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Graded Bourbaki ideals of graded modules

In this paper we study graded Bourbaki ideals. It is a well-known fact that for torsionfree modules over Noetherian normal domains, Bourbaki sequences exist. We give criteria in terms of certain attached matrices for a homomorphism of modules to induce a Bourbaki sequence. Special attention is given to graded Bourbaki sequences. In the second part of the paper, we apply these results to the Koszul cycles of the residue class field and determine particular Bourbaki ideals explicitly. We also obtain in a special case the relationship between the structure of the Rees algebra of a Koszul cycle and the Rees algebra of its Bourbaki ideal.


Introduction
The purpose of this paper is to study Bourbaki sequences and Bourbaki ideals. Throughout this section let R be a commutative Noetherian ring and M a finitely generated R-module. Then a Bourbaki sequence of M is a short exact sequence (1) 0 → F → M → I → 0 of R-modules, where F is a free R-module and I is an ideal of R. I is called a Bourbaki ideal of M. As a fundamental result, a Bourbaki sequence of M always exists if R is a normal domain and M is a finitely generated torsionfree R-module (see [3, Chapter VII, Section 4, 9. Theorem 6.]). If R is a standard graded normal domain over an infinite field, then a graded Bourbaki sequence of M also exists (Theorem 2.1, see also [9,Corollary 2.4]). One of the advantages of Bourbaki's theorem is the fact, that, by passing to a Bourbaki sequence, many properties of a module are inherited by those of its Bourbaki ideals. One can find applications of Bourbaki's theorem, for instance, to the vanishing of cohomologies, the study of the maximal Cohen-Macaulay modules over hypersurface rings, the Hilbert functions, and the Rees algebras of modules ( [2,7,8,9,11,12]). On the other hand, even though we know about the existence of a Bourbaki sequence, it is not easy to construct one explicitly. Actually, for a given homomorphism of modules, it is still difficult to check whether the map induces a Bourbaki sequence. Now let us explain how we organized this paper. In Section 2 we prepare general propositions to study Problem 1.1, and discuss the existence of graded Bourbaki sequences. In Section 3 we introduce an invariant, the Bourbaki number, obtained from a graded Bourbaki sequence over the polynomial ring. The Bourbaki number is an integer which only depends on the degree of the generators and invariants of M. It will be useful to find in Section 5 graded Bourbaki ideals for Koszul cycles.
In Section 4 we solve Problem 1.1 under some additional conditions, as described in the following theorem. (a) Suppose that M is reflexive and take an exact sequence 0 → M ι − → F → X → 0 so that F is a free R-module and X is a finitely generated torsionfree R-module. Then the following conditions are equivalent: (i) 0 → R r−1 ϕ − → M → Coker(ϕ) → 0 is a Bourbaki sequence; (ii) height(I r−1 (ι • ϕ)) ≥ 2.
As an application of the above theorem, we will illustrate the ubiquity of graded Bourbaki sequences (Theorem 4.4). Furthermore, we also give a method to compute a Bourbaki ideal for a given Bourbaki sequence (Theorem 4.7).
Let S = K[x 1 , . . . , x n ] be the polynomial ring. In Section 5 we apply the previous results to the Koszul cycles Z i of the residue class field K, and determine particular Bourbaki ideals explicitly in the cases i = 2, n−2, n−1. For i = n−1 and i = n−2, we can choose multigraded Bourbaki sequences. Hence the corresponding Bourbaki ideals are monomial ideals (Proposition 5.2 and 5.3). On the other hand, as shown in Theorem 5.4, multigraded Bourbaki sequences do not exist for 1 < i < n − 2 when n ≫ 0 or n ≤ 6. We expect it is also also the case for all n > 6.
In the last part of this paper we show that our Bourbaki ideal for Z n−2 has the property that its Rees algebra is normal and Cohen-Macaulay. Moreover, for n even it is Gorenstein and it is of Cohen-Macaulay type 2 if n is odd. The same properties are known for the Rees algebra of Z n−2 , see [10, Theorem 3.1 and Theorem 3.4]. Let us fix our notation throughout this paper. In what follows, let R be a commutative Noetherian ring. Let ϕ : F → G be an R-module homomorphism between finitely generated free R-module F and G. We then denote I t (ϕ) the ideal of R generated by the t-minors of a matrix representing ϕ. Q(R) denotes the total ring of fraction of R and the functor (−) * denotes the R-dual. For a finitely generated R-module M, we say that M is torsionfree (resp. reflexive) if the canonical map M → Q(R) ⊗ R M is injective (resp. the canonical map M → M * * is bijective).
