TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE
TO GORENSTEIN

arXiv:1705.05714v1 [] 16 May 2017

ANDREW R. KUSTIN AND ADELA VRACIU
A BSTRACT. Let S be a deeply embedded, equicharacteristic, Artinian Gorenstein
local ring. We prove that if R is a non-Gorenstein quotient of S of small colength,
then every totally reflexive R-module is free. Indeed, the second syzygy of the
canonical module of R has a direct summand T which is a test module for freeness
over R in the sense that if TorR+ (T, N) = 0, for some finitely generated R-module
N, then N is free.

In honor of Craig Huneke, on the occasion of his sixty-fifth birthday.

1. I NTRODUCTION .
Let R be a commutative Noetherian ring. A finitely generated R-module M is
called totally reflexive if there exists a doubly infinite complex of finitely generated
free R-modules
F : · · · → Fi+1 → Fi → Fi−1 → · · · ,
such that M is isomorphic to the cokernel of Fi+1 → Fi , for some i, and such that
both F and the dual complex HomR (F, R) are exact. Totally reflexive modules were
introduced by Auslander and Bridger [1], who proved that R is Gorenstein if and
only if every finitely generated R-module has a finite resolution by totally reflexive
R-modules. Over a Gorenstein ring, the totally reflexive modules are precisely the
maximal Cohen-Macaulay modules; hence, in particular, every singular Gorenstein
local ring has a non-free totally reflexive module.
Totally reflexive modules have been the object of extensive study ever since their
introduction; yet it remains an open problem to determine conditions on a ring R
that are necessary and sufficient for the existence of a non-free totally reflexive
R-module. We follow the lead of Takahashi [15] and call a commutative Noetherian local ring G-regular if every totally reflexive module over the ring is free.
Regular local rings are trivial examples of G-regular local rings. Avramov and
Martsinkovsky [2, 3.5.(2)] proved that every Golod local ring that is not a hypersurface is G-regular.
The main result of the paper is Theorem 7.5 where we prove that if R is a nonGorenstein quotient of small colength of a deeply embedded, equicharacteristic Artinian Gorenstein local ring, then R is G-regular. Roughly speaking, a local ring R
Date: May 17, 2017.
2010 Mathematics Subject Classification. 13A02, 13H10, 13D02, 18G20.
Key words and phrases. Canonical module, G-regular local ring, Gorenstein colength, Higher
matrix factorization, Summands of syzygies, Test modules, Teter ring, Totally reflexive modules.
Research partly supported by Simons Foundation collaboration grant 233597 (ARK) and NSF
grant DMS-1200085 (AV).
1

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A. R. KUSTIN AND A. VRACIU

is deeply embedded if R = P/I for some regular local ring (P, M) and I ⊆ Mn0 for
some large integer n0 ; see 2.5.(b) or [13, 4.1] for a more precise formulation.
We consider local rings (R, m) of the form R = S/(0 :S K) where (S, n) is an
Artinian Gorenstein local ring, and K is an ideal of S. Notice that K is also an Rmodule; furthermore, the canonical module of R, which is denoted by ωR , is equal
to K. Sometimes we assume S = P/a, where P is a standard graded polynomial
ring and a is a homogeneous ideal of P.
Our favorite method of showing that a ring R is G-regular is to exhibit a proj-test
module which is a direct summand of a syzygy of the canonical module of R. The
finitely generated R-module T is a proj-test module if the only finitely generated
R-modules N with TorR+ (T, N) = 0 are the projective R-modules; see 2.10. We
begin by summarizing our results that establish the existence of direct summands
of syzygies of the canonical module for certain classes of rings. See 2.2 for our use
of Fitt1 , 4.2 for the proof of Theorem 1.1, 6.1 for the proof of Theorem 1.2, and 6.5
for the proof of Theorem 1.3.
Theorem 1.1. Let (R, m,kk) be a local ring of the form R = S/(0 :S K) where (S, n)
is an Artinian Gorenstein local ring and K is an ideal of S. If
(1.1.1)

Fitt1S (K) = n,

and
(1.1.2)

nK :S n 6= K :S n,

then the second syzygy of the canonical module of R, viewed as an R-module, has a
direct summand isomorphic to R/m.
Theorem 1.2. Let P be a commutative Noetherian ring and A and B be ideals of P
with A ⊆ B2 , and P/A and P/B both Artinian Gorenstein local rings. Assume B is
generated by a regular P-sequence of length at least two.
Let R be the ring R = P/(A :P B). Then the second syzygy of the canonical module
of R, viewed as an R-module, has a direct summand isomorphic to B/B2 .
Theorem 1.3. Let k be a field of arbitrary characteristic, P = k [X1 , X2, X3 ,Y1 , . . . ,Ys ]
be a standard-graded polynomial ring over k in 3 + s variables for some nonnegative integer s, B′ be an ideal in P which is generated by five linearly independent
quadratic forms in the variables X1 , X2, X3 , B′′ be the ideal (Y1 , . . . ,Ys ) of P, B be
the ideal B′ + B′′ of P, and A ⊆ ([P]1)5 be a homogeneous ideal of P with P/A an
Artinian Gorenstein local ring. Assume [socle(P/B)]1 = 0.
Let R be the ring R = P/(A :P B). Then the second syzygy of the canonical module
of R, viewed as an R-module, has a direct summand isomorphic to B′ /BB′ .
In Section 3 we prove that for each ring under consideration in the main result,
Theorem 7.5, one of the above Splitting Theorems applies. We prove the Splitting
Theorems in Sections 4, 5, and 6. In Section 7 we prove that the summands produced by the Splitting Theorems are indeed proj-test R-modules. We complete the
proof by appealing to Observation 2.10.3.
The Splitting Theorems of Sections 4, 5, and 6 and the Test Module Theorems of
Section 7 are of independent interest and are stated with more general hypotheses
than those of the main result, Theorem 7.5.

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

2. N OTATION ,

3

CONVENTIONS , AND PRELIMINARY RESULTS .

In this paper k is always a field. If V is vector space over k , then dimk V is the
dimension of V as a vector space over k .
2.1. Let I be an ideal in a ring S, N be an S-module, and L and M be submodules of
N. Then
L :I M = {x ∈ I | xM ⊆ L}

and L :M I = {m ∈ M | Im ⊆ L}.

If L is the zero module, then we also use “annihilator notation” to describe these
“colon modules”; that is,
annS M = 0 :S M.
2.2. If M is a finitely generated module over a local ring S, and
φ

Sb −
→ Sµ(M) →
− M→0
is a presentation of M with µ(M) equal to the minimal number of generators of M,
then we write Fitt1S (M) for the ideal of S generated by the entries of φ. Notice that
Fitt1S (M) is equal to the usual Fitting ideal Fittµ(M)−1 (M) of M and does not depend
on the choice of φ.
2.3. If M is a matrix (or a homomorphism of free R-modules), then Ir (M) is the
ideal generated by the r × r minors of M (or any matrix representation of M). We
denote the transpose of a matrix M by M T .
2.4. If S is a ring and M is an S-module, then let λS (M) denote the length of M as
an S-module. If R is a quotient of S, then the colength of R as an S-module is
cS (R) = λS (S) − λS (R).
2.5. “Let (S, n,kk) be a local ring” identifies n as the unique maximal ideal of the
commutative Noetherian local ring S and k as the residue class field k = S/n.
(a) The embedding dimension of S is dimk (n/n2 ).
(b) The parameter v(S) is defined by
n
o
,
v(S) = inf i dimk (ni /ni+1 ) < (n−1)+i
i

where n is the embedding dimension of S. (This notation is introduced in [13,
(4.1.1)].) Notice that if n is minimally generated
by x1 , . . . , xn and i is an integer
(n−1)+i
with 0 ≤ i ≤ v(S) − 1, then the
monomials of degree i in the symbols
i
x1 , . . . , xn represent a basis for the k -vector space ni /ni+1 .
(c) We say that a local ring S is “deeply embedded” if v(S) is large.

2.6. Gorenstein duality. If (S, n,kk) is an Artinian Gorenstein local ring, then S is an
injective S-module, HomS (−, S) is an exact functor, and λS (M) = λS (HomS (M, S))
for all S-modules M of finite length. Furthermore, if A is an ideal of S, then (0 :S A)
is the canonical module of S/A. (See, for example, [3, 4].) The following statements
are immediate consequences of the above facts. We refer to them as Gorenstein
duality.
Proposition 2.6.1. If A and B are ideals of an Artinian Gorenstein local ring S,
then

4

(a)
(b)
(c)
(d)
(e)

A. R. KUSTIN AND A. VRACIU

λS (S/A) = λS (annS A),
annS (annS A) = A,
annS (A + B) = (annS A) ∩ (annS B),
annS (A ∩ B) = (annS A) + (annS B), and
SA
if A ⊆ B, then λS ( AB ) = λS ( ann
annS B ).

2.7. If R is a commutative ring and M is an R-module, then an R-module M ′ is a
second syzygy module of M if there are projective R-modules P1 and P2 and an
exact sequence of R-modules of the form
0 → M ′ → P2 → P1 → M → 0.
We denote the second syzygy module of the R-module M by syzR2 (M).
Remark 2.7.1. Recall from Schanuel’s Lemma that the notion of second syzygy
module is well-defined up to formation of direct sum with a projective module. In
other words, if M ′ and M ′′ are both second syzygy modules of M, then there are
projective R-modules P′ and P′′ with M ′ ⊕ P′ isomorphic to M ′′ ⊕ P′′ .
2.8. As always, let k be a field.
(a) If M is a graded module, then we write [M]i to represent the component of M
which consists of all homogeneous elements of degree i.
(b) If Q is a non-negatively graded ring over a localL
ring ([Q]0, m,kk ), then Q has a
unique maximal homogeneous ideal M = mQ + 1≤i [Q]i . If M is a Q-module,
then the socle of M is the vector space socle(M) = 0 :M M.
(c) Recall that if M is a finitely generated graded module over a graded polynomial
ring P = k [x1 , . . ., xn ], then the socle degrees of M may be read from the back
twists in a minimal homogeneous resolution
0 → Cn =

Ls

i=1 P(−βi ) → ... → C0

of M by free P-modules. Indeed, the computation of TorPn (M,kk) in each coordinate yields a graded isomorphism
socle M ∼
=
L

s
M
i=1

n

k

∑ deg x j

j=1




− βi .

(d) A graded ring S = 0≤i [S]i is called a standard-graded k -algebra, if S is a
commutative ring with [S]0 = k , S is generated as an [S]0-algebra by [S]1, and
[S]1 is finitely generated as an [S]0 -module.
(e) If (S, n,kk) is a local ring, then the associated graded ring of S is the standardgraded k -algebra
∞
M
ni
Sg =
.
i+1
n
i=0
2.9. Let S be a commutative Noetherian graded ring and M be a graded S-module.
(a) Define
topdeg(M) = inf{τ | [M] j = 0 for all j with τ < j},


L
maxgd(M) = inf τ S
j≤τ [M] j = M .

