CONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES ANDAPPLICATIONS TO COMMUTATIVE RINGSHIROKI MATSUI AND RYO TAKAHASHI
Abstract. In this paper, as an analogue of the spectrum of a tensor-triangulated category introducedby Balmer, we define a spectrum of a triangulated category which is not necessarily tensor-triangulated.
We apply it for some triangulated categories associated to a commutative noetherian ring.

1. Introduction
Classification of thick subcategories has been one of the main approaches in the studies of triangulated
categories for several decades. It has been studied so far in many areas such as stable homotopy theory,
modular representation theory, algebraic geometry, commutative/non-commutative algebra and so on;see [3–5,9,11,12,16,19–21] and references therein. In commutative algebra, Hopkins [11] and Neeman [16]classified thick subcategories of derived categories of perfect complexes over commutative rings via Zariski
spectra. The second author [20] classified thick subcategories of singularity categories of hypersurfaces
via singular loci, and it was extended to complete intersections by Stevenson [19].
On the other hand, there is a beautiful theory initiated by Balmer [2] which is called tensor-triangular
geometry. He introduced the concepts of thick tensor-ideals, radical thick tensor-ideals and prime thicktensor-ideals of an essentially small tensor-triangulated category T as analogues of ideals, radical ideals
and prime ideals of commutative rings. Then he defined a topology on the set Spc T of prime thicktensor-ideals of T , which is called the Balmer spectrum of T . He accomplished the following monumental
work in this theory, which enables us to do algebro-geometric studies of tensor-triangulated categories.
Theorem 1.1 (Balmer). Let T be an essentially small tensor-triangulated category. Taking the Balmer
supports gives a bijection between the radical thick tensor-ideals of T and the Thomason subsets of Spc T .
The Balmer theory works for arbitrary (essentially small) tensor-triangulated categories, but by definition, it does not (at least directly) work for those triangulated categories which are not tensor-triangulated.
Such triangulated categories include two of the most important triangulated categories in commutative
algebra: the bounded derived category Db (R) of finitely generated R-modules, and the singularity category Dsg (R) := Db (R)/ Dperf (R) of R, where R is a commutative noetherian ring. These triangulated
categories are not tensor-triangulated naturally, in the sense that the derived tensor product − ⊗L
R − isnot well-defined on them. This is why in the present paper we make a variant of Balmer’s theory for
general (i.e., not necessarily tensor-triangulated) triangulated categories T . Once we assign a class Cof thick subcategories of T , we can define the topological space EspC T whose underlying set is C. We
then call the elements of EspC T the prime thick subcateories. We prove the following analogous result
to Theorem 1.1.
Theorem 1.2 (Theorem 2.9). Let T be an essentially small triangulated category. Then there exists a
one-to-one correspondence
1-1
RadC T ←→ ParamC T .
Let us explain the notation used above: RadC T is the set of radical thick subcategories of T , which
are variants of radical thick tensor-ideals of a tensor-triangulated category, while ParamC T consists ofcertain subsets of EspC T , which parametrizes the radical thick subcategories as stated above.
2010 Mathematics Subject Classification. 13D09, 13H10, 18E30.
Key words and phrases. complete intersection, derived category, hypersurface, perfect complex, singularity category,
spectrum, (prime/radical/tame) thick subcategory, triangulated category.
HM was partly supported by JSPS Grant-in-Aid for JSPS Fellows 16J01067. RT was partly supported by JSPS Grantsin-Aid for Scientific Research 16K05098 and JSPS Fund for the Promotion of Joint International Research 16KK0099.
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HIROKI MATSUI AND RYO TAKAHASHI

Thus, an important and essential point is to find out a suitable class C for a given triangulated category.
In this paper, we do this for the bounded derived category Db (R) of finitely generated R-modules, thederived category Dperf (R) of perfect R-complexes, and the singularity category Dsg (R) of R, where R is
a commutative noetherian ring. For such triangulated categories T and a class of thick subcategories C,
we introduce a pair of mapss
! X ,Esp T "
C

S

T

where we set XDb (R) = XDperf (R) = Spec R and XDsg (R) = Sing R. We define tame thick subcategories of
T to be thick subcategories representable by the corresponding support. A tame thick subcategory isalways radical, and it is natural to consider when the converse holds. Our results include the following.
Theorem 1.3 (Theorems 3.10, 3.14, 3.17, 4.6, 4.10, 4.21 and Corollary 3.11). With the notation intoduced
above, the following statements hold.
(1) Let R be a commutative noetherian ring, and T = Dperf (R). Then the following hold true.
(i) The maps s and S are mutually inverse homeomorphisms.
(ii) The tame thick subcategories coincide with the radical ones.
Furthermore, ParamC Dperf (R) coincides with the set of Thomason subsets of EspC Dperf (R).
(2) Let R be a complete intersection local ring of positive Krull dimension. The following are equivalent.
(i) The topological spaces EspC Db (R) and Spec R have the same Krull dimension.
(ii) The topological spaces EspC Db (R) and Spec R are homeomorphic.
(iii) The maps s and S are mutually inverse homeomorphisms.
(iv) The tame thick subcategories coincide with the radical ones.
(v) The ring R is a regular local ring.
When one of these equivalent conditions holds, ParamC Db (R) coincides with the set of Thomason
subsets of EspC Db (R).
(3) Let R be an excellent complete intersection local ring. Then the following are equivalent.
(i) The topological spaces EspC Dsg (R) and Sing R have the same Krull dimension.
(ii) The topological spaces EspC Dsg (R) and Sing R are homeomorphic.
(iii) The maps s and S are mutually inverse homeomorphisms.
(iv) The tame thick subcategories coincide with the radical ones.
(v) The ring R is a hypersurface local ring.
When one of these equivalent conditions holds, ParamC Dsg (R) coincides with the set of Thomason
subsets of EspC Dsg (R).
We should remark that recently the first author [13] has also defined a spectrum of a triangulated
category, which is diﬀerent from a spectrum introduced in the present paper. More precisely, the spectrum
introduced in [13] is defined as the set of irreducible subcategories of T together with a topology, which is
uniquely determined by using the triangulated structure of T . The advantage of a spectrum introducedin the present paper is that there is flexibility in the choice of the underlying set; for a given triangulated
category, choosing an appropriate class C makes it more manageable.
This paper is organized as follows. In Section 2, following Balmer’s work [2], for each triangulated
category T and a given class C of thick subcategories, we define the topological space EspC T . Several
analogous notions to Balmer’s theory, such as supports and radical thick subcategories, are also introduced. Later sections are devoted to applications to commutative noetherian rings R. The spectra of
the derived categories Db (R), Dperf (R) and intermediate triangulated categories are explored in Section
3, and the spectrum of the singularity category Dsg (R) is investigated in Section 4 as well.
2. Spectra of triangulated categories
In this section, following Balmer’s theory [2], we define a topological space for each triangulated
category which is not necessarily tensor-triangulated, and by using it we give a classification of a certainclass of thick subcategories.
Throughout this section, let T be an essentially small triangulated category. Denote by Th T the setof thick subcategories of T . Fix a subset C of Th T . A thick subcategory of T which belongs to this set
C is called a prime thick subcategory. First we introduce a topology on the set C.

CONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES

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Definition 2.1 (cf. [2, Definition 2.1]). For a class E of objects of T , we put
Z(E) := {P ∈ C | P ∩ E = ∅}.
One can easily check that the following statements hold; note that the fact that thick subcategories areclosed under direct summands enables the last one to hold.
• Z(T! ) = ∅ and Z(∅)
" = C.
• i∈I Z(Ei ) = Z( i∈I Ei ) for a family {Ei }i∈I of classes of objects of T .
• Z(E)∪Z(E ′ ) = Z(E ⊕E ′ ) for classes E, E ′ of objects of T , where E ⊕E ′ := {X⊕X ′ | X ∈ E, X ′ ∈ E ′ }.
Thus we can put a topology on C by defining the collection of closed subsets to be Z(E). We then denote
this topological space by EspC T and call it the spectrum of T with respect to C.♣
For an object M ∈ T , we define its support by
Spp M := Z({M }) = {P ∈ EspC T | M ∕∈ P}.
!
It directly follows by definition that the equality Z(E) = M ∈E Spp M holds for each E. This shows thatthe family {Spp M }M ∈T of closed subsets forms a closed basis of EspC T . For a thick subcategory X of
T , we define its support by"
Spp X := M ∈X Spp M = {P ∈ EspC T | X ⊈ P},
which is a specialization-closed subset of EspC T .

For a full subcategory E of T , we denote by thick E the smallest thick subcategory of T containing E.
We provide some basic properties of supports, which says that the pair (EspC T , Spp) is a support data
for T in the sense of [13].
Remark 2.2. (1) Spp(0) = ∅.
(2) Spp(M [n]) = Spp M for M ∈ T and n ∈ Z.
(3) Spp(M ⊕ N ) = Spp M ∪ Spp N for M, N ∈ T .
(4) Spp M ⊆ Spp L ∪ Spp N for an exact triangle L → M → N → L[1] in T .
In particular, for a subset W of EspC T , the full subcategory
Spp−1 W := {M ∈ T | Spp M ⊆ W }of T is a thick subcategory, which implies that Spp X = Spp(thick X ) for each full subcategory X of T .
Proof. This is a direct consequence of the definition of a thick subcategory. Indeed, the fact that a thicksubcategory contains the zero object shows (1), the fact that it is closed under shifts shows (2), the factthat it is closed under finite direct sums and direct summands shows (3), and the fact that it satisfies the
2-out-of-3 property shows (4).
!
The following proposition is proved in [2] for Balmer spectra of tensor-triangulated categories. We canprove the same statement for our topological space EspC T and the proof is completely the same.
Proposition 2.3 (cf. [2, Proposition 2.9]). For any prime thick subcategory P of T , one has{P} = {Q ∈ EspC T | Q ⊆ P}.
In particular, EspC T is a T0 -space.
Proof. Let S be the right-hand side. Then P ∈ S = Z(T \ P). If Z(E) contains P, then P ∩ E = ∅, andQ ∩ E = ∅ for all Q ∈ S, which shows S ⊆ Z(E). Now the assertion follows.
!
Next, we define the height of a prime thick subcategory and the dimension of T .
Definition 2.4. For a prime thick subcategory P of T , we define the height ht P of P to be the largest
number n such that there exists a chain P0 ⊊ P1 ⊊ · · · ⊊ Pn = P of prime thick subcategories of T . We
define the dimension dimC T of T as the supremum of the heights of prime thick subcategories of T .
♣For a tensor-triangulated category T , the spectrum of T in our sense does not necessarily coincide with the Balmer
spectrum Spc T of T . To avoid confusion, instead of Spc T we adopt the notation Esp T , which comes from the Spanishtranslation espectro of spectrum.

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Remark 2.5. There is another notion of dimension for a triangulated category, which is nowadays called
the Rouquier dimension; see [18]. Our dimension dimC T of course depends on the choice of C and it
may or may not be equal to the Rouquier dimension.
Let X be a topological space. Recall that the Krull dimension dim X of X is defined as the supremum
of integers n ≥ 0 such that there exists a chain Z0 ⊊ Z1 ⊊ · · · ⊊ Zn of irreducible closed subsets of X.
Recall also that X is sober if every irreducible closed subset of X is the closure of exactly one point of X.
A typical example of a sober topological space is the Zariski spectrum Spec R of a commutative ring R.
Remark 2.6. By Proposition 2.3, for two prime thick subcategories P, Q of T one has {P} ⊊ {Q} if
and only if P ⊊ Q. This shows dim T ≤ dim EspC T . If EspC T is sober, then dim EspC T = dimC T .
Using the spectrum of T , we can classify a certain class of thick subcategories characterized by EspC T ,which we call radical thick subcategories.
Definition 2.7 (cf. [2, Lemma 4.2]). For a thick subcategory X of T , define the radical of X by
#
√
X :=
P.
P∈EspC T , X ⊆P

We say that a thick subcategory X is radical if
subcategories of T .