If R = n≥0 R n is a graded Noetherian ring over a field K = R 0 and M is a finitely generated graded R-module, t 0 (M) denotes the maximum degree of an element in a minimal homogeneous system of generators of M. software CoCoA ( [1]) and Singular ( [5]) for our computations. The authors would like to thank Ernst Kunz and Bernd Ulrich for discussions around Lemma 4.3.
This paper was written while the second author visited Essen for three months and Bucharest for two weeks. During the visits, Professor Herzog and Professor Stamate were extremely kind to him every day and he would like to express sincere gratitude for their hospitality. The third author is grateful to Professor Herzog and the Department of Mathematics of the Universität Duisburg-Essen for being excellent hosts, one more time.

Preliminaries
Let R be a Noetherian ring and M a finitely generated R-module. As a fundamental result, a Bourbaki sequence of M always exists if R is a normal domain and M is a finitely generated torsionfree R-module (see [3, Chapter VII, Section 4, 9. Theorem 6.]). We also have a graded version of Bourbaki sequences. Theorem 2.1. Let R = n≥0 R n be a standard graded Noetherian normal domain where R 0 is an infinite field and dim R ≥ 2. Let M = n∈Z M n be a finitely generated torsionfree graded R-module of rank r > 0.
Then for any integer k ≥ t 0 (M), there exists a graded Bourbaki sequence of M, for some integer m and I is a graded ideal of R.
Moreover, if R is a factorial ring, then there exists a Bourbaki sequence as in (2) with grade(I) ≥ 2.
Proof. By lack of good reference we outline the proof. Since R is standard graded and k ≥ t 0 (M), it follows that M ≥k is generated in degree k.
By [9,Corollary 2.4] we have an exact sequence where N ′ is a rank 1 torsionfree R-module. From this we construct a graded Bourbaki sequence of M. Indeed, consider the following commutative diagram where N denotes the cokernel of the composition R(−k) r−1 → M ≥k → M. Since M/M ≥k has finite length, N has rank one and Ass(N) ⊆ {0, m}, where m = R >0 is the graded maximal ideal in R. If m ∈ Ass(N), since depth R m ≥ 2, it follows that depth Rm M m = 0 and m ∈ Ass(M). This is a contradiction for the torsionfreeness of M. Whence Ass(N) = {0}. Therefore, N is torsionfree of rank 1. Let S be the multiplicatively closed set of non-zero homogeneous elements in R. It follows that N → S −1 R ⊗ R N is injective and S −1 R ⊗ R N ∼ = S −1 R. This implies that N ∼ = I(m) as a graded R-module, where I ⊂ R is a graded ideal and m is a suitable integer. Now assume in addition that R is a factorial ring. If I is of grade 1, then I = α·J for some graded ideal J with gcd(J) = 1. Therefore we may as well assume that gcd(I) = 1, and hence height(I) ≥ 2. Since R is a factorial domain, it is a normal ring and satisfies Serre's condition (S 2 ). Therefore, grade(J) ≥ 2.
Non-trivial Bourbaki sequences we only obtain when the grade of the Bourbaki ideal is 2. Indeed, we have Lemma 2.2. Let R be a Noetherian ring and M a finitely generated R-module with In particular, if M is not free and reflexive or M is an indecomposable module of rank ≥ 2, then grade(I) = 2.
In particular, R/I is Cohen-Macaulay on the punctured spectrum of R if M is locally free on the punctured spectrum of R.
Proof. For p ∈ Spec R, let be the localization of the above Bourbaki sequence. Then proj dim On the other hand, since grade(I) = 2, it follows that proj dim Rp IR p = 0 ⇔ I ⊆ p, and proj dim Rp IR p = 1 ⇔ IR p is perfect in the sense of [4,Definition 1.4.15], in other words, a Cohen-Macaulay ideal.

The Bourbaki number of graded torsionfree modules
In this section, S = K[x 1 , x 2 , . . . , x n ] is a polynomial ring of dimension n ≥ 2 over an infinite field K. For any finitely generated graded S-module M of positive dimension s, the Hilbert function of M, which is defined as H M (t) = dim K M t for all t ∈ Z, eventually agrees with a polynomial function of degree s − 1. Thus we may write  Proof. By using the additivity of the Hilbert function on the graded exact sequence we get that for t ≫ 0, Since the dimension of S/I is at most n−2, the degree of the polynomial H S/I (t+m) in t is at most n − 3. Hence m + e 1 (M) − k(r − 1) = 0, which gives the desired formula for m.  If a graded free resolution of M is known, e 1 (M) can be computed as follows. Proposition 3.3. Let M be a finitely generated graded S-module with dim M = dim S, and let 0 → F p → · · · → F 1 → F 0 → M → 0 be any finite graded free resolution of M.
and dim M = n, we have −e 1 (M) = p i=0 j∈Z (−1) i ·b ij ·(−j). Let us consider the graded Bourbaki ideals. When proj dim R M = 1, the ideal I in (4) can be described as an ideal of maximal minors of a certain matrix over S.
