Burch Index, Summands of Syzygies and Linearity in Resolutions

The Burch index is a new invariant of a local ring R whose positivity implies a kind of linearity in resolutions of R-modules. We show that if R has depth zero and Burch index at least 2, then any non-free 7th R-syzygy contains the residue field as a direct summand. We compute the Burch index in various cases of interest.


Introduction
Let (R, m, k) be a commutative, Noetherian local ring.Little is known about the matrices that can occur in the minimal free resolutions of a finitely generated R-module of infinite projective dimension.In this paper we study conditions on R, expressed by an invariant that we call the Burch index, that imply that these matrices contain many elements outside the square of the maximal ideal, or that the syzygies of M contain k as a direct summand.
The second condition is generally stronger than the first: by the main theorem of Gulliksen [8], the Koszul complex of the maximal ideal m of R is a tensor factor of the minimal R-free resolution of k, so, when dim k (m/m 2 ) ≥ 2, the entries of each matrix in the minimal free resolution of k generate m.It follows that the same holds for any module with k as a direct summand.
To exclude the phenomena associated with modules of finite projective dimension, it is useful to consider the case of rings of depth 0, and the Burch index of an arbitrary Noetherian local ring is defined in Section 1 by reducing to this setting.Suppose that R is a depth 0 local ring with residue field k, and that R = S/I, where (S, n, k) is a regular local ring.Note that n 2 ⊂ In : (I : n) ⊂ n.We define the Burch index of R (relative to S) to be Intuitively this measures the minimal generators of I that are multiples of elements in the socle of S/I.Our definition is inspired by ( [5]) in which a local ring (R, m, k) is said to be a that is not free, then the entries of each matrix in the minimal free resolution of syz R 5 (M) generate m.Furthermore, k is a direct summand of syz R n (M) for all n ≥ 7. See Theorem 3.1 for a stronger but more complicated result.Theorem 0.1 seems to be new even in the case of the rings R = k[x 1 , . . ., x d ]/(x 1 , . . ., x d ) n (n ≥ 2), which have Burch index d.Theorem 0.1 implies immediately that over a ring of Burch index at least 2, each module has either finite projective dimension or maximal curvature (that is, its Poincare series has the same radius of convergence as that of the residue field).See [1] for a survey of the vast literature on this subject.Also, any high enough (nonfree) syzygy M over such ring can be used to test for finite projective dimension, in the sense that Tor R i (M, N) = 0 for a single i > 0 forces N to have finite projective dimension.If M is a finitely generated R-module, we write J(M) for the ideal generated by the entries of a minimal presentation matrix for M, and syz R n (M) for the minimal n-th R-syzygy module of M. Without assuming that R has depth 0 (and using the definition in Section 1 we deduce that over rings of Burch index ≥ 2 the presentation matrices of high syzygies contain many linear forms: Corollary 0.2.Let (R, m, k) be a local ring and suppose Burch(R) ≥ 2. Let M be any finitely generated module over R which has infinite projective dimension.For n ≥ 5 + depth R, In Theorem 2.4 we draw a connection between the Burch index of an ideal I in a regular local ring S and the linear entries of matrices in its S-free resolution.Our results there suggest a connection between linearity of resolution of I over S and the resolution of modules over S/I.
We conclude the paper by computing the Burch index for several special classes of ideals: ideals in a regular local ring of dimension 2, ideals of general sets of points in the projective plane, ideals with almost linear resolution, and ideals of certain fibre products.We give examples of monomial ideals to illustrate our results.
The main results of this paper were suggested by extensive computations using the computer algebra system Macaulay2 [10].We wish to acknowledge the importance of this tool, primarily produced and constantly improved by Dan Grayson and Mike Stillman.