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

5

The expressions “topdeg” and “maxgd” are read “top degree” and “maximal
generator degree”, respectively. If M is the zero module, then topdeg(M) and
maxgd(M) are both equal to −∞.
(b) If S is non-negatively graded, [S]0 is Artinian, and M is finitely generated, then
(i) the Hilbert function of M is the function HFS (M, ), from the set of integers to the set of non-negative integers, with HFS (M, i) = λ[S]0 ([M]i),
and
(ii) the Hilbert series of M is the formal generating function
HSS (M,t) = ∑ HFS (M, i)t i.
i∈Z

If M = S, we often write HS(S,t) (or simply HS(S)) in place of HSS (S,t).
2.10. Test modules and G-regularity. Observation 2.10.3 is the starting point for
the present paper. The authors of [14] refer to [14, Lemma 3.2], which is a similar
result, as “well-known by the experts”. The present version uses the notion of projtest module. Related test modules have been used in [12, 5, 6].
Definition 2.10.1. Let R be a commutative Noetherian ring and T be a finitely
generated R-module. Then T is called a proj-test module for R, if the only finitely
generated R-modules N with TorRi (T, N) = 0 for all positive i are the projective
R-modules.
Observation 2.10.2. If (R, m,kk) is an Artinian local ring with canonical module
ωR and M is a finitely generated R-module, then ExtiR (M, R) and TorRi (M, ωR ) are
Matlis duals of one another.
Proof. Recall that Matlis dual is the functor (−)∨ = HomR (−, E), where E is the
injective envelope of the R-module k ; furthermore, in the present situation, E and
ωR are isomorphic R-modules. The ring R is complete; so, N ∨∨ ∼
= N for every
finitely generated R-module N.
Let F be a resolution of M by finitely generated free R-modules. Observe that
ExtiR (M, R) = Hi (HomR (F, R))
= Hi (HomR (F, HomR (ωR , ωR ))) because R∨∨ ∼
=R
i
∼
by the adjoint isomorphism theorem
= H (HomR (F ⊗R ωR , ωR ))
= Hi ((F ⊗R ωR )∨ )
∼
= (Hi (F ⊗R ωR ))∨

because (−)∨ is exact

= (TorRi (M, ωR ))∨ .

Observation 2.10.3. Let (R, m,kk) be an Artinian local ring with canonical module
ωR . If T is a proj-test module for R and T is a summand of a syzygy of ωR , then R
is G-regular.
Proof. If X is a totally reflexive R-module, then, by definition, ExtiR (X , R) = 0 for all
positive i and therefore, from Observation 2.10.2, TorRi (X , ωR ) = 0 for all positive i.
It follows that TorRi (X , T ) = 0 for all positive i; and therefore X is a free R-module
by the definition of proj-test module.


6

A. R. KUSTIN AND A. VRACIU

3. R INGS

OF SMALL

G ORENSTEIN

COLENGTH .

Let S be an equicharacteristic Artinian Gorenstein local ring. Assume that the
invariant v(S), of 2.5.(b), is large. In this section we prove that if R is a nonGorenstein quotient of S of small colength, then one of the Splitting Theorems of
Section 1 applies to R.
Theorem 3.1. Let (S, n,kk) be an equicharacteristic Artinian Gorenstein local ring
with embedding dimension at least two, and R be the ring R = S/J for some proper
non-zero ideal J of S. Assume that either
• 1 ≤ cS (R) ≤ 4, or
• cS (R) = 5 and S is a standard-graded algebra over a field [S]0.
If the parameter v(S) is sufficiently large, then one of the Theorems 1.1, 1.2, or
1.3 applies to R. In particular, the following statements hold, where k denotes the
minimal number of generators of n/(0 :S J).
(a) If k = 0, then Theorem 1.1 applies to R.
(b) If k = 1, then Theorem 1.2 applies to R, provided 2cS (R) ≤ v(S).
(c) If 2 ≤ k = cS (R) − 1, then Theorem 1.1 applies to R, provided 3 ≤ v(S).
(d) If k = 2, cS (R) = 4, and n3 ⊆ n(0 :S J), then Theorem 1.2 applies to R, provided
5 ≤ v(S).
(e) If k = 2, cS (R) = 4, and n3 6⊆ n(0 :S J), then Theorem 1.1 applies to R, provided
4 ≤ v(S).
(f) If k = 2, cS (R) = 5, and S is a standard-graded algebra over a field [S]0 , then
Theorem 1.1 applies to R, provided 4 ≤ v(S).
(g) If k = 3, cS (R) = 5, 3 ≤ maxgd(0 :S J), and S is a standard-graded algebra over
a field [S]0, then Theorem 1.1 applies to R, provided 3 ≤ v(S).
(h) If k = 3, cS (R) = 5, 2 ≤ dimk socle(S/0 :S J), and S is a standard-graded algebra
over a field [S]0, then Theorem 1.1 applies to R, provided 3 ≤ v(S).
(i) If k = 3, cS (R) = 5, maxgd(0 :S J) ≤ 2, dimk socle(S/(0 :S J)) = 1, and S is
a standard-graded algebra over a field [S]0, then Theorem 1.3 applies to R,
provided 5 ≤ v(S).
Assertions (a) – (e) of Theorem 3.1 are proven in 3.2; assertions (f) – (i) are
proven in 3.6. First we make some remarks about the statement and introduce some
notation.
Remark 3.1.1. The ring S of Theorem 3.1. is equicharacteristic and complete; so,
the Cohen structure theorem guarantees that S contains a copy of the residue field
k ; that is, there is a commutative diagram


k ❆

❆❆
❆❆
❆
∼
= ❆❆

/S
⑤
⑤
⑤⑤
⑤⑤ natural map
⑤
~
⑤

S/n.
Let n be the embedding dimension of S, U be an n-dimensional vector space over k ,
P be the polynomial ring Symk• (U ), and M be the maximal ideal of P generated by
∼
=
Sym1 U . Fix a vector space isomorphism U −
→ n/n2 . The ring S is Artinian; so, the

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

7


/ S and the isomorphism U ∼
inclusion k 
= n/n2 combine to induce a surjection
/ / S . Let A be the kernel of π. No harm is done if we write
π: P

P/A = S.
Recall from 2.5.(b) that A ⊆ Mv(S) ; but A 6⊆ Mv(S)+1 .
Remark 3.1.2. In the situation of of Theorem 3.1, let K be the ideal (0 :S J) of S.
Gorenstein duality (see 2.6) guarantees that
(0 :S K) = J.
Observe that
(a) K is an R-module,
(b) K is the canonical module of R, and
(c) λS (K) = cS (R).
Assertion (a) is obvious. For (b), observe that S and R have the same Krull dimension and S is a Gorenstein local ring which maps onto R. It follows that the
canonical module of R is
ωR = HomS (R, S) = HomS (S/J, S) = (0 :S J) = K.
Assertion (c) holds because λS (R) = λS (ωR ) = λS (K) = λS (S) − λS (S/K); hence,
λS (S/K) = λS (S) − λS(R) = cS (R).
Remark 3.1.3. In the situation of of Theorem 3.1, let B = π−1 (K) be the preimage
of K in P. It follows that
S
P/A
P
S
=
=
.
R= =
J (0 :S K) (A :P B)/A (A :P B)
Remark 3.1.4. In the situation of of Theorem 3.1, the inequality k + 1 ≤ cS (R)
always holds because


n
k = dimk K+n
≤ λS (n/K) = λS (S/K) − 1 = cS (R) − 1.
2
The final equality is established in Remark 3.1.2.(c).

Remark 3.1.5. The assertion of Theorem 3.1 does not hold if the ring R is Gorenstein; and for this reason the hypotheses of Theorem 3.1 exclude J = 0, exclude
cS (R) = 0, and exclude the case where n is a principal ideal. Indeed, if J were
equal to 0 or if cS (R) were equal to 0, then R would be equal the Gorenstein ring S.
Similarly, if n were a principal ideal, then S and R would both be Gorenstein rings
of the form k [X1 ]/(X1Ni ) for integers N1 and N2 .
Remark 3.1.6. The rings of Theorem 3.1.(a) are called Teter rings. It was already
shown in [14] that Teter rings are G-regular. In fact, the present paper is inspired
by [14].
Remark 3.1.7. All cases with 1 ≤ cS (R) ≤ 5 are covered in (a)–(g) of Theorem 3.1
because once (a), (b), and (c) are established, then, it follows from Remark 3.1.4,
that it is only necessary to consider 2 ≤ k ≤ cS (R) − 2 for cS (R) equal to 4 or 5.

8

A. R. KUSTIN AND A. VRACIU

3.2. Proof of assertions (a) – (e) of Theorem 3.1. Retain the notation P, U , M, A,
K, and B which is introduced in Remarks 3.1.1, 3.1.2, and 3.1.3. Choose a minimal
generating set x1 , . . ., xk , y1 , . . . , ys for n such that x1 , . . . , xk represents a minimal
generating set for n/K and y1 , . . . , ys are in K. Select Y1 , . . .,Ys , X1, . . . , Xk ∈ U to be
preimages of y1 , . . . , ys , x1 , . . . , xk , respectively. Observe that Y1 , . . . ,Ys , X1 , . . ., Xk is
a basis for the k vector space U ; indeed, P is the polynomial ring
P = k [Y1 , . . .,Ys , X1 , . . . , Xk ].
Let (x) and (y) denote the ideals (x1 , . . . , xk ) and (y1 , . . ., ys ) of S, respectively,
and let d denote the top degree of the associated graded ring
(S/K)g =

∞
M
ni + K
i=0

ni+1 + K

of S/K. In other words, d is the smallest positive integer with (x)d+1 ⊆ K. There
are strict inclusions
K ( K + (x)d ( · · · ( K + (x)2 ( K + (x) = n ⊆ S.

(3.2.1)

Notice that each quotient of consecutive terms in (3.2.1) is annihilated by n and the
Hilbert series of (S/K)g is
d

K + (x)i i
t;
HS (S/K)g ,t = ∑ dimk
K + (x)i+1
i=0

furthermore,
k = dimk

K + (x)
= the coefficient of t in HS((S/K)g ,t) and
2
K + (x)
d

K + (x)i
cS (R) = λS (S/K) = ∑ dimk
= HS((S/K)g , 1).
i+1
K + (x)
i=0
(a) The parameter k is equal to 0; consequently, n = K and y1 , . . . , ys is a minimal
generating set for n = K. The hypotheses that n is not principal and K is not zero
(that is, J is a proper ideal of S) guarantee that 2 ≤ s. Therefore, each yi appears in
a Koszul relation on y1 , . . . , ys . It follows that
n ⊆ Fitt1S (n) = Fitt1S (K) ⊆ n;
and therefore condition (1.1.1) of Theorem 1.1 is satisfied. Furthermore, condition
(1.1.2) is satisfied because 1 ∈ (K : n), but 1 6∈ (nK : n). We have shown that if
k = 0, then Theorem 1.1 applies to the ring R.
(b) We assume that k = 1 and 2cS (R) ≤ v(S). It follows that each quotient of consecutive terms in (3.2.1) has length one; thus, d + 1 = cS (R) and
(y1 , . . . , ys , x1d+1 ) ⊆ K.

(3.2.2)
On the other hand,

λS (S/(y1, . . . , ys , x1d+1 )) = d + 1 = cS (R) = λS (S/K);

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

9

and equality holds in (3.2.2). In this case, the ideal B of P from (3.1.3) is generated
by a regular sequence. The regular sequence has length at least two because n is
not zero and not principal. Furthermore, the hypothesis that
2cS (R) ≤ v(S)
forces
A ⊆ Mv(S) ⊆ M2cS (R) = M2(d+1) ⊆ B2 .
All of the hypotheses of Theorem 1.2 are satisfied by the ring R.
(c) We assume that 2 ≤ k = cS (R) − 1 and 3 ≤ v(S). It follows from (3.2.1) that
K + (x)2 = K; hence,
(y) + (x)2 ⊆ K.
In fact, equality holds, because
n
(y) + (x)2

and

n
K

both have length k.
The hypothesis 3 ≤ v(S) ensures
that K = (y) + (x)2 is minimally generated by

y1 , . . ., ys , together with the k+1
monomials of degree two in x1 , . . . , xk .
2

3.2.3. The ideal K has at least two minimal generators; so, any Koszul relation
involving any yi (with 1 ≤ i ≤ s) and any other minimal generator exhibits yi as an
element of Fitt1S (K).
Furthermore, the parameter k is at least two; so, if i and j are arbitrary with
1 ≤ i 6= j ≤ k,
then the relation xi (x2j ) − x j (xi x j ) exhibits as xi and x j elements of Fitt1S K. It is now
apparent that hypothesis (1.1.1) of Theorem 1.1 holds. It is clear that x1 n ⊆ K. The
hypothesis 3 ≤ v(S) also ensures that x1 n 6⊆ nK. We conclude that both assumptions
of Theorem 1.1 hold in this case.
3.2.4. The case (k, cS (R)) = (2, 4). We make some preliminary calculations before
dividing this case into the two sub-cases (d) and (e).
Examine the filtration (3.2.1) in order to see that HS((S/K)g ) = 1 + 2t + t 2 . It
follows, in particular, that
(K + (x1 , x2 )2 )/K

(3.2.5)

is a one-dimensional k -vector space. It also follows that (x1 , x2 )3 ⊆ K. The hypothesis 3 ≤ v(S) ensures that the k -submodule of S which is spanned by x21 , x1 x2 , x22 is
a three-dimensional vector space, which we call V . Select a basis q0 , q1 , q2 for V so
that q0 represents a basis for (3.2.5) and q1 and q2 are in K. It follows that
(y1 , . . ., ys , q1 , q2 ) + (x1 , x2 )3 ⊆ K;

(3.2.6)
and therefore,
(3.2.7)

3 = λS

n
K

≤ λS




n
.
(y1 , . . . , ys , q1 , q2 ) + (x1 , x2 )3

10

A. R. KUSTIN AND A. VRACIU

On the other hand, the module on the right side of (3.2.7) has length at most three
because this module is generated by the images of q0 , x1 , and x2 , and these generators give rise to a filtration of length three whose factors are vector spaces of
dimension at most one. We conclude that equality holds in both (3.2.7) and (3.2.6);
in particular,
(3.2.8)

(y1 , . . . , ys , q1 , q2 ) + (x1 , x2 )3 = K.

