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This is to certify that the
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Licci Gorenstein ideals of
deviation 2
presented by

Elias Manuel Lopez

has been accepted towards fulfillment
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Ph . D degree in Mathematics

 

 

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Date 4-28-88

 

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LICCI GORENSTEIN
IDEALS OF
DEVIATION TWO

By

Elias Manuel L0pez

A DISSERTATION

Submitted to
Michi; an State University
in partial fulfillment of the requirements
for the degree of

 

DOCTOR OF PHILOSOPHY

Department of Mathematics

1988

OOY‘JUVJ

ABSTRACT

LICCI GORENSTEIN IDEALS
OF DEVIATION TWO

By
Elias Manuel. Lopez

A new family of licci Gorenstein ideals of odd height and deviation 2, for
height at least 7, is introduced. A characterization, up to deformations and
specializations, of licci Gorenstein ideals of height 3 and deviation 2 is given in
terms of a minimal set of generators. Also, a family of licci Gorenstein ideals of
any even height larger or equal to 6 and deviation 2 defining a rigid algebra which

are not hypersurface sections is constructed.

A Maria Julia y
Maria Alejandra
Por todo lo que son

Por todo lo que serafi

111

ACKNOWLEDGMENTS

I express great thanks to Conicit, Caracas, Venezuela for their economic
support while I studied at MSU. I thank Bernd Ulrich for his more than patient
guidance during this work, for his encouragement and support. Andy Kustin gave
me some of his time to listen to my ideas when they were still not clear. He also
asked the question that precipitated the last chapter of this work. Byron
Drachman gave me a helpful hand with computers and Darlene Robel made my
manuscript readable. To all of them thanks. Many thanks and a life of gratitude
go to my wife and my daughter, for pushing me always and giving me their

constant belief.

1v

TABLE OF CON'I'l-IN'IS

INTRODUCTION
CHAPTER I: BASIC RESULTS AND NOTATIONS

CHAPTER II: LICCI GORENSTEIN IDEALS OF DEVIATION
TWO AND ODD HEIGHT

CHAPTER III: THE EVEN GRADE CASE
SUMMARY
BIBLIOGRAPHY

13
57
77
79

INTRODUCTION

Let (R, m) be a Gorenstein local ring, and let I be a proper ideal of R. The
deviation of I, d(I), is the difference between the minimal number of generators of
I and the grade of I. In some sense, it is a measure of the failure of I being a
complete intersection. Some authors refer to it as "complete intersection defect".

For perfect Gorenstein ideals of height 3 and deviation 2 we have the result
of Buchsbaum and Eisenbud [B—E, 2.1]. These ideals are generated by the 4x4
pfaffians of a 5x5 alternating matrix with entries in m. For perfect Gorenstein
ideals of height 4 and deviation 2, which are generically complete intersections, we
have the result of Vasconcelos and Villarreal [V—V, 1.1] (based on Herzog and
Miller [H—M, 1.7]). These ideals are hypersurface sections, namely they can be
written as (K, y), K a Gorenstein ideal of height 3 and deviation 2 and y a regular
element on R/ K.

In this dissertation, we give a structure theorem for licci Gorenstein ideals
of height 5 and deviation 2. Moreover, we construct a class of rigid algebras
defined by licci Gorenstein ideals of even height and deviation 2 which are not a
hypersurface section for every even height larger or equal to 6. By licci we
understand for an ideal to be in the linkage class of a complete intersection.

Deviation 2 is the smallest possible deviation for a perfect Gorenstein ideal
not being a complete intersection [Ku]. The first non trivial examples of licci
Gorenstein ideals of odd height and deviation 2, were constructed by Huneke and

Ulrich [H-U-l]. If R is local Gorenstein, n 2 2, X a 2nx2n generic alternating

2

matrix, Y a 2nxl generic matrix, then in R[X, Y](X,Y) the ideal Hn = (11(XY),
Pf(X)) is licci, Gorenstein, has deviation 2 and height 2n—1. Moreover it is not a
hypersurface section.

Our main tool is linkage of ideals. Linkage was first introduced by Peskine
and Szpiro [P-S], and it has been found to be very useful as a classifying tool, as a
method for proving results about Special varieties [K—M, H—U—2], and for studying
the divisor class group of rigid algebras [H—U—l]. Our main interest is for ideals I
and J which are linked and the linking sequence is part of a minimal set of
generators of I. In this case we write I .. J.

For the ideals of height 5 we are able to describe them, up to deformations
and specializations in terms of a minimal set of generators. To accomplish this, we
use induction on the number of steps needed to link such ideals to a complete
intersection.

We are able to show that every licci, Gorenstein ideal I of height 5 and
deviation 2 in a local Gorenstein ring is up to deformations and specializations,
either H3 or (H2, x, y), x and y regular modulo H:2 (theorem 2.22). To obtain this
result, we first characterize, up to deformations and specializations, all perfect
almost complete intersections of height 4 and type 2 in R (pr0position 2.16). Then
we compute all possible ideals J and K such that H3 -+ J 4 K and (H2, x,y) -+ J -+ K
(lemmas 2.17 to 2.21). It follows that K is either a hypersurface section or has a
common specialization with H3. Then, we prove our main result. As an
application, we find that if R = k[[xl,- - - ,xr]], then I is exactly of the desired form.

We also find a family Kn of licci Gorenstein ideals of height 211-1 and
deviation 2, and show that K 4 is not obtained from any Hn by deformations,
specializations and taking hypersurface sections.

This new family can be described as follows: with the same notation, let Xi'

I
be the pfaffian of the matrix obtained from X after deleting the i-th and j—th rows

3

and columns and let [[1- - - l2n]t = XY. Then the ideal

Kn=(/ x x

1" ° ° ’/2n-3’ x2n—l 2n’ Pf(X))

211—2 211'

2n—2 2n—l ’

is licci, Gorenstein, has height 2n—l, and for n 2 3, it has deviation 2. Also there is
a tight double link [K—M—2, 2.1] between HH and Kn'

These ideas are discussed in Chapter II.

In even grade, the only known result for Gorenstein ideals of deviation 2 is
due to Herzog and Miller, [H—M, 1.7] and Vasconcelos [V—V, 1.1]. They show
that, under some mild hypothesis, a perfect Gorenstein ideal of height 4 and
deviation 2 is a hypersurface section (this implies such ideals are licci). Kustin
once asked whether a Gorenstein ideal of even height and deviation 2 is a
hypersurface section. We answer this question negatively. We found a family En,
n 2 4, of licci rigid Gorenstein ideals of even height 2n-2 and deviation 2 which is
not a hypersurface section for any height (theorem 3.7). We obtain En from

(Hn-l’ x1, x2, x3), x1, x2, x3 regular module H by performing what Kustin

n-1’
and Miller call a semi—generic tight double link [K-M, 3.1]. The fact that En is
rigid follows from [K—M, 3.2 and K] and the fact that ED is not a hypersurface
section follows from the fact that En is contained in the square of the maximal
ideal.

We also find a. family Fn of ideals which are obtained from BD by
specialization. These ideals I“n are easily described. Let k be a field, X a
2n—lx2n-l generic alternating matrix, Y a (2n—1)x1 generic matrix, It _>_ 3. In
14x, Y](X.Y) consider the product [1,. . -,/2m_1]t = XY, and let Xil' , 'ir be the
pfaffian of the matrix obtained from X by deleting the rows and columns i1,- - "ir’
The” Fn = (l 1" ° "l 211-4, X2n—3 211—2 211-1, X2n—3’ X2n-2’ X2n—l) is "Cd,

Gorenstein, has deviation 2 and height 2n—2 (theorem 3.3). F3 is a hypersurface

4

section but P11 is not if n > 4 (corollary 3.10). We finish by showing that

le’Y](X,Y)/Fn is (R2) but not (R3) (proposition 3.11).

We want to close with a question that has been around for some while, but
we want to formulate it again. If I is a perfect Gorenstein ideal of R, and I has
deviation 2, is I licci? If the answer is yes, we would have to describe all

Gorenstein ideals of height 5 and deviatibn 2.

CHAPTER 1
BASIC RESULTS AND NOTATIONS

We will use as a general reference H. Matsumura's book Commutative
Algebra [Mat]. Unless otherwise stated, (R, m) will always denote a local
Gorenstein ring with identity.

We begin by establishing the notations needed for the manipulations of
pfaffians.

Let X be a nxn generic matrix of indeterminates (that is xi]. = “xji’ xii =
0). In R[X] = R[{xij/l S i,j 5 n}], the determinant of X is a perfect square. If n is
odd, this determinant is zero. For 11 even, the pfaffian of X, denoted by Pf(X) is a
uniquely defined square root of the determinant of X such that if X is specialized
to s = diagonal {S,- - -,S}, S = (—(l) (1)), then pf(s) = 1. [Jac, 6.4].

If r < n, we denote by Pfi . . . i (X) the pfaffian of the alternating matrix
1’ ’ r

obtained from X by deleting the rows and columns i1,- . "ir'

Let (i) denote the multi index i1,- - "ir' Define a(i) to be zero if (i) has a
repeated index, and otherwise to be the sign of the permutation that rearranges
i1,o - -,ir in ascending order (see [K]).

Let

and let

6

. 1 .
(1) x0) = (—1)"'+ a(1)Pf(i)(X)
If r = 11, let X“) = (—1)lil+la(i), and for r > n, let X0) = 0. The pfaffian of X

can be expressed in terms of lower order pfaffians as follows:

(2) Pf(X) = 233:, xijxij

while

(see [ALE], page 142).
Let Y be a nxl matrix of indeterminates over R[X], and write [I ,- ~ -,/n]t
= XY.

Consider the product

 

 

Then, on one hand this product is

7

while on the other, by(2 )it is [yj Pf(X) ]. Then

B) 2?_1mx1

xiii =3

Also, because X is alternating, we obtain
(4) {lyl +~ - -+ [nyn = 0.

Let a1,- . -,ar be elements of R. We denote by [al- - ~ar] the lxr matrix
whose entries are the ai's and by (a1,- - -,ar) the ideal they generate.

If I is a preper ideal of R, 11(1) denotes the minimal number of generators of
I, grade 1 its grade and ht(I) its height. d(I) = ”(D—grade I is the deviation of I. If
d(I) = 0, I is called a complete intersection and, in this case, I is generated by a
regular sequence. If d(I) 5 1, then I is called an almost complete intersection.

The ideal I is a hypersurface section if I = (K, y) where grade K= (grade
I) —1 and y is regular on R/K.

Let (R, m) be a Cohen—Macaulay ring with canonical module KR (for
definitions, see [H—K]). We define the type of R, r(R), as ”(KR).

If R = P/ I, where P is a local Gorenstein ring and I is a perfect ideal and if

04P“»~.4P1~P

is the minimal resolution of R, then r(R) = gn. We now list some basic facts of
linkage of ideals.

Definition 1.1. Let (R, m) be a local Cohen—Macaulay ring, and let I and J
be two ideals of R.

a) We say that I and J are linked, and write I ~ J, if there is a regular

8

sequence 5: = 0:1,- . -,as in I n J such that I = (a):J and J = (g):I.

b) I and J are gmmetrigally linked, if I and J are linked and have no
associated primes in common.

Definition 1.1 (a) is what Peskine and Szpiro [P-S] call algebraic linkage.
We drop the word algebraic in this work. Notice that I and J are not allowed to
be the unit ideal. Moreover I and J are unmixed ideals of height g.

Definition 1.2. An ideal I of R is generically a complete intersmtign if I is

unmixed and II) is a complete intersection in Rp, for all p E Ass(R/I).

R_emark 1.3. With the notations of 1.1 (b), if I and J are geometrically
linked, then they are generically complete intersections.

The proof of the following proposition can be found in Peskine and Szpiro
[P—S].

mm. Let I be an unmixed ideal of height g of the Gorenstein
local ring R, let 1; be a regular sequence inside I with (g) at I, and set J = (g):I.
Then

a) I = (g):J (i.e. I and J are linked)

b) R/ I is Cohen—Macaulay if and only if R/J is Cohen Macaulay

c) Let R/I be Cohen—Macaulay, then KB” 2 J/(g) and KR/J ; I/(g).

d) In addition to the assumptions of 1.4 (c), assume that proj. dimR(R/I)
is finite. Let F. be the minimal free resolution of R/I and K. the Koszul complex
of (9). Let U: K. -o F. be a morphism of complexes lifting the embedding (g) -+ I.
Then, the dual of the mapping cone of U, C(U*), is a resolution of R/ J .

Definition 1.5;. [H—U—2, 2.6] Let I and J be two ideals in R. We say that
J is minimally linkgl to I, and write I .. J, if I and J are linked with respect to the
regular sequence 51, and the elements of g form part of a minimal generating set of

I.

In general, minimal linkage is preferable to arbitrary linkage, partly

9

because of the following observation, which follows directly from 1.4.c.

W. [H—U-2, 2.7] In adition to the assumptions of 1.4.c, let k =
R/m.

a) r(R/J) = p(I/(o)) = p(I)— dimk((fl+mI)/mI) 2 d(I). In particular, if
R/ J is Gorenstein, I is an almost complete intersection.

b) r(R/J) = d(I) if and only if I -> J. In particular, if I -+ J and I is an
almost complete intersection, then R/J is Gorenstein.

Definition 1.7. Let I be an unmixed ideal of R.

a) The l_in_kege_c_le§§ (even linkage class) of I is the set of all R—ideals J
which can be obtained from I by a finite number of links (even number of links).

b) [H—U—2, 2.9] We say that I is Lied if I is in the linkage glass of a.
eomplete intersection.

Recall that if X is a nxm matrix of indeterminates over R, R[X] denotes the
polynomial ring R[{xij | 1 5 i 5 n, 1 5 j 5 m}], and (m, X) is the ideal of R[X]
generated by m and the entries of X.

Definition 1.3. Let R be a local Gorenstein ring. Let I be an ideal of R of
grade g, and let f = f1,- - - ,fu be a generating set of I, let X be a gxn generic matrix

over R, and consider the R[X] regular sequence o with
t t
[a] = Xlfl

a) [H-U—l, 2.3], L1(f) = (o)R[X]: IR[X] is called a first gegerie link of I.

