∗-transforms of acyclic complexes

arXiv:1307.1500v2 [] 12 Dec 2013

Taro Inagawa
Graduate School of Science, Chiba University,
1-33 Yayoi-Cho, Inage-Ku, Chiba-Shi, 263-8522, JAPAN

Abstract
Let R be an n-dimensional Cohen-Macaulay local ring and Q a parameter ideal
of R. Suppose that an acyclic complex (F• , ϕ• ) of length n of finitely generated free
R-modules is given. We put M = Im ϕ1 , which is an R-submodule of F0 . Then F•
is an R-free resolution of F0 /M . In this paper, we describe a concrete procedure to
get an acyclic complex ∗F• of length n that resolves F0 /(M :F0 Q).

1

Introduction

The ∗-transform of an acyclic complex of length 3 is introduced in [1]. The purpose
of this paper is to generalize it for acyclic complexes of length n ≥ 2. Let (R, m) be
an n-dimensional Cohen-Macaulay local ring and x1 , x2 , . . . , xn an sop for R. We put
Q = (x1 , x2 , . . . , xn )R. Suppose that an acyclic complex
ϕ1

ϕn

0 −→ Fn −→ Fn−1 −→ · · · −→ F1 −→ F0
of finitely generated free R-modules such that Im ϕn ⊆ QFn−1 is given. In this paper, we
describe an operation to get an acyclic complex
∗ϕ

∗ϕ

0 −→ ∗Fn −→ ∗Fn−1 −→ · · · −→ ∗F1 −→ ∗F0 = F0
n

1

such that Im ∗ϕ1 = Im ϕ1 :F0 Q and Im ∗ϕn ⊆ m · ∗Fn−1 , which is called the ∗-transform of
F• with respect to x1 , x2 , . . . , xn . As we give a practical condition for ∗Fn to be vanished,
it is possible to consider when the depth of F0 /(Im ϕ1 :F0 Q) is positive. This is useful
when we compute the saturation of ideals. In fact, in the subsequent paper [2], using
∗-transform we compute the saturation of the m-th power of the ideal generated by the
maximal minors of the following m × (m + 1) matrix
 α1,1

α
α
α
α1,m+1
x1
x2 1,2 x3 1,3 · · · xm1,m xm+1
 xα2,1 xα2,2 xα2,3 · · · xα2,m xα2,m+1 
1
m+1
4
3
 2α3,1

α3,2
α3,3
α3,m
α3,m+1 
x
x
x
·
·
·
x
x
3
4
5
1
2

,
 ..
..
..
..
.. 
 .
.
.
.
. 
αm,1
αm,2
αm,3
αm,m+1
αm,m
xm
xm+1 x1
· · · xm−2 xm−1
1

where x1 , x2 , x3 , . . . , xm , xm+1 is an sop and {αi,j }1≤i≤m,1≤j≤m+1 is a family of positive
integers.
Throughout this paper, R is a commutative ring, and in the last section, we assume
that R is an n-dimensional Cohen-Macaulay local ring. For R-modules G and H, the
elements of G ⊕ H are denoted by column vectors;
 
g
(g ∈ G, h ∈ H).
h
In particular, the elements of the forms
 
 
0
g
and
h
0
are denoted by [g] and hhi, respectively. Moreover, if V is a subset of G, then the family
{[v]}v∈V is denoted by [V ]. Similarly hW i is defined for a subset W of H. If T is a subset
of an R-module, we denote by R · T the R-submodule generated by T . If S is a finite set,
♯ S denotes the number of elements of S.

2

Preliminaries

In this section, we summarize preliminary results. Let R be a commutative ring.
Lemma 2.1. Let G• and F• be acyclic complexes, whose boundary maps are denoted
by ∂• and ϕ• , respectively. Suppose that a chain map σ• : G• −→ F• is given and
σ0−1 (Im ϕ1 ) = Im ∂1 holds. Then the mapping cone Cone(σ• ) :
Gp−2
G1
G0
Gp−1
ψp
ψ2
ψ1
· · · −→ ⊕ −→ ⊕ −→ · · · −→ ⊕ −→ ⊕ −→ F0 −→ 0
Fp−1
F2
F1
Fp
is acyclic, where
ψp =



∂p−1
0
p−1
(−1)
· σp−1 ϕp




for ∀p ≥ 2 and ψ1 = σ0 ϕ1 .