Assume M has rank r > 0 and let k ≥ t 0 (M). Let be a graded free resolution of M and a graded Bourbaki sequence with height(I) ≥ 2, respectively. Since S(−k) r−1 is a projective module, there exists a graded S-module homomorphism ψ 2 : S(−k) r−1 → F 0 such that π • ψ 2 = ϕ. Then I can be obtained as follows.
Proposition 3.4. Assume M is a finitely generated graded module which is torsionfree of rank r > 0 with proj dim M = 1. With the above notation we set is a graded free resolution of I(m), I = I α−1 (ψ 1 ψ 2 ), and Furthermore, if M is generated in degree k, then there exist a matrix A representing ψ and a (β + 1) × β submatrix A ′ of A such that I = I β (A ′ ).
Proof. Note that r = α − β. It is easy to check that is a graded exact sequence of S-modules. Since Ker(ψ 1 ψ 2 ) is a torsionfree S-module of rank β + (α − β − 1) − α + 1 = 0, it follows that Ker(ψ 1 ψ 2 ) = 0. Therefore, since grade(I) ≥ 2, the Hilbert-Burch theorem [4,Theorem 1.4.17] implies that I = I α−1 (ψ 1 ψ 2 ). The displayed formula for m follows by a simple computation from Theorem 3.1 and Proposition 3.3. Suppose that M is generated in degree k. Since ψ 2 : S(−k) α−β−1 → F 0 = S(−k) α is an injective graded map of degree 0, we may pick bases of S(−k) α−β−1 and of F 0 such that the matrix representing ψ 2 is We pick any free basis of F 1 and we denote A = (a ij ) the matrix representing ψ 1 in the fixed bases. Then

Characterization of Bourbaki sequences
In this section we consider maps ϕ : R s → M and enquire whether Coker(ϕ) is a torsionfree module. This applies, when R is a normal domain and M a torsionfree R-module of rank r, to characterize the maps ϕ : R r−1 → M which are part of a Bourbaki sequence like (1).
In the following lemma we present characterizations for torsionfree and for reflexive modules, respectively. For the convenience of the reader, we include the proof.
Lemma 4.1. Let R be a Noetherian ring and M a finitely generated R-module. Suppose that R is generically Gorenstein, that is, R p is Gorenstein for all p ∈ Ass(R). (a) The following conditions are equivalent: (i) M is torsionfree; The following conditions are equivalent: where F and G are finitely generated free R-modules.
Proof. Note that Ass(R) = Min(R) since R p is an Artinian Gorenstein ring for all p ∈ Ass(R). Let p ∈ Ass(R). Since R p is an Artinian Gorenstein ring, by Matlis duality we get that ϕ p is a bijective map. Thus Ass(R) ∩ Ass(Ker(ϕ)) = ∅. Since Ass(Ker(ϕ)) ⊆ Ass(M) ⊆ Ass(R), we obtain that the map ϕ is injective, which proves (iv).
For any p ∈ Ass(R), R p is an Artinian Gorenstein ring, hence Ext 1 Rp (X p , R p ) = 0. The latter implies N p = 0. Since N is torsionfree, arguing as before, we obtain that N = 0.
Hence, by applying Hom R (−, R) again, we obtain the commutative diagram with exact rows. The cokernel of the canonical map ϕ : M → M * * is isomorphic to the kernel of the map X → Coker(ϕ * * ), whence Coker(ϕ) is torsionfree. Since ϕ p is bijective for all p ∈ Ass(R), ϕ is bijective. It follows that M is a reflexive module.
Theorem 4.2. Let R be a normal domain of dimension ≥ 2 and M a finitely generated reflexive R-module, and let ϕ : be an exact sequence of R-modules with F free and X torsionfree. Then the following conditions are equivalent: Proof. (i) ⇐⇒ (ii): We can rephrase the condition that ϕ is injective as follows.
Hence we may assume that ϕ is injective. 9 Set N = Coker(ϕ) and C = Coker(ι • ϕ). Consider the commutative diagram Then where the first and third equivalence follow from Lemma 4.1 and the second equivalence follows from the inclusions On the other hand, by [4, Proposition 1.4.1(a)], C is torsionfree if and only if depth C p > 0 for all p ∈ Spec R with height(p) ≥ 1. Let p be a prime ideal of R. By (5), proj dim Rp C p ≤ 1. From the Auslander-Buchsbaum theorem we obtain If height(p) ≥ 2, since R is a normal ring it satisfies Serre's condition (S 2 ), hence depth R p ≥ 2. Thus depth Rp C p > 0.