It can be shown that this definition is essentially independent of the
generating set 1' of I [H-U—l, 2.4].

b) [H—U—Z, 2.12(a)] with the notation as in (a), set L0(I) = I, L1(I) =
L1(f), Li(I) = L1(Li—1(I))’ i > 0. Li(I) is an MEM-

Defioition 1.2. Let I, X, g and f as in 1.8 or let I = R, g an arbitrary

IO

integer. Denote by R(X) the local ring RIX]mR[X]

a) [H—U—2, 2.10(b)] If [o]t = X[f]t then L1(£) = (o)R(X):IR(X) is called a
first universal link of I.

It can be shown that this definition is essentially independent of t and g
[H—U-2, 2.11(b)].

b) [H-U—2, 2.12(b)] With the notations of (a), set L0(I) = I, L1(I) =
L1(t), Li(I) = Li(Li_1(I)), i > o. L1(l) is called an i—th universal link of 1.

Universal links have the following property.

Lemma 1.10. [H-U-3, 2.4] Set the notations as in 1.9, and assume R/m is
infinite. Then I is licci if and only if for some n > 0, Ln(I) is the unit ideal.

We will use the idea of deformation.

Definition 1.11. a) [H—U—2, 2.1.a] Let (R, I) and (S, J) be pairs of local
Gorenstein rings R and S, and ideals I C R and J C S. We say that (R, I) is
isomorphic to (S, J) if there is an isomorphism caR -1 S with tp(I) = J.

b) [H—U—2, 2.2.a] With the notation of (a), we say that (S, J) is a
deformation of (R, J) if there is a sequence 3,, regular on S and on S/J such that
(S/(e)), e+J/(e)) is isomorphic to (R, I). Equivalently, one says that (R, I) is a
smeialization of (S, J).

The following pr0positions give a connection between deformation and
linkage.

Promsition 1.12. [H—U—l, 1.12] Let I be a Cohen-Macaulay ideal of grade

g in R, o a regular sequence inside I with (o) it I and set J = (o):I. Let Q be a

sequence in R regular on R/(o), set R = R/(d), and set T, e, T the images of I,

(o), J in R. Then T = (e): I,T = (a): T and fl is a regular sequence on R/I and
R/J.
Proposition 1.13. [H—U—2, 2.16] Let (R, I) and (S, J) be pairs of local

11

Gorenstein rings R, S and Cohen—Macaulay ideals I of R, J of S, such that (S, J) is
a deformation of (R, I). Let I = I0 ~ - . ~ ~ In be a sequence of links in R.

Then, there is a sequence of links J = J0 ~ Jl ~ . - - ~ Jn in S such that
(S,Ji) is a deformation of (R, Ii) for all 0 S i 5 n.

Pronosition 1.14. [H—U—2, 2.17] Let I be a Cohen—Macaulay ideal of R,
and let I = I0 ~ - ~ - ~ In be a sequence of links in R0 = R. For 1 5 i 5 11. Consider
the i—th generic link Li(I) ; Ri’ where Ri is a polynomial ring over Ri-l' Let T =
Rn' Then, there is a q E Spec(T) such that m C q and (Tq, Li(I)Tq) is a
deformation of (R, Ii)’ 0 5 i S 11.

Remark 1.1§. Let I be a Cohen—Macaulay ideal of R, let (R, T) be a
specialization of (R, I), let T be as in 1.14, let q E Spec(T) such that m Q q, let S =

TqandletS= R®RS.

In S consider the sequence of localizations of generic links

IS~LfiUS~~~~LJUS

Then in S we have the sequence of links T S ~ L1(I)T ~ ~ Ln(I)S, where

Li(I)S = Li(T)S and (S, Li(I)S) is a Specialization of (S, Li(I)).

Proposition 1.16. [U—l] Let I be a licci ideal of R. Then, there is a
deformation (S, J) of (R, I) such that J is generically a complete intersection.

Qefinitiop 1.17. Let R be a formal power series ring over a field k, and I an
ideal of R. We say that the pair (R, 1) defines a r_igio algebra A = R/I if the first
upper cotangent module T1(A/k, A) vanishes (for definitions see [L—M]).

In this case (R, I) has the following preperty: let (S, J) be a deformation of
(R, I), with S also a formal power series ring over k. Then, there is a set of

indeterminates _Z_ such that (R[l Z | ], IR“ 1]] is isomorphic to (S, J).

12

For more details on deformation theory, see [Ar. M].

Let (R, I) and (S, J) be pairs of local Gorenstein rings R and S and ideals I
of R and J of S. We write (R, I) a (S, J) if there is a pair (T, K), T a local
Gorenstein ring, with char R = char S = char T, such that (T, K) is a deformation
of (R, I) and of (S, J).

Definition 1.18. [He, 2.2] The equivalence relation generated by z for the
pair (R, I) is called the Herzog glass of (R, I). Note that a need not be an
equivalence relation.

Definition 1.19. [He, 1.5] Let R be a formal power series ring over a field
and I a Cohen—Macaulay ideal of R such that R/I is reduced. I is called my
nonobstructed if I / I2 e R /I KR/I is Cohen-Macaulay.

Herzog [He] is able to show that if R = k[[xl,- - "er and if I is strongly
nonobstructed, Cohen-Macaulay and reduced, then (R, I) has a deformation (S,J),
S = k[[T1,o - -,Tn]] and S/J rigid such that the following holds: let (P, K) be any
pair in the Herzog class of (R, I), P = k[[zl,- - -,zm]], K reduced, then there is a
finite set of indeterminates w such that (P, K) is a specialization of (S[|w|],
JS[|_vg|]). In particular, all rigid algebras in the Herzog class of (R, I) are
isomorphic, after adjoining power series variables. We use these ideas together
with the following propositions.

Proposition 1.20. [B] Let R = k[[xl,- - "er and I a licci reduced ideal of R.
Then I is strongly nonobstructed.

Proposition 1.21 [U—l, Cor 2.2]. Let k be a field, R = k[[x1,...,xn]] and let
I be a licci ideal of R. Then, there is a pair (R', I') in the same Herzog class of
(R, I) such that R'/I' is reduced.

Finally, we will use the following result.

Prooosition 1.22 [H-U—3]. Let (R, m) be a local Gorenstein ring with

infinite residue class field and let I be a licci Gorenstein of R Then there is a

13

sequence of minimal links I = I0 -1 II -+ -+ In—l _. I n where In is a complete

intersection and I2i is Gorenstein for 0 3 2i 5 n.

CHAPTER II
LICCI GORENSTEIN IDEALS OF
DEVIATION TWO AND ODD HEIGHT

Let (R, m) be a local Gorenstein ring. In this chapter we will study licci
Gorenstein ideals of R of odd height and deviation 2.

For n 2 2, let X be a 2nx2n generic alternating matrix over R and Y a 2an
generic matrix over R[X]. Denote by {/1- .. [2nlt = XY. Set S = R[X, Y].
Consider the ideal Hn = ([1,- .. /2n’ Pf(X)). Using the ideas of Huneke and
Ulrich, one can show that in S(m, X, Y)’ HD is licci, Gorenstein, has deviation 2
and height 2n—1. [H—U—l, prop. 5.8 to 5.12], [K, introduction].

In this chapter we will produce a new family of licci Gorenstein ideals of
odd height and deviation 2, and we will establish a structure theorem for licci
Gorenstein ideals of height 5 and deviation 2.

We begin with the following sequence of lemmas.

Lemme 2.1 [J—P, th 2.3]. With the previous notations, in R[X, Y]

ht(Pf2r(X)) = [2n’§’+2] 0 3 2r 5 2n

where Pf2r(X) is the ideal generated by the 2rx2r pfaffians of X.
Proposition 2.2 [H, 1.1]. Let P be a Cohen Macaulay domain, and let M be

a P-module having a finite resolution

A

O—er——->Pn——-+M——+0

14

15

Denote by by S(M) the symmetric algebra of M and by It(A) the ideal generated
by the txt minors of A. Then, the following are equivalent

i) S(M) is a domain

ii) grade (It(A)) 2 m+2—t for 1 g t g m.

Recall that if M has a presentation of the form

Pm (aij) Pn

 

 

M 0

then S(M) is isomorphic to P[T1,o - -,Tn]/J where J is the ideal generated by the
m linear forms 231:1 aijTj'

With this notation, Huneke shows that if ii) holds, then ht J = m.

In addition to the previous notations, lets assume R is a domain. Then we
can now establish the following pr0position.

Proposition 2.3. The ideal (I ,- - °’/2n—2) is a prime ideal in R[X, Y] of
height 2n—2, n 2 3.

m. Let A be the matrix obtained from X by deleting the last 2 rows.
Then [I ,- - -,/2m_2]t = AY.

First, because R is a Cohen-Macaulay domain, R[X] is a Cohen—Macaulay

domain. Consider now the following complex:

2n—2 A 2n
0 ___. (R[X]) —-+ (R[X]) ————» M ———-» 0.

Because A is a matrix with maximum rank over R(X), the map from (R[X])2n_2

to (R[X])2n is injective. Moreover

16

SW R[XllYl .1 Ram _
half: nyj |1$i$2n—2) ‘(/1,..., /2n_2)

Thus, by preposition 2.2, it is enough to check that grade (It(A)) >

(2n—2)+2—t = 2n—t, 1 _<_ t 5 2n—2.

Because R[X] is Cohen—Macaulay, grade equals height. Let T be the
matrix formed with the first 2n—2 columns of A. Then T is alternating and
ht(It(T)) S ht(It(A)). We will show that for t 5 2n—4, ht(It(T)) 2 2n—t.

Buchsbaum and Eisenbud show [B-E, 2.6] that for T

12,0?) 912,41") 9 Pf2,(('r)) ; W .
Then, by proposition 2.1 and the definition of radical of an ideal
ht(12r(T))= ht(I2r_ 1(1‘)) = ht(Pf2r(T)) = [2‘3”] .
Hence, for 2r S 2n—4
ht(I2 r(T))_ (2n-2r)(2n—2r— 1) > 2n —2r.
Also, for 2r—1 5 2n—5
ht (I 21._1(T))= htI(2r(T)) = (n—r)(2n——2r-1) 2 (2n—2r—1)+2 = 2n—(2r—1).

Thus, we have shown that if t S 2n—4, then

his“ (2“x))2ht(I (T ”2211-1

17

Hence, we are left with the cases t = 2n—3 and t = 2n—2. For the case t = 2n—3,

consider the (2n—2)x2n matrix

 

 

_g -.’f12 "13 x14 0 ° 0
12.

‘xls. °

D= -x14 '

0 ' 6
3 ° . . . . '. x14

L 0 0 —x14 "x13 "x12 0 x12 X13,

We obtain D by specializing A. Here xii+j ——r X11+j’ 1 5 i 5 3, and xij—a 0

otherwise. Therefore

ht(I2n_3(D)) s ht(12n_3(A)).

Now, I2n__3(D) contains

det

 

det

 

and

 

 

18

X12 X13 x14 0---0 0
' . _ 2n—3
(161; . . = X12 mOd(X13, X14).
. x
* 14
X13
X12

 

 

Hence ht(12n_3(D)) 2 3 and therefore, ht(12n_3(A)) 2 3. For t = 2n-2, we
specialize x14 -—1 0 and let D be the matrix obtained from D in this way. Then

I2n_2(D) contains

 

 

,
x13 0° 0
det = xg—2
*
0
x13
and
“(12 x13 0 0
- 2n—2
det * 0 .-=x12 mod(x13)
x13
x12 ]

 

 

Then ht(I2n_2(D)) 2 2, and therefore ht(12n_2(A)) 2 2. It follows then that
ht(It(A)) 2 2n—t. I

Recall that S = R[X, Y], where R is a local Gorenstein ring domain, 11 2 2,

X is a 2nx2n generic alternating matrix over R, Y a 2nx1 generic matrix over

19

R[X]. We have the following corollaries.
Corolleg 2.4. (l ,- - "/2n-1) is a complete intersection in S(m, X, Y)‘
m. We first treat the case where R = D is a Cohen-Macaulay domain.
By 2.3 ([1,. - "/2n—2) is a prime ideal in D[X, Y] of height 2n—2. Denote by

"—" images in D[X, Y]/(Xi2n_la Xj2n1 i # 2n! j # 211-1). If [211-1 E (/ .,/

1’”
... . _ n-2 .

2n—2)’ the“ 72n—l — in—l 2n y2n E (71,” ’f2n-2) — (212:1 Xijyj’ 1 5 1 5

2n—2), which is false. Then (Z ,- - "/2n—1) is a complete intersection in D[X, Y].

We now prove how the claim follows for an arbitrary local
Cohen—Macaulay ring R. Let V be a prime ring of R localized at the contraction
of m, let p E Spec S(m,X,Y) with (11,- - °’/2n—1) C p, and let (7r) = p n V. We
have to show that dim S p 2 2n—1.

First consider the case where r is not regular on Sp. Then since R is
Cohen—Macaulay, dim Sp/ rSp = dim Sp. On the other hand, D = V/(r) is a field
and Sp/ «Sp is faithfully fiat over D[X’Ylan[X,Y]' Then by what we have shown

__ < . < . = . .
before, 2n 1.__ dlm D[X’Ylan[X,Y] __ dlm Sp/ «Sp dlm Sp

Now consider the case where r is regular on Sp. In this case take D = V.
Now D is a domain and SI) is faithfully fiat over D[X’Ylan[X,Y]' Again as
above, dim Sp 2 2n—1.

Corollary 2-§- The ideal (11,- - °’/2n—2’ Pf(X)) is a complete intersection
in S(m,X.Y)'

Prmf. It runs parallel to the proof of 2.4, after noticing that Pf(X)}

(l1,- - "/2n—2)‘ Grade, with the notations as 2.4, D[X, Y] by assigning deg xij =
0, deg yj = 1. Then Pf(X) has degree 0, but each [i has degree 1. Then we are
done.

Corollary 2.4 is the key of the following lemma. In it we prove all our

claims about Hn'

Lemma, 2.6. H11 is a licci Gorenstein ideal of height 2n—1 and deviation 2.