Hence, if G• and F• are complexes of finitely generated free R-modules, then Cone(σ• )
gives an R-free resolution of Im ϕ1 + Im σ0 .
Proof. See [1, 2.1].
Lemma 2.2. Let 2 ≤ n ∈ Z and C•• be a double complex such that Cp,q = 0 unless
0 ≤ p, q ≤ n. For any p, q ∈ Z, we denote the boundary maps Cp,q −→ Cp−1,q and
Cp,q −→ Cp,q−1 by d′p,q and d′′p,q , respectively. We assume that Cp• and C•q are acyclic for
0 ≤ p, q ≤ n. Let T• be the total complex of C•• and let d• be its boundary map, that is,
if ξ ∈ Cp,q ⊆ Tr (p + q = r), then
dr (ξ) = (−1)p · d′′p,q (ξ) + d′p,q (ξ) ∈ Cp,q−1 ⊕ Cp−1,q ⊆ Tr−1 .
Then the following assertions hold.
2

(1) Suppose that ξn ∈ Cn,0 and ξn−1 ∈ Cn−1,1 such that d′n,0 (ξn ) = (−1)n · d′′n−1,1 (ξn−1 )
are given. Then there exist elements ξp ∈ Cp,n−p for 0 ≤ ∀p ≤ n − 2 such that
ξn + ξn−1 + ξn−2 + · · · + ξ0 ∈ Ker dn
⊆ Tn = Cn,0 ⊕ Cn−1,1 ⊕ Cn−2,2 ⊕ · · · ⊕ C0,n .
(2) Suppose that ξn +ξn−1 +· · ·+ξ1 +ξ0 ∈ Ker dn ⊆ Tn = Cn,0 ⊕Cn−1,1 ⊕· · ·⊕C1,n−1 ⊕C0,n
and ξ0 ∈ Im d′1,n . Then
ξn + ξn−1 + · · · + ξ1 + ξ0 ∈ Im dn+1 .
In particular, we have ξn ∈ Im d′′n,1.
Proof. (1) It is enough to show that if 1 ≤ p ≤ n − 1 and two elements ξp+1 ∈ Cp+1,n−p−1,
ξp ∈ Cp,n−p such that
d′p+1,n−p−1(ξp+1 ) = (−1)p+1 · d′′p,n−p(ξp )
are given, then we can take ξp−1 ∈ Cp−1,n−p+1 so that
d′p,n−p(ξp ) = (−1)p · d′′p−1,n−p+1(ξp−1).
In fact, if the assumption of the claim stated above is satisfied, we have
d′′p−1,n−p (d′p,n−p(ξp )) = d′p,n−p−1(d′′p,n−p(ξp ))
= d′p,n−p−1((−1)p+1 · d′p+1,n−p−1(ξp+1))
= 0,
and so
d′p,n−p(ξp ) ∈ Ker d′′p−1,n−p = Im d′′p−1,n−p+1,
which means the existence of the required element ξp−1 .
(2) We set η0 = 0. By the assumption, there exists η1 ∈ C1,n such that
ξ0 = d′1,n (η1 ) = d′1,n (η1 ) + d′′0,n+1 (η0 ).
Here we assume 0 ≤ p ≤ n − 1 and two elements ηp ∈ Cp,n−p+1, ηp+1 ∈ Cp+1,n−p such that
ξp = d′p+1,n−p (ηp+1 ) + (−1)p · d′′p,n−p+1(ηp )
are fixed. We would like to find ηp+2 ∈ Cp+2,n−p−1 such that
ξp+1 = d′p+2,n−p−1(ηp+2 ) + (−1)p+1 · d′′p+1,n−p (ηp+1).
In fact, as we have
d′p+1,n−p−1(ξp+1 + (−1)p · d′′p+1,n−p (ηp+1))
= d′p+1,n−p−1(ξp+1) + (−1)p · d′p+1,n−p−1(d′′p+1,n−p(ηp+1 ))
= (−1)p+1 · d′′p,n−p(ξp ) + (−1)p · d′′p,n−p(d′p+1,n−p(ηp+1 ))
= (−1)p+1 · d′′p,n−p(ξp − d′p+1,n−p (ηp+1 ))
= (−1)p+1 · d′′p,n−p((−1)p · d′′p,n−p+1(ηp ))
= 0,
3

it follows that
ξp+1 + (−1)p · d′′p+1,n−p (ηp+1 ) ∈ Ker d′p+1,n−p−1 = Im d′p+2,n−p−1.
Thus we see the existence of the required element ηp+2 .
Lemma 2.3. Suppose that
ϕ

ψ

ρ

0 −→ F −→ G −→ H −→ L
is an exact sequence of R-modules. Then the following assertions hold.
(1) If there exists a homomorphism φ : G −→ F of R-modules such that φ ◦ ϕ = idF ,
then
∗ψ
ρ
0 −→ ∗G −→ H −→ L
is exact, where ∗G = Ker φ and ∗ψ is the restriction of ψ to ∗G.
(2) If F = ′F ⊕ ∗F , G = ′G ⊕ ∗G, ϕ(′F ) = ′G and ϕ(∗F ) ⊆ ∗G, then
∗ϕ

∗ψ

ρ

0 −→ ∗F −→ ∗G −→ H −→ L
is exact, where ∗ϕ and ∗ψ are the restrictions of ϕ and ψ to ∗F and ∗G, respectively.
Proof. See [1, 2.3].

