fu . When F is an IL -fiber we let MF = hxu : xu ∈ F i be the monomial ideal generated by the elements of F . From the properties of the lattice and the definition of lattice ideals we get the following: Proposition 2.2. If xv ∈ Fu then Fu = Fv . Moreover Fu = {xv : xv − xu ∈ IL }. If xu − xv ∈ IL then u − v ∈ L. We remark that Fu is a singleton if and only if there is no binomial 0 6= B ∈ IL such that FB = Fu . We note that F ⊂ MF and G(MF ) ⊂ F . The following proposition follows also from [22, Theorem 8.6]. Proposition 2.3. Let L ⊂ Zn be a lattice. The following are equivalent: (1) The lattice L contains a nonzero pure element. (2) All IL -fibers are infinite. (3) There exists an IL -fiber which is infinite.

4

H. CHARALAMBOUS, A. THOMA, AND M. VLADOIU

Proof. (1) ⇒ (2) Let F be an IL -fiber and suppose that 0 6= u ∈ L ∩ Nn . It is easy to see that if v ∈ Nn and F is the IL -fiber such that xv ∈ F , then xv+lu ∈ F for all l ∈ N. Thus F is infinite. (2) ⇒ (3) obvious. (3) ⇒ (1) Suppose that an IL -fiber F is infinite. Let xv ∈ F be such that xv ∈ / G(MF ). Note that since F is infinite such a v exists. Since xv ∈ MF , there exists a monomial xu ∈ G(MF ) such that xu |xv and thus xv = xw xu for 0 6= w ∈ Nn . Since xv , xu ∈ F , it follows that w = v − u ∈ L, therefore w ∈ L ∩ Nn .  Corollary 2.4. Let L ⊂ Zn be a lattice. The lattice L is positively graded (i.e. L ∩ Nn = {0}) if and only if G(MF ) = F where F is any IL -fiber. Proof. Suppose that L ∩ Nn = {0}. Since MF = hF i to prove that G(MF ) = F , it is enough to show that if xa 6= xb ∈ F then a, b are incomparable. Suppose otherwise. Then a − b ∈ L is pure, a contradiction. For the other direction suppose that G(MF ) = F . Thus F is finite and the conclusion follows from Proposition 2.3.  Notation 2.5. We let L+ = L ∩ Nn , σL = supp(L+ ) and Lpure be the subgroup of L generated by L+ . In the course of the proof of Proposition 2.3 we proved the following: Proposition 2.6. Let L ⊂ Zn be a lattice and let F be an IL -fiber. If G(MF ) = {xa1 , . . . , xas } then s [ {xai xw : w ∈ L+ } . F = i=1

The next proposition considers the support of the elements of L that belong to Lpure . Proposition 2.7. There exists an element w in L+ such that supp(w) = σL . For u ∈ L we have that supp(u) ⊂ σL if and only if u ∈ Lpure . Proof. The existence of w follows from the observation that if w1 , w2 ∈ L+ then w1 + w2 ∈ L+ and supp(w1 ) ∪ supp(w2 ) = supp(w1 + w2 ). Suppose now that u ∈ L and supp(u) ⊂ σL . Let w ∈ L+ be such that supp(w) = σL . It is clear that for l ∈ N, l ≫ 0, u + lw = w′ ∈ Nn . Since u, lw ∈ L it follows that w′ ∈ L and thus w′ ∈ L+ . Therefore u = w′ − lw ∈ Lpure .  Since Lpure is generated by the elements of L+ it is clear that supp(Lpure ) = σL . n

σL

Let u = (ui ) ∈ Z . By u we mean the vector (ui )i∈σ / L . The following is an immediate consequence of Proposition 2.7. Corollary 2.8. Let L ⊂ Zn be a lattice and u ∈ L. Then u ∈ Lpure if and only if uσL = 0. Definition 2.9. A nonzero vector u ∈ L is called L–primitive if whenever λu ∈ L where λ ∈ Q then λ ∈ Z. In other words, u is L–primitive if and only if Qu∩L = Zu. Equivalently u is L–primitive if it is the “smallest” element of L in the direction determined by u. Proposition 2.10. Let 0 6= v ∈ L. There is an L–primitive vector u ∈ L such that v = λu for λ ∈ Z.