20

Also, in S, a homogeneous resolution of Hn starts and ends like
0 .. S(—(4n—3)) -» ... .. s?“(—2) e S(—n) .. s.

Prmf. By induction on n, consider first the case n = 2. H2 can be viewed

as the 4x4 pfaffians of the following 5x5 generic matrix

 

 

Then H2 is Gorenstein of height 3 and deviation 2 [B—E, 2.1]. Also, [B-E, proof of
2.1] the projective dimension of S/H2 is three, and hence it is perfect. Therefore,
H2 is licci [W]. It also follows [B-E, proof of 2.1] that if deg xi. 2 deg yj = 1,

1
then a homogeneous resolution of S / H2 starts and ends like

0-»S(—5)-+---aS5(—2)-»S

and notice that 5 = 4(2)—3.

Assume now that the result is true for n—1. Let X' be a 2(n—1)x2(n—1)
generic alternating matrix over R, Y' a 2(n—1)x1 generic matrix over R[X']. Then
in S' = R[X', Y'], Hn—l = (I1(X’Y'), Pf(X')) is licci, Gorenstein has deviation 2
and height 2(n-1)—2 = 2n—3. Grade 8' by assigning deg xi. = deg y. = 1. A

J J
homogeneous resolution of Hn—l starts and ends like

0 a S'(4(n-1)—3) .. ... .. s'2n’2(—2) e S'(-(n—1)) .. s'.

21
Write [/',- - -,/én_2]t = X'Y'. If X(resp Y) is a 2nx2n generic matrix over R
(resp 2an generic matrix over R[X]), then we can view 8' as a subring of S = R[X,
Y], X'(resp. Y') obtained from X(resp. Y) by deleting the last 2 rows and columns
(resp. last 2 columns). Then H(n—1) can be viewed as a licci ideal of S, with same

height, deviation and resolution
2n—2
0 -1 S(—(4(n—l)—3)) -+ -—- -1 S (—2) e S(—(n—1))-> S.

Then Pf(X') = x Write [/ ,. . .,./2n]t = XY. Consider in s, the

2n—l 2n'
ideal (H(n-1)’ y2n—l’ Y2n) = (/ 1’”°’2n—2’ x2n—1 2n’ Y2n—1’ Y2n)° A

homogeneous resolution of this ideal is
2 2n—2
0 -1 S(—(4n—5)) -+ - - - -l S (—1)$ S (-2) 9 S(—(n—1)) -1 S.

Consider the ideal (/ 1,~-,/ 2n—2’ y2n), which, by (2.3), is a complete

intersection. The Koszul complex of [1,- - ~ ’/2n—2’ y2n is
2n—2
o a S(-(4n—3))-+ ...1 s (—2) e S(—1)-» s

and then a resolution of Ln = ([1,~ - "/2n—2’ y2n): (H(n_1), y2n—1’ yzn) starts

and ends like
0 —e S(—-(3n—2)) e S(-—(4n-4)) ——» --- __. S(—1) e s2n‘1(—2) ... 3.

Then Ln is generated by (l1,- - "/2n—2’ y2n-1) and an element of degree 2.

Consrder l2n_1. Because

22

n-2
/ E? 1 ’iyi‘l

2n—1y2n—1 = ‘ r= 2ny2n’
the“ /2n—1Y2n—1 E (11" ' °’/2n—2’ y2n) ’2n—1X2n-1,2ni“ V1" "1/2n—2’ Y2nl'
2n-1
Also, because Pf(X)y2n = 2 Xi211 [i then
i=1
2n—2
X2n--1 2n [2n-1 = Pf(X)y 2n 3:1 Xi2n {6 ((1 _ [2n—2’ y 2n) '

Hence /2n-1 6 Ln. A similar argument to the one used in 2.4, shows that /2n_1 E
([1’°”’/2n-2’ Y2n)' Then(/1"”’/2n—1’y2n)§ Ln and Ln/(/1"'°’/2n—2’
y2n) is minimally generated by one element of degree 2. Then (l1,- - "/2n—1’
y2n) = Ln. Hence in S(m,X,Y)’ LH and (Hn—l’ y2n—1’ y2n) are linked, Ln ——+
(H,_1.y2,,_1,y2,,) and (H(n_1),y2,,_,,, yzn) —~ L,- Then r(S/Ln) = 2, d(Ln) =
1, by (1.6).

Consider now the complete intersection (l1,---,/2n_1) (see 2.4). Its

homogeneous Koszul complex is
2n—1
o ... S(-(4n-2)) —-+ ... —» s (—2) .... s.
Then a resolution of ((1; . "/2n—1): Ln starts and ends like
0 —-+ S(—(4n—3)) —-+ ... —-+ 3211(4) e S(-n) __. 3.

Hence ([1

2 and one element of degree 11. Consider [2n and Pf(X). (3) and (4) in chapter
(I) show that ([1,- ~ gén, Pf(X)) = Hn g (/I,'°',/2n_1)2 Ln“ If

[2ne (’1"”’/2n—1

,0 - "/2n—1): Ln is generated by (l1,- - "/2n—1)’ one element of degree

), then [2n 5 0 (mod(xij, i _<_ 2n—1, j 5 2n—1, y2n) which is

23

not the case. As in 2.5, Pf(X) j! (l ,- - °’/2n)‘ Then Hn g ( [1,- - "/2n—1): LH
and (l , - ’/2n— 1): L [111/( - °"’/2n—1) is minimally generated by one element of

degree 2 and one element of degree n. Then Hn = (Z ,. - -,/2n_ 1):n L . Hence in
S(m,XY) Ln «Hn andHn ”Ln“ Then d(Hn )=2, r(S/Hn )=1, htHn =2n—1.
The last twist of a homogeneous resolution of H11 18 4n—3, and H11 double linked to

(H ), then HD is licci. Now the result follows. I

n—l’ y2n--l’ y2n
In some unpublished notes, Andrew Kustin obtains Hn from (Hn—l’ y, z),

y, 2 regular modulo Hn—l’ by what Kustin and Miller call semi generic tight
double link [K—M, 3.1]. In those notes, he proves, by induction, that Hn is licci,
Gorenstein, has height 2n-1 and deviation 2. But moreover, he shows that in
k[[X, Y]], k a field, Hn defines a rigid algebra (see also [K]). We will use this fact
several times.

Corollary 2.5 gives the following lemma.

Lemma 2.7. The ideal Jn = (l1,- . "/2n—2’ 2n—1 2n’
ideal of type 2, deviation 1 and height 2n—1 in S(m,X,Y)'

X Pf(X)) is a licci

11m. We will show Hn .. J n via the regular sequence

./

Q = / 2n—2’

1,. Pf(X).

Grade S by assigning deg xij = deg yj = 1. A homogeneous free resolution of Hn

(see 2.6)' 13
o e S(—(4n—3)) a . .- -+ s2“(—2) e S(—n) .. 3
while the Koszul complex of Q is

0 .. S(—(5n—-4)) -. ... -+ 52n—2(_2) e S(—n) e s.

24

Then a resolution of (5): H11 starts and ends like
2 2n—2
0 .. S (—(5n—6)) -+ ~-- 4 S (-2) 0 S(—(n—1)) 6 S(—n) -» S.

Therefore, (5): HH is generated by (E) and an element of degree n—1. Consider
X2n__1 2n’ which has the apprOpriate degree. By (3) in Chapter (I), we see that

_ n—2
X2n--12n [Zn-l — Pf(X)y 2n ‘ 2L1 Xi2n [i E (5)

and

' n—2
X2n—12n/"2n >32=1Xi2n—1/1_Pf()y2n—l 6 (Q)-

It follows then that J n g (5): Hn'
Suppose X2n_1 2n 6 (,6). Then, there would be homogeneous elements

-,a2n_2, b e S such that

X2n—1 2n = 31/1+'”+a2n—2/2n—2 + be(X)°
Because 11 = deg Pf(X) > deg X2n-l 2n = n — 1, b = 0.

If we assign now deg x.. =0 deg y. = 1, then X2n—1 2n would have degree

lJ ’ J
0, but the 4's would have degree 1. This is impossible. Therefore X2n—l 2n 1! (,5).
Hence Jn/(fi) is minimally generated by X2n—l 2n' Also (Q):Hn/(Q) ls
generated by an element of degree n—l. It follows then Jn = (Q): Hn.
Then, in S(m X Y),J =(Q): H11’ and by 1.4(a), Jn and HH are linked.

Moreover Hn -+ Jn and hence Jn has type 2, by 1.6(b). Thus, Jn is not a complete

25
intersection but by 1.6(a), d(Jn) 5 1. This implies d(Jn) = 1. JH is also licci
because so is Hn. Now we are done. I
Lemme 2.3. The ideal (1) = ([1’°°°’/2n-—3’ X2n—1 2n,Pf(X)), n 2 2 is a
complete intersection in S(m,X,Y)'

Proof. We use induction on n. For n = 2

,
0 x12 x13 x14 [Y1

—x 0 x y

x: 12 23 "24 ,Y____ 2
‘X13 ‘X23 0 x34 y3
_”‘14 ‘x24 "X34 0 , _y4,

 

 

 

 

and (1) is (x12y2 + x13y3 + x14y4, x12, xlzx34 + x14x23 — x13x24), which has
height 3.

Consider now the case n = 3, and let p be a prime ideal containing (1) such

that ht p 5 5. If p contains x12, x13 and x23, then p also contains

and hence ht p 2 6. We may then assume w.l.o.g, that x12 i p.

Then, there is a matrix A invertible over Sp such that

AtXA=P0 1 o

 

 

 

 

and the entries of X' are generic over R[X]-é, {xij’ i 5 2}]. If Y is replaced by

A—lY, then in Sp’ (1) becomes (yl, y2, I], xi2, Pf(X')), which has height 3+2 =
5.

26

Assume the result is true for n—l, n 2 4. Let p be a prime ideal of
S(m X Y) such that (1) C p and ht p 5 2n—1. Then p does not contain the ideal I
generated by the entries of the (2n—4)x(2n—4) matrix obtained from X, by deleting

the last 4 rows and columns. In efect, for n = 4, if I C p, then p contains

(E?=5 Xinjll $154),

which is a complete intersection [Ho], and the xij’

this ideal. Then ht(p) 2 6+4, which is impossible. For n 2 5, ht (p) 5 2n—1 <
(n-2) (211—5) = ht I.

1 5 i,j 5 4 are regular modulo

We may assume then, w.l.o.g, that x12 ¢ p. Then, there is a matrix A,

invertible over Sp, such that

0 1 0

 

0 X'

 

and the entries of X' are regular over R[{xi. J|i_ < 2}, x12]. If Y' = A-IY, and Y' is
obtained from Y" by deleting the first 2 rows, let [liu Z211 2_]t = X' Y'. Then

(1) becomes
91,er ’11 ° r/én 51”“

which, by induction hypothesis, has height (2n—3) + 2- — 2n—1. Then ht p= 2n-1
= ht(7) and 1 is a regular sequence in S(m X y). I

Theorem 2.2. In S(m X Y)’ for 11 2 3, the ideal
K11 = (11" ° "/2n-3’ X2n—2 2n’ x2n—2 2n—1’ x2n-1 2n’ Pf(X))

27

is a licci Gorenstein ideal of height 2n—1 and deviation 2.

flop; We first show that Jn —-1 Kn, where J n is the ideal of 2.7, via the
regular sequence (1) of 2.8.

Grade S as usual. A resolution of Jn is(proof of 2. 8)

o ... S2(—(5n—6)) ——+ ... .2 32‘1‘2(—2) e S(—(n—1)) e S(—-n) —-» 3,
while the Koszul complex of 1 is
o —-» S(—(6n—7)) —-. ... —-. szn‘3(-2) e S(—(n-1)) e S(—n) ——. s.
Then a resolution for (1): J 11 starts as
32“‘3(-2) e s3(—(n—1)) e S(—n) ... s.
Then (1): J n is generated by 1 and 2 elements of degree n—l.

Consider X2n_2 2n and X2n—2 2n—1’ which have degree n—1. By (3) in

chapter (I) we have

2n-3
[2n—2X2n-2 2n: y2n 21.. lxi2n/i ”X2n—1 2n [2n—1 E (1)
and
_ n-3
[2n—2X2n—2 2n—1 _ fy 2n 1 Zli=1 xi2n— 1/ i+x2n—1 2n [2n—1 E (1)'

Therefore Kn g (7): J 11' If X2n—2 2n 6 (1), then for homogeneous elements

- ,a2n_3, b, c in S, we have

28

X2n—2 2n = a1/1+” °+a211-3/2n—3 + bx2n—1 2n + “Pf(X),
and, by degree reasons, b E R, c = 0. Then

(1) X

2n-2 2n-bX2n—1 2n = 3‘1 [1+ ° ° ' +a2n—3 [2n—3'

If deg xij = 0 and deg yj = 1, then (1) holds if and only if a1 = - -~

a2n_3 = 0. Then

(2) x b x

2n—2 2n = 2n—1 2n'

If we specialize, by making xi 2n—2 = 0, the left hand side of (2) is nonzero, but
the right hand side is. Similarly, X2n_2 2n_1 E (1, in_2 211) if and only if

(3) x = bX b, c e R.

2n—2 2n—1 2n—2 2n + C x2n—1 2n’

If we specialize by assigning xi2n—2 = xi2n—1 = 0 the right hand side of (3) is
zero, but the left hand side is different from zero.

If we assign deg xi] = 1, deg yj = 0, then, say, [1 is in

(1

1,...,/

2n—3’ X2n—1 2n’ x2n-2 2n’ X2n—2 2n—1’ Pf(X))

[2114), which is contrary to 2.3.

ifand only if [1 E (l2,-~,

If deg xi]. = 0, deg yj = 1, then Pf(X) is in

(1

1,...,/

2n—3’ X X X

2n—2 2n’ 2n—12n’ 2n—2 2n-1)

29

if and only if

Pf(X) E (X2n—2 2n’ X2n-1 2n’ X2n-—2 2n—1)'

Specialize by assigning xij —-1 0 i, j 5 n and xi j —-1 xi]. otherwise. Then X

becomes

><l
ll

 

 

 

 

where A is a generic nxn matrix. Then det(X) = (det A)2 at 0 and hence, Pf(X)#0.
However, for i, j E {2n—2, 2n—1, 2n} (note 11 2 3), if we delete the i—th rows

and j-th columns, we get

 

 

where D is a(n—2)x(n—2) matrix. 80 we get that this matrix has determinant 0.