3

∗-transform

Let 2 ≤ n ∈ Z and let R be an n-dimensional Cohen-Macaulay local ring with the maximal
ideal m. Suppose that an acyclic complex
ϕn

ϕ1

0 −→ Fn −→ Fn−1 −→ · · · −→ F1 −→ F0
of finitely generated free R-modules such that Im ϕn ⊆ QFn−1 is given, where Q =
(x1 , x2 , . . . , xn )R is a parameter ideal of R. We put M = Im ϕ1 , which is an R-submodule
of F0 . In this section, transforming F• suitably, we aim to construct an acyclic complex
∗ϕ

∗ϕ

0 −→ ∗Fn −→ ∗Fn−1 −→ · · · −→ ∗F1 −→ ∗F0 = F0
n

1

of finitely generated free R-modules such that Im ∗ϕn ⊆ m · ∗Fn−1 and Im ∗ϕ1 = M :F0 Q.
Let us call ∗F• the ∗-transform of F• with respect to x1 , x2 , . . . , xn .
In this operation, we use the Koszul complex K• = K• (x1 , x2 , . . . , xn ). We denote
the boundary map of K• by ∂• . Let e1 , e2 , . . . , en be an R-free basis of K1 such that
∂1 (ei ) = xi for 1 ≤ ∀i ≤ n. Moreover, we use the following notation:
• N := {1, 2, . . . , n}.
• Np := {I ⊆ N | ♯ I = p} for 1 ≤ ∀p ≤ n and N0 := {∅}.
4

• If 1 ≤ p ≤ n and I = {i1 , i2 , . . . , ip } ∈ Np , where 1 ≤ i1 < i2 < · · · < ip ≤ n, we set
eI = ei1 ∧ ei2 ∧ · · · ∧ eip ∈ Kp .
In particular, for 1 ≤ ∀i ≤ n, ěi := eN \{i} . Furthermore, e∅ denotes the identity
element 1R of R = K0 .
• If 1 ≤ p ≤ n, I ∈ Np and i ∈ I, we set
s(i, I) = ♯ {j ∈ I | j < i}.
We define ♯ ∅ = 0, so s(i, I) = 0 if i = min I.
Then, for 0 ≤ ∀p ≤ n, {eI }I∈Np is an R-free basis of Kp and
X
∂p (eI ) =
(−1)s(i,I) · xi · eI\{i} .
i∈I

As x1 , x2 , . . . , xn is an R-regular sequence, K• gives an R-free resolution of R/Q. Hence,
for any R-module N, we have the following commutative diagram;
HomR (∂n ,N )

−→

HomR (K
 n−1 , N)
y∼
=
N ⊕n

n
HomR (K
 n , N) −→ ExtR (R/Q, N) −→ 0 (ex)
y∼
=

(x1 −x2 ··· (−1)n−1 ·xn )

−→
N
−→
N/QN
−→ 0 (ex),
which implies ExtnR (R/Q, N) ∼
= N/QN. Using this fact, we show the next result.
Theorem 3.1. (M :F Q)/M ∼
= Fn /QFn .
0

Proof. We put L0 = F0 /M. Moreover, for 1 ≤ ∀p ≤ n − 1, we put Lp = Im ϕp ⊆ Fp−1
and consider the exact sequence
ϕp−1

0 −→ Lp −→ Fp−1 −→ Lp−1 −→ 0,
where ϕ0 : F0 −→ L0 is the canonical surjection. Because
p
Extp−1
R (R/Q, Fp−1 ) = ExtR (R/Q, Fp−1 ) = 0,

we get

p−1
ExtpR (R/Q, Lp ) ∼
= ExtR (R/Q, Lp−1 ).
n−1
Therefore ExtR
(R/Q, Ln−1 ) ∼
= HomR (R/Q, F0 /M) ∼
= (M :F0 Q)/M. Next, we look at
the exact sequence
ϕn−1
ϕn
0 −→ Fn −→ Fn−1 −→ Ln−1 −→ 0,

which yields the following commutative diagram;
ϕ
f
n

n−1
n
0 −→ ExtR
(R/Q, Ln−1 ) −→ ExtnR (R/Q,
 Fn ) −→ ExtR (R/Q,
 Fn−1 ) (ex)
y∼
y∼
=
=