BINOMIAL FIBERS

5

k v where m 6= 1 and Proof. If v is not L–primitive there is v ′ ∈ L such that v ′ = m ′ gcd(k, m) = 1. Thus mv = kv and m divides all coordinates of v. Moreover there are t1 , t2 ∈ Z such that 1 = t1 k + t2 m. It follows that 1 1 v = t1 v ′ + t2 v =⇒ u = v ∈ L . m m We note that v is an integer multiple of u. If u is not L–primitive we repeat this procedure. Since v ∈ Zn this procedure has to end in a finite number of steps. 

Consider now any basis of L as a Z-module. The next theorem states that the elements of such a basis are necessarily L–primitive. Theorem 2.11. Let L ⊂ Zn be a lattice and let B be a basis of L as a Z-module. The elements of B are L–primitive. ∼ Zr, Proof. Since L is a sublattice of Zn , there exists an r ∈ N such that L = r = rank(L). Let B = {u1 , . . . , ur }. Suppose that for some i ∈ [r], ui is not L– primitive. By Proposition 2.10 it follows that there is v ∈ LP and m ∈ N, m 6= 1, 1 ui . Since B is a basis of L it follows that v = λj uj where λj ∈ Z such that v = m 1 for j ∈ [r]. Since the Q-coordinates of v are unique it follows that λi = m and λj = 0 for j ∈ [r] \ {i} and thus λi = m = 1, a contradiction.  We consider the usual Euclidean inner product in Zn : if a = (ai ), b = (bi ) we let P a · b = ai b i .

Theorem 2.12. Let L be a lattice and u1 an L-primitive vector. There exists a basis B of L such that u1 ∈ B.

Proof. We will do induction on r, the rank of L. If r = 1 we are done. Assume r > 1. Let w1 be a vector in Zn such that w1 · u1 = 0. Since r > 1 it is clear that there is u2 ∈ L such that w1 · u2 ∈ N. Choose u2 ∈ L to be such that w1 · u2 is positive and as small as possible. We will show that for u ∈ L there is a λ ∈ Z such that w1 · u = λ(w1 · u2 ). Suppose not. Then w1 · u = q(w1 · u2 ) + r, 0 ≤ r ≤ w1 · u2 . It follows that w1 · (u − qu2 ) = r which contradicts the choice of u2 . Notice also that u1 , u2 are linearly independent. We continue this way and obtain a sequence of linearly independent vectors u1 , . . . , ur and a sequence of vectors w1 , . . . , wr−1 that satisfy the following properties for 1 ≤ i ≤ r − 1: a) wi · uj = 0 for j = 1, . . . , i b) wi · ui+1 > 0 and c) if u ∈ L then there is a λ ∈ Z such that wi · u = λ(wi · ui+1 ). It is clear that {u1 , . . .P , ur } is a Q-basis of L. We show that {u1 , . . . , ur } is a Z-basis r of L. Let u ∈ L, u = i=1 λi ui where λi ∈ Q. Consider wr−1 · u. Then wr−1 · u = λr (wr−1 · ur ) .

P By c) above it follows that λr ∈ Z. Next consider u′ = u − λr ur = r−1 i=1 λi ui . Since wr−2 · u′ = λr−1 (wr−2 · ur−1 ) it follows as above that λ ∈ Z. In this way r−1 Pr we get that λr , . . . , λ2 ∈ Z. Consider now v = u − i=2 λi ui = λ1 u1 . Since v ∈ L and u1 is L–primitive it follows that λ1 ∈ Z.  Corollary 2.13. Let L be a lattice. There exists a basis of Lpure whose elements are in L+ and have support equal to σL .