Definitions and preliminary results
Throughout this paper, all rings are local and Noetherian and modules are finitely generated.We regard positively graded algebras over a field as local as well, with the results applying to graded modules only.Our definition of the (absolute) Burch index of a local ring uses the Cohen presentation, a residue field extension, and reduction modulo regular sequences to reduce to the case of an ideal of codepth 0 in a regular local ring.By a regular presentation of R, we mean a surjective local map (S, n, k) → (R, m, k) where S is regular.Such a presentation is called Definition 1.2.Let (R, m, k) be a local ring and let R denote the m-adic completion of R.
(1) If depth R = 0, let R = S/I be a minimal regular Cohen presentation of R. Define BI(R) to be BI S (I) R ∩ R and Burch(R) to be Burch S (I).These are well-defined by Theorem ) .We define Burch(R) to be the maximum, over all maximal regular sequences (x Recall that a ring R with Burch(R) > 0 is called a Burch ring in [5].
Theorem 1.3.Let R be a local ring of depth 0 and consider a regular presentation R = S/I.Then the R-ideal BI S (I)R and the number Burch S (I) are independent of the presentation of R.
Proof.We can assume R is complete and not a field.Consider any two presentations of R, S 1 ։ R and S 2 ։ R. Then S = S 1 × R S 2 is also complete local, and thus by the Cohen Structure Theorem it is a homomorphic image of another regular local ring T .Hence it is enough to prove the equality of our definitions for the pair (T, S 1 ) and since T maps onto S 1 , by induction we reduce to the case where our presentations are coming from (T, Suppose R = S/I.Let (J, m) be lifts of (I, n S ) to T respectively.By that we mean: the minimal generators of J, together with x, generate the preimage of I in T minimally, and similarly for m, n S .Write n for n T .
We need to show that (the LHS is BI S (I) and the RHS is BI T (I)).Observe that (J + (x))n = Jm + xn.Let A = T /Jm and (slightly abuse notations) x, n, H be the images of x, n, (J, x) : n in A. We are reduced to showing (x) : H = xn : H.By assumption R is not a field so H ⊂ n.Thus, if a ∈ (x) : H, then aH ⊂ (x) ∩ n 2 = (x)n, and we are done.
The following result shows in particular that to compute the Burch index for rings of positive depth one must use regular sequences consisting entirely of elements not in the square of the maximal ideal, that is, linear regular sequences: Proposition 1.4.Let S be regular and I, J be ideals such that IJ = I ∩ J (equivalently, Tor S i>0 (S/I, S/J) = 0).If S/I, S/J are both singular, then Burch(S/(I + J)) = 0.In particular, if R = S/(I + f ), where f is a nonzerodivisors in the square of the maximal ideal of S/I, then Burch(R) = 0.
Proof.Let R = S/(I + J).Then R is not Tor-friendly, in the language of [2].Namely, there are R modules M, N of infinite projective dimension such that Tor R i>0 (M, N) = 0. On the other hand, Burch rings are Tor-friendly, by [5,Theorem 1.4].
At this point, we have enough to give basic estimates and precise values for the Burch index of certain well known rings.See also Section 4. Proof.(1) The inequality is obvious if depth R = 0.When depth R > 0, by 1.4, to compute Burch(R), we just need to use a regular sequence in m − m 2 , thus the statement reduces to the depth 0 case, whence the assertion is obvious.
(3) We may assume R is Artinian, and thus has the presentation S/n 2 for some regular local ring (S, n), so we can apply (2) The inequality is obvious if depth R = 0.When depth R > 0, by 1.4, to compute Burch(R), we must use a regular sequence in m − m 2 , so the statement reduces to the depth 0 case.
(4) By the argument above, the computation reduces to the Artinian case, so we can assume R = S/(t a ) for some DVR (S, (t)) and some a > 1, for which the claim is clear.

Burch index and linearity
Let (S, n, k) be a local ring, let I ⊆ S an ideal and set R = S/I.Our main results link high values of Burch index to linear elements in the S-free resolution of R and the R-free resolutions of R-modules.Definition 2.1.Recall that we write J(M) for the ideal generated by the entries of a minimal presentation matrix for a finitely generated module M. We set M), and think of LinSp S n (M) as the vector space of linear forms in the n-th matrix in a minimal free resolution of M.
In the case where the ring S has depth ≥ 2, the following result gives another characterization of BI S (I).If depth S ≥ 2 then n contains an element that is a nonzerodivisor on S/tS, so (tI) : n = t(I : n), and the previous argument can be reversed.
Remark 2.3.Let (S, m) be a local ring and M an S-module.If t ∈ S is a regular element on M and S, then for each i > 0, If Burch S I ≥ 2 then let the associated primes of S be P 1 , . . ., P a .Let V 0 = (Burch S I) + n 2 )/n 2 and V i = (P i + n 2 )/n 2 .For each nonzero vector t ∈ n/n 2 outside of ∪ a j=0 V j we must have t + LinSp S i (I) = n/n 2 and each V j has codimension at least 2, so item (2) follows.

