On stable degenerations of Cohen-Macaulay modules over simple singularities of type ($A_n $)

0.23 MB·English

by  Naoya Hiramatsu

ON STABLE DEGENERATIONS OF COHEN-MACAULAY MODULES OVER
SIMPLE SINGULARITIES OF TYPE (A )
n
5
NAOYAHIRAMATSU
1
0
Dedicated toProfessor Yuji Yoshino on the occasion of his sixtiethbirthday.
2

y
Abstract. WestudythestabledegenerationproblemforCohen-Macaulaymodulesoversimple
a
M singularities of type (An). We prove that the stable hom order is actually a partial order over
theringandareabletoshowthatthestabledegenerationscanbecontrolledbythestablehom
order.
2
1

]
C
1. Introduction
A
The concept of degenerations of modules introduced in representation theory for studying the
.
h structureofthemodulevarietyoverafinitedimensionalalgebra[1,9,12,13,15,16,17]. Classically
t
a Bongartz[1]investigatedthe degenerationproblemofmodules overanartinianalgebrainrelation
m with the Auslander-Reiten quiver. In [16], Zwara gave a complete description of degenerations of
[ modules over representation finite algebras by using some order relations for modules known as
the hom order, the degeneration order and the extension order. Now a theory of degenerations is
3
consideredfornotonlymodulecategories,butderivedcategories[7]orstablecategories[14],more
v
generally, triangulated categories [10].
7
2 Let R be a commutative Gorenstein local k-algebra which is not necessary finite dimensional.
0 Yoshino [14] introduced a notion of the stable analogue of degenerations of (maximal) Cohen-
5 Macaulay R-module in the stable category CM(R). The notion of the stable degenerations is
0
closely related to the ordinary degenerations. In fact, the author [4] give a complete description
.
1 of degenerationsof Cohen-Macaulaymodules overa ring ofeven dimensionalsimple singularity of
0
type (A ) by using the description of stable degenerations over it. Hence it is also important for
n
5
the study of degeneration problem to investigate the description of stable degenerations.
1
The purpose of this paper is to describe stable degenerations of Cohen-Macaulay modules over
:
v
simple singularities of type (A ).
i n
X
k[[x ,x ,x , ,x ]]/(xn+1+x2+x2+ +x2).
r 0 1 2 ··· d 0 1 2 ··· d
a First we compare Auslander-Reiten theory on CM(R) with that on CM(R). We look into the
relation between AR sequences and AR triangles of Cohen-Macaulay modules (Proposition 2.2).
And we consider an order relation on CM(R) which is the stable analogue of the hom order
(Definition 2.6). We shallshowthatitisactuallyapartialorderifaringis offinite representation
type with certain assumptions.
In Section 4, we devote to describe stable degenerations of Cohen-Macaulay modules over the
ring above. We shall show that all stable degenerations can be controlled by the stable hom order
(Theorem 4.6). To show this, we use the stable analogue of the argument over finite dimensional
algebras in [16].
Date:May13,2015.
2000 Mathematics Subject Classification. Primary16W50;Secondary13D10.
Keywords and phrases. stabledegeneration, Cohen-Macaulaymodule,stablecategory.
1
2 NAOYAHIRAMATSU
2. Stable hom order on Cohen-Macaulay modules
Throughout the paper R is a Henselian Gorenstein local ring that is k-algebra where k is an
algebraically closed field of characteristic 0. For a finitely generated R-module M, we say that M
is a Cohen-Macaulay R-module if
Exti (M,R)=0 for any i>0.
R
We denote by CM(R) the category of Cohen-Macaulay R-modules with all R-homomorphisms.
AndwealsodenotebyCM(R)thestablecategoryofCM(R). TheobjectsofCM(R)arethesameas
thoseofCM(R),andthemorphismsofCM(R)areelementsofHom (M,N)=Hom (M,N)/P(M,N)
R R
for M,N CM(R), where P(M,N) denote the set of morphisms from M to N factoring through
∈
free R-modules. For a Cohen-Macaulay module M we denote it by M to indicate that it is an
object of CM(R). For a finitely generated R-module M, take a free resolution
d
F F M 0.
1 0
···→ −→ → →
WedenoteImdbyΩM. Wenotethatthisdefinesthefunctorgivinganauto-equivalenceofCM(R).
It is known that CM(R) has a structure of a triangulated category with the shift functor defined
by the functor Ω−1. We recommend the reader to [3, Chapter 1], [11, Section 4] for the detail.