6

H. CHARALAMBOUS, A. THOMA, AND M. VLADOIU

Proof. By Proposition 2.7 and Proposition 2.10 there is an L-primitive vector u1 ∈ L+ such that supp(u1 ) = σL . By Theorem 2.12 there exists a basis {u1 , . . . , ur } of Lpure . It is clear that for l ≫ 0, u′i = ui + lu1 ∈ L+ for i = 2, . . . , r. The set {u1 , u′2 , . . . , u′r } has the desired properties.  Bases of the lattice L are clearly important for the study of IL . We note though that it is well known that it is not enough to compute a basis for L to find a generating set of IL . Indeed we can associate to each u ∈ L the polynomial Bu = + − xu −xu where u = u+ −u− and u+ , u− ∈ Nn . Of course there are many binomials w1 w2 x − x such that u = w1 − w2 and they all belong to IL , but Bu and −Bu are the only ones with the property that its monomial terms are relatively prime: in some sense Bu is the lowest term binomial that corresponds to u. It is a basic fact that IL = hBu : u ∈ Li. Let E be a basis of L and define I(E) = hBu : u ∈ Ei. Clearly I(E) ⊂ IL but the equality does not have to hold. For example let L be the lattice of Z4 with basis E = {u1 = (1, −1, −1, 1), u2 = (1, −2, 2, −1)} and let B1 = xw − yz, B2 = xz 2 − y 2 w. The lattice L corresponds to the Macaulay curve:   4 3 1 0 L = ker 0 1 3 4 We see that u = (2, −3, 1, 0) ∈ L since u = u1 + u2 and that the polynomial B = x2 z −y 3 of k[x, y, z, w] is in IL . However B does not belong to I(E) = hB1 , B2 i since there is no way to create x2 z from the monomial terms of B1 and B2 . The problem is that the relation (2, −3, 1, 0) = (1, −1, −1, 1) + (1, −2, 2, −1) does not translate to a relation among the lowest term binomials. However the above relation on the elements of L translates to the following relation on elements of I(E): wB = xzB1 + yB2 . Indeed, as it is shown in [33, Lemma 12.2], for toric ideals the following relation holds: IL = I(E) : (x1 · · · xn )∞ , where for f ∈ k[x1 , . . . , xn ] and J an ideal of k[x1 , . . . , xn ], J : f ∞ = {g ∈ k[x1 , . . . , xn ] : gf r ∈ J, r ∈ N}. In the next two sections we are going to describe minimal generating sets of lattice ideals using properties of the IL -fibers and of the bases of Lpure . 3. Fibers and Markov bases of Lattice Ideals Let R = k[x1 , . . . , xn ] where k is a field, L ⊂ Zn a lattice. For simplicity of notation from now on we write I := IL ,

σ := σL .

If G ⊂ Tn and t ∈ Nn we let xt G := {xt xu : xu ∈ G}. Lemma 3.1. Let G, F be I-fibers. If there exists w1 ∈ Nn , xu ∈ G such that xw1 xu ∈ F then xw1 G ⊂ F . Moreover if xw2 F ⊂ G for w2 ∈ Nn then w1 + w2 ∈ L+ and supp(wi ) ⊂ σ, i = 1, 2.