The hypothesis 4 ≤ v(S) guarantees that the elements y1 , . . . , ys , q1 , q2 are the
beginning of a minimal generating set for K; furthermore,
(3.2.9)

(y) + (q1 , q2 ) = K ⇐⇒ n3 ⊆ nK.

We consider two sub-cases.
(d) Assume (k, cS (R)) = (2, 4), n3 ⊆ nK, and 5 ≤ v(S). Continue the discussion of
3.2.4. It follows from (3.2.8) and (3.2.9) that
(x1 , x2 )3 ⊆ (y1 , . . . , ys , q1 , q2 ).
Recall the polynomial ring P = Symk• U with P/A = S which was introduced
Remark 3.1.1. Select Y1 , . . .,Ys , X1, X2 ∈ U to be preimages of y1 , . . . , ys , x1 , x2 , respectively. Observe that Y1 , . . .,Ys , X1, X2 is a basis for the k vector space U . Select
Q1 and Q2 in the vector space k X12 ⊕ k X1 X2 ⊕ k X22 to be preimages of q1 and q2 ,
respectively. Notice that B, from Remark 3.1.3, is equal to (Y1 , . . . ,Ys , Q1 , Q2 ) + A.
Observe that Q1 , Q2 is a regular sequence in P. Otherwise, Q1 and Q2 have a
common factor and the k -submodule of P which is spanned by
{Xi Q j | 1 ≤ i, j ≤ 2}
is a vector space (which we call V ′ ) of dimension at most 3. On the other hand,
the image of the above V ′ in S contains the four dimensional vector space spanned
by the monomials of degree 3 in x1 , x2 . (Keep in mind that the monomials in y1 ,
. . . , ys , x1 , x2 of degree i represent a basis for ni /ni+1 for 0 ≤ i ≤ v(S) − 1.) This
contradiction establishes the claim that Q1 , Q2 is a regular sequence in P.
Now that we know that Q1 , Q2 is a regular sequence, a quick calculation shows
that
M5 ⊆ (Y1 , . . .,Ys , Q1 , Q2 )2 .
Therefore the hypothesis that 5 ≤ v(S) guarantees that
A ⊆ Mv(S) ⊆ M5 ⊆ (Y1 , . . .,Ys , Q1 , Q2 )2 .
Thus, B is generated by the regular sequence Y1 , . . . ,Ys , Q1 , Q2 ; this regular sequence
has length at least two; and A ⊆ B2 . All of the hypotheses of Theorem 1.2 are in
effect.
(e) Assume (k, cS (R)) = (2, 4), n3 6⊆ nK, and 4 ≤ v(S). Continue the discussion of
3.2.4. It follows from (3.2.8) and (3.2.9) that
(3.2.10)

(x1 , x2 )3 6⊆ (y1 , . . . , ys , q1 , q2 ).

We saw in the proof of (d) that the quadratic forms Q1 , Q2 of k [X1, X2 ] must have a
common linear factor. In other words, there are homogeneous linear forms L, L1 , L2
in k [X1 , X2] with Q1 = LL1 and Q2 = LL2 . If ℓ1 and ℓ2 are the images in S of L1 and
L2 , then ℓ1 q2 = ℓ2 q1 is a relation on a minimal generating set for K that exhibits ℓ1

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

11

and ℓ2 as elements of Fitt1S (K). The linear forms L1 , L2 must be linearly independent
in P; therefore, k (ℓ1 , ℓ2 ) = k (x1 , x2 ) as sub-vector spaces of S; hence, x1 and x2 are in
Fitt1S (K). Of course, the argument of (3.2.3) shows that all of the elements y1 , . . . , ys
are in Fitt1S (K). This verifies that Hypothesis (1.1.1) in Theorem 1.1 holds for R.
We verify Hypothesis (1.1.2) by showing that q0 ∈ K :S n but q0 6∈ nK :S n. The first
assertion holds because
q0 n ⊆ (y1 , . . ., ys ) + (x1 , x2 )3 ⊆ K.
For the second assertion, write (x1 , x2 )3 = (q0 , q1 , q2 )(x1 , x2 ). The ambient hypothesis n3 6⊆ nK guarantees that there is an element ℓ ∈ k x1 ⊕kk x2 with ℓq0 ∈ K 6∈ nK. 
Assertions (f) – (i) of Theorem 3.1 are proven in 3.6. We first prove some preliminary results.
Lemma 3.3. Let B′ be a homogeneous ideal of the standard-graded polynomial
ring P′ = k [X1 , X2]. If
(a) the embedding dimension of P′ /B′ is 2 and the length of P′ /B′ is 5, or
(b) the embedding dimension of P′ /B′ is 2, the length of P′ /B′ is 4, and the maximal
generator degree of B′ is at least 3,
then Fitt1P′ (B′ ) = (X1 , X2 ).
Proof. (a) The Hilbert series of P′ /B′ is either 1 + 2t + 2t 2 or 1 + 2t + t 2 + t 3 . We
first assume that HS(P′ /B′ ) = 1 + 2t + 2t 2 . Compare HS(P′ /B′ ) and HS(P′ ) to
see that HS(B′ ) = t 2 + 4t 3 + . . . . It follows that B′ is minimally generated by one
quadratic form and two cubic forms. Furthermore, we observe that
socle(P′ /B′ ) = [P′ /B′ ]2 ∼
= k (−2)2 .
The ideal B′ of P′ is perfect of grade two and the minimal homogeneous resolution
of P′ /B′ by free P′ -modules looks like
P′ (−2)1
→ P′ .
0 → P′ (−4)2 −−→ ⊕
P′ (−3)2
φ

(We used 2.8.(c).) The matrix φ has the form


φ1,1 φ1,2
(3.3.1)
φ = φ2,1 φ2,2  ,
φ3,1 φ3,2

where each φi, j is a homogeneous form in P′ and
(
2 if i = 1
deg φi, j =
1 if 2 ≤ i.
The Hilbert-Burch Theorem guarantees that B′ is generated by the maximal order
minors of φ; hence B′ is contained in the ideal generated by the degree one entries
of φ. The ideal B′ has grade 2; thus, the degree one entries of φ generate the entire
ideal (X1, X2 ).

12

A. R. KUSTIN AND A. VRACIU

We now assume that HS(P′ /B′ ) = 1 + 2t + t 2 + t 3. In this case, the Hilbert function of B′ is 2t 2 + 3t 3 + 5t 4 + · · · . The ideal B′ has two minimal quadratic generators; but these generators have a linear factor L in common; consequently, the
Hilbert function of the ideal generated by the two quadratics is 2t 2 + 3t 3 + 4t 4 + . . . .
We conclude that B′ is minimally generated by two quadratic forms and a homogeneous form of degree 4. The element of P′ /B′ represented by L is in the socle of
P′ /B′ and [P′ /B′ ]3 is contained in the socle of P′ /B′ . At this point we know that the
minimal homogeneous resolution of P′ /B′ by free P′ -modules looks like
P′ (−3)
P′ (−2)2
⊕
φ
→ P′ ,
0 → P′ (−5) −−→ ⊕
⊕
P′ (−4)1
F
for some homogeneous free P′ -module F. Rank considerations show that F is zero.
The matrix φ has the form (3.3.1), where each φi, j is a homogeneous form in P′ ,
φ3,1 = 0 and
(
1 if (i, j) equals (1, 1), (2, 1), or (3, 2)
deg φi, j =
3 if (i, j) equals (1, 2) or (2, 2).
Once again, the Hilbert-Burch Theorem guarantees that B′ is generated by the maximal order minors of φ; hence B′ ⊆ (φ1,1 , φ2,1 ). The ideal B′ still has grade 2; thus,
(φ1,1 , φ2,1 ) = (X1, X2 ), and the proof is complete.
(b) The ring P′ /B′ is standard-graded; consequently,
[P′ /B′ ]i = 0 =⇒ [P′ /B′ ]i+1 = 0.
It follows that HS(P′ /B′ ) = 1 + 2t + t 2 . The ideal B′ has two minimal generators
of degree two and, by hypothesis, at least one minimal generator of degree at least
three. The quadratic generators must have a linear factor L in common as in the
HS = 1 + 2t + t 2 + t 3 case. It follows follows that B′ has a minimal cubic generator and no further generators are needed. The element L represents an element of
degree one in the socle of P′ /B′ and [P′ /B′ ]2 is also contained in the socle. Rank
considerations show that the socle can not be any larger than k (−1) ⊕ k (−2) as in
the HS = 1 + 2t +t 2 +t 3 case. Thus, the minimal homogeneous resolution of P′ /B′
by free P′ -modules looks like
P′ (−2)2
P′ (−3)
φ
→ P′ ,
0 → ⊕ −−→ ⊕
P′ (−4)
P′ (−3)1
for some homogeneous free P′ -module F. The matrix φ has the form (3.3.1), where
each φi, j is a homogeneous form in P′ , φ3,1 = 0 and
(
1 if (i, j) equals (1, 1), (2, 1), or (3, 2)
deg φi, j =
2 if (i, j) equals (1, 2) or (2, 2).
Once again, the entries in the first column of φ must generate (X1, X2 ).



TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

13

Lemma 3.4. Let P′ be a standard-graded polynomial ring over the field k with
maximal homogeneous ideal M′ and let B′ be an M′ -primary homogeneous ideal
of P′ . If the top degree of P′ /B′ is less than the maximal generator degree of B′ ,
then
(B′ :P′ M′ ) 6= (M′ B′ :P′ M′ ).
Proof. Let d denote the top degree of P′ /B′ . First notice that maxgd(B′ ) is equal to
d + 1. Indeed, the inequality d + 1 ≤ maxgd(B′ ) is part of the hypothesis. On the
other hand, [P′ ]d+1 = [M′ ]1 [P′ ]d ⊆ B′ . (The equality holds because P′ is standardgraded.) It follows that if p ∈ [P′ ]d+ℓ , for some ℓ with 2 ≤ ℓ, then p ∈ M′ B′ (again
because P′ is standard graded) and p is not a minimal generator of B′ ; in other
words, maxgd(B′ ) ≤ d + 1.
Now observe that
[B′ ]d [M′ ]1 $ [B′ ]d+1 = [M′ ]d+1 = [M′ ]d [M′ ]1 .
The strict inclusion occurs because B′ requires a minimal generator in degree d + 1.
It follows that there exists an element θ of [M′ ]d with θ[M′]1 6⊆ [B′ ]d [M′ ]1 . This θ
is in (B′ :P′ M′ ) \ (M′B′ :P′ M′ ).