Fn /QFn

ϕn

−→

Fn−1 /QFn−1 ,

where ϕ
fn and ϕn denote the maps induced from ϕn . Let us notice ϕn = 0 as Im ϕn ⊆
QFn−1 . Hence
n−1
ExtR
(R/Q, Ln−1 ) ∼
= Fn /QFn ,
and so the required isomorphism follows.
5

e = Λ × N and take a family
Let us fix an R-free basis of Fn , say {vλ }λ∈Λ . We set Λ
{v(λ,i) }(λ,i)∈Λe of elements in Fn−1 so that
ϕn (vλ ) =

X

xi · v(λ,i)

i∈N

for ∀λ ∈ Λ. This is possible as Im ϕn ⊆ QFn−1 . The next result is the essential part of
the process to get ∗F• .
Theorem 3.2. There exists a chain map σ• : Fn ⊗R K• −→ F•
F ⊗∂

F ⊗∂

n
n
n
1
0 −→ Fn ⊗R Kn −→
Fn ⊗RKn−1 −→ · · · −→ Fn ⊗R K1 −→
Fn ⊗R K0
yσn−1
yσ1
yσ0
yσn

0 −→

Fn

ϕn

−→

−→ · · · −→

Fn−1

F1

ϕ1

−→

F0

satisfying the following conditions.
(1) σ0−1 (Im ϕ1 ) = Im(Fn ⊗ ∂1 ).
(2) Im σ0 + Im ϕ1 = M :F0 Q.
e
(3) σn−1 (vλ ⊗ ěi ) = (−1)n+i−1 · v(λ,i) for ∀(λ, i) ∈ Λ.

(4) σn (vλ ⊗ eN ) = (−1)n · vλ for ∀λ ∈ Λ.

Proof. Let us notice that, for 0 ≤ ∀p ≤ n, {vλ ⊗eI }(λ,I)∈Λ×Np is an R-free basis of Fn ⊗R Kp ,
so σp : Fn ⊗R Kp −→ Fp can be defined by choosing suitable element w(λ,I) ∈ Fp that
corresponds to vλ ⊗ eI for ∀(λ, I) ∈ Λ × Np . We set w(λ,N ) = (−1)n · vλ for ∀λ ∈ Λ and
e Then
w(λ,N \{i}) = (−1)n+i−1 · v(λ,i) for ∀(λ, i) ∈ Λ.
ϕn (w(λ,N ) ) = (−1)n · ϕn (vλ )
X
= (−1)n ·
xi · v(λ,i)
=

X

i∈N

(−1)s(i,N ) · xi · w(λ,N \{i}) .

i∈N

Moreover, we can take families {w(λ,I) }(λ,I)∈Λ×Np of elements in Fp for 0 ≤ ∀p ≤ n − 2 so
that
X
ϕp (w(λ,I) ) =
(−1)s(i,I) · xi · w(λ,I\{i})
(♯)
i∈I

for 1 ≤ ∀p ≤ n and ∀(λ, I) ∈ Λ ×Np . If this is true, an R-linear map σp : Fn ⊗R Kp −→ Fp
is defined by setting σp (vλ ⊗ eI ) = w(λ,I) for ∀(λ, I) ∈ Λ × Np and σ• : Fn ⊗R K• −→ F•
becomes a chain map satisfying (3) and (4).

6

In order to see the existence of {w(λ,I) }(λ,I)∈Λ×Np ,
F• ⊗R K• .
..
.
y
· · · −→ Fp ⊗R Kq
yFp ⊗∂q

· · · −→ Fp ⊗RKq−1
y
..
.

ϕp ⊗Kq

−→

ϕp ⊗Kq−1

−→

let us consider the double complex
..
.
y

Fp−1 
⊗R Kq −→ · · ·
yFp−1 ⊗∂q

Fp−1 ⊗R Kq−1 −→ · · ·
y
..
.

We can take it as C•• of 2.2. Let T• be the total complex and d• be its boundary map.
In particular, we have
Tn = (Fn ⊗R K0 ) ⊕ (Fn−1 ⊗R K1 ) ⊕ · · · ⊕ (F1 ⊗R Kn−1 ) ⊕ (F0 ⊗R Kn ).
For ∀I ⊆ N, we define
t(I) =

For a while, we fix λ ∈ Λ and set
ξn (λ) = (−1)
ξn−1 (λ) = (−1)

n(n+1)
2
(n−1)n
2

X

(i − 1) if I 6= ∅,


 i∈I



0

if I = ∅.