√X = X . Denote by RadC T the set of radical thick

The following proposition tells us that the support cannot distinguish a thick subcategory and its
radical. Thus to classify thick subcategories by their supports we focus on the radical thick subcategories.
√
Proposition 2.8. (1) For M ∈ T one has Spp√
M = ∅ if and only if M ∈ 0.
(2) For each X ∈ Th T there is an equality Spp X = Spp X .
√!
Proof. The first assertion is straightforward from the equality 0 = P∈Esp T P. To show the second, fix√√
P ∈ EspC T . Then P ∈ Spp X if and only if X ∕⊆ P, if and only if X ∕⊆ P, if and only if P ∈ Spp X . !
We define the parameter set ParamC T as the set of supports of thick subcategories of T :ParamC T := {Spp X | X ∈ Th T } ⊆ 2EspC T .

The reason why we call this so is that it parametrizes the radical thick subcategories of T as follows.
Theorem 2.9. There is a one-to-one correspondence
RadC T "

Spp
Spp−1

! Param T .
C

Proof. Let X ∈ Th T and M ∈ T . The condition Spp M ⊆ Spp X is equivalent to saying that for eachP ∈ EspC T √, if P contains X , then P also contains M . Therefore we have equalities Spp−1 (Spp X ) =!
X . This shows that Spp−1 : ParamC T → RadC T is a well-defined map, and that it is
X ⊆P P =a retraction of the map Spp√ : RadC T → ParamC T . Applying Proposition 2.8(2), we get equalities
Spp(Spp−1 (Spp X )) = Spp X = Spp X , which completes the proof of the theorem.
!
A subset of a topological space X is said to be Thomason if it is the union of some closed subsets
of X whose complements are quasi-compact. We denote by Thom X the set of Thomason subsets of X.
Theorems 2.9 and 1.1 naturally lead us to the following question.
Question 2.10. When does the equality ParamC T = Thom(EspC T ) hold?
We will give answers to this question in Sections 3 and 4.
3. Spectra of derived categories
In this section, we consider what we got in the previous section for derived categories of commutative
noetherian rings, that is, we discuss those spectra.
Throughout this section, let R be a commutative noetherian ring, and let D be a triangulated subcategory of the bounded derived category Db (R) of finitely generated R-modules containing the derived
category Dperf (R) of perfect R-complexes. The purpose of this section is to investigate EspC D for a

CONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES

5

certain class of thick subcategories C. To begin with, we recall the notion of a homological support andits basic properties for later use.
Definition 3.1. For each object M ∈ D, the homological support of M is defined by
Supp M := {p ∈ Spec R | H(M )p ∕= 0} = {p ∈ Spec R | Mp ∼∕ 0}.
=
For a full subcategory X ⊆ D, the homological support of X is defined by$
Supp X :=Supp M.
M ∈X

For an ideal I of R, let K(I) stand for the Koszul complex of a system of generators of I. We caneasily verify that the homological support satisfies the following properties.
Remark 3.2. (1) For an ideal I of R it holds that Supp K(I) = V(I).
(2) For M ∈ D the set Supp M is Zariski-closed (i.e. closed in Spec R).
(3) For M ∈ D one has Supp M = ∅ if and only if M ∼= 0 in D.
(4) For M ∈ D one has Supp(M [1]) = Supp M .
(5) For M, N ∈ D there is an equality Supp(M ⊕ N ) = Supp M ∪ Supp N .
(6) For an exact triangle L → M → N → L[1] in D there is an inclusion Supp M ⊆ Supp L ∪ Supp N .
In particular, for a subset W of Spec R, the full subcategorySupp−1
D (W ) := {M ∈ D | Supp M ⊆ W }of D is a thick subcategory of D.

For a partially ordered set S, let max S (resp. min S) be the set of maximal (resp. minimal) elementsof S. We always regard a set of sets as a partially ordered set with respect to the inclusion relation.
To define the spectrum of D, we have to fix a class of prime thick subcategories. We adopt the following
definition of prime thick subcategories of D.
Definition 3.3. For X ∈ Th D, we define the set I(X ) of ideals of R byI(X ) := {I ⊆ R | K(I) ∕∈ X }.

This is well-defined since the condition K(I) ∕∈ X is independent of the choice of a system of generatorsof I; see [7, Proposition 1.6.21]. We say that a thick subcategory P of D is prime if I(P) has a uniquemaximal element, i.e., # max I(P) = 1. When this is the case, we denote by s(P) the maximal elementof I(P). Let C denote the set of prime thick subcategories of D. We thus obtain the spectrum EspC D of
D with respect to C by Definition 2.1.
To avoid complication, omitting the subscripts C, we simply write Esp D, Rad D and Param D.
Remark 3.4. The motivation of this definition of a prime thick subcategory of D comes from [14,Proposition 3.7], which states that I(P) has a unique maximal element for a prime thick tensor-ideal Pof the right bounded derived category D- (R) of finitely generated R-modules.
Next, we give a typical example of prime thick subcategories. For each prime ideal p of R, we define
the full subcategory S(p) of D by
S(p) := {M ∈ D | Mp ∼= 0}.
Lemma 3.5. Let p ∈ Spec R. Then S(p) is a prime thick subcategory of D with s(S(p)) = p.
Proof. It is easy to check that
(3.5.1)

S(p) = Supp−1
D {q ∈ Spec R | q ⊈ p},which especially says that S(p) is a thick subcategory of D. For an ideal I of R, one has K(I) ∕∈ S(p) if
and only if K(I)p ∼∕ 0, if and only if p ∈ Supp K(I) = V(I). Hence I(S(p)) = {I ⊆ R | p ∈ V(I)}, and we=
see that max I(S(p)) = {p}.
!
Lemma 3.5 says that S assigns to each prime ideal of R a prime thick subcategory of D. Conversely,s also assigns to each prime thick subcategory of D a prime ideal of R.

Lemma 3.6. For a thick subcategory X of D every maximal element of I(X ) is a prime ideal of R. In
particular, s(P) is a prime ideal of R for any prime thick subcategory P of D.

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Proof. Take a maximal element p of I(X ). Note then that p ∕= R. Consider elements a, b ∈ R with ab ∈ p.
By the octahedral axiom, there is an exact triangle K(a) → K(ab) → K(b) → K(a)[1] in D. Tensoring
K(p) with this, we obtain an exact triangle K(p + (a)) → K(p + (ab)) → K(p + (b)) → K(p + (a))[1] in
D. Since p + (ab) = p, the complex K(p + (ab)) does not belong to X . Therefore, either K(p + (a)) orK(p + (b)) is outside X . The maximality of p implies that a ∈ p or b ∈ p.
!
Lemmas 3.5 and 3.6 lead us to the construction of a pair of maps of topological spacess : Esp D ⇄ Spec R : S.