Lemma 3.5. Let P be a standard-graded polynomial ring over the field k with
maximal homogeneous ideal M, and let A ⊆ B be M-primary ideals of P, with
P/A a Gorenstein ring. Denote P/A by S, B/A by K, and S/(0 :S K) by R. Let
Y1 , . . . ,Ys , X1, . . . , Xk be a basis for [P]1 and P′ be the subring k [X1, . . . , Xk ] of P. Denote the maximal homogeneous ideal of P′ by M′ . Suppose that B = (Y1 , . . . ,Ys , B′ P)
for some ideal B′ of P′ . If
(a) Fitt1P′ B′ = M′ ,
(b) maxgd(B′ ) < topdeg(P′ /B′ ), and
(c) topdeg(P′ /B′ ) + 1 ≤ v(S),
then Theorem 1.1 applies to R.
Proof. Let d denote the top degree of P′ /B′ and n denote the maximal homogeneous
ideal of S. Hypothesis (b) enables us to apply Lemma 3.4 in order to learn that there
is an element θ′ ∈ [P′ ]d with θ′ M′ ⊆ B′ but θ′ [M′ ]1 6⊆ [M′B′ ]d+1 . Observe that θ′ is
also in [P]d with
θ′ M ⊆ B but θ′ [M]1 6⊆ [MB]d+1 .
Apply the natural quotient map π : P → S. Let θ ∈ [S]d denote π(θ′ ). Hypothesis
(c) guarantees that
π : [MB]d+1 → [nK]d+1
is an isomorphism; thus θ ∈ [S]d with
θn ⊆ K

but θ[n]1 6⊆ [nK]d+1 ,

and Hypothesis (1.1.2) of Theorem 1.1 holds for R.
Now we verify that Hypothesis (1.1.1) of Theorem 1.1 holds for R. Let f1 , . . . , fc1
be a minimal homogeneous generating for B′ and
φ′

h
i
f = f 1 , . . . , f c1

→ (P′ )c1 −−−−−−−−−−→ B′
(P′ )c2 −

14

A. R. KUSTIN AND A. VRACIU

be a presentation of B′ . It follows that
h

i

f Y1 . . . Ys
Pc2 +sc1 +( ) →
− Pc1 +s −−−−−−−−−−−−→ B,
s
2

with

φ


φ′ Y1 Ic1 Y2 Ic1 · · · Ys Ic1


−f
−Y2




−
f
Y
1


φ=
,
.
.


.
···



−Ys 
−f
Ys−1
is a presentation B by free P-modules with f1 , . . ., fc1 ,Y1 , . . . ,Ys a minimal generating set for B. Hypothesis (a) guarantees that I1 (φ′ ) = M′ . It follows that


I1 (φ) = M.

(3.5.1)

Apply − ⊗P S to obtain the complex
(3.5.2)

Sc2 +sc1

h

i

f Y1 . . . Ys ⊗P S
φ⊗P S
−−−→ Sc1 +s −−−−−−−−−−−−−−→ K.

Observe that π1 ( f1 ), . . ., π( fc1 ), π(Y1), . . . , π(Ys) is still a minimal generating set for
K because Hypothesis (c) ensures that v(S) is larger than the maximal generator
degree of K. (See the proof of Lemma 3.4, if necessary.) The fact that (3.5.2) is a
complex ensures that I1 (π(φ)) ⊆ Fitt1S (K). On the other hand, (3.5.1) ensures that
I1 (π(φ)), which is equal to π(I1(φ)), is equal to π(M) = n. Thus, Fitt1S (K) = n and
Hypothesis (1.1.1) of Theorem 1.1 holds for R.

3.6. Proof of assertions (f) – (i) of Theorem 3.1. Throughout this proof, the ring
S of Theorem 3.1 is standard-graded over [S]0 = k , the polynomial ring P of Remark 83.1.1 is standard-graded over [P]0 = k , the ideals K in S of Remark 3.1.2
and A and B in P of Remarks 3.1.1 and 3.1.3 are homogeneous. The ring S/K is
standard-graded and is equal to the associated graded ring (S/K)g .
Recall the elements x1 , . . . , xk , y1 , . . ., ys in S and the corresponding elements X1,
. . . , Xk ,Y1 . . . ,Ys of P from the beginning of 3.2. Let P′ be the subring
P′ = k [X1 . . . , Xk ]
of P. Observe that B = (Y1 , . . . ,Ys ) + B′ P for some homogeneous ideal B′ of P′ ;
furthermore, there is a natural isomorphism (P′ /B′ ) ∼
= S/K, which is induced by
(3.6.1)

P′ 


/

P

π //

S.

(f) We assume that (k, cS (R)) = (2, 5) and 4 ≤ v(S). There are two possible Hilbert
series for S/K which are consistent with the hypothesis (k, cS (R)) = (2, 5); that is,
HS(S/K) is equal to 1 + 2t + t 2 + t 3 or 1 + 2t + 2t 2. Recall the standard-graded
polynomial P′ = k [X1, X2 ], the homogeneous ideal K ′ of P′ and the isomorphism
P′ /B′ ∼
= S/K of (3.6.1). Apply Lemma 3.3 to see that Fitt1P′ (B′ ) is equal to the maximal homogeneous ideal M′ = (X1 , X2 ) of P′ . The proof of Lemma 3.3 identifies
topdeg(P′ /B′ ) and maxgd(B′ ) for both choices of HF(P′ /B′ ). In each case, the inequality topdeg(P′ /B′ ) < maxgd(B′ ) holds; furthermore in each case topdeg(P′ /B′ )

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

15

is at most 3. All of the hypotheses of Lemma 3.5 are in effect. It follows that Theorem 1.1 applies to R.
3.6.2. The case (k, cS (R)) = (3, 5). Examine the filtration (3.2.1) in order to see
that HS(S/K) = 1 + 3t + t 2. It follows, in particular, that
(3.6.3)

(K + (x1 , x2 , x3 )2 )/K

is a one-dimensional k -vector space and that (x1 , x2 , x3 )3 ⊆ K. Thus,
K = (y1 , . . . , ys , q1 , . . ., q5 ) + (x1 , x2 , x3 )3 ,
where q1 , . . . , q5 are linearly independent quadratic forms in (x1 , x2 , x3 ). Furthermore, y1 , . . . , ys , q1 , . . ., q5 is the beginning of a minimal generating set for K. We
show that Hypothesis (1.1.1) of Theorem 1.1 is satisfied. Consider the ten products
(3.6.4)

{xi q j ∈ S | i ∈ {1, 2} and 1 ≤ j ≤ 5},

which is a subset of the ten-dimensional vector space generated by the monomials
of degree 3 in x1 , x2 , x3 . (We have used the hypothesis that 3 ≤ v(S).) The monomial
x33 is not in the vector space spanned by (3.6.4); so the elements of (3.6.4) must be
linearly dependent. This gives rise to a non-trivial relation ∑i qi ℓi = 0, with each
ℓi a linear form in x1 and x2 . The vector space generated by ℓ1 , . . . , ℓ5 must have
dimension more than one because qℓ = 0 is not possible in S with q a non-zero
quadratic form in x1 , x2 , x3 and ℓ a non-zero linear form in x1 , x2 , x3 . (Again, we
have used the hypothesis that 3 ≤ v(S).) Thus, (x1 , x2 ) ⊆ Fitt1S (K). A permutation
of the subscripts yields that x3 is in Fitt1S (K), as well. We conclude that Hypothesis
(1.1.1) holds for R.
At this point we have identified the beginning, y1 , . . . , ys , q1 , . . . , q5 , of a minimal
generating set for K and the beginning, [S/K]2, of the socle of S/K. The exact
argument we use to finish the proof depends on whether more generators are needed
to generate all of K and whether more socle elements are needed to generate all of
socle(S/K).
(g) Assume (k, cS (R)) = (3, 5) and 3 ≤ maxgd(K). We proved in 3.6.2 that Hypothesis (1.1.1) is satisfied. We now consider Hypothesis (1.1.2). The inclusion
(x1 , x2 , x3 )3 ⊆ K guarantees that (x1 , x2 , x3 )2 ⊆ (K :S n). On the other hand, the
hypothesis that some element of (x1 , x2 , x3 )3 is a minimal generator of K ensures
that
(x1 , x2 , x3 )2 6⊆ (nK :S n);
so, Hypothesis (1.1.2) is satisfied and Theorem 1.1 applies to R in this case.
(h) Assume (k, cS (R)) = (3, 5) and 2 ≤ dimk socle(S/K). We proved in 3.6.2 that
Hypothesis (1.1.1) is satisfied. We now consider Hypothesis (1.1.2). In this case,
there must be an element ℓ of [S]1 which represents an element of the socle of S/K.
It is clear that ℓ ∈ (K :S n). On the other hand, ℓn contains some of the minimal
quadratic generators of K; so ℓ 6∈ (nK :S n). Thus, Hypothesis (1.1.2) is satisfied
and Theorem 1.1 applies to R.

16

A. R. KUSTIN AND A. VRACIU

(i) Assume (k, cS (R)) = (3, 5), maxgd(K) ≤ 2, dimk socle(S/K) = 1, and 5 ≤ v(S).
We use the notation introduced in 3.6.2. The hypotheses ensure that K is minimally generated by y1 , . . . , ys , q1 , . . ., q5 and the socle of S/K is equal to the onedimensional vector space [S/K]2. The ideal B of Remark 3.1.3 is equal to
(Y1 , . . . ,Ys , Q1 , . . . , Q5 ) + A,
where P is the standard-graded polynomial ring k [Y1 , . . . ,Ys , X1 , X2, X3 ] and Qi is the
quadratic form in X1 , X2, X3 which represents qi . The hypothesis 5 ≤ v(S) implies
that
A ⊆ Mv(S) ⊆ M5 ⊆ (Y1 , . . . ,Ys , Q1 , . . . , Q5 ),
where M is the maximal homogeneous ideal of P; consequently,
B = (Y1 , . . .,Ys , Q1 , . . ., Q5 ).
Observe that A ⊆ M5 , S = P/A is an Artinian Gorenstein local ring, and
[socle(P/B)]1 = [socle(S/K)]1 = 0.
Theorem 1.3 applies to P/(A :P B), which according to Remark 3.1.3, is equal to R.
This completes the proof of Theorem 3.1

4. T HE

FIRST SPLITTING THEOREM .

Although not explicitly stated there, Theorem 1.1 was shown in [14]. The proof
we give is essentially the dual of the earlier proof.
Lemma 4.1. Let (S, n,kk) be a local ring, J be an ideal of S, and d : F → G be a
homomorphism of free S-modules of finite rank. Suppose that I1 (d) ⊆ n and
(J :S n) · I1 (d) 6⊆ nJ.
Then k is a direct summand of the SJ -module ker(d ⊗S SJ ).
Proof. Let R denote SJ . If f is an element of F, then let f¯ denote the image of f in
F
F ⊗S R, which is equal to JF
.
According to the hypothesis, there are elements s ∈ S, θ ∈ F, and α ∈ G∗ such
that sn ⊆ J and sα(d(θ)) is a minimal generator of J. Observe first that sθ is in
ker(d ⊗S R). Indeed,
d(sθ) ∈ s im d ⊆ snG ⊆ JG.
Now observe that sθ is a minimal generator of ker(d ⊗S R). Otherwise, there are
elements si ∈ n and θi ∈ F with sθ − ∑i si θi ∈ JF and dθi ∈ JG. In this case,


sα(d(θ)) = α ∑ si d(θi ) + an element of nJ · G
i

= α(an element of nJ · G) ∈ nJ,

which is a contradiction.
Let sθ, φ2 , . . . , φm be a minimal generating set for ker(d ⊗S R). Observe that as
an R-module, ker(d ⊗S R) is equal to the direct sum
Rsθ ⊕ R(φ2 , . . . , φm ).
It is clear that Rsθ + R(φ2 , . . . , φm ) = ker(d ⊗S R). Furthermore, if
rsθ ∈ R(φ2 , . . . , φm ),

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

17

for some r in R, then the definition of the φ’s ensures that r is in the maximal ideal
of R and rsθ = 0 in F/JF.
We have shown that Rsθ is a non-zero direct summand of ker(d ⊗S R). The proof
is complete because ns ⊆ J; and therefore Rsθ is isomorphic to k .