In particular

= x2n—2 2n = x2n—2 2n—1 = 0'

X2n—1 2n

Then Kn is minimally generated by (1, X211_2 2n’ X2n—2 2n—1) and it is
contained in (1): JD. Also (1): J n/ (1) is generated by 2 elements of degree n—l.
Then Kn = (1): Jn'
Then, in S(m,X,Y)’ Kn .... JH and JH —-o Kn‘ By 1.6(b), then

30

d(Kn) = r(S(m,X,Y)/Jn) = 2

and

r(S(m,X,Y)/Kn) = d(Jn) = 1
as claimed. Also, since Jn is licci, so is Kn' I

Theorem 2.9 presents a family of Gorenstein ideals of deviation 2 and any
odd height larger or equal to 5. K3 and H3 have height 5, and if deg xi j = deg yj
= 1, they are generated by 6 elements of degree 2 and one element of degree 3
(later we will see they have a common specialization).

We turn our attention to the case n = 4.

In this case set R = Q, S = k[X, Y], X an 8x8 generic alternating matrix

over k, Y a 8 x 1 generic matrix over k[X].

Then, the minimal free resolution of H 4 [K, 5.5 and 6.1] starts as
-s99——rs37—-r89—-+s

and in particular the second Betti number of H 4 is 37.
Several applications of the mapping cone produce a (not necessarily

minimal) resolution of K 4 that starts as

d

.—-.s ——»sg—-»s

and d 2 44. That means then that H 4 and K 4 are in different Herzog classes.

The minimal resolution for H3 is

31

o=s=s7=sz2=322=s7=s.
If x, y are regular on S/H3, their Koszul complex is
o ... s —. s2 -+ s

and hence a resolution for (H3, x, y) starts as

and hence K4 is not in the same Herzog class as (H3, x, y). The minimal

resolution for H2 is

o—rs—rs5—rs5—rs.

If x, y, z, w are regular on S/H2, their Koszul complex is

0—»s—es4-es6—rs4—es

and the resolution for (H2, x, y, z, w) starts as

...—as3l—es9-es.

We see then that (H2, x, y, z, w) and K 4 are in different Herzog classes.

So, it is still left to determine if for some n > 4, (K4, t1,---,t ) and

r
(Hn,x1,--o,xe) are in the same Herzog class, where t1,---,tr form a regular

sequence modulo K 4 and x1,v - ~,xe form a regular sequence modulo Hn. The

32

answer is in the following sequence of lemmas.
LemmeI 2.19. Let (R, m) be a local Gorenstein ring, and let I be a
hypersurface section in R. Then, any specialization of I is a hypersurface section.
P_ro_ot. Write I = (K, y), y regular on R/K. If o is a regular sequence on R
and on R/I, then the specialization via o is (R/(o), (I,(a))/(o)). Now

(I, e) = (Kilns)
(e) (e)

 

and y, o are regular on R/K. Because R is local, or, y are regular on R/K and

hence

(K1152) (K1300
7217—: W

with y regular on R/ (K, o). I

Lemma 2.11. Let (R, m) and I as in 2.10. Assume R/m is infinite. If I is
licci, so is K.

M. Write I = (K, y), y regular on R/K. By 1.10, there is an n such
that Ln(I) = L“(K, y) = (1). Because y e m, it is enough to show Ln(K, y) =
(Ln(K),y) since (Ln(K), y) = (1) if and only if Ln(K) = (1), and by induction on
n, we only need to consider the case n = 1.

Let g = grade I, 1‘ = f1,- - -,fr generators of K. Then (11,- - °’fr’ y) = I. Let
X be a gx(r+l) generic matrix over R. Then in R(X) = R[leR[X]’ xgr+1 is

invertible, and after row and column Operations X becomes

X' 0
0 l

 

 

33

with X' generic over R[{x , 1 5 i 5 g, 1 5j 5 1+1}, x _1]. Then y is an

ir+1’ xgj
element of the regular sequence o given by

rg+1

[elt= x'] o f

 

0'1 y

and then L1(K, y) = (L1(K), y). |

Lemma 2.12. Let k be a field, 8 = k[[T1,- - "Tr“ and let I be an ideal of S
such that S/I is rigid. If I is a hypersurface section, then I t m2.

M. Write I = (K, y), y regular on S/ K. Let x be an indeterminate over
S. Then (S[[x]], (K, x)) is a deformation of (S, (K,y)) via the regular sequence
x—y. Because (S, I) is rigid, for some of indeterminate z, (S[[x]], (K, x)) is
isomorphic to (S[[z]], (K, y) S[[z]]). If I = (K, y) 9 (mS[[z]])2’ then (K,x) Q
(msllxll)2 and hence x e (ms[[x]])2’ which is impossible.

The following proposition is important by itself. It gives a criterion for a
licci ideal to be a hypersurface section. '

Proposition 2.13. Let k be an infinite field and S a formal power series ring
over k. Let I be a licci ideal of S, and let (T, J), T = k[[_x]], be in the same Herzog
class as (S, I). Then I is a hypersurface section if and-only if J is.

Em. Assume I is a hypersurface section, I = (K, y), y regular on S/ K.
Because I is licci, then by lemma 2.11, K is licci. By proposition 1.21, there is a
deformation (S, K) of (S, K) which is reduced. Also, by 1.13, K is licci and hence
strongly nonobstructed by 1.20.

By the discussion following definition 1.19, there is a pair (S, K), S a power

a a
series ring over k, such that (S, K) is rigid, reduced, and for some indeterminate Y,

(S[[Yl], (K, Y)) is a rigid deformation of (S, I). Let (T, J) be a pair in the same

Herzog class of (S, I). Then J is licci [U—2], and hence (T, J) has a deformation

34

(T, J) with T/J reduced. By the same discussion following definition 1.19, (T, J)

is a specialization of

(S[[Y, on, (R, Y) S[[Y, w”) for a finite set of indeterminates y. Then J, and
hence J, is a hypersurface section, by 2.10. I

Another criterion for an ideal I being a hypersurface section is given by
Herzog and Miller [H-M, 1.3]. They show that if k is an algebraic closed field and
I is a Gorenstein ideal of deviation 2 of R = k[[x1,...,xe]] such that I is generically
a complete intersection and Ill2 is a R/I — Cohen—Macaulay module, then I = (J,
t1,...,tr), t1,...,tr regular on R/J if and only if 11(1) - p(H1(I)) 2 r, where H1(I) is
the first Koszul homology module of I.

We come back now to our question prior to lemma 2.10, and assume the
answer is yes. Then repeated applications of proposition 2.13 would imply that H n
is in the same Herzog class as (K 4,t1,- - °’tr) or K 4 is in the same Herzog class as
(Hn’ x1,- - - ,xe) for n = 2, 3 or 4. But we saw this is impossible.

We summarize our findings in the next result.

Theorem 2.14. (8, (K4, t1,- ~ -,tr)) (t1,- - -,tr a regular sequence on S/K4)
is not in the same Herzog class as (T, (H 11’ x1,- - -,xe)), (x1,- - -,xe regular
sequence on T/Hn) for any n 2 2.

We now want to study the case n = 3 more closely. In this case ht K3 = ht

3 — .

We begin with the following lemma due to Brodman.

Lemme 2.15. [Br, 3. Satz] Let R be a local noetherian ringjwith infinite
residue class field, and let I be an ideal of R of grade g > 0 which is generically a

complete intersection. Then there are x1, - - . ,x g in I such that

i) x1,- - - ,x g form a regular sequence.

.. = . . . I.
11) (x1, ,xg)p Ip for all mlnlmal prlmes p of R/

35

iii) {x1,- . -,xg} is part of a minimal set of generators of I.

We now recall our notation: (R, m) is always a local Gorenstein ring. We
will assume also that 2 is a unit in R, and that R/m is infinite.

PrOposition 2.16. Let I be a perfect almost complete intersection ideal of R
of height 4 and type 2. Further assume (R, I) has a deformation which is
generically a complete intersection. Then there is a pair (S, J) in the same Herzog
class as (R, I) such that J is licci and either:

i) J = 11(AB), A a 5x5 alternating matrix, B a 5x1 matrix, aij’ bj 6 ms’ or

ii) J is a hypersurface section.

m. We can assume, after deforming, that I is generically a complete
intersection.

Let K be a minimal geometric link of I, which exists by 2.15. Then K is
perfect, Gorenstein, generically a complete intersection (1.3) and d(K) 5 2.

If d(K) = 0, K is a complete intersection, say, K = (a1, a2, a3, a4). If x1,

x2, x3, x4 are indeterminates over R, then
(R', K.) = (R[X1,X2,X3 x4](Xl,X2,X3,X4)’ (X1,X2,X3,X4))

is a deformation of (R, K) via the regular sequence ,3 = [xi-aim 5 i 5 4}. This

sequence is regular on R'/K' if and only if
(X11 "2’ "3’ X41 "1‘31’ "2‘32’ "3’a3’ ’91—‘14)

is a complete intersection in R'. But this later ideal is

(x1. x2, x3, x4. a1. a2, a3, a4)

36

which is a complete intersection in R'. Notice also that R'/(,Q) ; R and
K'+(5)/(Q) 3"- K.
Let K'CZ) J' where (R', J') is a deformation of (R, J) (1.13). Since

r(R'/J') = 2 = MIC/(1)),

two elements which are part of a minimal set of generators of K', say x1 and x2
are in (1) Q J '. Then J ' is a hypersurface section.

Now assume K is not a complete intersection. Because K is Gorenstein and
perfect, then, [Ku], d(K) 9% 1.

If d(K) = 2, then K -» I (1.6(b)). Because K is generically a complete
intersection, then K is a hypersurface section [V—V, 1.1]. Namely, K = (L, t), L a
Gorenstein ideal of height 3 and deviation 2 and t regular on R/ L. It follows then
I is licci.

Thus, there is a 5x5 alternating matrix D = [dij] with entries in m such
that L is generated by the 4x4 pfaffians of D, [B—E, 2.1], i.e. L = (Di’ 1 5 i 5 5)
where Di are the pfaffians (with signs) of D.

We now compute all possible minimal links of K. Let (Q) = (01, 02, a3,
04) be the regular sequence defining K —-1 I. Then, there is a 4x6 matrix T with

entries in R such that

...;
J

 

 

al ' D
02 = T D2
“3 D3
(14 . D 4
D5

_ t

 

 

37
Because the link is minimal, a maximal minor of T is invertible. Thus, if "—"
denotes the reduction modulo m, T has rank 4 over R. Therefore, the matrix

formed with the first 5 columns of T has rank at least 3 over R. Then, after

performing row Operations, one can assume that

 

 

31‘11‘1

s q r

T: 13 2 2 2
33 ‘13 1'3

01x3 84. ‘14 r4

 

 

and because the link is minimal, one of the element of the last row is a unit.

Let

'10 0—81 qll
010—82q2
MI: 0 01-—s3q3
0 0 0 10
[0 0 0 0 0 _

 

 

then M is invertible over R, MtDM is alternating and for 1 5 i 5 3
_. t _
D] — (M DM)i — Di—siD4+qiD5.

Hence, we can perform colulrm operations in the first 5 columns of T and after

doing so, T becomes

38

 

 

 

0 0 r1”
I3 0 0 r2
0 0 r3
0 0 0 s4 q4 r4‘

, where Z = [Zij] is a

we now deform (R, K) to (S. P), where S = R[Z, Wl(m z W)

5x5 generic matrix over R, w an indeterminate over R[Z] and P = (Z1, Z2, Z3, Z 4,
Z5, w). We deform here using the regular sequence zij_dij’ w—t, 1 5 i, j 5 5.

Let now J =--- (o):P. Then (S, J) is a deformation of (R, I) (1.13).

If r4 is a unit (or if one of the ri's is a unit) then w+S4Z4+q425 E (or) Q J
(or w+Zi E (o) C J) and hence J is a hypersurface section.

If none of. the ri's is a unit, then we can assume s4 is a unit (the case q4 a
unit is symmetric). We deform then r1, r2, r3 and 14 to undeterminates over S.
By (1.13) the link also deforms. Now specialize r1, r2, r3 to zero. Then, after

column operations T becomes '

3 03x3

 

 

['IX3 101'.

and o = Z1, Z2, Z3, Z4 + rw, which is a regular sequence, because 1 and w are
regular modulo (Z1, Z2, Z3). Then (by 1.12) the linkage also specializes. Our
ambient ring is now S = R[Z, w, r]. Grade S by assigning degree 1 to all
indeterminates.

The homogeneous minimal resolution of P is [B—E]
o ——. S(—6)—-1~-—-+ s5(—2) e S(-—1)-—1 3

while the Koszul complex of or is

39

0-vS(—8)-—+-~—->S4(—2)—-18

and hence a homogeneous resolution of (o):P starts as -4 85(—2) —-1 (S) and
hence (o):P/(o) is generated by an element of degree 2.
Consider z 45. Because 245Z4 and z 45Z5 are in (Z1, Z2, Z3), then

245rw = z45(Z4+rw) — Z45Z4 E (or)

and

z45rZ5 = r(z45Z5) E (or)

and it follows that (o, 2451') <_: (o):P.

If for homogeneous elements a1, a2, a3, a4 in R[Z, w, r]
z45r = a1Zl + a2Z2 + a3Z3 + a4(Z4+rw)
then by degree reasons, ai E R, 1 5 i 5 4. But this is impossible.

Then J = (o):P = (21’ Z2, Z3, Z4 + rw, Z451'), (R[Z, r,w](
deformation of (R, 1). Moreover J = (11(AB)), where

m,Z,r,w)’ J) is a

0 ’214 234 0 zl3 “225
z 0 z 0 z z
A: 14 24 12 B: 35
""34 “‘24 0 0 z23 z15
o o 0 o —r w
[‘le Z12 ’223 r 0 J . Z45,

 

 

 

 

as claimed. I
Notice that if J is a hypersurface section, almost complete intersection of

height 4 and type 2, then J = (P, t) where P is an almost complete intersection of

40

height 3 and type 2, which are described in [B—E].