If height(p) = 1, since R satisfies the condition (R 1 ) we get that R p is a regular local ring. Consequently, depth Rp Hence, the torsionfreeness of C is equivalent to saying that C p is R p -free for any height one prime ideal p of R. By [4, Proposition 1.4.10] the latter condition is equivalent to saying that (ι • ϕ) p is a split monomorphism for any height one prime p, which in turn is equivalent to I s (ι • ϕ) p = R p for any height one prime p.
Therefore, N is a torsionfree module if and only if height(I s (ι • ϕ)) ≥ 2.
Assume R is graded and M is a graded R-module of rank r. We will prove that under mild assumptions on R and M, given k ≫ 0 and a K-basis m 1 , . . . , m α of M k , then any r − 1 generic K-linear combinations of m 1 , . . . , m α generate a free submodule of M which is the beginning of a graded Bourbaki sequence of M. Lemma 4.3. Let R = n≥0 R n be a graded Noetherian ring such that R 0 = K is an algebraically closed field. Let T = R[z 1 , . . . , z m ] be the polynomial ring over R. We regard T as a graded ring by using the grading of R and deg(z i ) = 0 for 1 ≤ i ≤ m. Let I be a graded ideal of T such that I ⊆ n>0 T n and set S = T /I.
is a Zariski open subset of K m for any integer e.
By semicontinuity of fiber dimension (see for example [6,Theorem 14.8,b]), for any integer e, there exists an ideal J e of R ′ such that Combining Theorem 4.2 and Lemma 4.3 yields the following result.
Theorem 4.4. Let R = n≥0 R n be a graded Cohen-Macaulay normal domain such that R 0 = K is an algebraically closed field. Let M = n∈Z M n be a finitely generated reflexive graded R-module of rank r > 0. Suppose that M has no free summands. Let With these assumptions and notation, the set Proof. Our assumptions on M imply that there exists a graded exact sequence 0 → M ψ − → H → X → 0 of R-modules such that H is a free R-module generated in single degree of rank β and X is torsionfree. Let h 1 , . . . , h β be a basis of H.
Let T = R[z ij ] 1≤i≤α,1≤j≤r−1 be the graded polynomial ring over R with deg(z ij ) = 0, and let A denote the matrix representing ψ • ι • π with respect to the bases g 1 , . . . , g α and h 1 , . . . , h β . Note that I = I r−1 (A·(z ij )) is a graded ideal of T such that I ⊆ n>0 T n since M has no free summands.
If M p is not free, then considering the mapping cone of ϕ p we obtain that Using the Auslander-Buchsbaum formula again, we get that depth Rp M p = depth Rp N p . On the other hand, since M is torsionfree, one has depth Rp M p ≥ 1. Therefore, we have the following chain of equivalences: where the second equivalence follows from the above argument. For the third equivalence we use that when height(q) = 1 the ring R q is regular, so proj dim The next corollary is an immediate consequence of the previous theorem.
The next theorem tells us how to compute a Bourbaki ideal, once we have its relation matrix.
With respect to the canonical bases, B T represents ψ * , and let the vector u = (u 1 , . . . , u α ) T represent the map γ. Then B T u = 0, and the elements u 1 , . . . , u α generate the ideal I.
Let ∆ i (C T ) be the determinant of the matrix which is obtained by deleting the ith column of C T , and set for a unique nonzero element c ∈ Q(R). As R is a factorial ring, the greatest common divisor of f 1 , . . . , f α exists. Let g = gcd{f 1 , . . . , f α }. Then I = cgJ where J = (1/g)(f 1 , f 2 , . . . , f α ) is an ideal in R of grade ≥ 2. Since both ideals I and J have grade ≥ 2, it follows that cg is a unit element in R. The desired conclusion follows with x = 1/c.

Bourbaki ideals of Koszul cycles
In this section we study Bourbaki ideals of Koszul cycles, and compute them explicitly in some cases. Let S = K[x 1 , x 2 , . . . , x n ] be a polynomial ring of dimension n ≥ 2 over a field K, which is not necessarily infinite. Let m = (x 1 , x 2 , . . . , x n ) be the maximal graded ideal of S and be the Koszul complex on the sequence x 1 , x 2 , . . . , x n , which is a linear free graded resolution of S/m ∼ = K.
We set Z i = Im ∂ i for 1 ≤ i ≤ n, which are torsionfree graded S-modules generated in degree i. Clearly, Z 1 ∼ = m and Z n ∼ = R(−n). Hence we focus on the Bourbaki ideals of Z i where 2 ≤ i ≤ n − 1.