· (−1)t(N ) · w(λ,N ) ⊗ e∅ ∈ Fn ⊗R K0 ,
X
·
(−1)t(N \{i}) · w(λ,N \{i}) ⊗ ei ∈ Fn−1 ⊗R K1 .
i∈N

It is easy to see that
ξn (λ) = vλ ⊗ e∅
since t(N) = (n − 1)n/2 and n2 + n ≡ 0 (mod 2). Moreover, we have
X
ξn−1 (λ) = (−1)n ·
v(λ,i) ⊗ ei
i∈N

since t(N \ {i}) = (n − 1)n/2 − (i − 1). Then
(ϕn ⊗ K0 )(ξn (λ)) = ϕn (vλ ) ⊗ e∅
X
=(
xi · v(λ,i) ) ⊗ e∅
i∈N

=

X

v(λ,i) ⊗ xi

i∈N

= (Fn−1 ⊗ ∂1 )(

X

v(λ,i) ⊗ ei )

i∈N

= (−1)n · (Fn−1 ⊗ ∂1 )(ξn−1 (λ)).
7

Hence, by (1) of 2.2 there exist elements ξp (λ) ∈ Fp ⊗ Kn−p for 0 ≤ ∀p ≤ n − 2 such that
ξn (λ) + ξn−1 (λ) + ξn−2 (λ) + · · · + ξ0 (λ) ∈ Ker dn ⊆ Tn ,
which means
(ϕp ⊗ Kn−p )(ξp (λ)) = (−1)p · (Fp−1 ⊗ ∂n−p+1 )(ξp−1(λ))
for 1 ≤ ∀p ≤ n. Let us denote N \ I by I c for ∀I ⊆ N. Because {eI c }I∈Np is an R-free
basis of Kn−p , it is possible to write
ξp (λ) = (−1)

p(p+1)
2

·

X

(−1)t(I) · w(λ,I) ⊗ eI c

I∈Np

for 0 ≤ ∀p ≤ n − 2 (Notice that ξn (λ) and ξn−1 (λ) are defined so that they satisfy the
same equalities), where w(λ,I) ∈ Fp . Then we have
(ϕp ⊗ Kn−p )(ξp (λ)) = (−1)

p(p+1)
2

·

X

(−1)t(I) · ϕp (w(λ,I) ) ⊗ eI c .

I∈Np

On the other hand,
(−1)p · (Fp−1 ⊗ ∂n−p+1 )(ξp−1(λ))
= (−1)p · (−1)

(p−1)p
2

X

·

{(−1)t(J) · w(λ,J) ⊗ (

X

c

(−1)s(i,J ) · xi · eJ c \{i} )}.

i∈J c

J∈Np−1

Here we notice that if I ∈ Np , J ∈ Np−1 and i ∈ N, then
I c = J c \ {i}

⇐⇒

I = J ∪ {i}.

Hence we get
(−1)p · (Fp−1 ⊗ ∂n−p+1 )(ξp−1(λ))
= (−1)

p(p+1)
2

·

X

I∈Np

{(

X

(−1)t(I\{i})+s(i,I

c ∪{i})

· xi · w(λ,I\{i}) ) ⊗ eI c }.

i∈I

For ∀I ∈ Np and ∀i ∈ I, we have
t(I \ {i}) = t(I) − (i − 1),
s(i, I) + s(i, I c ∪ {i}) = s(i, N) = i − 1,
and so
t(I \ {i}) + s(i, I c ∪ {i}) = t(I) − s(i, I)
≡ t(I) + s(i, I) (mod 2).
Therefore we see that the required equality (♯) holds for ∀I ∈ Np .
8

Let us prove (1). We have to show σ0−1 (Im ϕ1 ) ⊆ Im(Fn ⊗ ∂1 ). Take ∀ηn ∈ Fn ⊗R K0
such that σ0 (ηn ) ∈ Im ϕ1 . As {ξn (λ)}λ∈Λ is an R-free basis of Fn ⊗R K0 , we can express
X
X
ηn =
aλ · ξn (λ) =
aλ · (vλ ⊗ e∅ ),
λ∈Λ

λ∈Λ

where aλ ∈ R for ∀λ ∈ Λ. Then we have
X
X
aλ · σ0 (vλ ⊗ e∅ ) = σ0 (ηn ) ∈ Im ϕ1 .
aλ · w(λ,∅) =
λ∈Λ

λ∈Λ

Now we set
ηp =

X

aλ · ξp (λ) ∈ Fp ⊗R Kn−p

λ∈Λ

for 0 ≤ ∀p ≤ n − 1. Then
ηn + ηn−1 + · · · + η1 + η0 =

X

aλ · (ξn (λ) + ξn−1 (λ) + · · · + ξ1 (λ) + ξ0 (λ))

λ∈Λ

∈ Ker dn ⊆ Tn .
Because
η0 =

X

aλ · ξ0 (λ)

λ∈Λ

=

X

aλ · (w(λ,∅) ⊗ eN )