In the following sequence of lemmas, all ideals J and K with H3 e J e K or
(H2, y, z) 4 J 4 K, y, z regular modulo H2, are computed, up to deformations and
specializations. J turns out to be an almost complete intersection of type 2, while
K is a perfect Gorenstein ideal of deviation at most 2. We will observe that K is
either a hypersurface section, or it is obtained from H3 by deformations and
specializations. Figure 1 summarizes our findings. Arrows mean minimal links
and the numbers associated to the arrows indicate the number of the lemma in
which the link is computed, h.s.s. means hypersurface section, while s.h.h.s
(d.h.s.s.) means single hypersurface section (double hypersurface secton). The

other notations are standard.

h.s.s.
2.17 H
H3 / 3
2.1x 2.19. {h.s.s.

([1’12’l31/4’ X56,Pf(x))___, H
3

h.s.s.
s.h.s.s. we { H
220/uh.s.s. 3

(H2’ y, Z) d.h.s.s. fle {h-Ifii-S-
3

(4,12, 13.14,x56, Pf(X) ___.. H3
Figurel

Lemme 2.17. Let X be a 6x6 generic alternating matrix over R, Y a 6x1

generic matrix over R[X]. In S = R[X, Y](m,X,Y) consider the ideal H = H3
1:

generated by {1, l2, 6, l4, l5, [6 and Pf(X), where [[1 l2 l3 l4 [5 16] = XY.

41

Let JV be an ideal of S such that H —-1 J. Then there is a pair (8', J ') in the same
Herzog class as (S, J) such that either

i) J ' is a hypersurface section or

ii) There is a 6x6 alternating matrix A and a 6x1 matrix B, aij’ bj E mS'
such that if [t1 t6]t = AB, then T = (t1, t2, t3, t4, A56, Pf(A)).

Epoof. Because H is Gorenstein and has deviation 2, r(R/J) = 2 and d(J) =
1. Consider the regular sequence 3 = 71, 72, 73, 74, 75 defining H -» J. It is

given by the product

 

 

 

 

_ , ’1
71 42
72 ’3
73 =T ’4
74 ’5
75 ’6

.Pf(X).

where T is a 5x7 matrix with entries in S. Because H ... J, one maximal minor of

T is invertible. Thus, if "-——" denotes the reduction modulo the maximal ideal
of S, then T has rank 5 over S, and therefore the matrix formed with the first 6

columns of T has rank at least 4 over S. Hence, after row Operations, T becomes

 

 

 

 

d1 Q1 81

I4 d2 ‘12 8’2
d3 q3 33

d4 ‘14 S4.

0 0 0 0 d5 q5 85‘

and one Of the elements Of the last row is a unit.

As in 2.16, we can perform column Operations in the first 6 columns of T.

42

After doing so, T becomes

 

 

 

 

81

I4 04x2 82
33

S4

0 d5 q5 85]

If neither (15 nor q5 are units we can deform them to indeterminates and 35

is a unit. We get the regular sequence
1' = {/i+si Pf(x), d5/5+q5/6+s5 Pf(X) | 1 5 i 5 4}.

By Specializing d5 and q5 to zero, this sequence becomes 7 = {/1,/2,/3,/4,Pf(X)}
which is regular, by corallary 2.5.

Then d5, q5 are regular on S/(1’), and hence by 1.12 (1'): H specializes tO
(1):H and by the proof of 2.7, this link is ([1, 6, 6, 4, X56’ Pf(X)).

If d5 is a unit (the case q5 a unit is symmetric), then by column Operations

in the first 6 columns Of T, this matrix becomes

 

 

 

 

S1
14 04x2 32
33
34
01x4 1 0 s5]

and 1= {/i + sin(X) i 5 i 5 5}.
If none Of these si's is a unit, we can deform them to variables, and then

specialize to zero. 1 becomes {ll/2, [3,14,/5}, which is a regular sequence, by

43

2.4, and 1:H is a hypersurface section, by the proof of 2.6.
Otherwise, w.l.O.g., 35 is a unit.
After row Operations and column Operations in the first 6 columns of T, we

can assume that T is

5 04x2
0 1

and 1 = {[11/21/31141/5 + Pf(X)}.

By (3) in chapter I we see that

[5(X56+y6) = (Pf(X)+/5)y6 “ i=1 Xi6/i

and by (3) and (4) in chapter I,
/6(X56+y6) = ‘(Pf(x)+/5)y5 + zi=1(xi5—yi)/i

and hence y6+X56 E (1):H and then the link is a hypersurface section. I

Lemme 2.13. Let X be a 5x5 generic alternating matrix over R, Y a 5x1
generic matrix over R[X], v an indeterminate over R[X, Y]. In S = R[X, Y,
V](m,X,Y,v)’ consider the ideal J = (11/2,! ,/4,/5, v), where [Zr/5]t = XY,
and let J -—1 K. Then K is a licci, Gorenstein ideal Of height 5 and deviation at
most 2, and there is a pair (8', K') in the same Herzog class as (S, K) such that
either

i) K’ is a hypersurface section or

ii) K' = H3.

44

£11m. Observe first that (S, (ll,/2,/3,/4,/5)) is in the same Herzog class
as the pair Obtained in 2.16 (i). Actually, such a pair is a specialization of
(s. (4.- «15».

Then (11,- . -,/5) is licci, (2.16 and 1.13), is an almost complete
intersection, has height 4 and type 2.

Because v is regular on S/(/ ,o - -,/5), then J is licci of type 2, deviation 1

and height 5.
Hence if J -—-» K, then K is licci, Gorenstein Of height 5 and d(K) 5 2.

Let 1 = 71, 72, 73, 74, 75 be the regular sequence defining the link J ... K.

Then, for some 5x6 matrix T with entries in S we have

 

 

 

 

[ 71 . . [1 .
72 ’2
73 = T ’3
74 ’4
l
[75, 5
L V .1
Because J ... K, a maximal minor Of T is invertible, and if "—" denotes the

reduction modulo the maximal ideal of S, then T has rank 5 on S, and therefore

the matrix formed with the first 5 columns of T has rank at least 4 over S.
Then, after row and column Operations, in the first 5 columns Of T, T

becomes

'1"!

"I
011$me

4 04XI

 

 

"8

O!
H

 

 

45

and one Of the elements of the last row is a unit.

If one Of the ri's is a unit, then v+/i or v+d5 [5 is in (1) Q K, and hence K

is a hypersurface section.

Otherwise, none Of the ri's is a unit. Then we can deform them to variables
and then 1 becomes

7'=(/i+riv |15i55).

Consider, then the following generic matrices over R

 

 

 

 

 

 

F r11
r2 [Y1
x: x 1r3 Y'=
r4 17
r5
_—r1—r2—r3—r4—r5 0 .

Then, if [/',- - .,/é]t = X'Y', z; = /i+riv, 1 g i g 5. Moreover,

(R[X', Y'](m,XU,Yl)1(/ia ° 14.5))

is (S, 1') and

(R[X't Y'](m,x',Y')’ (41"‘1/5’ V”

is a deformation of (S, J), modulo the regular sequence r1 - . o,15.
Moreover (lip-J5): (IE-owls, v) = H3 (proof Of 2.6) and we are
done. I

Lemme 2.19. Set the notations as in 2.17, in S consider the ideal

46

J = ([1,/2,/ 1/41 X561 Pf(X)),

and let J ——1 K. Then K is licci Of height 5 and deviation at most 2 and there is a
pair (8', K') in the same Herzog class as (S, K) such that either

i) K' is a hypersurface section or

ii) K' = H3. _

gm. That K is licci, Gorenstein of height 5 and d(K) 5 2 is clear (1.6 and
2.7). Let 1 be the regular Sequence defining J —-1 K. Then there is a 5x6 matrix T

with entries in S such that

[2] =T

\\\‘\
euro

56
.Pf(X).

 

 

By standard arguments (see the proof of'2.16 to 2.18) T becomes

f

t—i
OD

COO

Heft?

 

f.-

4,-0-
019me
CD
00

HH
01$

 

 

2X3

and at least one element of the fourth row and one element of the fifth row, in
different columns, are units.
Assume first r 4 is a unit (the case 15 a unit is symmetric). Then, after row

operations T becomes.

47

 

 

 

 

F t1 S1
I4 t2 32

t3 S3

t4 34

01x4 t5 35]

and one element of the last row is a unit. If t5 is a unit, we may assume then that
1 = {li-l-si Pf(X), X56 + 35 Pf(X) | 1 5 i 5 4},
and because

(y6+ 23‘le xi6si + [53)Pf(X)=L¢11___1(/.+s. Pf(X))Xi6+ /5(X56+s5 Pf(X))

is in (1), y6 + 231:1 X1631 + [535 E (7):J and the link is a hypersurface section.
If t5 is not a unit, then 35 is a unit and we get

1 = {/i+ti x Pf(X)-l-t5X56 | 1 s i s 4}-

56’

Since t 5 is not a unit, we may deform it to a variable and specialize to 0. Then

1: {Ii-Hi X56, Pf(X) | 15154}.

If none of the ti's is a unit, again we may deform them to variables and specialize
to 0. Then 1 = {/1,/2,/3, l4, Pf( X()} and 1:J = H3 by 2.7.
If, say t4 is a unit, we may assume then 1 = l1,/2,/3,/4+X56, Pf(X).

Notice our ground ring is still 8. Assuming deg xi. = deg yj = 1. A resolution for

11
J, which was Obtained in the proof Of 2.7, is

48

o ... s2(—9) _. ... ... s5(—2) e S(—3) ... 3
while the Koszul complex Of 1 is
o ... S(—11)—-1 ... ... s4(—2) e S(—3) ... s.
Then a resolution of 1. J starts as
56(—2) e S(—3) ... 5.

Consider /5—X 46 and 6+X45, which have degree 2. Notice that, because Of (3) in

chapter I
[4(/5_X46) = ‘prfm + 27i=1 xiii/i + (X56+ ’4) [5 E (1)
and
/4(/6+X45) = 3’53“) " i=1 X15 Z1+(X56+ ’4) ’6 E (1)
then

K = (l1,/2,/3,/4+X56, /5_X46’/6+x45’ Pf(X)) Q (1):J.
By setting yi = 0, these elements minimally generate K if and only if X46’ X56’
X 45, Pf(X) minimally generate some ideal, which was the case of the proof of 2.9.
Then K = (1):J and d(K) = 2.

Specialize (S,K) via the regular sequence

49

* _ _
— {y3+"121 y2 X131 Y1+x23’ y51x25“26”‘349‘451x46 x56}'
We obtain a pair (8', K'), where
K' = (X14y4+x16y 6’ X2414 x315" 612(X14x23‘x13x24)12(x15x23+"12X35)1
2("16"23‘LXI2X36)’ x16‘24X35‘X15X24X36)’
and then, because 2 is a unit, we get

' _ _ .
K _ (x14y4+x16y6’ x24y4, x36y6, x14x23 x13x24, X15X23+X12X351

X16x23+x12x36’ x16X24x35‘x15X24x36)'

That is K' = (11(AB), Pf(A)) where

F 0 X12 X13 X14 x15 x16 “X231
"x12 0 x23 x24 0 0 x13

A: “x13 x23 0 0 x35 x36 B: x12
x14 —x24 0 0 0 0 y4
—x15 0 -x35 0 0 0 0
L—XIG 0 —x36 0 0 0 y6

 

 

 

 

Here we use the fact that 2 is a unit. We now have to Show that (S', K') is a
specialization of (S, K). It is enough to Show that ht K' = 5.

Let p :1 K', be a prime ideal in S'. If x23 is in p, K' contains also
x14y4+x16y6, x36y6, x12x35 and x13x24 and then, ht p 2 5. If x23 1! p, then 1n
Sp, K' becomes

50

x13"24 "12x35

(x14y4+x16y6,x24y4,x36y6,x14— x23 ”‘15”r x23

1

X12x36

1

x16+ 23 x16x24 X35‘x15x24x36)

x x x x x x
I: 24, x15 + I: 35, x16+ I: 36
23 23 23

x24y4, x36y6. Then ht K' 2 5, and then, because K' is a specialization Of an ideal

which contains 3 linear forms, x14- , and

of height 5, then ht K' = 5. Also notice that K' is a specialization of H3.
Finally, assume that neither 14 and 15 are units. Then row and column

Operations in the first 4 columns, T becomes

 

 

1—1
00
C
COO
COCO

 

 

where r ('1, r5 are still not units. Then

1 = {/1,/ ,/3, X56+rl]/4, Pf(X) + r5/5}.
Now we deform r ('1 and r5 to indeterminates and specialize them to zero. 1 then
becomes {ll/2,13, X56’
linkage specializes (1.12). Moreover (1):J = (Iv/2,43, X45, X46’ X56, Pf(X))

Pf(X)}, which is a regular sequence by 2.8. Then, the

(2.9). If we apply the specialization * to (S, K), we obtain the pair (S', K')
Obtained before. Now we are done. I

We now turn our attention to the following case: let (R, m) as usual, and

51

let X' a 4x4 generic alternating matrix Over R, Y' a, 4x1 generic matrix over R[X']
and y, z indeterminates over R[X', Y']. Set S = R[X', Y', y, 2]. We are interested

in all ideals J and K Of S(m,X',Y' such that (H2, 3’2) -—-+ J —-t K. Because H2

.152)
is a deviation 2 Gorenstein ideal of height 3, we may assume it is generated by the

4x4 pfaffians Of a 5x5 generic alternating matrix X[B—E, 2.1]. We write [i for
(—1)i+1 xi. Write then s = R[X, y, 2].
Lemma 2.211. With the previous notation, let I = (H2, y, z) ... J. Then

there is a pair (8', J ') in the same Herzog class as (S( , J) such that either

m,X,y,z)
i) J ' is a hypersurface section or

ii) there is a 6x6 alternating matrix A and a 6x1 matrix B aij’ bj 6 ms,

such that if [t1,- . ~16]t = AB, then J = (11,- - -,t4, A Pf(A).