These maps are basic tools to investigate the structure of Esp D by comparison with Spec R.
It is hard to check from the definition whether a given thick subcategory is prime or not. We give a
useful characterization of prime thick subcategories in terms of their homological supports.
Theorem 3.7. Assume either (a) D = Dperf (R), or (b) R is a local complete intersection. Then for a
thick subcategory P of D it holds that
P is prime⇐⇒

Supp P = {q ∈ Spec R | q ∕⊆ p} for some p ∈ Spec R.

When the latter condition holds, one has p = s(P).

Proof. We prove the theorem step by step.
(1) Let W be a specialization-closed subset of Spec R, and let p be a prime ideal of R. Then it isstraightforward that max(W ∁ ) = {p} if and only if W = {q ∈ Spec R | q ∕⊆ p}.
(2) Let X be a thick subcategory of D. Then max(I(X ) ∩ Spec R) = max I(X ). In fact, Lemma 3.6
implies that there are inclusions max I(X ) ⊆ I(X ) ∩ Spec R ⊆ I(X ). It remains to note the general factthat for two partially ordered sets A, B with max B ⊆ A ⊆ B one has max A = max B.
(3) For each thick subcategory X of D, the equality I(X ) = {I ⊆ R | V(I) ∕⊆ Supp X } holds. In fact,case (a) follows from [16, Theorem 1.5]. Case (b) will follow if we show that K(I) ∈ X if and only if
V(I) ⊆ Supp X for an ideal I of R and a thick subcategory X of D. Taking the homological supports
shows the ‘only if’ part. For the ‘if’ part, as there exist only finitely many minimal primes of I, we canfind a complex M ∈ X such that V(I) ⊆ Supp M . Since R is assumed to be a local complete intersection,
it follows from [17, Theorem 5.2] that M is proxy small, and by [8, Proposition 4.4] the Koszul complex
K(I) is in thick M , and hence it belongs to X .
(4) For X ∈ Th D one has I(X ) ∩ Spec R = (Supp X )∁ . Indeed, for a prime ideal p of R it holds that
V(p) ⊈ Supp X if and only if p ∈ (Supp X )∁ . Using (3), we deduce the equality.
(5) Now let us show the assertion of the theorem. Let P be a thick subcategory of D, and let p be aprime ideal of R. Then Supp P = {q ∈ Spec R | q ⊈ p} if and only if max((Supp P)∁ ) = {p} by (1), if andonly if max(I(P) ∩ Spec R) = {p} by (4), if and only if max I(P) = {p} by (2), if and only if the thicksubcategory P is prime with s(P) = {p}.
!
Remark 3.8. The same characterization of prime thick tensor-ideals of Dperf (R) holds. Indeed, it follows
from [2, Corollary 5.6], every prime thick tensor-ideal of Dperf (R) is of the form {M ∈ Dperf (R) | Mp ∼= 0}
for some p ∈ Spec R. Therefore the thick subcategory P of Dperf (R) is prime if and only if Supp P ={q ∈ Spec R | q ∕⊆ p} for some p ∈ Spec R.
Here we consider composition and continuity of the maps s and S.

Corollary 3.9. (1) The maps S and s are order-reversing with s · S = 1. Moreover, S is continuous.
(2) Assume either that D = Dperf (R), or that R is a local complete intersection. Then s is also continuous, and P ⊆ S(s(P))) = Supp−1
D (Supp P) for any prime thick subcategory P of D.
Proof. (1) It is directly verified that S, s are order-reversing. Lemma 3.5 shows s · S = 1. The equality(3.9.1)S −1 (Spp M ) = Supp M

holds for each M ∈ D, which shows that S is a continuous map.
(2) Fix an ideal I of R and a prime thick subcategory P of D. If K(I) is not in P, then I belongsto I(P) and is contained in s(P). Conversely, if K(I) is in P, then V(I) is contained in Supp P. The
latter set consists of the prime ideals that are not contained in s(P) by Theorem 3.7. In particular,s(P) does not contain I. This shows that s−1 (V(I)) = Spp K(I), whence s is continuous. The equality
S(s(P))) = Supp−1!
D (Supp P) follows by Theorem 3.7 and (3.5.1).

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We may ask when the maps s and S are mutually inverse bijections. Here is an answer to this question.
Theorem 3.10. (1) One has mutually inverse homeomorphisms
Esp Dperf (R) "sS! Spec R.

(2) Let R be a local complete intersection. Then the following mutually inverse bijections are induced.
max(Esp D) "sS! min(Spec R).

(3) Let R be a local complete intersection of positive Krull dimension. Then the following are equivalent.
(i) The maps s : Esp D ⇄ Spec R : S are mutually inverse bijections.
(ii) The maps s : Esp D ⇄ Spec R : S are mutually inverse homeomorphisms.
(iii) One has D = Dperf (R).