λ∈Λ

=(

X

aλ · w(λ,∅) ) ⊗ eN

λ∈Λ

∈ Im(ϕ1 ⊗ Kn ),
we get ηn ∈ Im(Fn ⊗ ∂1 ) by (2) of 2.2.
Finally we prove (2). Let us consider the following commutative diagram
F ⊗∂

n
1
Fn ⊗R K1 −→
Fn ⊗R K0 −→ Fn /QF
 n −→ 0 (ex)
yσ1
yσ0
yσ0

F1

ϕ1

−→

−→ F0 /M

F0

−→ 0 (ex),

where σ0 is the map induced from σ0 . For ∀λ ∈ Λ and ∀i ∈ N, we have
xi · w(λ,∅) = ϕ1 (w(λ,{i}) ) ∈ M,
which means w(λ,∅) ∈ M :F0 Q. Hence Im σ0 ⊆ M :F0 Q, and so Im σ0 ⊆ (M :F0 Q)/M.
On the other hand, as σ0−1 (Im ϕ1 ) = Im(Fn ⊗ ∂1 ), we see that σ0 is injective. Therefore
we get Im σ0 = (M :F0 Q)/M since (M :F0 Q)/M ∼
= Fn /QFn by 3.1 and Fn /QFn has a
finite length. Thus the assertion (2) follows and the proof is complete.

9

In the rest, σ• : Fn ⊗R K• −→ F• is the chain map constructed in 3.2. Then, by 2.1
the mapping cone Cone(σ• ) gives an R-free resolution of M :F0 Q, that is,
Fn ⊗R Kn−1
Fn ⊗R Kn−2 ′
Fn ⊗R Kn−3
ϕn−1
ψn
⊕
⊕
⊕
0 −→ Fn ⊗R Kn −→
−→
−→
Fn
Fn−1
Fn−2
ψn+1

Fn ⊗R Kn−4
Fn ⊗R K1 ∗ Fn ⊗R K0 ∗
ϕ1
ϕ2
−→ F0
−→
⊕
−→
⊕
−→ · · · −→
⊕
F1
Fn−3
F2

∗ϕ

n−2

is acyclic and Im ∗ϕ1 = M :F0 Q, where






Fn ⊗ ∂n
Fn ⊗ ∂n−1
0
Fn ⊗ ∂n−2
0
′
ψn+1 =
, ψn =
, ϕn−1 =
,
(−1)n · σn
(−1)n−1 · σn−1 ϕn
(−1)n−2 · σn−2 ϕn−1



Fn ⊗ ∂p−1
0
∗
ϕp =
for 2 ≤ ∀p ≤ n − 2 and ∗ϕ1 = σ0 ϕ1 .
p−1
(−1)
· σp−1 ϕp
Because σn : Fn ⊗R Kn −→ Fn is an isomorphism, we can define
n

φ = 0 (−1) ·

σn−1

 Fn ⊗R Kn−1
⊕
:
−→ Fn ⊗R Kn .
Fn

Then φ ◦ ψn+1 = idFn ⊗R Kn and Ker φ = Fn ⊗R Kn−1 . Hence, by (1) of 2.3, we get the
acyclic complex
′ϕ

′ϕ

∗ϕ

∗ϕ

∗ϕ

0 −→ ′Fn −→ ′Fn−1 −→ ∗Fn−2 −→ ∗Fn−3 −→ · · · −→ ∗F2 −→ ∗F1 −→ ∗F0 = F0 ,
n

n−1

n−2

2

1

where
Fn ⊗R Kn−2
Fn ⊗R Kp−1
∗
⊕
⊕
Fn = Fn ⊗R Kn−1 , Fn−1 =
, Fp =
for 1 ≤ ∀p ≤ n − 2
Fn−1
Fp


Fn ⊗ ∂n−1
′
.
and ϕn =
(−1)n−1 · σn−1

′

′

Although Im ′ϕn may not be contained in m · ′Fn−1 , removing non-minimal components
from ′Fn and ′Fn−1 , we get free R-modules ∗Fn and ∗Fn−1 such that
∗ϕ

∗ϕ

∗ϕ

0 −→ ∗Fn −→ ∗Fn−1 −→ ∗Fn−2 −→ · · · −→ ∗F1 −→ ∗F0 = F0
n

n−1

1

is acyclic and Im ∗ϕn ⊆ m · ∗Fn−1 , where ∗ϕn and ∗ϕn−1 are the restrictions of ′ϕn and ′ϕn−1 ,
respectively. In the rest of this section, we describe a concrete procedure to get ∗Fn and
∗
Fn−1 . For that purpose, we use the following notation. As described in Introduction, for
any ξ ∈ Fn ⊗R Kn−2 and η ∈ Fn−1 ,
 