56’
Proof. Let 1 be the regular sequence defining I —-» J. Then there is a 5x7

matrix with coefficients in S( ) such that

m,X,y,z

M\ 1—1\

[1] =T

 

 

N '~<
m\ uh\ w\

Because I —-t J, a maximal minor of T is invertible. By using standard arguments,

after row Operations and column Operations in the first 5 columns of T, T becomes

U)

 

 

 

1 d1

I3 03.2 S2 d2
S3 d3

02x3 t4 1’4 34 d4
t5 r5 35 d5,

52

and at least one element of the fourth row and one element Of the fifth row, in
different columns, are units.

If 35 or d5 are units (and likewise for any other si or di) y+t5 l4+r5 [5 or
z+t5 [4+15 [5 are in ( 1) and hence 1.1 is a hypersurface section.

Then, we may assume that T is:
31 d1
32 d2
5 33 d3
34 d4
35 d5 )

 

 

 

and none of the si's, di's are units 1 5 i 5 5. Therefore we may deform all Of them

to indeterminates and specialize all of them, but s4, d5, to 0. Then
1 = {111/21 l3,/4+sy, Z5 + dz}
and ht( 1) = 5.
If we assign deg xi]. = deg y = deg z = deg s = deg d = 1, then, a resolution
for I is
o -e S(—7) ... .. - ... s5(—2)es2(—1) _. 3,
while the Koszul complex of 1 is

o ... S(—10)—-1 ... ... s5(-2) ... s.

Then, a resolution of ( 1): I starts as

53

s5(—2) e S(-3) ... 3.
Consider x45 sd, which has degree 3. Because

x45 sdy = x45d(/4+sy) - x45d/4 e (1)
and

x45 sdz = x45 s(/5+dz) - X45S/5 E (1)

it follows that x453d e (1): I. If deg xij = 1, deg y = deg z = 2, deg d = deg s = 0,
then ( 1) is generated by homogeneous elements of degree 2, but x45 sd has degree
1. Then x45 sd i (1). Thus J' = (1):I = (l1,/2,/3,/4+sy, /5+dz, x45 sd) which
can be viewed as (t1, t2, t3, t 4, A56’ Pf(A)) where, if

0 x45 x24 ”‘25 0 0‘ ’x13
““45 0 x34 ”‘35 0 0 x12
A: ““24 ‘x34 0 “‘23 ‘d 0 13: x15
"25 x35 x23 0 0 ‘3 "14
0 o d 0 o o
0 0 0 s 0 0. y

 

 

 

 

then the ti's are as in (ii). I

In lemma 2.20 (i), J is a hypersurface section, namely J = (N, t), t regular
on S/N and N is a type 2 almost complete intersection Of height 4, which is licci
because J is licci (2.11). By 2.16, after deforming and specializing if needed, N is
either of the form 11(AB), A a 5x5 alternating matrix, B a 5x1 matrix, or N is a
hypersurface section. In the first case, if J --v K, then K is described in 2.18. We
now study the second case.

Lemma 2.21. Let J be a licci, almost complete intersection ideal of R Of

54

height 5 and type 2 which is a double hypersurface section. If J —-l K, K is licci,
Gorenstein, has height 5 and deviation at most 2, and K is a hypersurface section.

m. Write J = (N, y, z), y, 2 regular on R/N, N a height 3, type 2
almost complete intersection. Let Y, Z be indeterminates over R. Then (R', J ') =
(R[Y, Z](m,Y,Z)’ (111, Y, Z)) is a deformation Of (R, J). Let J' ~ K' such that
(R',K‘) is a deformation of (R,K) (such K' exists by 1.13). Hence J ' --» K' and we
will show that K' is a hypersurface section.

Let 1 be the regular sequence defining J ' —-1 K'. Then, for some matrix T

with coefficients in R', we have

[2] =

 

 

where N = (a1, a2, a3, a 4). By standard arguments, we may assume that T is

Ti

 

OOOOH

 

 

Hence 1 contains Y+t12Z+Zfi1=1 t1 i +2 ai and hence K' is a hypersurface section.
I

We are now ready to prove the main result of this chapter.

Theorem 2.22. Let (R, m) be a local Gorenstein ring in which 2 is a unit,

and assume the residue class field of R is infinite. Let I be a licci, Gorenstein ideal

55

of R Of height 5 and deviation 2. Then, there is a pair (R', I') in the same Herzog
class as (R, I) for which either

i) I is a double hypersurface section or

ii) there is a 6x6 alternating matrix A and a 6x1 matrix B, such that I' =
(I1(AB), Pf(A)), aij’ bj 6 ms..

floot. By the conditions on R and because I is licci and Gorenstein, there
is a sequence of links

[:1

—-)Il——9-oo—-11

0 n

where In is a complete intersection and I2i is Gorenstein 0 5 2i 5 n (1.22).

Therefore 12 is Gorenstein of height 5 and deviation at most 2 (1.6). If I2
has deviation 0, then it is a complete intersection. If not, because I2 is perfect,
then by [Ku], d(12) = 2, and I2 is linked in two less steps to a complete
intersection. Therefore, by induction hypothesis, we may assume that either I2 is
a complete intersection or after deformations and specializations, I2 is as (i) or (ii).

Assume first d(12) = 2. Then, because d(I2) = r(R/Il) = 2, I2 ... I1, and
because d(Il) = r(R/IO) = 1, Il —e 10.

If L1(I2) is a first generic link Of I2 in some polynomial ring R[Z], and if
L202) is any second generic link of I2 in some polynomial ring R[Z, W], then for
some p in Spec(R[Z, W]), m C p, (R[Z, W]p, Li(12)R[Z, W]p) is a deformation Of
(R, I2—i)’ 0 5 i 5 2 (prOposition 1.14).

Therefore we can replace R by R[Z, W]p, I2—i by Li(12)R[Z, W]p. Change
notations and call this ring R, and the ideals 12, I1 and 10. Then R is Gorenstein
local, 2 is a unit, I2 .... II -—+ I0 and these links deform and Specialize (proposition

1.13 and remark 1.15). Therefore, and by induction hypothesis, we may assume 12

is actually equal to the ideal described in i) or ii) and we can also assume all

56

matrices are generic.

Assume I2 is as (i). Then I2 is a double hypersurface section. Then
lemmas 2.20, 2.16, 2.18, 2.21 and 2.19 describe I.

If I2 is as (ii), then lemmas 2.17, 2.16, 2.18, 2.21 and 2.19 describe I.

SO, we now assume I2 is a complete intersection, and I —-1 Il ... 12. For II
we know d(Il) = 1 and because r(R/Il) = 1, II -+ 1. Hence I2~I1 -—+ I.

Let (1) be the linking sequence from I2 to 11. Then since r(R/Il) = 2 =
11(12/ (1)), three elements Of I2 which are part Of a minimal set of generators Of I2
are in (o).

Deform (R, 12) to

(R[T, T2 ,T

T4,T T,.TTTT5))

l ( 1 . ,
5(T ,5,T2.T3.T4.T) 2 3 4

3,

Then (or) contains 3 of the Ti's and hence I1 is a double hypersurface section.
Lemma 2.21 describes I. Now we are done. I

Corollary 2.23. Let (R, I) be as in 2.22. Then the minimal resolution Of
R/ I is either

i) o—eR—eR7—eR22-eR22—eR7—ea
01'
ii) o—eR-eR7—eR13—eR13—eR7—en.

Proof. By 2.22, (R, I) is in the same Herzog class of (R[X, Y](m X y), H3)
for a 6x6 generic alternating matrix X and a 6XI generic matrix Y, or in the same
Herzog class as (R[Z, t, w](m,Z,t,w)’ (H2, t, w)) for some 5x5 generic alternating
matrix Z and indeterminates t and w. A resolution for H3 is (i) and a resolution

for (H2, t, w) is (ii) [K]. NOw we use the fact that the Betti numbers are

57

invariants of the Herzog class.

Corollery 2.24. Let J be a licci almost complete intersection ideal of R Of
height 5 and type 2. Then there is a pair (R', J ') in the same Herzog class as (R,
J) such that either:

i) J is a hypersurface section or

ii) There is a 6x6 alternating matrix A and a 6x1 matrix B, aij’ bj E mR"
such that if [/ ,...,/6]‘ = AB, 1' = (1122,12 , A56, Pf(X)).

£1991. Let J —-1 I. Then J is licci, Gorenstein, has height 5 and deviation
at most 2.

If d(I) = 0, I is a complete intersection and then, by the proof of 2.22, J is a
hypersurface section, and it is described in 2.16.

Otherwise d(I) = 2 [Ku]. Then, there is a pair (R', I') in the same Herzog
class Of (R, I) such that I' is as (i) or (ii) of theorem 2.22. If I' is as i, lemmas 2.20
and 2.16 describe J. If I' is as (ii), lemmas 2.17 and 2.16 describe J. I

Corollary 2.25. Let R = k [[x1,- - °’xr]]’ where k is an infinite field with
char k # 2 and let I be a licci Gorenstein ideal of R of height 5 and deviation 2.
Then, either

a) I is a double hypersurface section or

b) I = (11(AB), Pf(A)), where A is a 6x6 alternating matrix, B a 6x1
matrix, aij’ bj not units.

M- By (2.22), (R, I) is in the same Herzog class as (R[[X, Y]]), H3) or
(R[[Z,t,w]], (H2,t,w)), where X is a 6X6 generic alternating matrix, Y a 6x1 generic
matrix. But since R[[X,Y]]/H3 and R[[Z,t,w]]/(H2,t,w) are rigid [H-U—l, K], it
follows from the discussion following (1.19) that (R, I) is a specialization Of

(R[[X,Y,w]], H3 R[[X,Y,w]]) or (R[[Z,t,w,v]], (H2,t,w) R[[Z,t,w,v]]). I

CHAPTER III
THE EVEN GRADE CASE

In this chapter, we construct a family of licci, Gorenstein ideals of deviation
2 and even grade that are not hypersurface sections in any even grade larger or
equal to 6.

The only positive result known on even grade is the result of Herzog and
Miller [H—M, th. 1.7]. They show that if R is a local Gorenstein ring in which 2 is
a unit, and if I is a perfect, Gorenstein, generically a complete intersection ideal Of
R of height 4 and deviation 2 such that I/I2 is Cohen—Macaulay, then I is a
hypersurface section.

Vasconcelos and Villarreal show that the condition of I/I2 being
Cohen—Macaulay follows from the other assumptions [V—V, th. 1.1].

We now construct a family of licci Gorenstein ideals of even height and
deviation 2.

Let k be a field, 11 an integer, larger or equal to three, X a (2n—1)x(2n—1)
generic alternating matrix, Y a (2n—1)x1 generic matrix Over k[X]. Let R =

k[X,Y], S = R(X,Y) and in this ring consider the ideal

/

I1n = (/ 2n-2’

1,..., x

2n—1’y2n—l) = (Hn—l’ y2n—1)’

where [/ ,- - -,/2n_1]t = XY.
Notice In is licci, Gorenstein, has even height 2n—2 and deviation 2.

Lemma 3.1. With the above notations,

58

59

Gn=(/ / x x

1M 2n—4’ 211—3 2n—2 2n—1’ 2n—l’y2n—l)

is a licci ideal of deviation 1, height 2n—2 and type 2 in S.

Proof. This is essentially lemma 2.7.

 

Lemma 3.2. on = ([1’”"/2n—4’ X2n_3 2n—2 2n—1’ X2n_1) is a complete
intersection in S.

Prmf. By induction on n. For n = 3,

 

 

 

 

F 0 "12 "13 "14 "15 y1
"12 0 "23 "24 "25 y2

x: "13 ”‘23 0 "34 "35 Y: ya
”‘14 ”‘24 ”‘34 0 "45 y4

"15 "25 ”‘35 "45 0 (yo

and

(93) = ("12’ "13y 3 + "14y4 + "153’5’ "23y3 + "24Y 4 + "253’5’ "13"24”‘14"23)'

If T = S/(x12), then T is a Cohen-Macaulay domain. A direct application Of (2.2)
shows that (x13y3 + x14y4 + x15y5, x23y3 + x24y4 + x25y5) is a prime ideal of
height 2 (it actually defines Sym(12(Z)) where Z is a 2x3 generic matrix) (see also
[Ho]) and x13x24—x14x23 is not in this ideal. Hence (o) is a complete intersection.

For 11 2 4, let p e Spec(S) be such that ht p 5 2n—2 and (on) Q p. Then, as
in 2.8, p does not contain the ideal generated by the entries of the matrix formed
with the first 2n—4 rows and 2n—4 columns of X. Then, we can assume, w.l.o.g,,
that x12 is invertible in Sp. Hence, there is a matrix A, invertible over Sp, such

that

 

 

 

 

and the entries of X' are generic over k[xIé, {xij’ i 5 2}]. If Y" = A_1 Y, then
(an) becomes (yl, y2, an_1), which by induction hypothesis, has height (2n—4)+2
= 2n-2. I

We can establish now the following proposition.

Proposition 3.3. With the previous notations,

Fn = ([1”“’/2n—4’ x2n—1’ x2n—2’ X‘2n-3’ X2n—3 2n—2 2n—1)

is a licci Gorenstein ideal of height 2n—2 and deviation 2.

Proof. Grade R by assigning deg xij = deg yj = 1. A resolution for Gr1

starts and ends like (2.7)
2 2n-4
0 —-1 R (-(5n—10)) -—->- . .... R(—1)e R (-2) e R(—(n—2)) e R(—(n—1)) ——+R
while the Koszul complex for (on) is
2n—4
0 ——» R(—-(6n-11)) —-+' - 0—» R (-2) e R(—(n-2)) 0 R(—(n—1)) —-1 R.
Therefore a resolution for (on): Gn starts as
. - --» R2n’4(-2) a R(—(n-2)) e R3(—(n-1)) -—+ R.

Consider X2n—2 and X2n_3, which have degree (n—l). Let X' (reSp. Y') be the

matrix obtained from X (resp Y) by deleting the (2n—2)th row and column (resp.