Proof. (1) We use Corollary 3.9. It suﬃces to show that S is surjective. Let P ∈ Esp Dperf (R). Then we
have S(s(P)) = Supp−1
Dperf (R) (Supp P) = P, whose last equality follows from [16, Theorem 1.5].
(2) Again, we use Corollary 3.9. We have only to show that S : min(Spec R) → max(Esp D) is awell-defined surjection. First, let p ∈ min(Spec R). Take a prime thick subcategory P of D containing
S(p). We have s(P) ⊆ s(S(p)) = p, and p = s(P) by the minimality of p. Therefore P ⊆ S(s(P)) = S(p),and we get P = S(p). Thus the map S : min(Spec R) → max(Esp D) is well-defined. Next, takeP ∈ max(Esp D). The inclusion P ⊆ S(s(P)) and the maximality of P show that P = S(s(P)). Choose aminimal prime p of R contained in s(P). Then P = S(s(P)) ⊆ S(p). Again the maximality of P implies
P = S(p). Thus S : min(Spec R) → max(Esp D) is surjective.
(3) It follows from the assertion (1) that (iii) implies (ii), while it is trivial that (ii) implies (i). Let us
show that (i) implies (iii). Denote by m the maximal ideal of R. Since dim R > 0, there is a prime ideal
p of R strictly contained in m. Set P = thick{K(q) | q ∈ Spec R, q ⊈ p}. Then we can verify that theequality Supp P = {q ∈ Spec R | q ∕⊆ p} holds. Theorem 3.7 implies that P is a prime thick subcategory
perfwith s(P) = p, whence P = S(s(P)) = S(p). We have Supp−1(R).
D {m} ⊆ S(p) = P ⊆ D
Take any X ∈ D. Taking a truncation of a projective resolution of X gives rise to an exact triangle
P → X → M [n] ⇝ in Db (R), where P is a perfect complex, M is a maximal Cohen–Macaulay R-moduleand n is an integer. Let x be a maximal regular sequence on R. Then x is also a regular sequence onperf
M , and M/xM ∈ Supp−1(R). This shows that M/xM has finite projective dimension as an
D {m} ⊆ D
R-module. It follows from [7, Exercise 1.3.6] that M ∈ Dperf (R), and we conclude that X ∈ Dperf (R). !
Applying Theorem 3.10(3) to D = Db (R), we immediately obtain the following result.
Corollary 3.11. Let R be a local complete intersection with dim R > 0. Then the following are equivalent.
(1) The maps s : Esp Db (R) ⇄ Spec R : S are mutually inverse bijections.
(2) The maps s : Esp Db (R) ⇄ Spec R : S are mutually inverse homeomorphisms.
(3) The ring R is regular.
Remark 3.12. If R is an artinian complete intersection local ring, then the maps s : Esp Db (R) ⇄Spec R : S are always mutually inverse homeomorphism. Indeed, write Spec R = {m}. Let P be a prime
thick subcategory of Db (R). Then Theorem 3.7 implies that Supp P = {p ∈ Spec R | p ∕⊆ m} = ∅. Thisimplies P = 0, and thus Esp Db (R) = {0}. It remains to note that S(m) = 0 and s(0) = m.
From now on, we discuss the relationship of radical thick subcategories of D with tame thick subcategories, which are defined below.
Definition 3.13. (1) A thick subcategory X of D is called tame if there is a subset W of Spec R suchthat X = Supp−1
D W . In this case, W can be taken as a specialization-closed subset; in fact, it holds that−1
−1
Supp−1W=
Supp
D
D (Supp(SuppD W )). Denote by Tame D the set of tame thick subcategories of D.
−1tame(2) We put X:= SuppD (Supp X ) for each full subcategory X of D. This is the smallest tame thicksubcategory of D containing X . In this sense, we call it the tame closure of X .
For a topological space X, we denote by Spcl X the set of specialization-closed subsets of X, that is,
the unions of closed subsets of X. The theorem below complements the bijection given in Theorem 2.9.

8

HIROKI MATSUI AND RYO TAKAHASHI

Theorem 3.14. There is a diagram of maps of sets
Spp

Rad
$ D"
inc

Spp

−1

()tame! Param D
$
s−1

S −1

#
#
Supp
! Spcl(Spec R),
Tame D "Supp−1D

where inc stands for the inclusion map. The horizontal maps are mutually inverse bijections, and thecompositions of the maps ending at bottom sets are commutative (in particular, the composition of thetwo vertical maps starting from each bottom set is the identity). Moreover, Rad D = Tame D if and only
if the maps s : Esp D ⇄ Spec R : S are mutually inverse bijections.
Proof. Let us prove the theorem step by step.
(1) We have already got the mutually inverse bijections (Spp, Spp−1 ) in Theorem 2.9.
(2) Using Remark 3.2(1), we see that W = Supp(Supp−1
D W ) for a specialization-closed subset W ofSpec R. This yields the mutually inverse bijections (Supp, Supp−1).
D!
(3) It is easy to check that there is an equality Supp−1
W=Dp∈W ∁ S(p) for any subset W of Spec R.
This shows Tame D ⊆ Rad D, and we have ()tame · inc = 1.
(4) It is seen from (3.9.1) that S −1 (Spp X ) = Supp X for a full subcategory X of D, which shows that−1
S is well-defined. We claim that for a specialization-closed subset W of Spec R there are equalities
s−1 (W ) = Spp{K(p) | p ∈ W } = Spp(thick{K(p) | p ∈ W }).
Indeed, the second equality follows from Remark 2.2. To show the first equality, pick any P ∈ s−1 (W )and put p := s(P) ∈ W . Then p ∈ I(P), which implies K(p) ∈/ P and hence P ∈ Spp K(p). Conversely,
pick any p ∈ W and P ∈ Spp K(p). Then K(p) ∈/ P, which implies p ∈ I(P). The unique maximalityof s(P) shows that s(P) contains p, and we get s(P) ∈ W since W is specialization-closed. Therefore
P belongs to s−1 (W ). Thus the claim follows, which shows that s−1 is well-defined. Since s · S = 1 by
Corollary 3.9(1), we see that S −1 · s−1 = 1.
(5) We have S −1 (Spp X ) = Supp X = Supp(X tame ) for a full subcategory X of D, which implies−1
S · Spp = Supp ·()tame . Using this equality and the fact that the horizontal maps are bijections which−1
we have already seen in (1) and (2), we can easily deduce ()tame · Spp−1 = Supp−1.
D ·S
Finally, let us prove the last assertion of the theorem. The ‘if’ part is easily deduced from thecommutative diagram. To show the ‘only if’ part, assume Rad D = Tame D. Then every prime thicksubcategory of D is tame, and hence it is in the image of S by Corollary 3.9(2).
!
Now we give a partial answer to Question 2.10.
Corollary 3.15. There are equalities and a one-to-one correspondence
Th Dperf (R) = Tame Dperf (R) = Rad Dperf (R) "

Spp
Spp

−1

! Param Dperf (R) = Thom(Esp Dperf (R)).