Since R is Gorenstein, by the definition of a triangle, L M N L[1] is a triangle in CM(R)
if and only if there is an exact sequence 0 L M′→ N→ 0→in CM(R) with M′ = M in
→ → → → ∼
CM(R), that is, M′ is isomorphic to M up to free summand. Since R is Henselian, CM(R), hence
CM(R), is a Krull-Schmidt category,namely eachobject can be decomposed into indecomposable
objects up to isomorphism uniquely.
In the paper we use the theory of Auslander-Reiten (abbr. AR) sequences and triangles of
Cohen-Macaulaymodules. Letus recallthe definitions ofthosenotions. See [11]forARsequences
and [3, 8] for AR triangles.
Definition 2.1. Let X, Y and Z be Cohen-Macaulay R-modules.
f
(1) A short exact sequence Σ :0 Z Y X 0 is said to be an AR sequence ending
X
→ → −→ →
in X (or starting from Z) if it satisfies
(AR1) X and Z are indecomposable.
(AR2) Σ is not split.
X
(AR3) If g : W X is not a split epimorphism, then there exists h : W Y such that
→ →
g =f h.
◦
f w
(2) We also say that a triangle Σ : Z Y X Z[1] is an AR triangle ending in X (or
X
→ −→ −→
starting from Z) if it satisfies
(ART1) X and Z are indecomposable.
(ART2) w =0.
6
(ART3) If g : W X is not a split epimorphism, then there exists h : W Y such that
→ →
g =f h.
◦
f
Proposition 2.2. Let Σ : 0 Z Y X 0 be an AR sequence ending in X. Then
X
→ → −→ →
f w
Σ :Z Y X Z[1] is an AR triangle ending in X.
X
→ −→ −→
Proof. We shall show that Σ satisfies (ART1), (ART2) and (ART3).
(ART1) It is obvious.
(ART2) If w is zero, then Σ is split. Thus there exists g : X Y such that f g = 1 . Note
X
→ ◦
that f g radEnd (X) since Σ is not split. It yields that f g =1 radEnd (X). This is a
◦ ∈ R ◦ X ∈ R
contradiction and w must be non zero.
ON STABLE DEGENERATIONS OF COHEN-MACAULAY MODULES 3
(ART3) Let g : W X be not a split epimorphism. Then g : W X is also not a split
→ →
epimorphism. By (AR3), we have a morphism h:W Y such that f h=g. We conclude that
f h=g. → ◦ (cid:3)
◦
WesaythatCM(R)(resp. CM(R))admitsARsequences(resp. ARtriangles)ifthereexistsan
ARsequence(reap. ARtriangle)endinginX (resp. X)foreachindecomposableCohen-Macaulay
R-module X. We also say that (R,m) is an isolated singularity if each localization R is regular
p
for each prime ideal p with p = m. If R is an isolated singularity, CM(R) admits AR sequences
6
(cf. [11, Theorem 3.2]). As a corollary of Proposition 2.2, CM(R) admits AR triangles if R is an
isolated singularity.
Corollary 2.3. If R is an isolated singularity, we have an 1-1 correspondence between the set of
isomorphism classes of AR sequences in CM(R) and that of AR triangles in CM(R).
Proof. According to Proposition 2.2, we can define the mapping from the set of AR sequences
to the set of AR triangles by taking its triangles. Note that AR triangles (resp. AR sequences)
ending in X (resp. X) is unique up to isomorphism of triangles (resp. sequences) for a giving
indecomposable X (resp. X) (see [3, 8]). Hence it follows from Proposition 2.2 that the mapping
is surjective. The injectivity of the mapping is clear. (cid:3)
For an AR triangle Z Y X Z[1], we denote Z (resp. X) by τX (resp. τ−1Z). For an
→ → →
AR sequence 0 Z Y X 0, Z (resp. X) is also denoted by τX (resp. τ−1Z) (see [11,
→ → → →
Definition2.8].). ByvirtueofProposition2.2,τX =τX foreachindecomposableCohen-Macaulay
∼
R-module X.
Remark 2.4. ReitenandVandenBergh[8]showthataHom-finitek-lineartriangulatedcategory
T
admitsARtrianglesifandonlyif hasaSerrefunctor. WecanshowthatCM(R)isaHom-finite
T
triangulated category which has a Serre functor if R is an isolated singularity. Actually CM(R)
has a Serre functor τ( )[ 1] = ( )[d 1] where d = dimR. (Cf. [11, Lemma 3.10]. See also [6,
− − ∼ − −
Corollary2.5.].) Hereτ isanARtranslation(seeabove). ThereforewecanalsoshowthatCM(R)
admits AR triangles by [8].