61

(2n—2)th row).

Let [/',.. '/2n- 3 2n_1]t = X'Y', and notice X2 n—2 = -Pf(X'). Then
X2n—2y2n—1 = ‘y2n—1 ”(K')
= 2?:‘3 Xi2n—2 2n—1 /
=22ni4 Xi2n—2 2n—1 1+ +X2n—3 2n-2 2n—1 [2n—3
— (2?:{3 X12n- 2 2n—1xi2n—2) y 2n—2
=23? Xi2n-2 2n—l [1+X2n-3 2u-2 2n—l Z2n—3”‘2n—1 y2n—2
6 (an).

Similarly, X2n—3 y2n_1 E (on), and then Fn Q (o): G If in_3 E (Q), then for

n'
homogeneous elements a1,- ~ -,a2n_4, b, c in R,

n-4
X 22: 31 [i +bX2n—3 2n—2 2n—1+° X2n—l'

2n—3

Then deg b = 1. If we set deg Xijz 0, deg yi = 2, then a1 = — -a2n_4= 0.

Therefore

X2n—3 = b x2n-3 2n-2 2n—1 + ‘ X2n—1'

Specialize, by assigning xi2n—3 = 0. Then in_1 becomes zero but

62

X =bX

2n—3 2u-3 2n—2 2n—1 1‘ 0’

but this is impossible, because in_3 is an irreducible element of R and neither b
nor X2n—3 2n—2 2n—1 are units in R.

Similarly, X2n_2 E (o, X2n_3) only if, for homogeneous elements

X =aX

2n—2 2n—3 2n—2 2n—l + b x2n—3 + c X2n—1'

Then deg a = 1. If we Specialize by assigning xi 2 = 0, we obtain

n—2

x2n—2 = a X2n—3 2n—2 2n—1 " 0’

which is impossible. In the same way

X2n--1 ¢ (/ ’° ° °’/2n—4, x2n—3 2n—2 2n—1’ x2n—3’ X2n—2)

Finally, as in 2.9, 11 E (l / X X X X f

2""’ 2n—4’ 2n—3 2n-2 2n—1’
and only if [1 e ( [2"”’/2n—4)’ which is contrary to 2.4.

211—3, 2n—2’ 2n—l) ‘

Then Fn Q (_o_n):Gn and (on):Gn/(on) is minimally generated by 2 elements
of degree n—l. Then Fn = (on): Gn’ and on form part of a minimal generating set
of En

Hence in S = k[X, Y](X,Y)’ Fu and GH are linked FI1 -—> GH and GI1 ... Fn.
Then r(S/Gn) = d(Fn) = 2 and r(S/Fn) = d(Gn) = l. I

Proposition 3.3 produces a family Of licci, Gorenstein ideals of every height
larger or equal to 4 and deviation 2.

When n = 3, F has height 2(3)—2 = 4 and then by [H—M] and [V—V], it is

3
a hypersurface section. We ask if PH, 11 2 4, is a hypersurface section.

63

Before we answer this question we need some definitions and propositions.

Qefinitiop 3.4. [K—M—2, def. 3.1] Let I be a grade g Gorenstein ideal of the
local Gorenstein ring P. Let e be a 1xn vector which generates I, t an integer with
05t5g—1, Y a (g—1)x(n—t) matrix of indeterminates, and v an indeterminate.

Define 12 by the product

lbl‘= __‘_ Y [alt-

Let y be an element of IP[[Y]] with o, y regular and 11(1/(12, y)) 2 2. If w is any
element Of P[[Y]] such that J = (o, y): I is generated by (b, y, w) and K = (o,
w+vy): J, then P[[Y, v]]/K is called a semi neri i h uble link of P/I.

Kustin and Miller show that K is a grade g. Gorenstein ideal, and if P =
k[[x1,- - -,xn]] and if I is a grade g ideal of P such that P/I is rigid, then any semi
generic tight double link of P/ I is also rigid [K—M, prop. 3.2 and Cor. 3.7]. .

We now begin the construction of our example.

Set the notations as in the beginning of this chapter, let n 2 4, and let
X'(reSp. Y') be the matrix obtained from X (resp. Y) by deleting the last 3 rows
and columns. Let P" = k[[X', Y]], P' = k[[X, Y]].

In P" consider the ideal Ln = (/"”"/2n—4’ f', y2n_3, y2n—2’ y2n—1)
where [Ann/1114f = x' Y', r = —Pf(X'), (actually it is x2n_3 21H 2n_l).

Then, Ln has height 2n—2 and deviation 2, it is Gorenstein, licci and P"/Ln

is rigid ([H-U—l], [K]). Let h be defined by

64

 

 

 

 

 

x1 2n—3 "1 2n—2 "1 2n—1 z]
lbl‘ = 1M 11.1.4.
x2n—4 2n—3 x2n—4 2n—2 x2n—4 2n—1 f.
x2 x x y 2n—3
_ n—2 2n—1 211-3 2n—2 2n—3 2n—1 . y2n_2
_ y 2n—l 4

That is

_ I _
bl ‘ ’1 + "l2n—3 y2n—3+ "12n-2Y2n—2+ "12n—1 y2n-1 " i

b2n-4= ’2n—4+"2n—4 2n—3y2n—3+x‘2n—4 2n—2y2n-2+x2n-4 2n-1Y2n—l = [2n—4

b2n—3= f'1”‘2n--2 2n-1 y2n—3+x2n-3 2n—2 y2n—2+x2n—3 2n—1 y2n—1

. t .
Then, we notlce that [/ ""’/2n—1] = XY and that y2n_1 E LnP' is such that (1),,
y2n_1) is a complete intersection contained in Ln P', by 2.5. Also 11(Ln/ (o,y2n_1)
= 2, being generated by the images Of y2n_3 and y2n_2.
Le‘ w = X2n—l ‘ "2n—2 2n—1 /2n-2 + "2n—3 2n—2 /2n-3'
Le_mr_n_e_3._5. With the previous notations, Tn = (o, y9n_l, w) is a licci,
type 2, deviation 1 ideal of height 2n—2 in P'.

Prmt. With C = (11, y2n_1): LH and denote by "——" reduction modulo

(1) = (x2n-3 2n—1’ x2n—3 2n—2’ x2n—2 2n—l)’

which is a complete intersection in S'.
Then (_5, y2n_1) = ([1’/2”"’/2n—4’ f', y2n-l)’ which by corollary 2.5 is a

regular sequence in P". Then (1) is a regular sequence on P'/(_b, y2n_1) and

65
therefore C = (5, y2n_1): In = (E, y2n_1): Ln’ and ( 1) is also a regular sequence
on 9/0 (1.12).
In S'” = k[X, Y]/(1) assign deg xij = deg yj = 1. Then a resolution of
LnS‘" in S'” is ([K]), or (2.6)
o ... S"'(——(4n — 8)) ... ... s"'2"‘4(—2) e S"'(—(n—2)) e s"'3(—1) ... 5...

while the Koszul complex of o, y2n_1 is

o ... S"'(—(5n—9)) .... . .__. s"'2“‘4(—2) e S'”(—(n—2)) e s"'(—1)—e s'".

Then a resolution of C S'” in S'” starts as

31112n_4(_2) 9 Sill(_(n_2)) 9 Sill(_(n_1)) 9 SIII(_1) __, Sill

Now x2n—1 has degree n—1, and it is in C S'” but not in (E, y2n_1). Then

‘6' 8'" = (B, an-.. Kan—.7) 8'"

and therefore

6 = (5' 3’-2n--1’ x2n—1) P = T;-

Consider now y2n_2 w and y2n_3 w.

First let X" (resp Y") be the matrix obtained from X(resp. Y) by deleting
the (2n—l)th row and column (reSp (2n—1)th row). Let [l",...,/'2'n_2]t = X"Y".
Since X2n—1 = Pf(X"), by (3) and (4) in chapter (I) we get

66

/

2n— 2+X2n—3 2n—2 /

(X 2n—3)

Y2n—2‘” = y2n—2 2n—1 "2n—2 2n— 1

/

2n—2+y2n—2 "2n—3 2n—2 /

X 2n—3

= y2n—2 2n—1—y2n-2X2n—2 2n—1

n-4(X
=22: X—2n 1) i2n —-2’i X2n--3 2n—2 2n—1 ’2n—3

/

2n—3'

/

‘ y2n—2"2n—2 2n—1 2n—2 + y2n—2"2n—3 2n—2

Now we add and subtract

22n—3((X

i=1X2n-1) i 2n-2"l 2rr--1)y 2n—l'

Notice then that

(x 4-

1 _.
2n—1)i 2n—2"i 2n—1y2n—1+(X2n-1)i2n-2 ’1 " (X2n-1)i 2n—2

Hence, we obtain

/

_ n-4
_ 2i=1( + X 2n—3

(X2n—1)i2n-2 / 2n—3 2n—2 2n—1

_(22n-—3 2n—1

i=1 (X2n—1)i 2n—2xi 2n—1)y2n—1 + 121 4y i x2n—2 2n—1

i # 2n—2

/

.+ "2n—3 2n—2 y2n—2 2n—3

2n—4
=Zi_ 1[(x2n-1 i2n —2 y 1x2n-2 2n—11/i

67

+ ’2n—3IX2n-3 2n—2 2n—1+y2n—3x2n—2 2n—1

+X2n—3 2n—2y2n—2+x2n—3 2n—1y2n—1]

_[22= n-3(X
X)2n-1 i 2n—2 xi 2n—1
+ [2n—1x2n—2 2n—1+x2n—3 2n—l [Zn—3ly2n—1 E (12’ y 2n—1)

Also

y 2n—3W = y 2n—3(X2n—1Tx2n—2 2n—1/2n—2 + x2n—3 2n—2 l2n—3)
= y2n—3 X2n—1x2n-2 2n—ly2n—3 ’2n—2 + "2n—3 2n -2y2n—3 ’2n—3
= 22n-4(X2n—1)i 2n—3 ’i"(X2n—1)2n—3 2n—2 ’2'n—2
‘ "2n—2 2n—ly2n—3 ’2n—2+"2n-3 2n-2y 2n—3 ’ 211—3
We now add and subtract
23:? X2n—1)i 2n—2xi 2n—1)y 2n—1
We obtain then
= 2h?“ 2n—1)i 2n—3’i ’X2n—3 2n—2 2n—1/2n—2
‘ (2?:3 (X2n—l)i 2n-2 "i 2n—1) y2n—l”‘2n—2 2n—1y2n—3 ’2n—2

68

—/

2n—2(X

2n—3 2n—2 2n—l+x‘2n—2 2n—1y2n—3+x2n—3 2n—2y2n—2

+x2n—3 2n-1y2n—1)

2n—2
(21:1 (X2n—l)i 2n-3 xi 2n—1+/2n—1x2n-3 2n—2

—x2n-3 2n-1 [2n—2) e (9’ y2n—1)'

On the other hand, we had seen that Tn = ‘0- ( "—" denote reduction module
(1) = (X2n—2 2n-1’ x2n-3 2n—2’ X2n—3 2n—1) and that 7 is. a regular sequence on
P' / C. Then also Tn = C. Moreover Ln ——-1 TH and hence Tn has type 2, deviation
1 and height 2n-2. I

fimark 3.5. Let R' = k[X, Y] and grade R' by assigning deg xi j = 1, degyj
= n—3. Then Ln R' has a presentation

2n—3 3
R' (-(n-2)) 9 R' (—(n-3)) -1 R',

and hence a homogeneous resolution of TH R' starts and ends as

2 2n—3
0 -+ R' (-(n-2)(2n-3)) 2' ° '2 R' (-(n-2))‘PR'(-(n-3))9R'(-(n-1))-’ R'.

Let now v be an indeterminate over P' and let P = P'[[v]]. In this ring, we

69

consider, with the above notations, the complete intersection (h, w+vy2n_l)

’7 = X211—2 ' Vy2n—2 ‘ x2n—2 2n—l [211-1 + x211—3 2n—1 [211—3
and
C= x2n—3 + vy211—3 " "2n—3 2n—l [211—2 “ x211—3 211—2 [211—1

with the above notations, we have the following theorem.

Theorem 3.7. Let n 2 4. In P, the ideal En = (b, w+vy2n_1, 17, C) is a
licci, Gorenstein ideal of height 2n—2 and deviation 2, P /En is rigid and En is not
a hypersurface section.

M- We will show Tn -+ En and En ... Tu and that En is a
semi—generic tight double link of In Then the conditions on the height, deviation
and type will be satisfied. Because Ln is rigid [H—U—1][K], then P/En is also
rigid, En being a semi generic tight double link of Ln' Also, En is contained in the
square of the maximal ideal, and hence, by (2.12), En is not a hypersurface section.

Therefore we only have to compute those links. Set deg x.. and deg y. as in

1] J
3.6, and set deg v = 2. A resolution for (b, y2n—1’ w) R' was found in 3.6

2 2n—3
0 -’ R' (-(2n-3)(n-2)) —+- ° °-’ R' (-(n-2)) ‘9 R'(-(n-3))

e R'(-(n-1)) —-+ 11'

while the Koszul complex of (b, y2n_lv+w) R' is

2 -3
0 —~ R'(—<2n—3)(n—2>—<n—1)) —-» . ... R' n (an—2110 R'(-(n-1)) -—+ R'

and then a resolution for (b, w+y2n—1V) R':(b, y2n_1, w) [1' starts as

2n—3 3
R' (—(n—2)) e R' (—(n—1)) —-+ R'.