For 2 ≤ i ≤ n − 1, we denote r i = rank Z i . By Theorem 2.1 and Lemma 2.2, there exist a graded ideal I and a graded exact sequence such that height(I) = 2, where the integer m i is computed in Theorem 3.1.

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Furthermore, Proposition 2.3 implies that S/I is Cohen-Macaulay on the punctured spectrum of S.
Let e 1 , . . . , e n be a basis of the free S-module K 1 . For any subset {i 1 , i 2 , . . . , i k } of {1, 2, . . . , n}, we denote the wedge product e i 1 ∧ e i 2 ∧ · · · ∧ e i k by e i 1 i 2 ...i k . For any k ≥ 1, the elements e i 1 i 2 ...i k with 1 ≤ i 1 < · · · < i k ≤ n form a basis for K k . For convenience, we denote by e i 1 i 2 ...i k the unique canonical basis element in K n−k such that e i 1 i 2 ...i k e i 1 i 2 ...i k = e 12...n .
Considering the free resolution of Z i which results from the Koszul resolution of K we obtain that rank(Z i ) = i−1 k=0 (−1) k n i−1−k . The desired result follows then by induction on i.

5.1.
Bourbaki ideals of Z n−1 and Z n−2 . In this subsection we investigate the Bourbaki ideals of Z n−1 and Z n−2 . In this case, we show that we can choose a Bourbaki sequence so that F is a submodule of K i generated by a part of the canonical basis of K i . It will follow that the Bourbaki ideal We first treat the case i = n − 1. By Lemma 5.1, r n−1 = rank(Z n−1 ) = n − 2. Note that proj dim Z n−1 = 1, hence we may apply Proposition 3.4.
Proof. Let F be the submodule of K n−1 generated by the elements e k with k ∈ {1, 2, . . . , n} \ {i, j}. We denote ϕ : F → Z n−1 the restriction of ∂ n−1 to F .
We claim that ϕ(F ) ∼ = S(−n + 1) n−2 . Indeed, suppose that k∈{1,2,...,n}\{i,j} where f k ∈ S. On the left hand side of the above equation, the coefficient of e ik is f k x i up to sign, hence f k = 0 for all k. Then from the chain of isomorphisms we derive the graded Bourbaki sequence as desired.
Let F be the submodule of K n−2 generated by the elements e ij with 1 ≤ i < j ≤ n and (i, j) ∈ A. Let r = r n−2 . Note that rank(F ) = r − 1.
Let ι : Z n−2 → K n−3 be the inclusion map and ϕ : F → Z n−2 the restriction of ∂ n−2 to F . Then, by Theorem 4.2, ϕ is injective and Coker(ϕ) is a torsionfree module if and only if the height(I r−1 (ι • ϕ)) ≥ 2.
By our choice of the basis of F , the map ι • ϕ : F → K n−3 is a multigraded Smodule homomorphism. So, if D is the matrix representing ι • ϕ with respect to the canonical bases, then I r−1 (D) is a monomial ideal. To conclude that height(I r−1 (D)) is at least two, it suffices to show that no x i divides all the monomial generators of I r−1 (D). By symmetry, it is enough to show this for i = n. In other words, we have to show that there exists an (r − 1) × (r − 1) submatrix of D such that its determinant is not divisible by x n .

Hence the determinant of
x 2 x 3 · · · x n−2 , which is not divisible by x n . Therefore, the sequence 0 → F ϕ − → Z n−2 → Coker(ϕ) → 0 is exact and Coker(ϕ) is a torsionfree module of rank one.
Set N = Coker(ϕ). Then, since N ∼ = K n−2 /(Im ∂ n−1 + F ), for N we have the graded free presentation gives the map with respect to the canonical bases of K n−1 and (i,j)∈A S e ij . Recall that I is the ideal generated by the monomials x/x i x j with (i, j) ∈ A. We denote ψ : Hence N ∼ = I since the Kernel of ψ is torsionfree of rank zero. Therefore, we have the graded Bourbaki sequence 0 → F ϕ − → Z n−2 → I → 0, as desired.

5.2.
When does Z i have a multigraded Bourbaki sequence? For i = n − 1 or i = n−2, in the previous subsection we found free submodules F ⊂ K i such that 0 → is a graded Bourbaki sequence and F is generated by a subset of the canonical basis of K i . In this situation, Z i admits a multigraded Bourbaki sequence with respect to the following multigrading: we assign to the variable x i the multidegree mdeg(x i ) = e i = (0, . . . , 0, 1, 0, . . . , 0) ∈ Z n , for i = 1, . . . , n, and we let mdeg(e i 1 ,...,i j ) = e i 1 + · · · + e i j when 1 ≤ i 1 < · · · < i j ≤ n.