Lemma 2.5. [11, Lemma 3.9] Let M and N be finitely generated R-modules. Then we have a
functorial isomorphism
Hom (M,N)=TorR(TrM,N).
R ∼ 1
Here TrM is an Auslander transpose of M.
According to Lemma 2.5, Hom (M,N) is of finite dimensional as a k-modules for M, N
R ∈
CM(R) if R is an isolated singularity. Thus the following definition makes sense.
Definition 2.6. For M, N CM(R) we define M N if [X,M] [X,N] for each X
hom
∈ ≤ ≤ ∈
CM(R). Here [X,M] is an abbreviation of dim Hom (X,M).
k R
Now let us consider the full subcategory of the functor category of CM(R) which is called the
Auslander category. We give a brief review of the Auslander category (see [11, Chapter 4 and 13]
for the detail). The Auslander category mod(CM(R)) is the category whose objects are finitely
presentedcontravariantadditivefunctorsfromCM(R)tothecategoryofAbeliangroupsandwhose
morphisms are natural transformations between functors. The following lemma will be a key of
our result in this section.
Lemma 2.7. [11, Theorem 13.7] A group homomorphism
γ :G(CM(R)) K (mod(CM(R))),
0
→
defined by γ(M)=[Hom ( ,M)] for M CM(R), is injective. Here G(CM(R)) is a free Abelian
R
group Z X, whereX runsthroughall∈isomorphism classes ofindecomposable objects inCM(R).
·
L
4 NAOYAHIRAMATSU
We denote by mod(CM(R)) the full subcategory mod(CM(R)) consisting of functors F with
F(R) = 0. Note that every object F mod(CM(R)) is obtained from a short exact sequence in
∈
CM(R). Namely we have the short exact sequence 0 L M N 0 such that
→ → → →
0 Hom ( ,L) Hom ( ,M) Hom ( ,N) F 0
R R R
→ → → → →
isexactinmod(CM(R)). SinceF mod(CM(R))isasubfunctorExt1(,L)forsomeL CM(R),
∈ R ∈
F(X)hasfinite lengthforeachX CM(R)ifRisanisolatedsingularity. Thereforewecandefine
∈
a group homomorphism associated with X in CM(R)
(2.1) ϕ :K (mod(CM(R))) Z ; [F] dim F(X).
X 0 k
→ 7→
If 0 Z Y X 0 is an AR sequence in CM(R), then the functor S defined by an exact
X
→ → → →
sequence
0 Hom ( ,Z) Hom ( ,Y) Hom ( ,X) S 0
R R R X
→ → → → →
is a simple object in mod(CM(R)) and all the simple objects in mod(CM(R)) are obtained in this
way from AR sequences. We say that R is of finite representation type if there are only a finite
number ofisomorphismclassesof indecomposableCohen-MacaulayR-modules. We note that if R
is of finite representationtype, then R is an isolatedsingularity (cf. [11, Chapter 3.]). It is proved
in [11, (13.7.4)] that
foreachobjectF inmod(CM(R)),thereisafiltrationbysubobjects0 F F F =F
1 2 n
⊂ ⊂ ⊂···⊂
such that each F /F is a simple object in mod(CM(R)) if R is of finite representation type.
i i−1
We also remark that, since CM(R) is a Krull-Schmidt category,
k if X =Y,
S (Y)= ∼
X
(0 if X 6∼=Y.
for an indecomposable module Y CM(R). See [11, (4.11)] for instance.
∈
Lemma 2.8. If R is of finite representation type, then we have the equality in K (mod(CM(R)))
0
[Hom ( ,M)]= [X ,M] [S ]
R − i · Xi
Xi∈inXdCM(R)
for each M CM(R).
∈
Proof. For F =Hom ( ,M), F(R)=0, so that F mod(CM(R)). Since R is of finite represen-
R − ∈
tation type, F has a filtration by simple objects S (see above). Hence we have the equality in
Xi
K (mod(CM(R))):
0
[F]= c [S ].
i· Xi
Xi∈inXdCM(R)
By using homomorphism in (2.1), we see that
[X ,M]=ϕ ([F])=dim c dim S (X )=c .
j Xj k i· k Xi j j
Xi∈inXdCM(R)
Therefore we obtain the equation in the lemma. (cid:3)
Theorem2.9. LetRbeoffiniterepresentationtypeandM andN beCohen-MacaulayR-modules.