70

Then ED is generated by (b, y2n_1v+w) and 2 elements of degree n—l.
Consider

/ /

'7 = x2n—2 — vy2n—2 — x2n—2 2n—1

2n—1 + x211—3 2n—1 211—3
and
C = X2n—3 + Vy211—3 + x211—3 2n—1 [211—2 " x211-3 2n—2 [211—1

Let X" (resp. Y") the matrix obtained from X (resp. Y) be deleting the (2n—2)th
row and column (resp. (2n—2)th row). Then X2n-2 = —Pf(X"). Let also

U I 0 t _ II II
M""’ 2n—3 2n—1] - X Y

then, by (1), (3) and (4) in chapter (I), we have

77y2n—1 = (X2n—2 ' vy211—2 ‘ X211—2 2n-1/2n-1+ x211—3 2n—1 / 211—3)y 211—1 =

l

/ 2n—3y2n—1

X 2n—1y2n—1 +x2n—3 2n—l

211—2y 2n—l—vy 211—2y 2n—l ”‘2 n—2 2n—1

which by (3) in Chapter 1 becomes

__ n-4

=1 (X2n-2)i 2n—1 Z'1' '( 1'

X 2n—3 — vy2n-2y2n—1

2n—2)2n—3 2n—1

/

_ x2n—2 2n—1 6n—1y 2n-1 + x2n-—3 2n—1 2n-3 y 2n—1

Now, we add and substract

2n-3
(21:1 Xi 211—2 2n—1xi 2n—2)y2n-2'

71

We obtain then, by also setting the indices in increasing order

/+X

n—4
i=1 Xi 2n—2 211—1 i 2n—3 2n—2 2n—1/2n-3_X2n—1y2n—2

/

2n—1y2n—1 +x2n—3 2n—l /

_vy2n—2y2n—1_X2n—2 2n—1 2n—3y2n—1'

And by (4) in Chapter I, we get

2?:4X /i+X —X

i 211—2 2n—1 2n—3 211—2 211—1 / 211-3 211—1y 2n—2

2n-2

-Vy2n—2y2n—1+x2n—2 2n—1 121 [13’ i+x2n-3 2n—1/2n-3y2n—1

59‘ A1 = Xi 211—2 2n—1+"2n—2 2n-1y 1’ 1 5 i 5 211—4.
Then, the last expression becomes

2n-4
21 =1 Ai/i+ x211—3 2n—2 2n—l [211—3 ‘ X2n—ly2n—2 ' vy2n—ly2n—2

/

/ 2n—2y2n—2

+ x2n--2 2n—l 2n—3y 211—3+ 9‘2n-2 211-1

/

+ X2n—3 2n—1 2n—3y2n—1 =

2n-4
21:1 Ai’fi [2n—3(X2n—3 2n—2 2n—1+"2n—2 2n—1y 211—3

+ x2n-3 2n—1y2n—1)

’ y2n_2()(2n_1+vy2n_1 ‘ x211—2 211—1 [211—2)

72

_ n—4
" Zi=1 Ai/i +

/

2n-—3(x

211-3 211-2 211-1+ x211-2 211-1y 211-3 + "211-3 211—2y 211—2

+ x211—3 211-13’ 211-1)

/

211-2+ x211-3 211-2/

(X 2n—3)

_ y 2n—2 2n—1+vy2n—1—x2n—2 2n—1

E (b, y2n_lv+w).

Let now X (reSp. if) the matrix obtained from X (resp. Y) by deleting the

(2n—3)th row and column [resp. (2n-3)th row)]. Then X n—3 = Pf(X). Let

2

~ ~ ~ ~ ~ ~

[ll’m’l2n—4 [2n-2/2n—1] : X Y’

Therefore, by (1), (3) and (4) in Chapter I, we have

(y2n-1 =

l

(X 2n—1+ x2n—3 2n—1/2n-2)y2n-1 =

2n—3+ ”211-3“ x211-3 211-2

x211-3y 211-1+ VY 211-3 y211-1“ x211-3 211-2 / 211-1y 211-1
+ x211-3 211-1 / 211-2y 211-1'

By (3) in chapter (I) this expression becomes

2n—4 - -

121 (X2n—3)i 211-1/1+X211-3 211-2 211-1 [211-2 + W 211-3y 211-1

73

/

‘ x211-3 211-2/211-1y211-1 + x211-3 211-1 211-2y 211-1-
Add and subtract
2n—2
(1.21 (x2n-3)i 2n—1Xi 2111-3)y 211-3
we obtain then
23214(X211-3h 211-1/1 + X211-3 211—2 211-1 1211-2+"Y211-3y211-1
"(XE—12(X211-3h 211-1"i 211-3 ) y211—3""211-3 211-2/2n-1y211-1
+x2n—3 2n-1 [2n—2y2n—l'

Which by (2) and (4) in chapter I becomes

22n-4

i=1 (X211-3)i 211-1/1’r X211-3 211-2 211-1/

211-2+Vy 211-3y 211-1
+X +(22n'2/ )x +x /
211-1Y211-3 i=1 iyi 211-3 211-2 211-3 211-1 2n—2y2n—1'

set Bi = (X2n—3)i 211-1 + yi"211-3 211-2’ 1 3 i 5 “‘4'

Then we obtain

2n—4
2i =1 Bi (1+[211-2(X211-3 211-2 211—1 +y 211-1 x211-3 211-2

+ x211-3 211-2Y211-2) ‘ y211-3(X211-1 + Vy211-1+ [2n—3x2n—3 211—2)

74

_ 2n-4

‘ 21:1 314 +

[Zn—2(X2n—3 2n—2 2n—1 + y 2n—1x2n—3 2n—1 + x2n—3 2n—2y2n—2
+ X211-2 211-1y211—3)

" y211-3(X211-1 + vy211-1 + ’2n-3x2n-3 211-2 ‘X211-2 211-1 1211—2)

E (b, w + y2n-l v).

Therefore, (b, w + y2n_lv, n, C) R' Q (h, w + y2n_lv) R':(b, w, y2 n_1)R'. If 17 E
(b, w + y2n_1 v), then the image of n is in the image of (b, w + y2n_1v) in

R.
(V1 x211-3 211—21 x211-3 211—1' x211—2 211-1)'

 

Let "-——" denote images in this ring. Suppose that
C = X211-3 E (/ 1""1/211-41 X2n-3 211-2 211-11X211-1)

where X is the matrix such that 3?, j = xi j for

(i, j) z {(2n—2, 2n—1), (2n—3, 2n—2), (2n—3, 2n—1)}

and 3c.” = 0 if (i, j) is in this set.

As usual it would follow that X21?3 is in (in_3 2n—2 2n—1’ X2n—1) (see

75

proof of 2.9). But that says that X2n—3 E 0 mod({x1jl2 S j S 2n—4} U {Xi 211-3”

and this is false. If n E (b, w+y2n_lv, C) then, as usual,

" = x211-2 E (X211-3 211—2 211-1X211-3’ X211-1)

and hence

X2n—2 E 0 mod({x1j|2 S j S 2n—4} U {Xi 2n—2})

and this is false.

/

If [1 E (/ 2n—4’ w+y2n_lv, n,

2,..., C), then in IT‘- I E (l l ) and

1 2""’ 2n—4
this is false.

Thus, we have seen (9, w+y2n_lv, n, C) R' is not a complete intersection.
Moreover n and C are homogeneous of degree n—1 and part of a minimal system of

generators of

(Q: W+y2n_1V)R" :(ba W, Y2n_1R.)
(9,111+):2 n_1v)R'

 

On the other hand, we had seen that this module is generated by two homogeneous

elements of degree n—l. Then

(h, W+y2n_1V)R':(h, W, y2n_1)R' = (1)., + Y2n_1va 77$ C)R'

Hence in particular, (b, w+y2n__lv)P: T n = En' Now we are done and all our

claims follow. I

76

We are interested in the relation between En and En Before studying it,

we need a lemma.
Lemma 3.8. Let X be the matrix obtained in 3.7, let n 2 3 and let Y as
usual. In R = [X, Y] the ideal

X

Fn = (Z / 2n—2’ XZn—l’

1,... 2n-4’ X2n—3’ X

2n-3 2n-2 2n—l)

has height 2n—2.

Prmf. Induct on 11. Notice first that i3 = F3 and hence ht F3 = 2(3)—2

For larger 11 notice that the ideal generated by the entries of the matrix
formed with the first 2n—4 rows and columns of X has height (n—2) (Zn—5) (2.1).

Let p be in Spec(k[X, Y]) be such that ht p 5 2n—2 and FR E p. Then (see the

proof of 2.8) one of those entries, w.l.o.g, x12, is not in p.
Then x12 is invertible in k[X, Y] p' Hence there is a matrix A, invertible

over k[X, Y] p such that

ATXA =

 

o x' x"

 

 

 

 

. t
(‘X') 03x3

and the entries of X' and X" are generic over k[{xijli 5 2}, x3]. Replace Y by

A_1Y. Then in k[X, Y]p, 7511 is (yl, y2, EH) which, by induction has height
2(n—l)—2+2 = 2n—2. I

77

Theorem 3.9. (S, En) and (S, F11) are in the same Herzog class.

 

Proof. In k[X, Y], both pairs are (k[X, Y], Pu), and grade En = grade PD
= grade Fn, by 3.3. |

Now we can answer the question posed after 3.3.
Corollary 3.10. Fn is not a hypersurface section if n 2 4.
Proof. Theorem 3.9, theorem 3.7 and pr0position 2.13. I

We close this chapter with the following proposition about Fn.
Promoition 3.11. Let S = k[X, “(X Y) and let n 2 3. Then S/Fn is (Ro)
but not (R3). In particular, S/Fn is a normal domain, but it is not rigid.

Proof. For n = 3

 

 

 

 

0 x12 x13 x14 x15 y1
"12 0 "23 x24. x25 y2
X: “x13 “x23 0 x34 x35 Y: y3
‘x14 ’X24 ”‘34 0 x45 Y 4
”‘15 "x25 ”‘35 x45 0 , 3’5

and

F 3 = (X121 x133’3 + x143'4 + x15’51"23y3 + x24y4+ x25y5’

x14"25 " 11151124, 11131124-11141123, 11131125-11151123).

Let p be a prime ideal of S of height at most 6 containing F3. Because x12
6 p, {xiin 5 2, 3 5 j 5 5} E p, and therefore one of those indeterminates is not in p.
Since each one of them appears in 3 generators of F3, one can see then that F38p

has at least 4 elements, part of a regular system of parameters of S p. Hence

78
(R/F3) is (32).

If q = ({x12} U {xijli = 1,2, j = 1,2,3}). Then ht q = 7 and F3 C q.
However F33 q has at most 3 elements part of a regular system of parameters of
S q' Hence S/F3 is not (R3).

For n 2 4, let p e Spec(S) such that ht p 5 2n. Let X' be the matrix formed
with the first 2n—4 rows and 2n—4 columns of X. Then Il(X)' I p. Otherwise, for
n = 4, p would also contain 21?:5 xijyj’ 1 5 i S 3 and then ht p _>_ 9. For larger 11
Il(x') I p, since ht(Il(X')) > (n-2)(2n—5), (see 2.1) and (n—2)(2n—5) > 2n if n >
4.

Then, w.l.o.g., x12 1! p. Hence, in Sp there is an invertible matrix A such

that

ATXT =

 

0 X '

 

 

 

l

and the entries of X' are generic over k[xfé, {xi j Ii 5 2}]. If we replace Y by A—lY,
then in Sp, we have FnSp = (yl, y2, Fn—l)’ n 2 4, and then, by induction S/Fn is
(112).

To show that R/Fn is not R3, localize at the prime

p = ({xij/Zn-5 _<_ i, 2n—5 51)}

and observe that (R/Fn)p is obtained from R/F3 by a purely trascendental

extension of the residual class field of R/F3.

SUMMARY
We keep the notation corresponding to each result.

(1) The ideal (11""’/2n—2) is a prime ideal of height 2n—2 in R[X, Y] where R

is a Cohen—Macaulay domain (proposition 2.3).

(2) The ideal Kn = (l / X X X Pf(X)) is

1’”" 2n-3’ 2n-l 2n’ 2n—2 2n’ 2n—2 2n-l’
a licci, Gorenstein ideal of height 2n—1 and deviation 2 (theorem 2.9).

(3) Let (s, I) be a pair where s = k[[wn and I a licci ideal of 3. Let (s, i) be
in the same Herzog class as (S, I). Then I is a hypersurface section if and only if I

is. (proposition 2.13)

(4) (S, K4) is not in the same Herzog class as (S, Hn) for any n 2 2 (theorem

2.14).

(5) Let I be a perfect almost complete intersection ideal of height 4 and type 2
in R. Then, there is a pair (S, J) in the same Herzog class as (R, I) such that
either:
i) J is a hypersurface section or
ii) There is a 5x5 alternating matrix A and a 5x1 matrix B, aij’ by 6 ms’
such that J = 11(AB) (pr0position 2.16).

(6) If I is a licci Gorenstein ideal of R of height 5 and deviation 2, then there is
a pair (R', I') in the same Herzog class as (R, I) such that either:

i) I' is a double hypersurface section or

79

80

ii) there is a 6x6 alternating matrix A and a 6x1 matrix B such that I' =
(11(AB), Pf(A)), aij’ bj E mR.. (theorem 2.22).

(7) With the notations of (6) if J is a licci type 2 almost complete intersection
of height 5. Then, there is a pair (R',J') in the same Herzog class as (R,J) such
that either:

i) J ' is a hypersurface section or

ii) JJ = (/ ,...,/ , A56’ Pf(A)) (corollary 2.24).
(8) Let R = k[[x1,...,xn]], and let J be a licci Gorenstein ideal of R of height 5
and deviation 2. Then either I is a specialization of H3 or a specialization of (H2,
x, y) (corollary 2.25).
(9) FB is a licci Gorenstein ideal of deviation 2 and height 2n—2 in k[X,
Y](X,Y)' (pr0position 3.3).
(10) En is a licci Gorenstein ideal of height 2n—2 and deviation 2 is S, such that
S/En is rigid and En is not a hypersurface section (theorem 3.7).
(11) (S, En) and (S, F11) are in the same Herzog class, for n 2 4 (theorem 3.9).

(12) P11 is not a hypersurface section, 11 2 4 (corollary 3.10).
(13) R/Fn is (R2) but not (R3) (proposition 3.11).

[Ar.E]

[Ar.M]

[Br]

[311]

[13-13]

[He]
[H-M]

[H-K]

[H0]
[HUI

[J ac]
[H-U—H

[H—U—2]

[H—U—3]

lJ-P]

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N STRTE UNIV

9 . LIBRARIES
lIHHII lllllll] Ill]lllllllllllllll
1111 1

29300112

nrcurc
111]]
:31