4.2. Proof of Theorem 1.1. We apply Lemma 4.1 to the Artinian Gorenstein local
ring (S, n,kk). Take J to be the ideal (0 :S K), which defines R, from the statement of
Theorem 1.1, and take d to be a minimal presentation matrix for K as an S-module.
It follows that I1 (d) = Fitt1S (K), see 2.2. According to Gorenstein duality, see 2.6,
the hypothesis (J :S n) · I1(d) 6⊆ nJ of Lemma 4.1 is equivalent to

(0 :S nJ) 6⊆ 0 :S (J :S n) · I1(d)


⇐⇒ ((0 :S J) :S n) 6⊆ 0 :S (J :S n) :S I1 (d)
⇐⇒ (K :S n) 6⊆ (0 :S (0 :S nK)) :S I1 (d)

⇐⇒ (K :S n) 6⊆ (nK :S I1 (d)).
In Theorem 1.1 the hypotheses Fitt1S (K) = n and K :S n 6⊆ nK :S n are both in effect;
hence Lemma 4.1 applies and the proof of Theorem 1.1 is complete.

5. T HE

METHOD FOR THE SECOND AND THIRD SPLITTING THEOREMS .

The main results in this section are Lemmas 5.4 and 5.5. In each result we identify a particular summand of a second syzygy. These results are used in Section 6
to prove Theorems 1.2 and 1.3.
Setup 5.1. Let P be a commutative Noetherian ring, A ⊆ B be ideals of P with
A ⊆ B2 , and ∆ be an element of P with
(a) (A :P B) = (A, ∆), and
(b) (A :P ∆) = B.
Let S denote the ring S = P/A, and R denote the ring R = P/(A, ∆). Let
¯ denote − ⊗P S

and

denote − ⊗P R.

We study BS as an R-module.
Remark 5.2. If Setup 5.1 is in effect and S is Artinian, Gorenstein, and local, then
BS is the canonical module of R because S and R have the same Krull dimension
and S is a Gorenstein local ring which maps onto R. It follows that the canonical
module of R is


P
P
ωR = HomS (R, S) = HomP/A (A,∆) , A = A:AP ∆ = AB = BS.
Assumption 5.3. In the notation of Setup 5.1, let
b

b

2
1
B2 −→
B1 −→
B→0

be a presentation of B by free P-modules. Assume that there exists a P-module
homomorphism L : B1 → B2 with
b2 ◦ L ≡ ∆ · idB1

mod AB1 .

18

A. R. KUSTIN AND A. VRACIU

Lemma 5.4. Assume that the notation and hypotheses of 5.1 and 5.3 are in effect.
If B is generated by a regular sequence and B · I1(L) ⊆ (A, ∆), then the R-module
B/B2 is a direct summand of the R-module syzR2 (BS).
Remark. Our application of Lemma 5.5 occurs when B′′ is generated by variables
which are not involved in B′ . One can read the statement of Lemma 5.5 with this
application in mind; indeed, the statement is meaningful (and easier to digest) if B′′ ,
B′′1 , and B′′2 all are zero and b1 = b′1 , b2 = b′2 and L = L′ .
Lemma 5.5. Assume that the notation and hypotheses of 5.1 and 5.3 are in effect.
Suppose that there are free P-module decompositions of B1 and B2 of the form:
B1 = B′1 ⊕ B′′1

and B2 = B′2 ⊕ B′′2

and corresponding decompositions of the homomorphisms b1 , b2 , and L of the form:
 ′

 ′

 ′

b2 b′′′
L
0
′′
2 , and L =
b1 = b1 b1 , b2 =
,
0 b′′2 ,
0 L′′
where

b′1 : B′1 → B,

b′′1 : B′′1 → B,

′′
′
b′′′
2 : B2 → B1 ,

L′ : B′1 → B′2 ,

b′2 : B′2 → B′1 ,

b′′2 : B′′2 → B′′1 ,

L′′ : B′′1 → B′′2

and

are homomorphisms of free P-modules. Let B′ and B′′ be the ideals b′1 (B′1 ) and
b′′1 (B′′1 ) of P, respectively. Suppose further that
(a) B · I1(L′ ) ⊆ (A, ∆),
(b) L′ ◦ b′2 ≡ ∆ · idB′2 mod AB′2 ,
(c) I1 (L′ ◦ b′′′
2 ) ⊆ (A, ∆), and
b′

b′

2
1
(d) the complex B′2 −→
B′1 −→
B′ → 0 is exact.

Then the R-module B′ /BB′ is a direct summand of the R-module syzR2 (BS).
Lemmas 5.4 and 5.5 are proved in 5.10 and 5.11, respectively. We first identify
syzR2 (BS) in terms of homomorphisms than can be readily calculated from the given
data; see (5.7.1). This part of the argument is fairly routine and uses Assumption 5.3
but does not use the extra hypotheses of either Lemma 5.4 or 5.5.
Further notation 5.6. Assume the notation and hypotheses of Setup 5.1 and Assumption 5.3. Let
(5.6.1)

...
/

A2

a2

/

A1

c2



...
/

B2

a1

/

0

/

0

/ A
_

c1
b2

/



B1

b1


/

B

be a commutative diagram of P-modules with exact rows and each Ai and Bi free.
(Take the bottom row of (5.6.1) to be a Koszul complex on a regular sequence which
generates B, whenever this is possible.) The hypothesis that A ⊆ B2 guarantees that
we may choose c1 : A1 → B1 with
(5.6.2)

c1 (A1 ) ⊆ BB1 .

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

Let µ :

V2

19

B1 → B2 be a P-module homomorphism with
(b2 ◦ µ)(β1 ∧ β′1 ) = b1 (β1 ) · β′1 − b1 (β′1 ) · β1.

Define φ : A1 ⊗ B1 → B2 to be the composition
natural quotient map V2

c ⊗1

1
A1 ⊗ B1 −−
−→ B1 ⊗ B1 −−−−−−−−−−−→

µ

B1 −
→ B2 .

Let
B3
B2
⊕
δ2,ℓ
A2 −−→ ⊕ ,
A1
⊕
A1 ⊗ B1

B2
B1 −−→ ⊕ ,
A1
δ2,r

B3
⊕
B2
A2
δ2
⊕ −→ ⊕ ,
A1
A1 ⊗ B1
⊕
B1

and

B2
δ1
⊕ −→
B1
A1

represent the P-module homomorphisms


 




b3 c2
φ
L
δ2,ℓ =
, δ2,r =
, δ2 = δ2,ℓ δ2,r , and δ1 = b2 c1 .
0 −a2 1 ⊗ b1
0

Proposition 5.7. If the notation and hypotheses of 5.6 are in effect, then the following statements hold.
(a) The homomorphisms
B̄3
B̄2
⊕
δ̄2,ℓ
b̄1
δ̄1
Ā2 −−−−→ ⊕ −−−
→ BS → 0
→ B̄1 −−−
Ā1
⊕
Ā1 ⊗ B̄1
form an exact complex of S-modules.
(b) The homomorphisms
B̄¯ 3
⊕
B̄¯ 2
Ā¯ 2
b̄¯ 1
δ2
δ1
⊕ −−−→ ⊕ −−−
→ BS → 0
→ B̄¯ 1 −−−
¯
¯
¯
Ā1
Ā1 ⊗ B̄1
⊕
B̄¯ 1
form an exact complex of R-modules and

(5.7.1)

syzR2 (BS) = ker δ1 = im δ2 .

(Recall from 2.7 that “syzR2 (BS)” is meaningful up to formation of direct sum
with a R-projective module.)

20

A. R. KUSTIN AND A. VRACIU

Proof. In each case the indicated maps form a complex. We show that the complexes are exact. The mapping cone

(5.7.2)

"

#

"

#

b3 c2
b4 c3
h
i
0 −a3 B3 0 −a2 B2 b2 c1
b1 B
. . . −−−−−−−−→ ⊕ −−−−−−−−→ ⊕ −−−−−−→ B1 −→
= BS → 0
A
A1
A2

of (5.6.1) is a resolution of BS by free P-modules.
(a) Apply − ⊗P S to (5.7.2) to see that
B2 δ1 = b2 c1
b1
⊕ −−−−−−−−→ B1 −→
BS → 0
A1
i

h

is an exact sequence of S-modules. We compute ker(δ1 ). Suppose θ ∈ B2 ⊕ A1
and δ1 (θ) = 0 in B1 . In this case, δ1 (θ) = ∑i a1 (α1,i ) · β1,i for some α1,i ∈ A1 and
β1,i ∈ B1 . Observe that




0
∑ µ(c1 (α1,i ) ∧ β1,i )




0

(δ1 ◦ δ2,ℓ ) 
 = b2 c1  i b (β ) · α
∑ 1 1,i
1,i
∑ α1,i ⊗ β1,i
i

i

= ∑ a1 (α1,i ) · β1,i = δ1 (θ).
i

Thus,

0


0
θ − δ2,ℓ 
 ∈ ker(δ1 ).
∑ α1,i ⊗ β1,i


i

We see from (5.7.2) that



b3 c2
ker(δ1 ) = im
.
0 −a2
It follows that θ is in the image of δ2,ℓ . The completes the proof of (a).
(b) The proof of (b) is essentially the same as the proof of (a). The right exactness
of tensor product yields the exact sequence
B 2 δ1 = b c
2
1
b1
⊕ −−−−−−−−→ B1 −→
BS → 0.
A1
h

i

We compute ker(δ1 ). Suppose θ ∈ B2 ⊕ A1 and δ1 (θ) = 0 in B1 . In this case,
δ1 (θ) ≡ ∆β1 mod AB1 for some β1 ∈ B1 . Apply Assumption 5.3 and observe that
(δ1 ◦ δ2,r ) (β1 ) = (b2 ◦ L)(β1 ) ≡ ∆β1 ≡ δ1 (θ) mod AB1 .
Thus θ − δ2,r (β1) ∈ ker(δ1 ) = im(δ2,ℓ ), by part (a); and therefore, θ is in the image
of δ2 . This completes the proof that the complex of (b) is exact.

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

21

The exact complex of (b) gives rise to the exact sequence of R-modules
 δ1
b1
0 → ker δ1 → B2 ⊕ A1 −→
B1 −→
BS → 0

with B2 ⊕ A1 and B1 free. It follows from Definition 2.7 that ker(δ1 ) = syzR2 (BS).
The exactness of the complex of (b) also yields that ker(δ1 ) = im(δ2 ). Assertion
(5.7.1) has been established and the proof of Proposition 5.7 is complete.

Proposition 5.8. If the notation and hypotheses of Lemma 5.4 and Notation 5.6 are
in effect, then the following statements hold:
BB1
(a) ker L = (A:B)B
, and
1
(b) im L ∼
= B/B2 .

Proof. We prove (a) by showing that
(5.8.1)

{β1 ∈ B1 | L(β1 ) ∈ (A, ∆)B2} = BB1 .

The inclusion “⊇” is guaranteed by the hypothesis B · I1 (L) ⊆ (A, ∆). We prove
“⊆”. Let β1 be an element of B1 with L(β1 ) ∈ (A, ∆)B2 . It follows that there is an
element β2 ∈ B2 with Lβ1 ≡ ∆β2 mod AB2 . Apply b2 to both sides and then use
Assumption 5.3 to obtain
∆β1 ≡ (b2 ◦ L)(β1 ) ≡ ∆b2 (β2 )

mod AB1 .