 
ξ
0
′
[ξ] :=
∈ Fn−1 and hηi :=
∈ ′Fn−1 .
0
η
10

In particular, for any (λ, I) ∈ Λ × Nn−2 , we denote [vλ ⊗ eI ] by [λ, I]. Moreover, for a
subset U of Fn−1 , hUi := {hui}u∈U .
e and a subset U of Fn−1 so that
Now, let us choose a subset ′Λ of Λ
{v(λ,i) }(λ,i)∈′Λ ∪ U

is an R-free basis of Fn−1 . We would like to choose ′Λ as big as possible. The following
almost obvious fact is useful to find ′Λ and U.
e and a subset U of V
Lemma 3.3. Let V be an R-free basis of Fn−1 . If a subset ′Λ of Λ
satisfy
(i) ♯ ′Λ + ♯ U ≤ ♯ V , and
(ii) V ⊆ R · {v(λ,i) }(λ,i)∈′Λ + R · U + mFn−1 ,
then {v(λ,i) }(λ,i)∈′Λ ∪ U is an R-free basis of Fn−1 .
Let us notice that
{[λ, I]}(λ,I)∈Λ×Nn−2 ∪ {hv(λ,i) i}(λ,i)∈′Λ ∪ hUi
is an R-free basis of ′Fn−1 . We define ∗Fn−1 to be the direct summand of ′Fn−1 generated
by
{[λ, I]}(λ,I)∈Λ×Nn−2 ∪ hUi.
Let ∗ϕn−1 be the restriction of ′ϕn−1 to ∗Fn−1 .
e itself as ′Λ, then
Theorem 3.4. If we can take Λ
∗ϕ

∗ϕ

0 −→ ∗Fn−1 −→ ∗Fn−2 −→ · · · −→ ∗F1 −→ ∗F0 = F0
n−1

1

is acyclic. Hence we have depthR F0 /(M :F0 Q) > 0.
e there exists a homomorphism φ : ′Fn−1 −→ ′Fn such that
Proof. If ′Λ = Λ,
φ([λ, I]) = 0 for any (λ, I) ∈ Λ × Nn−2 ,

e
φ(hv(λ,i) i) = (−1)i · vλ ⊗ ěi for any (λ, i) ∈ Λ,
φ(hui) = 0 for any u ∈ U.
Then φ ◦ ′ϕn = id′Fn and Ker φ = ∗Fn−1 . Hence, by (1) of 2.3 we get the required
assertion.
e and put ∗Λ = Λ
e \ ′Λ. Then, for any
In the rest of this section, we assume ′Λ ( Λ
(µ, j) ∈ ∗Λ, it is possible to write
X (µ,j)
X
v(µ,j) =
a(λ,i) · v(λ,i) +
b(µ,j)
· u,
u
u∈U

(λ,i)∈′Λ

11

(µ,j)

(µ,j)

where a(λ,i) , bu

(µ,j)

∈ R. Here, if ′Λ is big enough, we can choose every bu

(µ,j)
bu

′

from m. In

′

fact, if
∈
/ m for some u ∈ U, then we can replace Λ and U by Λ ∪ {(µ, j)} and
U \ {u}, respectively. Furthermore, because of a practical reason, let us allow that some
terms of v(λ,i) for (λ, i) ∈ ∗Λ with non-unit coefficients appear in the right hand side, that
is, for any (µ, j) ∈ ∗Λ, we write
X (µ,j)
X
v(µ,j) =
a(λ,i) · v(λ,i) +
b(µ,j)
· u,
u
e
(λ,i)∈Λ

where
(µ,j)

a(λ,i) ∈



R



u∈U

if (λ, i) ∈ ′Λ,
and b(µ,j)
∈ m.
u

m if (λ, i) ∈ ∗Λ

Using this expression, for any (µ, j) ∈ ∗Λ, the following element in ′Fn can be defined.
X
(µ,j)
∗
v(µ,j) := (−1)j · vµ ⊗ ěj +
(−1)i−1 · a(λ,i) · vλ ⊗ ěi .
e
(λ,i)∈Λ

Lemma 3.5. For any (µ, j) ∈ ∗Λ, we have
X
X
(µ,j)
′
ϕn (∗v(µ,j) ) = (−1)j · [vµ ⊗ ∂n−1 (ěj )] +
(−1)i−1 · a(λ,i) · [vλ ⊗ ∂n−1 (ěi )] +
b(µ,j)
· hui.
u
e
(λ,i)∈Λ

u∈U

As a consequence, we have ′ϕn (∗v(µ,j) ) ∈ m · ∗Fn−1 for any (µ, j) ∈ ∗Λ.
Proof. By the definition of ′ϕn , for any (µ, j) ∈ ∗Λ, we have
′

ϕn (∗v(µ,j) ) = [(Fn ⊗ ∂n−1 )(∗v(µ,j) )] + h(−1)n−1 · σn−1 (∗v(µ,j) )i.