However, the next result shows that multigraded Bourbaki sequences only exist for special values of i and n.
Theorem 5.4. Let 2 ≤ i ≤ n − 1 and F a free submodule of K i generated by a subset of the canonical basis of K i such that the sequence is a graded Bourbaki sequence of Z i . Then Proof. To show the conclusion, we prove that i is at least each of the terms on the right hand side of the inequality (8). For 1 ≤ j ≤ n we set I j = {(ℓ 1 , . . . , ℓ j ) | 1 ≤ ℓ 1 < · · · < ℓ j ≤ n}. Then {e p | p ∈ I j } is the canonical basis of K j . We assume that H = {e p 1 , . . . , e p r−1 } is an S-basis of F , where r = rank(Z i ) = n−1 i−1 and p 1 , . . . , p r−1 ∈ I i . Let ϕ : F → K i−1 be the composition of the inclusion F → K i and ∂ i . Let A be a matrix representing ϕ with respect to the specified bases of F and K i−1 .
Since (7) is a Bourbaki sequence of Z i , Theorem 4.2 implies that height I r−1 (ϕ) ≥ 2. As whence a k ≤ |{q ∈ I i−1 | k ∈ q}| = n−1 i−2 . Adding these relations for all k we obtain n i=1 a k ≤ n n−1 i−2 , from which we get that i ≥ i n−1 i−1 − n n−1 i−2 . To verify the second inequality subsumed by (8) we let ψ : K i+1 → K i /F be the composition of ∂ i+1 and the canonical map K i → K i /F . Then ψ is a multigraded S-module homomorphism.
Since (7) is a Bourbaki sequence, the module Coker ψ is torsionfree. Now, using the free presentation K i+1 Arguing as in the first part of this proof, for all k = 1, . . . , n we find p On the other hand, the kth component of s Proof. It is easy to check that 2 < 2 n−1 1 − n n−1 0 = n − 2, when n ≥ 5, and that n − 3 < 3 n−1 n−3 − n n−1 n−2 when n ≥ 8. Then one applies Theorem 5.4. We formulate the following. When n = 5, the answer is positive, by Theorem 5.4. When n = 6 the answer is also positive: the case i = 2 is covered in Proposition 5.6, and the case i = 3 is treated by ad-hoc methods in Proposition 5.12.

5.3.
Bourbaki ideals of Z 2 and Z 3 . In this section we construct a graded Bourbaki sequence and determine explicitly a Bourbaki ideal of Z 2 for arbitrary n.
When n = 6, we show in Proposition 5.12 that Z 3 does not have a multigraded Bourbaki sequence. Nevertheless, we describe a graded one for it in Proposition 5.13. Proposition 5.8. Let F be the submodule of K 2 generated by the elements e i,i+1 − e i+1,i+2 with 1 ≤ i ≤ n − 2 and let ϕ = ∂ 2 | F : F → Z 2 be the restriction of ∂ 2 . Then Proof. Set f i = e i,i+1 − e i+1,i+2 for 1 ≤ i ≤ n − 2. They are a free basis for F and rank(F ) = n − 2 = rank(Z 2 ) − 1. Let ι : Z 2 → K 1 denote the inclusion map. Let A n be the matrix representing ι • ϕ with respect to the bases f 1 , . . . , f n−2 and e 1 , . . . , e n . Then We prove that height(I n−2 (A n )) ≥ 2 by induction on n ≥ 3. If n = 3, then , whence its height is two. Assume our assertion holds for  Let p be any height one prime ideal of S. If I n−2 (A n ) ⊆ p, then x n−1 I n−3 (A n−1 ) ⊆ p. By the induction hypothesis, height(I n−3 (A n−1 )) ≥ 2, hence x n−1 ∈ p. Arguing similarly, from x 2 I n−3 (B n−1 ) ⊆ p we derive that x 2 ∈ p. Then (x 2 , x n−1 ) ⊆ p, which is a contradiction. This shows that height(I n−2 (A n )) ≥ 2. Therefore, 0 → F ϕ − → Z 2 → Z 2 /∂ 2 (F ) → 0 is a graded Bourbaki sequence of Z 2 , by Theorem 4.2.
Next we compute the Bourbaki ideal of Z 2 determined by the embedding ϕ in Proposition 5.8, keeping the notation from there.