It follows that ∆(β1 − b2 (β2 )) ∈ AB1 and
β1 − b2 (β2 ) ∈ (A :P ∆)B1 = BB1 .
The hypothesis that B is generated by a regular sequence ensures that the bottom
row of (5.6.1) is a Koszul complex; and therefore, b2 (β2 ) ∈ BB1 . We conclude that
β1 ∈ BB1 ; hence, (a) is established. The proof of (b) follows in a routine manner:
B1 ∼ B1 ∼ B
im L ∼
=
= 2.
=
ker L BB1 B
The middle isomorphism is due to (a) and the final isomorphism is induced by the
surjection b1 : B1 → B.

Proposition 5.9. If the notation and hypotheses of Lemma 5.5 and Notation 5.6 are
in effect, then the following statements hold:
(a) ker L′ =
(b) im L′ ∼
=

b′2 (B′2 )+B·B′1
,
(A:P B)B′1
B′ /BB′ .

and

Proof. The proof makes use of the following statements:
(c) {β′1 ∈ B′1 | L′ (β′1 ) ∈ (A, ∆)B′2} = b′2 (B′2 ) + B · B′1, and
(d) (A :P B)B′1 ⊆ b′2 (B′2 ) + B · B′1.
We prove (c). Hypotheses (b) and (a) of Lemma 5.5 ensure that

(A, ∆)B′2 ⊇ L′ b′2 (B′2 ) + B · B′1 ;

and this establishes the inclusion “⊇”. Now, we prove “⊆”. Suppose β′1 is in B′1
with
L′ (β′1 ) ≡ ∆β′2 mod AB′2 ,

22

A. R. KUSTIN AND A. VRACIU

for some β′2 ∈ B′2 . Use Hypothesis 5.5.(b) to write
L′ (β′1 ) ≡ (L′ ◦ b′2 )(β′2)

mod AB′2 .

Apply b′2 to obtain
(b′2 ◦ L′ )(β′1 − b′2 β′2 ) ≡ 0

mod AB′1 .

Employ Assumption 5.3 to see that
∆(β′1 − b′2 β′2 ) ∈ AB′1 ;
hence,
β′1 − b′2 β′2 ∈ (A :P ∆)B′1 = BB′1 ,
by 5.1.(b), and β′1 ∈ b′2 (B′2 ) + BB′1 . This completes the proof of (c).
We prove (d). Recall from 5.1 that (A :P B) = (A, ∆) and A ⊆ B. Thus, it suffices
to consider ∆B′1 . Assumption 5.3 and the decompositions of Lemma 5.5 ensure that
∆ · B′1 ≡ (b2 ◦ L)(B′1 ) = (b′2 ◦ L′ )(B′1 ) ⊆ b′2 (B′2 ).
The proof of (a) is now straightforward:
{β′1 ∈ B′1 | L′ (β′1 ) ∈ (A, ∆)B′2} + (A :P B)B′1
(A :P B)B′1
b′ (B′ ) + B · B′1 + (A :P B)B′1
= 2 2
,
(A :P B)B′1
b′2 (B′2 ) + B · B′1
,
=
(A :P B)B′1

ker L′ =

by (c),
by (d).

The proof of (b) is also straightforward:
im L′ ∼
=

B′1
ker L′

=

B′1
(A:P B)B′1
b′2 (B′2 )+B·B′1
(A:P B)B′1

=

B′1
B′
∼
.
=
b′2 (B′2 ) + B · B′1 BB′

The final isomorphism is induced by the surjection b′1 : B′1
pothesis (d) of Lemma 5.5

//

B′ and uses Hy

5.10. The proof of Lemma 5.4. We use all of notation and hypotheses of 5.1, 5.4,
and 5.6. We proved in Proposition 5.8.(b) that
im(δ2,r ) ∼
= B/B2 .
We complete the proof of Lemma 5.4 by showing that im(δ2,r ) is a direct summand
of the R-module syzR2 (BS). We accomplish this goal by showing that
im(δ2,ℓ ) ⊕ im(δ2,r ) = syzR2 (BS).
We know from (5.7.1) that
im(δ2,ℓ ) + im(δ2,r ) = syzR2 (BS).
We show that im(δ2,ℓ ) ∩ im(δ2,r ) = (0). Suppose
β 1 ∈ B1

and

θ ∈ B3 ⊕ A2 ⊕ (A1 ⊗ B1 )

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

23

with
(5.10.1)

δ2,r (β1 ) − δ2,ℓ (θ) in (A, ∆)B2 ⊕ (A, ∆)A1 .

We complete the proof by showing that
L(β1 ) ∈ (A, ∆)B2.

(5.10.2)

We are told in (5.10.1) that



 

L(β1 )
∆β2
an element of AB2
− δ2,ℓ (θ) =
+
,
0
∆α1
an element of AA1


for some β2 ∈ B2 and α1 ∈ A1 . Apply δ1 = b2 c1 to see that

(5.10.3)

b2 Lβ1 − δ1 δ2,ℓ (θ) = ∆b2 β2 + ∆c1 α1 + an element of AB1 .

We know from 5.7.(a) that δ1 δ2,ℓ (θ) ∈ AB1 . The choice c1 (A) ⊆ BB1 , which was
made in (5.6.2), guarantees that ∆c1 α1 ∈ ∆BB1 ⊆ AB1 . Of course, Assumption 5.3
guarantees that b2 Lβ1 ≡ ∆β1 mod AB1 . Thus, (5.10.3) yields
∆(β1 − b2 β2 ) ∈ AB1 ;
and therefore, β1 − b2 β2 ∈ BB1 . Thus,
β1 ∈ b2 (B2 ) + BB1 ⊆ BB1 .
(The hypothesis that B is generated by a regular sequence ensures that the bottom
row of (5.6.1) is a Koszul complex; and therefore, b2 (B2 ) ⊆ BB1 and the final inclusion holds.) Apply the hypothesis BI1 (L) ⊆ (A, ∆) to conclude L(β1 ) ∈ (A, ∆)B2,
which establishes (5.10.2) and completes the proof.

5.11. The proof of Lemma 5.5. We use all of notation and hypotheses of 5.1, 5.5,
and 5.6. We proved in Proposition 5.9.(b) that
im(L′ ) ∼
= B′ /BB′ .
We complete the proof of Lemma 5.5 by showing that im(L′ ) is a direct summand
of the R-module syzR2 (BS). We accomplish this goal by showing that


′′
im(δ2,ℓ ) + im L ⊕ im(L′ ) = syzR2 (BS).
We know from (5.7.1) that


im(δ2,ℓ ) + im L′′ + im(L′ ) = syzR2 (BS).


′′
We show that im(δ2,ℓ ) + im L ∩ im(L′ ) = (0). Suppose
β′1 ∈ B′1 ,

β′′1 ∈ B′′1 ,

and

θ ∈ B3 ⊕ A2 ⊕ (A1 ⊗ B1 )

with
(5.11.1)

L′ (β′1 ) − L′′ (β′′1 ) − δ2,ℓ (θ) in (A, ∆)B′2 ⊕ (A, ∆)B′′2 ⊕ (A, ∆)A1 .

We complete the proof by showing that
(5.11.2)

L′ (β′1 ) ∈ (A, ∆)B′2 .

24

A. R. KUSTIN AND A. VRACIU

We are told in (5.11.1) that



 

L(β′1 − β′′1 )
∆β2
an element of AB2
− δ2,ℓ (θ) =
+
,
0
∆α1
an element of AA1


for some β2 ∈ B2 , and α1 ∈ A1 . Apply δ1 = b2 c1 to see that

(5.11.3)

(b2 ◦ L)(β′1 − β′′1 ) − δ1 δ2,ℓ (θ) = ∆b2 β2 + ∆c1 α1 + an element of AB1 .

We know from 5.7.(a) that δ1 δ2,ℓ (θ) ∈ AB1 . The choice c1 (A) ⊆ BB1 , which was
made in (5.6.2), guarantees that ∆c1 α1 ∈ ∆BB1 ⊆ AB1 . Of course, Assumption 5.3
guarantees that b2 L ≡ ∆ mod AB1 . Thus, (5.11.3) yields
∆(β′1 − β′′1 − b2 β2 ) ∈ AB1 ;
and therefore,
(5.11.4)

β′1 − β′′1 − b2 β2 ∈ (A :P ∆)B1 = BB1 .

Separate (5.11.4) into the components B′1 ⊕ B′′1 = B1 . Recall that b2 has the form

 ′
b2 b′′′
2
;
b2 =
0 b′′2
and write β2 = β′2 + β′′2 with β′2 ∈ B′2 and β′′2 ∈ B′′2 . Conclude that
 ′   ′
 ′  

β1
b2 b′′′
β2
an element of BB′1
2
−
=
.
−β′′1
0 b′′2 β′′2
an element of BB′′1
It follows that
′′′
′
β′1 − b′2 β′2 − b′′′
2 β2 ∈ BB1 .

Apply L′ and invoke Hypotheses (a), (b), and (c) of Lemma 5.5 to conclude L′ β′1 is

in (A, ∆)B′2 , which establishes (5.11.2) and completes the proof.
6. C OMPLETE

INTERSECTIONS AND FIVE QUADRATICS IN THREE VARIABLES .

In this section, we prove Theorems 1.2 and 1.3 which were promised in Section 1. The proof of Theorem 1.2 is based on Lemma 5.4 and is given in 6.1. The
proof of Theorem 1.3 is given in 6.5 after we collect the results that ensure that the
hypotheses of Lemma 5.5 are satisfied.
6.1. The proof of Theorem 1.2. We apply Lemma 5.4 and Remark 5.2. Gorenstein
duality guarantees that (A :P B)/A is a cyclic P/A-module. Let ∆ be an element of
P with (A :P B) = (A, ∆). Gorenstein duality also guarantees that (A :P ∆) = B.
Let θ1 , . . ., θn be a regular sequence in P which generates B and let
b

b

1
2
B→0
B1 −→
B2 −→

be the beginning of the Koszul complex on this generating set. We are now in the
situation of Setup 5.1. It remains to establish the existence of a homomorphism
L : B1 → B2 which satisfies
(a) (b2 ◦ L) ≡ ∆ · idB1 mod AB1 , and
(b) B · L(B1 ) ⊆ (A, ∆)B2.

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

25

Property (b) is satisfied provided




L(B1 ) ⊆ (A, ∆) : B) B2 = (A : B) : B) B2 = (A : B2 )B2 .

Consequently, it suffices to prove that



2



(A : B)B1 ⊆ b2 (A : B )B2 + AB1 .
We think of B1 as Pn and of the surjection B1 → B as the map given by the matrix
[θ1 , . . . , θn ]. We show that


(A : B)
 0 
 .  ⊆ Pn
 .. 

0


is in b2 (A : B2 )B2 + AB1 . One completes the argument by symmetry. Every
Koszul relation on θ1 , . . ., θn is in the image of b2 ; so, in particular, each column of


θ2
···
θn
−θ1 In−1

is in the image of b2 , where In−1 is the (n − 1) × (n − 1) identity matrix. Let Θ be an
element of (A : B). We show that there exist elements r2 , . . . , rn in (A : θ1 ) ∩ (A : B2 )
with ∑ni=2 ri θi ≡ Θ, mod A. In other words, it suffices to show that


Θ ∈ (A : θ1 ) ∩ (A : B2 ) (θ2 , . . . , θn ) + A.

Let Θ roam over (A : B). It suffices to show that

(A : B) ⊆ (A : (θ1 , B2 ))(θ2, . . . , θn ) + A.
By Gorenstein duality, it suffices to show
(6.1.1)

(θ1 , B2 ) : (θ2 , . . . , θn ) ⊆ B.

The hypothesis that B is a complete intersection of grade at least two ensures that
(6.1.1) holds and the proof is complete.

We turn our attention to the proof of Theorem 1.3. Some preliminary results are
needed.
Observation 6.2. Let k be a field of arbitrary characteristic, P′ be a standardgraded polynomial ring in three variables over k , and B′ be an ideal in P′ which is
generated by a 5-dimensional subspace of [P′ ]2 . Assume that
(6.2.1)

[socle(P′ /B′ )]1 = 0.