Furthermore, the maps in the diagram in Theorem 3.14 are all bijections for D = Dperf (R).
Proof. The Hopkins–Neeman theorem [16, Theorem 1.5] says that all the thick subcategories of Dperf (R)are tame. It follows from Theorem 3.14 that every tame thick subcategory of Dperf (R) is radical. Hence
we obtain the equalities Th Dperf (R) = Tame Dperf (R) = Rad Dperf (R).
Let us show the equality Param Dperf (R) = Thom(Esp Dperf (R)). The homeomorphisms s, S induce
mutually inverse bijections between the Thomason subsets of Esp Dperf (R) and those of Spec R. Notehere that a subset of Spec R is Thomason if and only if it is specialization-closed. Therefore, for each
T ∈ Thom(Esp Dperf (R)) we have s(T ) ∈ Thom(Spec R) = Spcl(Spec R), and T = s−1 (s(T )) belongs
to Param(Dperf (R)) by Theorem 3.14. Conversely, for any U ∈ Param(Dperf (R)) we have S −1 (U ) ∈Spcl(Spec R) = Thom(Spec R) by Theorem 3.14 again, whence U = S(S −1 (U )) ∈ Thom(Esp Dperf (R)).
We now conclude that Param Dperf (R) = Thom(Esp Dperf (R)). Combining Theorem 2.9, we are done. !

CONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES

9

Remark 3.16. Recall that the derived category Dperf (R) has the structure of a tensor-triangulated
category, so that Balmer’s theory can be applied to it. The one-to-one correspondence in Corollary 3.15
is identified with Theorem 1.1 for T = Dperf (R). Indeed, there are equalities
Esp Dperf (R) = Spc Dperf (R),Spp = supp,

where supp stands for the Balmer support.
Let us show these equalities. The latter one follows from the former one and the definitions of Spp
and supp. The former equality follows from (i) and (ii) below; let P be a thick subcategory of Dperf (R).
(i) Suppose that P belongs to Esp Dperf (R). Then P ∕= Dperf (R). As any thick subcategory of Dperf (R)is a tensor-ideal, so is P. Setting p := s(P), we have P = S(p) by Theorem 3.10(1). Let X, Y ∈ Dperf (R)be objects such that X ⊗R Y ∈ P. Then Xp ⊗Rp Yp ∼= 0 in Dperf (Rp ), which implies either Xp ∼= 0 or
∼Yp = 0. Hence either X ∈ P or Y ∈ P holds, and we conclude that P ∈ Spc Dperf (R).
(ii) Suppose that P ∈ Spc Dperf (R), and let I, J be maximal elements of I(P). Then neither K(I) nor
K(J) is in P. Since P ∈ Spc Dperf (R), we have K(I +J) = K(I)⊗R K(J) ∈/ P. Hence I +J ∈ I(P), and the
maximality of I, J shows I = I +J = J. Thus I(P) has a unique maximal element, and P ∈ Esp Dperf (R).
We close this section by proving a theorem on the dimension of D, which improves Theorem 3.10(3).
Theorem 3.17. (1) There is an inequality dim D ≥ dim R.
(2) Consider the following conditions:(a) dim D = dim R, (b) dim Esp D = dim R, (c) Esp D ∼= Spec R, (d) D = Dperf (R).
Then the implications (d) ⇒ (c) ⇒ (b) ⇒ (a) hold. If R is a complete intersection local ring withpositive Krull dimension, then the implication (a) ⇒ (d) holds as well.
Proof. (1) The assertion follows from Corollary 3.9(1).
(2) Theorem 3.10(1) shows the implication (d) ⇒ (c), while (b) ⇒ (a) is shown by Remark 2.6 and(1).
To show (c) ⇒ (b), let f : Esp D → Spec R be a homeomorphism. Then Esp D is a sober space since
so is Spec R, and we obtain dim Esp D = dim D by Remark 2.6. Note that f ({P}) = {f (P)} for eachP ∈ Esp D. Using Proposition 2.3, we see that P ⊆ Q in Esp D if and only if f (P) ⊇ f (Q). It follows
that dim D = dim R, and consequently, dim Esp D = dim R.
Finally, we show the implication (a) ⇒ (d), assuming that (R, m) is a local complete intersection with
d := dim R > 0. Take a chain p0 ⊊ · · · ⊊ pd = m of prime ideals of R. Applying S gives rise to a chain
0 = S(m) = S(pd ) ⊊ S(pd−1 ) ⊊ · · · ⊊ S(p0 )of prime thick subcategories of D. Since D has dimension d, we have ht S(pd−1 ) = 1. It follows from
Corollary 3.9(2) and the fact s(0) = m that s−1 (pd−1 ) = {S(pd−1 )}. Replacing p with pd−1 in the proofof Theorem 3.10(3), we obtain D = Dperf (R).
!
4. Spectra of singularity categories
In this section, we investigate the spectra of singularity categories of commutative noetherian rings.
The flow will go similar to the previous section, while the results which will be obtained in this section
are completely independent.
Throughout this section, we assume that R is a commutative noetherian ring. Recall that the singularity category Dsg (R) of R is defined as the Verdier quotientDsg (R) := Db (R)/ Dperf (R).
By definition Dsg (R) is a triangulated category, and we can define its spectrum EspC Dsg (R) for a given
C. The purpose of this section is to explore the structure of the topological space EspC Dsg (R).
We denote by Sing R the singular locus of R, that is, the set of prime ideals p of R such that the local
ring Rp is singular (i.e. nonregular). We equip Sing R with the relative topology to regard it as a subspaceof Spec R. First of all, let us recall the definition of a singular support, which plays an important role inthis section.
Definition 4.1. For an object M ∈ Dsg (R), we define the singular support of M by
Supp M := {p ∈ Sing R | Mp ∼∕ 0 in Dsg (Rp )} = {p ∈ Sing R | pd Mp = ∞}.
=
Rp

10

HIROKI MATSUI AND RYO TAKAHASHIFor a full subcategory X ⊆ Dsg (R), we define the singular support of X by"Supp X := M ∈X Supp M.

Here we state some basic properties of singular supports, which correspond to Remark 3.2.

Remark 4.2. (1) For p ∈ Sing R it holds that Supp(R/p) = V(p).
(2) For M ∈ Dsg (R) the set Supp M is Zariski-closed.
(3) For M ∈ Dsg (R) one has Supp M = ∅ if and only if M ∼= 0.
(4) For M ∈ Dsg (R) one has Supp(M [1]) = Supp M .
(5) For M, N ∈ Dsg (R) the equality Supp(M ⊕ N ) = Supp M ∪ Supp N holds.
(6) For an exact triangle L → M → N → L[1] in Dsg (R) it holds that Supp M ⊆ Supp L ∪ Supp N .
In particular, for each subset W of Sing R the full subcategorySupp−1 W := {M ∈ Dsg (R) | Supp M ⊆ W }.

of Dsg (R) is a thick subcategory.