Burch index and k-direct summands
As we explained in the introduction, the condition that a module M has k as a direct summand is stronger than the condition that lin i (M) is the maximal ideal.Our main theorem strengthens Theorem 0.1.Theorem 3.1.Let (R, m, k) be a local ring of depth 0 and embedding dimension ≥ 2. For every non-free R-module M: (1) If Burch(R) ≥ 2 then k is a direct summand of syz R i (M) for some i ≤ 5 and for all i ≥ 7.
(2) If Burch(R) ≥ 1, and k is a direct summand of syz R s (BI(R)) for some s ≥ 1, then k is a direct summand of syz R i (M) for some i ≤ s + 4 and for all i ≥ s + 6.Thus, if R is Burch, then k is a direct summand of all high syzygies of non-free R-modules if and only if k is a direct summand of some R-syzygy of BI(R).
Remark 3.2.The case of depth 0 and embedding dimension 1 is different, but easy to analyze completely: in this case R = S/t n for some discrete valuation ring (S, t), so BI S (R) = t.Any indecomposable R-module has the form M = R/t m and the syzygies of M have k as direct summand in alternate steps iff m = 1 or m = n − 1.
The following Propositions will be used in the proof of Theorem 3.1.Recall the convention that the ideal of entries of a matrix with no columns is R. We first show that the ideals J(syz R s−1 (M)) + J(syz R s (M)), for any R-module M, are an obstruction to socle summands: Proposition 3.3.Let M be a module over the local ring (R, m).If J(syz R s−1 (M)) + J(syz R s (M)) ⊆ K m for some ideal K of R then k is not a summand of syz p (K) for any 1 ≤ p ≤ s − 2.
Example 4.5.Corollary 4.4 cannot be improved.There are Artinian Gorenstein ideals with almost linear resolution in each generating degree d > 1 and number of variables n > 2, see for instance [6,Example 3.2].Since the quotients are Gorenstein and not hypersurfaces, the Burch index is 0 (the syzygies of k will never have k summands, as they are indecomposable).

Definition 1 . 1 (
relative Burch index ).Let (S, n, k) be a local ring and I ⊆ S an ideal.We define the Burch ideal of an ideal I ⊂ S to be BI S (I) := In : (I : n).If I = 0 or S = S/I has positive depth, we set Burch S (I) = 0. Otherwise n 2 ⊆ BI S (I) ⊆ n and we set Burch S (I) = dim k (n/ BI S (I)).

Proposition 2 . 2 .
Let t ∈ n be a nonzerodivisor.If t / ∈ BI S (I) then k is a direct summand of I/tI.If depth S ≥ 2 then the converse also holds.Proof.Since t ∈ n is regular, tI : n ⊆ tI : (t) = I, and thus soc(I/tI) = ((tI : n) ∩ I)/tI = (tI : n)/tI.Thus I/tI has a k-summand if and only if (tI) : n ⊆ nI.If t / ∈ BI S (I) = nI : (I : n), then t(I : n) ⊆ nI, and since t(I : n) ⊆ (tI) : n, I/tI has a k-summand.
M), and otherwise lin S/tS i (M/tM) = lin S i (M) − 1.The matrices in the S-free resolution of ideals with positive Burch have many linear entries: Theorem 2.4.Let (S, n, k) be a local ring of embedding dimension n, and let I ⊂ n be an ideal.If 0 < i < projdim I, then: (1) If Burch S I ≥ 1 and S has positive depth, then lin S i (I) ≥ n − 1. (2) If Burch S I ≥ 2 and dim(P + n 2 )/n 2 ≤ n − 2 for each associated prime P of S then lin S i (I) = n.Proof.If t / ∈ BI S (I) is a nonzerodivisor, then by 2.2 the module I/tI has a k-summand, so LinSp S/tS i (I/tI) = n/(n 2 + (t)), and thus LinSp S i (I) + t = n/n 2 by Remark 2.3, hence lin S i (I) ≥ n − 1.If Burch S I ≥ 1 and S has positive depth then, by prime avoidance, we can choose such an element t, proving (1).

Corollary 4 . 6 .
Let (S, m) be a polynomial ring and I ⊂ m 2 an ideal with linear resolution.Then Burch(S/I) = dim S − depth S/I.