Suppose that [X,M]=[X,N] for each X CM(R). Then M ΩM =N ΩN.
∈ ⊕ ∼ ⊕
Proof. Underthecircumstances,weseethat[Hom ( ,M)]=[Hom ( ,N)]inK (mod(CM(R))),
R − R − 0
hence in K (mod(CM(R))). Note that Hom ( ,M) = Ext1( ,ΩM) for each M CM(R) (cf.
0 R − ∼ R − ∈
[2]). We have the resolution in mod(CM(R)):
0 Hom ( ,ΩM) Hom ( ,P ) Hom ( ,M) Hom ( ,M) 0,
→ R − → R − M → R − → R − →
ON STABLE DEGENERATIONS OF COHEN-MACAULAY MODULES 5
where P is a free R-module. Thus we have
M
[Hom ( ,M)]+[Hom ( ,ΩM)] [Hom ( ,P )]=[Hom ( ,N)]+[Hom ( ,ΩN)] [Hom ( ,P )].
R R R M R R R N
− − − − − − − −
Hence,
[Hom ( ,M)]+[Hom ( ,ΩM)]+[Hom ( ,P )]=[Hom ( ,N)]+[Hom ( ,ΩN)]+[Hom ( ,P )].
R R R N R R R M
− − − − − −
According to Lemma 2.7, we get
M ΩM P =N ΩN P .
⊕ ⊕ N ∼ ⊕ ⊕ M
Therefore M ΩM =N ΩN. (cid:3)
⊕ ∼ ⊕
It immediately follows from the theorem that
Corollary 2.10. Let R be of finite representation type and M and N be Cohen-Macaulay R-
modules. Suppose that U = U[ 1] for each indecomposable Cohen-Macaulay R-module U. Then
∼ −
[X,M]=[X,N] for each X CM(R) if and only if M =N. Particularly, is a partial order
∈ ∼ ≤hom
on CM(R).
Example 2.11. Let R be a one dimensional simple singularity of type (A ), that is R =
n
k[[x,y]]/(xn+1 +y2). If n is an even integer, one can show that X is isomorphic to ΩX up to
free summed for each X CM(R), so that X =X[ 1]. See [11, Proposition 5.11]. Thus is
∈ ∼ − ≤hom
a partial order on CM(R) if n is an even integer.
If n is an odd integer, we have indecomposable modules X CM(R) such that X =X[ 1]. In
∈ ∼6 −
fact, let N = R/(x(n+1)/2 √ 1y). Then N (resp. N ) is isomorphic to ΩN (resp. ΩN ),
± + − − +
± −
so that N+ 6∼= N+[−1] (resp. N− 6∼= N−[−1]). Although we can also show that ≤hom is a partial
order on CM(R) even if n is an odd integer.
Proposition 2.12. Let R=k[[x,y]]/(xn+1+y2). Then [X,M]=[X,N] for each X CM(R) if
∈
and only if M =N.
∼
Proof. We show the case when n is an odd integer. Let I = (xi,y) be ideals of R for 1
i
≤
i (n 1)/2 and N = R/(x(n+1)/2 √ 1y). Then I , I ,N is a complete list of
± 1 (n−1)/2 ±
≤ − ± − { ··· }
non free indecomposable Cohen-Macaulay R-modules. Note that I = ΩI up to free summand
i ∼ i
for i = 1, (n 1)/2. See [11, Paragraph (9.9)]. Now let M, N CM(R) and suppose that
··· − ∈
[X,M] = [X,N] for each X CM(R). Set M = (n−1)/2I mi N m+ N m− and N =
∈ i=1 i ⊕ + ⊕ −
(i=n−11)/2Iini ⊕N+n+ ⊕N−n−. By Theorem 2.9, M ⊕LΩM ∼=N ⊕ΩN . Thus
L (n−1)/2 (n−1)/2
Ii2mi ⊕N+m++m− ⊕N−m++m− ∼= Ii2ni ⊕N+n++n− ⊕N−n++n−.
i=1 i=1
M M
Hence we have equalities:
m =n , m +m =n +n .
i i + − + −
Here we remark that
HomR(N±,N∓)∼=Ext1R(N±,N±)=0.