We adopt the following definition of prime thick subcategories of Dsg (R). We should compare thiswith Definition 3.3. The way to define the set I(X ) is quite diﬀerent, but the definition of a prime thicksubcategory is similarly done.
Definition 4.3. For each X ∈ Th(Dsg (R)) we define the set I(X ) of ideals of R byI(X ) := {I ⊆ R | V(I) ⊆ Sing R, R/I ∕∈ X }.

We say that a thick subcategory P is prime if I(P) has a unique maximal element. Denote by s(P)the maximal element of I(P) and by C the set of prime thick subcategories of Dsg (R). We obtain a
topological space EspC Dsg (R) as defined in 2.1.
Similarly to the previous section, we simply write Esp Dsg (R), Rad Dsg (R) and Param Dsg (R) by omitting the subscripts C. For each p ∈ Sing R we define the full subcategory S(p) of Dsg (R) by
S(p) := {M ∈ Dsg (R) | Mp ∼= 0 in Dsg (Rp )}.
Analogous statements to Lemmas 3.5 and 3.6 hold true, although the proof is rather diﬀerent.

Lemma 4.4. (1) Let p ∈ Sing R. Then S(p) is a prime thick subcategory of Dsg (R) with s(S(p)) = p.
(2) For a thick subcategory X every maximal element of I(X ) belongs to Sing R. In particular, s(P)belongs to Sing R for any prime thick subcategory P of Dsg (R).
Thus one obtains a pair of mapss : Esp Dsg (R) ⇄ Sing R : S.
Proof. (1) We easily verify the following, which shows that the subcategory S(p) of Dsg (R) is thick.

S(p) = Supp−1 {q ∈ Sing R | q ⊈ p}.
Since p is in Sing R, we have (R/p)p ∼∕ 0 in Dsg (Rp ). Hence p belongs to I(S(p)). Any ideal I belonging=
to I(S(p)) satisfies R/I ∕∈ S(p), which implies I ⊆ p. Thus, p is a unique maximal element of I(S(p)).
(2) Pick a maximal element p of I(X ); note p ∕= R. Let a, b ∈ R \ p be elements with ab ∈ p. Then theideals p + (a) and p : a strictly contain p. Therefore R/p + (a) and R/(p : a) ∼= p + (a)/p belong to X by
the maximality of p. The exact sequence 0 → p + (a)/p → R/p → R/p + (a) → 0 implies R/p ∈ X , which
is a contradiction. Thus p is prime. As p ∈ I(P), we have V(p) ⊆ Sing R, which implies p ∈ Sing R.
!
(4.4.1)

The following proposition should be compared with Corollary 3.9. They are similar, but there aresome diﬀerences. In particular, the inclusion relations between P and S(s(P)) are opposite.
Proposition 4.5. (1) The maps S and s are order-reversing with s · S = 1, and S is continuous.
(2) The map s is also continuous, if either s is injective and Sing R is closed, or Sing R is finite.
(3) Suppose that R is a Gorenstein local ring. For any P ∈ Esp Dsg (R) there is an inclusion S(s(P)) ⊆ P.
Proof. (1) It is straightforward that S, s are order-reversing maps, while Lemma 4.4(1) implies s · S = 1.
(2) First of all, we claim that if Sing R is closed and so is s−1 (V(p)) for each p ∈ Sing R, then s is
continuous. Indeed, write Sing R = V(I) with an ideal I of R. Each closed subset of Sing R is of the form
V(J) ∩ Sing R = V(I + J), where J is an ideal of R. Let p1 , . . . , pn be the minimal primes of I + J. ThenCONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES

11−1
for each i the prime ideal"n pi is in the singular locus of R, and s (V(pi )) is closed by assumption. Hences−1 (V(J) ∩ Sing R) = i=1 s−1 (V(pi )) is also closed, and the claim follows.
Now, assume that s is injective. Then s is a bijection with s−1 = S. Fix P ∈ Esp Dsg (R). We have

Supp P = Supp S(s(P)) = {q ∈ Sing R | q ∕⊆ s(P)}by (4.4.1) and Remark 4.2(1). Take any p ∈ Sing R. Then P ∈ s−1 (V(p)) if and only if p ⊆ s(P). Also,
P ∈ Spp(R/p) if and only if R/p ∈/ P, and in this case p ∈ I(P) and hence p ⊆ s(P). If p ⊆ s(P) andR/p ∈ P, then p ∈ V(p) = Supp(R/p) ⊆ Supp P, which gives a contradiction. Consequently, we obtains−1 (V(p)) = Spp(R/p) for all p ∈ Sing R. The above claim shows that s is continuous.
Next, assume that Sing R is a finite set. Then, in particular, Sing R is a closed subset of Spec R, andby the above claim it suﬃces to show that for each p ∈ Sing R one has"
s−1 (V(p)) = q∈V(p) Spp(R/q)

because by assumption the union is finite and hence it is closed. Note that each q ∈ V(p) is in the singularlocus of R. A prime thick subcategory P of Dsg (R) belongs to the right-hand side of the above equality
if and only if R/q ∈/ P for some q ∈ V(p), if and only if there exists a prime ideal q of R such thatp ⊆ q ∈ I(P), if and only if p ⊆ s(P), if and only if P belongs to the left-hand side of the above equality.
(3) Consider the thick subcategory X := thick{R/p | p ∈ Sing R, p ∕⊆ s(P)} of Dsg (R). We observethat X is contained in P and that the equality Supp X = {p ∈ Sing R | p ∕⊆ s(P)} holds. Therefore, R/pbelongs to X for every p ∈ Supp X . Since R is a Gorenstein local ring, we can apply [20, Corollary 4.11]
to get the equality X = Supp−1 (Supp X ), whose right-hand side coincides with S(s(P)) by (4.4.1).
!
Let R be a local ring with residue field k. For M ∈ Db (R), we define its complexity by

cxR M := inf{c ∈ Z≥0 | dimk ExtnR (M, k) ≤ rnc−1 for some r ∈ R and for all n ≫ 0}.