BINOMIAL FIBERS

7

′

Proof. Suppose that xw1 xu = xv ∈ F and let xu ∈ G. Since u′ − u ∈ L it follows ′ that (w1 + u′ ) − v ∈ L and thus xw1 +u ∈ F . If in addition xw2 F ⊂ G it follows that xw1 +w2 F ⊂ F . Since xw1 +w2 xv , xv ∈ F it follows that w1 + w2 ∈ L ∩ Nn .  Definition 3.2. Let F , G be I-fibers. We say that F ≡I G if there exist u, v ∈ Nn such that xu F ⊂ G and xv G ⊂ F . It is immediate that F ≡I G is an equivalence relation among the I-fibers. We denote the equivalence class of F by F . Thus F = {G : G is an I -fiber, G ≡I F } . We note that F ≡I G implies that the cardinality of F is equal to the cardinality of G. Lemma 3.3. If Lpure = {0} and F is an I-fiber then F = {F }. Proof. By Proposition 2.3, |F | < ∞. Let G be an I-fiber, G ≡I F . There are u, v ∈ Nn such that xu F ⊂ G and xv G ⊂ F . Since |F | = |xu F | = |G| = |xv G| it follows that xv xu F = F and xv = xu = 1.  Next we want to investigate the number of equivalent fibers inside each equivalence class when Lpure 6= {0} and σ 6= ∅. If u ∈ Zn we let uσ = (ui )i∈σ . If s = |σ| we can assume that uσ ⊂ Zs and then consider the sublattice (Lpure )σ of Zs generated by the vectors uσ , u ∈ Lpure . First we note the following: Remark 3.4. Let σ 6= ∅, F an I-fiber and u ∈ Nn such that supp(u) ⊂ σ. If G is an I-fiber with the property xu F ⊂ G then G ∈ F . Proof. Let w ∈ L+ be such that supp(w) = σ. There exists l ≫ 0 such that lw − u ∈ Nn . Since lw ∈ L it follows that xlw F ⊂ F . Let xv ∈ G such that xv = xu xp for xp ∈ F . It follows that xlw−u xv = xlw xp ∈ F and thus xlw−u G ⊂ F by Lemma 3.1.  Proposition 3.5. Let Lpure 6= {0} and F an I-fiber. The cardinality of F is equal to |Zs /(Lpure )σ |, where s = |σ|. Proof. For every G ∈ F choose uG ∈ Nn such that xuG F ⊂ G. Let φ : F → Zs /(Lpure )σ , φ(G) = (uG )σ + (Lpure )σ . The definition of φ is independent of the choice of uG . Indeed suppose that u, v ∈ Nn are such that xu F ⊂ G and xv F ⊂ G. This implies that u − v ∈ L. By Lemma 3.1 it follows that uσ = v σ = 0 and by Proposition 2.7 it follows that u − v ∈ Lpure and uσ − vσ ∈ (Lpure )σ . We will show that ϕ is a bijection: the only part needing proof is the surjection of ϕ. Let u′ + (Lpure )σ be an element of Zs + (Lpure )σ . First we remark that we can assume without loss of generality that u′ ∈ Ns . Indeed, let w ∈ L+ be such that | supp(wσ )| = s. It is clear that for l ≫ 0, lwσ + u′ ∈ Ns and thus u′ + (Lpure )σ = (lw + u′ ) + (Lpure )σ . Let u ∈ Nn be such that uσ = 0, uσ = u′ and let G be the I-fiber such that xu F ⊂ G. By Remark 3.4 it follows that G ∈ F , and thus φ(G) = u′ + (Lpure )σ . 

8

H. CHARALAMBOUS, A. THOMA, AND M. VLADOIU

Examples 3.6. (a) We consider the lattice ideal I = h1 − xyi ⊂ k[x, y] where L = h(1, 1)i ⊂ Z2 . There are infinitely many I-fibers: for any c ∈ Z the set Fc = {xi y j : i−j = c} is an I-fiber. All I-fibers are infinite and belong to the same equivalence class: the cardinality of this equivalence class is |Z|. Indeed Lpure = L, and Z2 / L ∼ = Z. (b) If we consider the lattice ideal I = h1 − xy, 1 − x5 i ⊂ k[x, y], where L = h(1, 1), (5, 0)i ⊂ Z2 then there are exactly five infinite I-fibers: Fk = {xi y j : i − j ≡ k

mod 5},

0≤k≤4,

which are all equivalent. Hence we have only one equivalence class F0 = {F0 , . . . , F4 } which has five equivalent fibers. Indeed Lpure = L and Z2 /L ∼ = Z5 . We define the relation “≤I ” among the equivalence classes of I-fibers. Definition 3.7. Let F , G be I-fibers. We say that F ≤I G if there exists u ∈ Nn such that xu F ⊂ G. It is immediate that “≤I ” is well defined and is a partial order among the equivalence classes of I-fibers. For simplicity of notation we occasionally write F ≤I G if F ≤I G and F I F k >I F k+1 >I · · · a chain of equivalence classes of fibers with no least element. Choose a representative Fi , i ∈ N for each class. Next consider the corresponding ascending chain of monomial ideals: MF1 ⊂ · · · ⊂ MF1 + · · · + MFk ⊂ MF1 + · · · + MFk+1 ⊂ · · · The chain stabilizes at some step, say s, so that MF1 + · · · + MFs = MF1 + · · · + MFs+1 . Let x ∈ G(MFs+1 ). By the above equality it follows that xa ∈ MFi for some 1 ≤ i < s + 1 and xa = xu xb where xb ∈ G(MFi ). Since xu xb ∈ Fs+1 it follows that xu Fi ⊂ Fs+1 . This leads to a contradiction since F s+1