Let B be the matrix representing ψ with respect to the canonical basis of K 3 , and {e ij | (i, j) ∈ H}, respectively. Let N = rank(K 2 /F ) = n 2 − (n − 2). In order to apply Theorem 4.7, we will describe an N × (N − 1) submatrix of B of rank N − 1.
(a) For i = j, the entries of the first column of C ij are zero.
With notation as above, there exist a graded ideal of height two isomorphic to Z 2 /∂ 2 (F ) and which we explicitly describe as follows. Proof. By Lemma 5.9, the matrix C has the following form It follows by Theorem 4.7 that there exists a unique element b ∈ S such that I = (1/b)I N −1 (C).
For (i, j) ∈ H, let ∆ ij (C) be the determinant of the matrix which is obtained from C by deleting the row corresponding to e ij . Then b = gcd(∆ ij | (i, j) ∈ H).

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Note that for Expanding this matrix with respect to the column corresponding to e i,i+1,i+2 , we see that . It follows that the greatest common divisor of ∆ 1,3 (C), . . . , ∆ n−2,n (C) is a = n−2 i=2 x n−1−i i . Clearly, b divides a. On the other hand, I is generated in degree n − 2, by Lemma 5.1(b). Hence It follows that deg(a) = deg(b), thus I = (1/a)I N −1 (C).
Example 5.11. We explain the previous constructions for n = 5. Then After computing its maximal minors with CoCoA ([1]), we find that a Bourbaki ideal of Z 2 is . Proposition 5.12. Suppose n = 6. Let F be a free submodule of K 3 generated by a subset of the canonical basis of K 3 and let ϕ = ∂ 3 | F : F → Z 3 be the restriction of ∂ 3 . Then the sequence is not a graded Bourbaki sequence of Z 3 .
Proof. Suppose that (9) is a graded Bourbaki sequence of Z 3 . Then rank(F ) = rank(Z 3 ) − 1 = 9, by Lemma 5.1. Let B be a basis of F which is part of the canonical basis for K 3 . If e m 2 ,m 3 ,m 4 = 0, then e m 2 ,m 3 ,m 4 ∈ Im ∂ 4 + F . Since Im ∂ 4 is generated in degree four, we get that e m 2 ,m 3 ,m 4 ∈ F , which is not the case. So, e m 2 ,m 3 ,m 4 is a nonzero torsion element in K 3 /(Im ∂ 4 + F ) ∼ = Z 3 /∂ 3 (F ) which contradicts the fact that (9) is a Bourbaki sequence.
Thus, we conclude that for any distinct m 1 , m 2 , m 3 , m 4 ∈ The 9 subsets of {1, . . . , 6} which index the elements in B use 27 indices, so by the pigeon hole principle, there exists one index which is used at least 5 times. Let G be the graph on the vertex set [5] and edges E(G) = {(ij) : e ij6 ∈ B}. Note that there is no cycle of length 3 in G. Indeed, if (ij), (jk), (ik) ∈ E(G), then e ij6 , e jk6 , e ik6 ∈ B, which is false by the key observation above.
If |E(G)| ≥ 7, then the complementary graph G has 5 vertices and at most 3 edges, so there exists j an isolated vertex in G and i, k ∈ [5] \ {j} so that (i, k) is not an edge in G. This implies that ijk is a 3-cycle in G, which is false.
If B is the matrix representing ψ with respect to the canonical bases of K 4 and (i,j,k)∈H Se ijk , each of these bases in the natural order, then Since C is a submatrix of B of full rank, there exists a unique element a ∈ S such that I = (1/a)I 10 (C). For 1 ≤ s ≤ 11, let ∆ s (C) be the determinant of the matrix which is obtained by deleting the sth row of C. Then, by computing (with CoCoA [1]) the greatest common divisor of ∆ 1 (C), . . . , ∆ 11 (C), we see that a = x 4 1 .
6. The Rees algebra of the Bourbaki ideal in Proposition 5.3 In this section we consider the Rees algebra of the Bourbaki ideal I of Z n−2 described in Proposition 5.3. We show that it is a normal Cohen-Macaulay ring, and it is Gorenstein if n is even. It turns out that the Rees algebra of Z n−2 has the same properties as the Rees algebra of I, see [10, Theorem 3.1 and Theorem 3.4]. At present we do not know whether our result can be directly deduced from [10].
Let, as before, S = K[x 1 , x 2 , . . . , x n ] be a polynomial ring of dimension n ≥ 2 over a field K.
x i . Therefore, to prove our assertion, it is enough to show that R(I) is a normal toric ring, by Hochster's theorem (see [4,Theorem 6.3.5]).