Then the ring P′ /B′ is Gorenstein and has Hilbert series HS(P′ /B′ ) = 1 + 5t + t 2 ;
furthermore, the ideals (B′ )2 and ([P′ ]1 )4 of P are equal.
Proof. Let φ : [P′ ]2 → k be a k -module homomorphism with [B′ ]2 = ker φ. If the
characteristic of k is different than two, then use standard results about symmetric

26

A. R. KUSTIN AND A. VRACIU

bilinear forms over a field (essentially Gram-Schmidt orthogonalization) to choose
a basis x, y, z for [P′ ]1 so that the matrix
 2

φ(x ) φ(xy) φ(xz)
(6.2.2)
T = φ(xy) φ(y2 ) φ(yz)
φ(xz) φ(yz) φ(z2 )

is a diagonal matrix. The non-degeneracy hypothesis (6.2.1) ensures that the entries
on the main diagonal of T are units in k , say u1 , u2 , u3 . The same conclusion holds
in characteristic two; but the argument is more subtle. The argument we offer was
taken from some expository notes written by Keith Conrad [7, Exercise 5.4].
Assume k has characteristic two. The non-degeneracy hypothesis (6.2.1) ensures
that there is an element z1 ∈ [P′ ]1 with z1 2 ∈
/ B′ . (Otherwise, if x1 , y1 , z1 is any basis
′
of [P ]1 , then φ(y1 z1 )x1 + φ(x1 z1 )y1 + φ(x1 y1 )z1 represents a non-zero element of
[socle(P′ /B′ )]1 .) If there is an element y1 with y1 z1 ∈ B′ and y21 ∈
/ B′ , then the usual
argument maybe used to complete the proof. On the other hand, if every element
of B′ :[P′ ]1 z1 squares to an element of B′ , then let x1 , y1 be a basis of B′ :[P′ ]1 z1 with
φ(x1 y1 ) = 1. Consider the basis
x = φ(z21 )x1 + z1 ,

y = (1 + φ(z21))x1 + y1 + z1 ,

and z = x1 + y1 + z1

for [P′ ]1 . Observe that the matrix (6.2.2) is a diagonal matrix. In this case also, we
take the entries on the main diagonal of T to be the units u1 , u2 , u3 of k .
It is now clear that
(6.2.3)

B′ = (xy, xz, yz, u2 x2 − u1 y2 , u3 x2 − u1 z2 ).

It is also clear that the row vector of signed maximal order Pfaffians of the alternating matrix


0
u3 x
−u1 u3 y −z
0
 −u3 x
0
u1 u3 z
0
y 


′

0
u3 x −u2 x
(6.2.4)
b2 = u1 u3 y −u1 u3 z

 z
0
−u3 x
0
0 
0
−y
u2 x
0
0

is

(6.2.5)

b′1 = [u3 xy, u2 xz, yz, u2 u3 x2 − u1 u3 y2 , u23 x2 − u1 u3 z2 ].

Observe that the image of b′1 is equal to B′ and that B′ has grade 3. It follows that
(6.2.6)

′

b′1 T

b′

b′

2
1
0 → P (−5) −−→ P′ (−3)5 −→
P′ (−2)5 −→
P′

is a minimal homogeneous resolution of P′ /B′ ; hence, P′ /B′ is a Gorenstein ring
with Hilbert series 1 + 5t + t 2 . One can check the assertion B′ 2 = ([P′ ]1 )4 by hand
or one can read the minimal homogeneous resolution
0 → P′ (−6)10 → P′ (−5)24 → P′ (−4)15 → P′
of P′ /B′ 2 from [11, page 36] in order to conclude that the socle degree of P′ /B′ 2 is
three; thus, ([P′]1 )4 ⊆ B′ 2 . The inclusion B′ 2 ⊆ ([P′ ]1 )4 is obvious.


TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

27

Lemma 6.3. Let k be a field of arbitrary characteristic, P = k [X1 , X2, X3 ,Y1 , . . . ,Ys ]
be a standard-graded polynomial ring over k in 3 + s variables for some nonnegative integer s, B′ be an ideal in P which is generated by five linearly independent
quadratic forms in the variables X1 , X2, X3 , B′′ be the ideal (Y1 , . . . ,Ys ) of P, B be
the ideal B′ + B′′ of P, and A ⊆ ([P]1)5 be a homogeneous ideal of P with P/A an
Artinian Gorenstein local ring. Assume [socle(P/B)]1 = 0. Then there is a presentation
b

b

2
1
B2 −→
B1 −→
B→0

(6.3.1)

of B by free P-modules, a P-module homomorphism L : B1 → B2 , and direct sum
decompositions
 ′

 ′


 ′
b2 b′′′
L 0
′
′′
′
′′
′′
2
, and L =
,
B1 = B1 ⊕B1 , B2 = B2 ⊕B2 , b1 = b1 b1 , b2 =
0 b′′2 ,
0 L′′

which satisfy Assumption 5.3 and conditions (a), (b), (c), and (d) of Lemma 5.5.

Proof. Let P′ be the polynomial ring k [X1 , X2, X3 ]. View P′ as a subring of P. Notice
that P/B = P′ /(P′ ∩ B′ ); so we may apply Observation 6.2 and pick a basis x, y, z
of [P′ ]1 so that the generators of (P′ ∩ B′ ) are given in (6.2.3) and the minimal
homogeneous resolution of P′ /(P′ ∩ B′ ) by free P′ -modules is given in (6.2.6) with
b′1 and b′2 given in (6.2.5) and (6.2.4), respectively.
Apply the functor P ⊗P′ − to (6.2.6) to obtain the minimal homogeneous resolution
′

B :

0 → B′3

b′1 T

b′

b′

1
2
B′0 = P
B′1 = P(−2)5 −→
= P(−5) −−→ B′2 = P(−3)5 −→

of P/B′ . (This establishes Hypothesis (d) of Lemma 5.5.) Let B′′1 be a free P-module
of rank s, b′′1 : B′′1 → P be the homomorphism
V

h

i

Y1 · · · Ys
P(−1)s −−−−−−−−−→ P,

and • B′′1 be the Koszul complex
associated to b′′1 . The minimal homogeneous
V• ′′
′
resolution of P/B is B ⊗P B1 . In this language, the presentation (6.3.1) of B by
free P-modules is
V
B′2 ⊗P 0 B′′1
V
h
i
⊕
B′1 ⊗P 0 B′′1 b1 = b′ b′′
V
1
1
b
2
⊕
−−−−−−−−−−→ B = B′ + B′′ → 0,
B′1 ⊗ 1 B′′1 −−−→
V
⊕
B′0 ⊗ 1 B′′1
V2 ′′
′
B0 ⊗ B1
where


b′2 −1 ⊗ b′′1
0
b2 =
.
0 b′1 ⊗ 1 1 ⊗ b′′1
To complete the promised decomposition of (6.3.1) let
(6.3.2)



V

B′1 ⊗ 1 B′′1
⊕
B′′2 =
V
′
B0 ⊗ 2 B′′1 ,



′′ 0
−1
⊗
b
b′′′
=
2
1

and,



b′′2 = b′1 ⊗ 1 1 ⊗ b′′1 .

28

A. R. KUSTIN AND A. VRACIU

The homogeneous ideals A and B satisfy A ⊆ B and both ideals define Gorenstein
rings. Gorenstein duality ensures the existence of a homogeneous element ∆ of P
with
(A :P B) = (A, ∆) and (A :P ∆) = B.

Let S denote P/A, and ¯ denote − ⊗P S. Let x,y,z,Yi1 ,...,Ys denote the set of monomials in P of degree i in x, y, z,Y1 , . . . ,Ys . The hypothesis A ⊆ ([P]1)5 guarantees
that

{m | m ∈ x,y,z,Yi1,...,Ys }
(6.3.3)

is a basis for [S]i for 0 ≤ i ≤ 4. Let d denote the socle degree of S and let α1 be an
element of [P]d with the property that α1 is a basis for the socle [S]d of S. For each
integer i, with 0 ≤ i ≤ d, the multiplication map
(6.3.4)

[S]i × [S]d−i → [S]d

is a perfect pairing. For 1 ≤ i ≤ 3, select {αm ∈ [P]d−i | m ∈

{αm | m ∈ x,y,z,Yi1 ,...,Ys }

x,y,z,Y1 ,...,Ys 
}
i

so that


is a basis for [S]d−i which is dual to {m | m ∈ x,y,z,Yi1 ,...,Ys } in the sense that if

m, m′ ∈ x,y,z,Yi1 ,...,Ys , then
(
α1 , if m = m′ , and
m′ α m =
0,
if m 6= m′ .


It follows from (6.3.4) that if m ∈ x,y,z,Yi1 ,...,Ys and m′ ∈ x,y,z,Yi1′ ,...,Ys for some i and
i′ with 0 ≤ i, i′ ≤ 3, then
(
αm′′ , if m′ m′′ = m, and
m′ α m =
0,
if m′ does not divide m.
The generators of B are Y1 , . . . ,Ys , together with the five quadratics given in
(6.2.3). Use the perfect pairing (6.3.4), with i = 2, to observe that
[A :P B]d−2 = [A]d−2 +kk (u1 αx2 + u2 αy2 + u3 αz2 );
hence, we may take the ∆ of (6.3.3) to be
(6.3.5)

∆ = u1 αx2 + u2 αy2 + u3 αz2 .

Let L′ be the 5 × 5 matrix of Table 1 with entries in P. Recall the matrix b′2 from
(6.2.4). A straightforward calculation shows that b′2 L′ and L′ b′2 are both congruent
to ∆I5 , modulo A. The conclusion b′2 L′ ≡ ∆I establishes half of Assumption 5.3.
(The other half is studied at (6.3.7).) The conclusion L′ b′2 ≡ ∆I demonstrates that
condition Lemma 5.5.(b) holds. Observe that
(6.3.6)

′′
′
I1 (b′′′
2 ) · S ⊆ B · S ⊆ annS I1 (L );

and therefore condition Lemma 5.5.(c) holds. (The second inclusion is easily read
from the definition of L′ .) To prove condition Lemma 5.5.(a), it suffices to show

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN




















Let L′ be the matrix
− uu13 αx3
0
u2 αy2 z
+u3 αz3
0
0
− uu1 u3 2 αx2 y

0
0

− uu1 u2 3 αy3
−u1 αx2 z
−u3 αz3
− uu1 u2 3 αx2 z

1
u1 αz3
u22
u3 αy3

u2
u3 αxy2

u2 αy3

− uu12 αx3

u23
u2

0

− uu13 αx3

0
−u1 αx2 y
−u2 αy3
−u3 αyz2
0

−u1 u2 αxyz −u1 u3 αxyz
0

0

− uu32 αxz2

− αz3

TABLE 1. The matrix L′ from the proof of Lemma 6.3.

29











.









that B2 · I1(L′ ) · S = 0. On the other hand,
2

B2 · I1 (L′ ) · S ⊆ (B′ + B′′ )I1 (L′ ) · S
2

= B′ I1 (L′ ) · S

see (6.3.6)

⊆ [S]4 · [S]d−3 · S ⊆ [S]d+1 · S = 0.
6.3.7. To complete the proof, we must establish the rest of Assumption 5.3; that is,
we must exhibit a P-module homomorphism L′′ : B′′1 → B′′2 with

  
0
0
mod AB1 .
(6.3.8)
b2 ′′ ≡
∆ · idB′′1
L
The next calculation is the main step in that direction.
Claim 6.3.9. The element ∆ of P is in the ideal B′ · (A :P B′′ ) + A.
Proof of Claim 6.3.9. By Gorenstein duality, 2.6, it suffices to show that
A :P (B′ · (A :P B′′ ) + A) ⊆ A :P (A, ∆).
The module on the right is B. The module on the left is
(A :P (A :P B′′ )) :P B′ = (A, B′′) :P B′ .
To prove Claim 6.3.9 it suffices to show that
(6.3.10)

(A, B′′) :P B′ ⊆ B.