Using this, we have
(n−1)/2 (n−1)/2
[N ,M]= m [N ,I ]+m [N ,N ]= n [N ,I ]+n [N ,N ]=[N ,N].
+ i + i + + + i + i + + + +
i=1 i=1
X X
This equality show that m+ =n+, so that m− =n−. Consequently M ∼=N. (cid:3)
6 NAOYAHIRAMATSU
Remark 2.13. The stable hom order is not always a partial order on CM(R) even if the
hom
≤
base ring R is a simple singularity of type (A ). Let R = k[[x,y,z]]/(x3 +y2 +z2). Then R
n
is a two dimensional simple singularity of type (A ). And let I (resp. J) be an ideal generated
2
by (x,y) (resp. (x2,y)). Note that the set I,J is a complete list of non free indecomposable
{ }
Cohen-MacaulayR-modules(see[11,Chapter10]forinstance). Thenitiseasytoseethat[I,J]=
[J,I]=1. However I =J. Thus is not a partial order on CM(R).
6∼ ≤hom
In what follows, we always assume that R is an isolated singularity. We denote by µ(M,Z) the
multiplicity of Z as a direct summand of M. On an AR triangle, we have the following.
Proposition 2.14. Let Σ : Z Y X Z[1] be an AR triangle in CM(R). Then the
X
→ → →
following statements hold for each U CM(R).
∈
(1) [U,X]+[U,Z] [U,Y]=µ(U,X)+µ(U,X[ 1]).
− −
(2) If U is periodic of period 2, that is, U =U[2], then
∼
[U,X]+[U,Z] [U,Y]=[U[ 1],X]+[U[ 1],Z] [U[ 1],Y]
− − − − −
Proof. (1) Apply Hom (U, ) to the triangle Σ , we have a long exact sequence as follows:
R − X
Hom (U,g[−1])
Hom (U,Z[ 1]) Hom (U,Y[ 1]) R Hom (U,X[ 1])
R − −−−−→ R − −−−−−−−−−−→ R − −−−−→
Hom (U,f) Hom (M,g)
Hom (U,Z) R Hom (U,Y) R Hom (U,X)
R −−−−−−−→ R −−−−−−−−→ R −−−−→
Hom (U,f[1])
Hom (U,Z[1]) R Hom (U,Y[1]) Hom (U,X[1]) .
R −−−−−−−−−→ R −−−−→ R −−−−→
Since End (X)/radEnd (X) = k for each non free indecomposable Cohen-Macaulay module X
R R ∼
and by the property of an AR triangle (ART3), we have
Ker Hom (U,f[i+1])=Coker Hom (U,g[i])=kµ(U,X[i])
R ∼ R ∼
for all i Z. In particular
∈
0 kµ(U,X[−1]) Hom (U,Z) Hom (U,Y) Hom (U,X) kµ(U,X) 0
→ → R → R → R → →
Therefore we obtain the required equation.
(2) Note from (1) that the equations
[U,X]+[U,Z] [U,Y]=µ(U,X)+µ(U,X[ 1])
− −
and
[U[ 1],X]+[U[ 1],Z] [U[ 1],Y]=µ(U[ 1],X)+µ(U[ 1],X[ 1])
− − − − − − −
hold. Sincetheshiftfunctor( )[1](hence( )[ 1])isanautofunctor,µ(U,X)=µ(U[ 1],X[ 1]).
− − − − −
Moreover µ(U[ 1],X) = µ(U,X[ 1]) for U[ 1][ 1] = U[ 2] = U. Consequently we get the
assertion. − − − − ∼ − ∼ (cid:3)
Remark 2.15. By using Proposition 2.14, one can show that is a partial order on CM(R)
hom
≤
without the assumption that R is of finite representation type.
3. Hom order and Grothendieck group of CM(R)
For later reference we state some results on the stable hom order among the same class in the
GrothendieckgroupofCM(R). The GrothendieckgroupofCM(R)(moregenerallyatriangulated
category) is defined by
K (CM(R))=G(CM(R))/<X +Z Y There is a triangle Z Y X Z[1] in CM(R)>,
0
− | → → →
where G(CM(R)) is a free Abelian group Z X. We refer the reader to [3, Chapter
X∈indCM(R) ·
3] for the details. Since R is Gorenstein, one can show that [M]=[N] in K (CM(R)) if and only
L 0
if [M P]=[N Q] in K (CM(R)) for some free R-modules P, Q.