Let e 1 , . . . , e n and f 1 , . . . , f n in Z n+1 be the exponent vectors of the monomials and s = s 1 + · · · + s n . Let D be the set of lattice points a = (a 1 , . . . , a n+1 ) T ∈ Z n+1 in the rational cone which is obtained by intersecting the half spaces of equations: a 1 + · · · + a n ≥ (n − 2)a n+1 .
By Gordan's lemma (see [4, Proposition 6.1.2 (b)]) we obtain that D is a normal affine semigroup. We prove that C = D, which implies that R(I) is a normal ring.
It is straightforward to check the inclusion C ⊆ D. Assume that C ⊇ D and take an element a = (a 1 , . . . , a n+1 ) T ∈ D \ C so that a 1 + · · · + a n+1 is as small as possible. By the observation in (10), we have a n+1 > 0. Claim 6.2. a 1 > 0, . . . , a n > 0.
Proof of Claim 6.2. Suppose that a i 1 = a i 2 = 0 for some 1 ≤ i 1 < i 2 ≤ n. If i 2 − i 1 = 1 mod n, by (12), a i 1 + a i 2 ≥ a n+1 > 0, which is a contradiction. Hence i 2 − i 1 = 1 mod n. By symmetry of a 1 , . . . , a n , we may assume that i 1 = 1 and i 2 = 2. Then, by (12), a i = a 1 +a i ≥ a n+1 for 3 ≤ i ≤ n−1 and a n = a 2 +a n ≥ a n+1 . Hence which is a contradiction. Hence 0 appears at most once among a 1 , . . . , a n . Assume a 1 = 0. Then a 2 > 0 and we have a i = a 1 + a i ≥ a n+1 for 3 ≤ i ≤ n − 1 and a 2 + a n ≥ a n+1 by (12). If a 2 + a n > a n+1 , then, all of the inequalities (11), (12), and (13) appearing a 2 are strict. It follows that a − e 2 ∈ D \ C, which is a contradiction for the minimality of n+1 i=1 a i . Hence a 2 + a n = a n+1 . Then a = a 2 f n + a n f 1 + This is also a contradiction. Hence a 1 > 0. By the symmetry of a 1 , . . . , a n , we have a 1 > 0, . . . , a n > 0.
We choose a subset and v is as large as possible.
For 1 ≤ w ≤ n we define its (circular) predecessor to be pred(w) = w − 1, if w > 1 and pred(1) = n. Similarly, its (circular) successor is succ(w) = w + 1, if w < n and succ(n) = 1. Proof of Claim 6.3. Assume that, for all w ∈ J ′ , pred(w) ∈ J. Then we may replace J ′ with the set J ′′ = {pred(w)|w ∈ J ′ } which satisfies (14) and |J ′′ | = |J ′ |. If J ′′ still does not have the desired property, we take predecessor sets until one finds a good substitute for J ′ . Indeed, this process must terminate in at most n steps. Otherwise, it means that |J| = n, i.e. a = (a, . . . , a) T , which is false.
We pick w ∈ J ′ so that pred(w) / ∈ J.
Proof of Claim 6.4. The vector a − f w satisfies (13), and also (11) since a i > 0 for all i. We now verify (12).
If {i 1 , . . . , i ℓ } ∩ {w, succ(w)} = ∅, since the wth and the succ(w)-th components of f w are zero, it follows that the corresponding inequality (12) for a − f w and the indices i 1 , . . . , i ℓ holds.
Clearly, a − f w is not in the semigroup C, because otherwise a ∈ C, which is false. So, the vector a − f w ∈ D \ C has the sum of its components less than n+1 i=1 a i , which is false. Consequently, C = D and the normality of the Rees algebra R(I) is now fully proven. Proof. Let R = R(I). In view of Proposition 6.1, keeping the notation from its proof, R is a toric ring generated by the monomials whose exponent is in the affine semigroup C ⊂ Z n+1 , which satisfies C = R + C ∩ Z n+1 .
Let ω R be the ideal (x F |F ∈ Z n+1 ∩ relint R + C)R, where relint R + C denotes the relative interior of the cone R + C. Then ω R is the canonical module of R ([4, Theorem 6.3.5(b)]).
Let F ∈ Z n+1 . Then F is in the semigroup C if and only if its coordinates satisfy the weak inequalities (11), (12), (13); and moreover F ∈ relint R + C if and only if none of the latter inequalities becomes an equation.

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Note that F 1 − F 2 has negative entries and F 2 − F 1 does not satisfy (13), so F 1 − F 2 / ∈ C and F 2 − F 1 / ∈ C. Hence (x F 1 )R = (x F 1 , x F 2 )R = (x F 2 )R. We conclude that when n is odd, ω R is minimally generated by x F 1 and x F 2 , so the Cohen-Macaulay type of R is two.