Because
(Fn ⊗ ∂n−1 )(∗v(µ,j) ) = (−1)j · vµ ⊗ ∂n−1 (ěj ) +

X

(µ,j)

(−1)i−1 · a(λ,i) · vλ ⊗ ∂n−1 (ěi )

e
(λ,i)∈Λ

and
X

σn−1 (∗v(µ,j) ) = (−1)j · σn−1 (vµ ⊗ ěj ) +

(µ,j)

(−1)i−1 · a(λ,i) · σn−1 (vλ ⊗ ěi )

e
(λ,i)∈Λ

= (−1)n−1 · v(µ,j) + (−1)n ·
= (−1)n−1 · (v(µ,j) −
= (−1)n−1 ·

X

X

e
(λ,i)∈Λ

· u,
b(µ,j)
u

u∈U

we get the required equality.
12

X

(µ,j)

a(λ,i) · v(λ,i)

e
(λ,i)∈Λ
(µ,j)

a(λ,i) · v(λ,i) )

Let ∗Fn be the R-submodule of ′Fn generated by {∗v(µ,j) }(µ,j)∈∗Λ and let ∗ϕn be the
restriction of ′ϕn to ∗Fn . By 3.5 we have Im ∗ϕn ⊆ ∗Fn−1 . Thus we get a complex
∗ϕ

∗ϕ

0 −→ ∗Fn −→ ∗Fn−1 −→ · · · −→ ∗F1 −→ ∗F0 = F0 .
n

1

This is the complex we desire. In fact, the following result holds.
Theorem 3.6. (∗F• , ∗ϕ• ) is an acyclic complex of finitely generated free R-modules with
the following properties.
(1) Im ∗ϕ1 = M :F0 Q and Im ∗ϕn ⊆ m · ∗Fn−1 .
(2) {∗v(µ,j) }(µ,j)∈∗Λ is an R-free basis of ∗Fn .
(3) {[λ, I]}(λ,I)∈Λ×Nn−2 ∪ hUi is an R-free basis of ∗Fn−1 .
Proof. First, let us notice that {vλ ⊗ ěi }(λ,i)∈Λe is an R-free basis of ′Fn and
vµ ⊗ ěj ∈ R · ∗v(µ,j) + R · {vλ ⊗ ěi }(λ,i)∈′Λ + m · ′Fn
for any (µ, j) ∈ ∗Λ. Hence, by Nakayama’s lemma it follows that ′Fn is generated by
{vλ ⊗ ěi }(λ,i)∈′Λ ∪ {∗v(µ,j) }(µ,j)∈∗Λ ,
e = ♯ ′Λ + ♯ ∗Λ. Let ′′Fn be the Rwhich must be an R-free basis since rankR ′Fn = ♯ Λ
′
submodule of Fn generated by {vλ ⊗ ěi }(λ,i)∈′Λ . Then ′Fn = ′′Fn ⊕ ∗Fn .
Next, let us recall that
{[λ, I]}(λ,I)∈Λ×Nn−2 ∪ {hv(λ,i) i}(λ,i)∈′Λ ∪ hUi
is an R-free basis of ′Fn−1 . Because
′

ϕn (vλ ⊗ ěi ) = [vλ ⊗ ∂n−1 (ěi )] + (−1)i · hv(λ,i) i,

we see that
{[λ, I]}(λ,I)∈Λ×Nn−2 ∪ {′ϕn (vλ ⊗ ěi )}(λ,i)∈′Λ ∪ hUi
is also an R-free basis. Let ′′Fn−1 = R · {′ϕn (vλ ⊗ ěi )}(λ,i)∈′Λ . Then ′Fn−1 = ′′Fn−1 ⊕ ∗Fn−1 .
It is obvious that ′ϕn (′′Fn ) = ′′Fn−1 . Moreover, by 3.5 we get ′ϕn (∗Fn ) ⊆ ∗Fn−1 . Therefore, by (2) of 2.3, it follows that ∗F• is acyclic. We have already seen (3) and the first
assertion of (1). The second assertion of (1) follows from 3.5. Moreover, the assertion (2)
is now obvious.

References
[1] K. Fukumuro, T. Inagawa and K. Nishida, On a transform of an acyclic complex of
length 3, J. Algebra, 384 (2013), 84–109.
[2] K. Fukumuro, T. Inagawa and K. Nishida, Saturations of powers of certain determinantal ideals, Preprint (2013).

13