Let p be a homogeneous element in (A, B′′ ) :P B′ . Write p = p′ + p′′ for homogeneous elements p′ and p′′ , with p′ in the subring k [x, y, z] of P and p′′ in the ideal B′′
of P. Of course, p′′ is in (A, B′′) :P B′ and p′′ ∈ B; consequently, p′ is in (A, B′′ ) :P B′
and it suffices to prove that p′ ∈ B. We are told that p′ B′ ⊆ (A, B′′). There are nonzero elements of
[p′ B′ ∩kk [x, y, z]]deg p′ +2 .

30

A. R. KUSTIN AND A. VRACIU

On the other hand, the ideal A is contained in ([P]1 )5 , by hypothesis, so
[(A, B′′) ∩kk [x, y, z]]i = 0

for i ≤ 4.

Therefore, 5 ≤ deg p′ + 2; hence 3 ≤ deg p′ and p′ ∈ B′ ⊆ B; and this completes the
proof of Claim 6.3.9. (Recall that the Hilbert series of P′ /B′ is given in Observation 6.2.)
Claim 6.3.9 guarantees that there is an element c in B′1 with
(6.3.11)

b′1 (c) ≡ ∆

and B′′ I1 (c) ⊆ A.

mod A

Recall from (6.3.2) that B′′2 = (B′1 ⊗ B′′1 ) ⊕

V2 ′′
B1 . Define L′′ : B′′1 → B′′2 by

L′′ (β′′1 ) = c ⊗ β′′1 ∈ B′1 ⊗ B′′1 ⊆ B′′2 ,

for β′′1 ∈ B′′1 .

Observe that


 
 ′

 ′′ ′′  

0
′′
0
b2 −1 ⊗ b1
0
−b
(β
)
·
c
0
′′
′′
1 1
c ⊗ β1  =
b2 ′′ (β1 ) =
≡
,
L
0 b′1 ⊗ 1 1 ⊗ b′′1
b′1 (c) · β′′1
∆ · β′′1
0

mod A. The final congruence is due to (6.3.11). This completes the proof of (6.3.8)
and also the proof of Lemma 6.3.

Remark 6.4. We used Macaulay2, [9], to solve a system of 250 non-homogeneous
linear equations in 300 unknowns in order to produce the matrix L′ of Table 1.
The matrix L′ is 5 × 5; so, it has 25 entries,
and each entry is a linear combination
x,y,z
of the ten elements {αm | m ∈ 3 }. Altogether, 25 × 10 coefficients must be
determined. Each entry of b′2L′ and each entry of L′ b′2 is a linear combination of the
six elements {αm | m ∈ x,y,z
2 }. Altogether (25 + 25) × 6 equations must be solved.
It is amusing to notice that, according to the computer, the vector space of solutions
of the corresponding homogeneous equations has dimension 86, over Q and also
over Z/(2); so, there are many choices for L′ . The fact that we used the computer
to identify one choice for L′ is essentially irrelevant because it is easy to check by
hand that the matrix of Table 1 does satisfy b′2 L′ = L′ b′2 = ∆I.
6.5. The proof of Theorem 1.3. Apply Lemma 6.3 to see that all of the hypotheses
of Lemma 5.5 hold. Lemma 5.5 yields the result.

7. T EST

MODULES

In Theorem 7.1 and Proposition 7.3 we prove that many modules are proj-test
modules in the sense of Definition 2.10.1. The modules “B′ /BB′ ” of Theorem 1.3
are covered by Theorem 7.1 and Remark 7.2. The modules “B/B2 ” of Theorem 1.2
are covered in Proposition 7.3 and Remark 7.4. We state and prove the main result
in the paper as Theorem 7.5.
Theorem 7.1. Let (R, m,kk) be a local Noetherian ring with 3 ≤ v(R). If T is a
finitely generated R-module with m2 T = 0, then T is a proj-test module for R, in the
sense of Definition 2.10.1.
Proof. Let x1 , . . . , xn be a minimal generating set for m. The assumption about v(R)
guarantees that the n+1
monomials of degree 2 in the symbols x1 , . . ., xn represent
2
k
a basis for the -vector space m2 /m3 ; see 2.5.(b).

TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

31

Fix finitely generated R-modules T and M with m2 T = 0 and TorR+ (T, M) = 0.
Let
d2
d1
F : · · · −→
Rb1 −→
Rb0
be the minimal resolution of M. The matrix di is well-defined for all non-negative
integers i; if necessary, di is allowed to be the zero matrix. Furthermore, each entry
of each di is in m. Write
n

di =

∑ Di, j x j ,

j=1

where each Di, j is a bi−1 × bi matrix with entries in R. Use di di+1 = 0 and the
hypothesis 3 ≤ v(R) to see that
(7.1.1)
(7.1.2)

Di, j Di+1, j ≡ 0
Di, j1 Di+1, j2 + Di, j2 Di+1, j1 ≡ 0

mod m,
mod m,

for all j, and
for all j1 =
6 j2 .

The hypotheses m2 T = 0 and TorR+ (T, M) = 0 guarantee that
(mT ) ⊗ Rbi ⊆ ker(T ⊗ di ) = im(T ⊗ di+1 )
Fix a positive index i. If u is an arbitrary vector in (mT )bi , then
n

u=

(7.1.3)

∑ Di+1, j0 x j0u j0 ,

j0 =1

for some u 1 , . . .,uun ∈ T bi+1 . Each x j0 u j0 is in (mT )bi+1 ; therefore
x j0 u j0 ∈ im(T ⊗ di+2 ) and
n

x j0 u j0 =

(7.1.4)

∑ Di+2, j1 x j1u j1 , j0 ,

j1 =1

for some u j1 , j0 ∈ T bi+2 . Combine (7.1.3) and (7.1.4) to obtain
n

u=

n

∑ ∑ Di+1, j0 Di+2, j1 x j1 u j1, j0 .

j0 =1 j1 =1

In its turn, x j1 u j1 , j0 ∈ (mT )bi+2 ⊆ im(T ⊗ di+3 ). When the above procedure is iterated k times, one obtains
n

(7.1.5)

u=

n

n

∑ ∑ · · · ∑ Di+1, j0 Di+2, j1 · · · Di+k+1, jk x jk u jk ,..., j1, j0 ,

j0 =1 j1 =1

jk =1

for some u jk ,..., j1 , j0 ∈ T bi+k+1 .
Observe that if n < k, then
(7.1.6)

Di+1, j0 Di+2, j1 · · · Di+k+1, jk x jk ≡ 0,

mod m

for each such product which appears in (7.1.5). Indeed there must be two j’s which
are the same, say ji1 = ji2 for some i1 and i2 with 0 ≤ i1 < i2 ≤ k. If i2 = i1 + 1,
then equation (7.1.1) shows that (7.1.6) holds. If not, use equation (7.1.2) to bring
the equal j’s closer together.

32

A. R. KUSTIN AND A. VRACIU

It now L
follows that (mT )bi = 0, and therefore mT = 0. In other words, T is a vector space k . The hypothesis that Tor1 (T, M) is zero now implies that Tor1 (kk, M) =
0; and therefore, M is a free R-module.

Remark 7.2. Theorem 7.1 may be applied to the module B′ /BB′ of Theorem 1.3.
Indeed, B′ /BB′ is annihilated by (X1 , X2, X3 )2 because
2

(X1, X2 , X3)2 B′ ⊆ (X1, X2 , X3)2 (X1, X2 , X3)2 = (X1 , X2, X3 )4 = B′ ⊆ BB′ .
The right most equality is established in Observation 6.2.
Proposition 7.3. Let P be a regular ring, M be a maximal ideal of P, I ⊆ M be an
ideal of P which is generated by a regular sequence, and R be P/J, where J is an
M-primary ideal of P and v(R) is sufficiently large. Then T = R/IR is a proj-test
module for R in the sense of Definition 2.10.1.
Proof. It does no harm to localize P at M and to inflate the residue field of PM in
order to assume that the inflated residue field is infinite; see, for example [10, Ch.
0, 10.3.1]. We alter the notation to make it less cumbersome. In the new notation,
(P, M,kk) is a regular local ring with k infinite, I is an ideal of P which is generated
by a regular sequence, and R is an Artinian quotient of P with v(R) sufficiently
large.
The Artin-Rees Lemma guarantees the existence of an integer n0 such that
I ∩ Mn0 ⊆ IM.
In this proof, we insist that n0 < v(R).
Let M be a finitely generated R-module with TorRi (M, T ) = 0 for all positive i.
The ring R and the R-module M both have depth zero; consequently, according to
the Auslander-Buchsbaum formula, to show that M is a free R-module, it suffices
to show that M has finite projective dimension as an R-module. We assume that M
has infinite projective dimension and we draw a contradiction.
Let
F:

d

d

d

3
2
1
. . . −→
F2 −→
F1 −→
F0

be a collection of P-module homomorphisms of free P-modules with the property
that F ⊗P R is the minimal R-resolution of M. The hypothesis about the vanishing
of TorR+ (M, T ) guarantees that (F ⊗P R) ⊗R T = F ⊗P T is the minimal resolution
of M ⊗R T by free T -modules.
The ring T is a complete intersection, the ring P is regular local, and the residue
field k of P is infinite; hence, the Eisenbud-Peeva [8] theorems apply to F ⊗P T . In
particular, if i is sufficiently large, then the syzygy module im(di ⊗P T ) in F ⊗P T
is a Higher Matrix Factorization module [8, Theorem 1.3.1]; and therefore, when
suitable bases are chosen in F, some of the entries of the product matrix di+1 di+2
are minimal generators of I. The hypothesis that n0 < v(R) guarantees that every
minimal generator of I represents a non-zero element of R. On the other hand,
F ⊗P R is a complex and di+1 di+2 ⊗P R = 0. We have reached a contradiction
0 6= di+1 di+2 ⊗P R = 0;
and this completes the proof.



TOTALLY REFLEXIVE MODULES OVER RINGS THAT ARE CLOSE TO GORENSTEIN

33

Remark 7.4. Proposition 7.3 may be applied to the module B/B2 of Theorem 1.2.
Indeed, the fact that the ideal B of P is generated by a regular sequence ensures that
B/B2 is a free P/B module of rank c, where c is the minimal number of generators
of B; hence, as an R-module,
 c
 c
B
R
B
P
⊗
R
=
.
=
⊗
R
=
P
P
B2 B2
B
BR
Theorem 7.5 is the main result of the paper. Recall that if R = S/J is a quotient of
the ring S, then cS (R) is the length of J as an S-module. Also, recall the parameter
v(S) from 2.5.(b).
Theorem 7.5. Let (S, n,kk) be an equicharacteristic Artinian Gorenstein local ring
with embedding dimension at least two, and R be the ring R = S/J for some proper
non-zero ideal J of S. Assume that either
• 1 ≤ cS (R) ≤ 4, or
• cS (R) = 5 and S is a standard-graded algebra over a field [S]0.
If the parameter v(S) is sufficiently large, then R is G-regular.
Proof. Apply Theorem 3.1 to see that one of the Theorems 1.1, 1.2, or 1.3 applies to
R. These theorems ensure that either k , “B/B2 ”, or “B′ /BB′ ” is a direct summand of
syzR2 (ωR ). The module k is the prototype of a proj-test module. It is shown in Theorem 7.1 (together with Remark 7.2) and Proposition 7.3 (together with Remark 7.4)
that “B/B2 ” and “B′ /BB′ ” are proj-test modules. Apply Observation 2.10.3 to complete the proof.

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A NDREW R. K USTIN , D EPARTMENT OF M ATHEMATICS , U NIVERSITY OF S OUTH C AROLINA ,
C OLUMBIA , SC 29208, U.S.A.
E-mail address: kustin@math.sc.edu
A DELA V RACIU , D EPARTMENT OF M ATHEMATICS , U NIVERSITY
C OLUMBIA , SC 29208, U.S.A.
E-mail address: vraciu@math.sc.edu

OF

S OUTH C AROLINA ,