0
⊕ ⊕
ON STABLE DEGENERATIONS OF COHEN-MACAULAY MODULES 7
Lemma3.1. IfRisoffiniterepresentation type, wehavetheequalityasasubgroupofG(CM(R)):
<X+Z Y There is a triangle Z Y X Z[1] in CM(R)>
− | → → →
=<X +Z Y There is an AR triangle Z Y X Z[1] in CM(R)>.
− | → → →
Proof. It follows from Corollary 2.3, [11, Theorem 13.7] and the definition of triangles in stable
categories. (cid:3)
Hence, if R is of finite representation type, [M] = [N] in K (CM(R)) if and only if there exist
0
AR triangles Z Y X Z [1] and Z′ Y′ X′ Z′[1] with the equality
i → i → i → i j → j → j → j
n m
(3.1) [M] [N]= c ([X ]+[Z ] [Y ]) d ([X′]+[Z′] [Y′]).
− i· i i − i − j · j j − j
i j
X X
Now we consider the following condition:
( ) For an AR triangle Z Y X Z[1], [X]+[Z] [Y]=[X[ 1]]+[Z[ 1]] [Y[ 1]] in
∗ → → → − − − − −
G(CM(R)).
Remark 3.2. (1) The condition ( ) says that for AR triangles Z Y X Z[1] and
Z′ Y′ X′ Z′[1], if [U∗,X]+[U,Z] [U,Y] = [U,X′]+→[U,Z′→] [U,→Y′] for each
inde→compo→sable→U then [X]+[Z] [Y]=[X−′]+[Z′] [Y′] in G(CM(R)−). Since it follows
fromProposition2.14thatµ(X,X−)+µ(X,X[ 1])=−µ(X,X′)+µ(X,X′[ 1]),weseethat
X =X′ or X =X′[ 1]. If X =X′ the equati−on is obvious and assume th−at X =X′[ 1].
∼ ∼ − ∼ ∼ −
Then by ( ),
∗
[X]+[Z] [Y] =[X′[ 1]]+[Z′[ 1]] [Y′[ 1]]
− =[X′]−+[Z′] [Y−′] − −
−
in G(CM(R)).
(2) We also remark that, under the condition ( ), X = X[ 1] holds if X = τX. For an AR
∗ ∼ − ∼
triangle Σ :τX E X τX[1], we have
X X
→ → →
X τX E [ 1]=X[ 1] τX[ 1] E .
⊕ ⊕ X − ∼ − ⊕ − ⊕ X
Assume that X = X[ 1]. Then X τX = E . However, this never happen (see Propo-
6∼ − ⊕ ∼ X
sition 2.14).
(3) Thecondition( )holdswhenR=k[[x,y]]/(xn+1+y2)(cf. [11,Proposition5.11,Paragraph
∗
(9.9)]).
Proposition 3.3. Let R be of finite representation type and M and N be Cohen-Macaulay R-
modules with [M]=[N] in K (CM(R)). Suppose that each AR triangle satisfies the condition ( ).
0
∗
Then M N if and only if there exist AR triangles Z Y X Z [1] with the equation
hom i i i i
≤ → → →
n
(3.2) [M] [N]= c ([X ]+[Z ] [Y ])
i i i i
− · −
i
X
in G(CM(R)).
Proof. It is easy to see that
[U,N] [U,M] = nc ([U,X ]+[U,Z ] [U,Y ]) md ([U,X′]+[U,Z′] [U,Y′])
− i i· i i − i − j j · j j − j
= nc µ(U,X )+µ(U,X [ 1]) md µ(U,X′)+µ(U,X′[ 1])
Pi i·{ i i − }−Pj j ·{ j j − }
for each indecomposabPle module U CM(R). Assume thaPt M N. If d = 0, there exist i
hom j
∈ ≤ 6
suchthatX′ =X orX′[ 1]=X withd =c . AsmentionedinRemark3.2(1),bythecondition
j ∼ i j − ∼ i j i
( ), we can omit such constituents in the expression. Repeating this procedure, we obtain the
e∗quation. (cid:3)
8 NAOYAHIRAMATSU
Under the circumstance of Proposition 3.3, we say that the expression (3.2) is irredundant if
each X is not isomorphic to each other up to a shift. Since R is of finite representation type,
i
we can always take the expression irredundant. We say that R satisfies ( ) if each AR triangle
∗
satisfies the condition ( ).
∗
Lemma 3.4. Let R be of finite representation type which satisfies ( ) and M and N be Cohen-
∗
Macaulay R-modules with M N and [M] = [N] in K (CM(R)). And let U be a non free
hom 0
≤
and indecomposable direct summand of N. Suppose that [U,M] = [U,N] and U = U[2]. Then U
∼
is also a direct summand of M.
Proof. By virtue of Proposition 3.3, since M N, there exist AR triangles Z Y X
hom i i i
≤ → → →
Z [1] with
i
n
[N] [M]= c ([X ]+[Z ] [Y ])
i i i i
− · −
i
X
in G(CM(R)). Thus we have
n n
N Y ci =M (X Z )ci.
⊕ i ∼ ⊕ i⊕ i
i i
M M
Now we assume that U is not a direct summand of M. Then there exists i such that U = X or
∼ i
U =Z and we can show the inequality
∼ i
[U,X ]+[U,Z ] [U,Y ]=µ(U,X)+µ(U,X[ 1])>0
i i i
− −
holds. If U =X , it is clear. If U =Z , since U is periodic of period at most 2,
∼ i ∼ i
X =τ−1U =U[d]=U or U[1].
i ∼ ∼ ∼
See Remark 2.4. Hence we have the inequality above. However the inequality never happen since
[U, Y ci]=[U, (X Z )ci].
i i i
⊕
i i
M M
Therefore, U is a direct summand of M. (cid:3)
Proposition3.5. LetRbeoffiniterepresentationtypewhichsatisfies( )andM andN beCohen-
∗
Macaulay R-modules with M N and [M] = [N] in K (CM(R)). Suppose that [U,M] =
hom 0
≤
[U,N] and U =U[2] for each indecomposable direct summand U of N. Then M =N.
∼ ∼
Proof. ItfollowsfromLemma3.4thatN is isomorphictoadirectsummandofM. SinceM
hom
N, we have M =N. ≤ (cid:3)
∼
4. Stable degenerations of Cohen-Macaulay modules
In this section, we shall describe the stable degenerations of Cohen-Macaulay modules over
simple singularities over type (A ), namely
n
(4.1) k[[x ,x ,x , ,x ]]/(xn+1+x2+x2+ +x2).
0 1 2 ··· d 0 1 2 ··· d
First let us recall the definition of stable degenerations of Cohen-Macaulay modules.
Definition 4.1. [14, Definition 4.1] Let M,N CM(R). We say that M stably degenerates to
∈
N if there exists a Cohen-Macaulay module Q CM(R V) such that Q[1/t] = M K in
∈ ⊗k ∼ ⊗k
CM(R K) and Q V/tV =N in CM(R).
⊗k ⊗V ∼
Itisknownthatthering(4.1)isoffiniterepresentationtype,sothatitisanisolatedsingularity.
If a ring is an isolated singularity, there is a nice characterizationof stable degenerations.
ON STABLE DEGENERATIONS OF COHEN-MACAULAY MODULES 9
Theorem 4.2. [14,Theorem5.1,6.1] Consider the following threeconditions for Cohen-Macaulay
R-modules M and N:
(1) M P degenerates to N Q for some free R-modules P, Q.
⊕ ⊕
(2) M stably degenerates to N.
(3) There is a triangle
Z M Z N Z[1]
−−−−→ ⊕ −−−−→ −−−−→
in CM(R).
If R is an isolated singularity, then (2) and (3) are equivalent. Moreover, if R is artinian, the
conditions (1),(2) and (3) are equivalent.
Remark 4.3. In general, the implications (1) (2) (3) hold. And it is required that the
⇒ ⇒
endmorphism of Z in the triangle in (3) is nilpotent. However if R is an isolated singularity, we
do not need the nilpotency assumption (cf. [14, Lemma 6.5.]). The theorem also saysthat M and
N give the same class in the Grothendieck group of CM(R) if M stably degenerates to N.
We state order relations with respect to stable degenerations and triangles.
Definition 4.4. [4, Definition 3.2., 3.3.] Let M and N be Cohen-Macaulay R-modules.
(1) WedenotebyM N ifN isobtainedfromM byiterativestabledegenerations,i.e. there
st
≤
is a sequence of Cohen-Macaulay R-modules L ,L ,...,L such that M = L , N = L
0 1 r ∼ 0 ∼ r
and each L stably degenerates to L for 0 i<r.
i i+1
≤
(2) We say that M stably degenerates by a triangle to N, if there is a triangle of the form
U M V U[1] in CM(R) such that U V =N. We also denote by M N if N
→ → → ⊕ ∼ ≤tri
is obtain from M by iterative stable degenerations by a triangle.
Remark 4.5. It has shown in [14] that the stable degeneration order is a partial order. Moreover
if there is a triangle U M V U[1], then we can show that M stably degenerates to U V
→ → → ⊕
(cf. [4, Remark 3.4. (2)]). Hence M N induces M N. It also follows from Theorem 4.2
tri st
≤ ≤
that M N induces that M N.
st hom
≤ ≤
The rest of the paper, we devote to describe stable degenerations over the one dimensional
simple singularity of type (A ). Namely we consider the ring of the form:
n
R=k[[x,y]]/(xn+1+y2).
As stated in [11, Proposition 5.11], if n is an even integer, the set of ideals of R
I =(xi,y) 1 i n/2
i
{ | ≤ ≤ }
is a complete list of isomorphic indecomposable non free Cohen-Macaulay R-modules. On the
other hand, if n is an odd integer, then
I =(xi,y) 1 i (n 1)/2 N =R/(x(n+1)/2+√ 1y), N =R/(x(n+1)/2 √ 1y)
i + −
{ | ≤ ≤ − }∪{ − − − }
is a complete list of the ones (cf. [11, Paragraph(9.9)]).
In this section, we shall show
Theorem 4.6. Let R=k[[x,y]]/(xn+1+y2). Then the stable hom order coincides with the stable
degeneration order. Particularly, we have the followings.
(1) If n is an even integer,
0 I I I .
st 1 st 2 st st n/2
≤ ≤ ≤ ···≤
10 NAOYAHIRAMATSU
(2) If n is an odd integer,
0 I I I N N .
st 1 st 2 st st (n−1)/2 st + −
≤ ≤ ≤ ···≤ ≤ ⊕
and
N N I N I N N N (double sign corresponds).
± st ± 1 st st ± (n−1)/2 st ± + −
≤ ⊕ ≤ ···≤ ⊕ ≤ ⊕ ⊕
To show this, we use the stable analogue of the argument in [16].
The lemma below is well known for the case in an Abelian category(cf. [16, Lemma 2.6]). The
samestatementfollowsin anarbitraryk-lineartriangulatedcategory,notnecessaryCM(R). (The
author thanks Yuji Yoshino for telling me this argument.)
Lemma 4.7. Let
f1
Σ :N (cid:18)v(cid:19) L N (u,g1) L N [1]
1 1 1 2 2 1
−−−−→ ⊕ −−−−→ −−−−→
and
f2
Σ :M (cid:18)w(cid:19) N M (v,g2) N M [1]
2 1 1 2 2 1
−−−−→ ⊕ −−−−→ −−−−→
be triangles in a k-linear triangulated category. Then we also have the following triangle.
M L M L M [1].
1 1 2 2 1
→ ⊕ → →
Proof. We consider the following triangle associated with Σ :
2
−f1fw2◦f2! 10 fv1 g02
M L N M (cid:18) (cid:19) N M [1].
1 1 1 2 2 1
−−−−−−−→ ⊕ ⊕ −−−−−−−−−→ −−−−→
We remark that the left morphism is given by
f f 1 f 0 0
1 2 1
− ◦ −
f = 0 1 0 f ,
2 2
    
w 0 0 1 w
    
to make the diagram below:
f1
N (cid:18)v(cid:19) L N (u,g1) L N [1]
1 1 2 2 1
−−−−→ ⊕ −−−−→ −−−−→
1 f1 0
0 v g2
(cid:18) (cid:19)
(cid:13) x x (cid:13)
N(cid:13)1 L1 N1 M2 L1 M2 N1(cid:13)[1]
(cid:13) −−−0−→ ⊕ ⊕ −−−−→ ⊕ −−−−→ (cid:13)
1
0!
−f1◦f2
f2
w !
x x
M1 M1.
 
By the octahedral axiom, we obtain the required triangle. (cid:3)
Remark 4.8. Combining (the Abelian version of) Lemma 4.7 with Lemma 2.8, the dimension of
HomcanbecalculateeasilyfromdatumofARsequences. Forinstance,letR=k[[x,y]]/(xn+1+y2)
where n is even. The AR sequences are
0 I I I I 0
i i−1 i+1 i
→ → ⊕ → →