We refer the reader to [1, §4.2] for fundamental properties of this numerical invariant. Since cxR P = 0
for P ∈ Dperf (R), the complexity is well-defined on the isomorphism class of each object of Dsg (R). Now
we state and prove the following theorem, which should be compared with Theorem 3.10.
Theorem 4.6. Let R be a local ring with closed singular locus and admitting a complex of finite positive
complexity (e.g., let R be a singular excellent local complete intersection). The following are equivalent.
(1) The maps s : Esp Dsg (R) ⇄ Sing R : S are mutually inverse bijections.
(2) The maps s : Esp Dsg (R) ⇄ Sing R : S are mutually inverse homeomorphisms.
(3) The ring R is a hypersurface.
Proof. The implication (2) ⇒ (1) is clear.
We show the implication (1) ⇒ (3). Let C be the full subcategory of Dsg (R) consisting of complexes
with finite complexity. Then C is a thick subcategory, and C ∕= 0 by assumption. If C does not contain theresidue field k of R, then I(C) contains the maximal ideal m of R, that is, C is a prime thick subcategory
with s(C) = m. Hence C = S(s(C)) = S(m) = 0, which gives a contradiction. Thus k is in C, and R is acomplete intersection by [10, Theorem 2.3]. There is a finitely generated R-module M with complexity 1
by [6, Proposition 2.2]. The full subcategory C ′ consisting of objects of Dsg (R) with complexity at most
1 is thick and nonzero. An analogous argument as above shows k ∈ C ′ . Therefore, R is a hypersurface.
Finally, we prove the implication (3) ⇒ (2). In view of (1) and (2) of Proposition 4.5, it suﬃces toshow that S is surjective. We first claim that
S(p) = {X ∈ Dsg (R) | R/p ∕∈ thick X}

for each p ∈ Sing R. Indeed, for each X ∈ Dsg (R), one has X ∕∈ S(p) if and only if Xp ∼∕ 0, if and only if=
V(p) ⊆ Supp X, if and only if Supp(R/p) ⊆ Supp X. Since R is a local hypersurface, the last condition isequivalent to saying that R/p ∈ thick X by [20, Theorem 6.8]. Thus the claim follows.
Let P be a prime thick subcategory of Dsg (R). Setting p = s(P), we have S(p) = S(s(P)) ⊆ P by
Proposition 4.5(3). Take any object X ∈ P. Then thick X is contained in P and R/p = R/s(P) does
not belong to P, which yields R/p ∈/ thick X. The above claim implies that X belongs to S(p), and we
obtain S(p) = P. It follows that S is surjective, which completes the proof of the assertion.
!

12

HIROKI MATSUI AND RYO TAKAHASHI

Remark 4.7. There is a case where R is not a hypersurface but s, S give mutually inverse homeomorphisms. For instance, let R be a Cohen–Macaulay local ring with quasi-decomposable maximal ideal (inthe sense of [15]) which is locally a hypersurface on the punctured spectrum. Then, by [15, Theorem
4.5] every thick subcategory of Dsg (R) is of the form Supp−1 W with W a specialization-closed subset ofSing R, even if R is not a hypersurface. Hence, the same argument as above proves what we want. For
such a ring R, the existence of complexes of finite positive complexity fails.
Next, as we did in the previous section, we introduce tame thick subcategories of Dsg (R) and relatethem with radical ones. Most of the arguments in the previous section does work for Dsg (R) just by replacing D, SuppD , Spec, K(p) with Dsg (R), Supp, Sing, R/p respectively. We will give definitions, properties,
results, and proofs that are essentially diﬀerent from the ones given in the previous section.
Definition 4.8. A thick subcategory X of Dsg (R) is said to be tame if there exists a subset W of Sing R
such that X = Supp−1 W . Denote by Tame Dsg (R) the set of tame thick subcategories of Dsg (R). We
put X tame := Supp−1 (Supp X ) for each full subcategory X of Dsg (R). Note that X tame is the smallest
tame thick subcategory of Dsg (R) containing X , and we call X tame the tame closure of X .
!
Proposition 4.9. (1) For W ⊆ Sing R one has Supp−1 (Supp(Supp−1 W )) = Supp−1 W = p∈W ∁ S(p).
(2) For a full subcategory X of Dsg (R) it holds that S −1 (Spp X ) = Supp X = Supp(X tame ).
(3) For a specialization-closed subset W of Sing R the equality s−1 (W ) = Spp(thick{R/p | p ∈ W }) holds.
If R is a Gorenstein local ring, then the equality s−1 (W ) = Spp(Supp−1 W ) also holds.
Proof. The only statement essentially diﬀerent from what we got in the previous section is the latter
assertion of (3), so we only give a proof of it. Let P be a prime thick subcategory of Dsg (R), and set
p := s(P). If P ∈ s−1 (W ), then p ∈ W and Supp(R/p) = V(p) ⊆ W as W is specialization-closed. Hence
R/p ∈ Supp−1 W . We have R/p ∈/ P since p ∈ I(P). Thus P ∈ Spp(R/p) ⊆ Spp(Supp−1 W ). Conversely,−1
suppose P ∈ Spp(Supp W ). Then P ∈ Spp X for some X ∈ Supp−1 W . If Xp ∼= 0, then by Proposition
4.5(3) we get X ∈ S(p) = S(s(P)) ⊆ P, which implies P ∈/ Spp X, a contradiction. Therefore Xp ∼∕ 0,=which gives p ∈ Supp X ⊆ W . Thus s(P) = p ∈ W , and we obtain P ∈ s−1 (W ). Now we conclude thatthe equality s−1 (W ) = Spp(Supp−1 W ) holds.
!
We should compare the following Theorem 4.10 with Theorem 3.14.
Theorem 4.10. Suppose that R is a Gorenstein local ring. Then there is a diagram of maps of sets
Spp

! Param D (R)Rad Dsg (R) "
$
$ sg
Spp−1
inc

()tame

s−1

S −1

Supp
#
#
! Spcl(Sing R)
Tame Dsg (R) "Supp−1

The horizontal maps are mutually inverse bijections, and the compositions of the maps ending at bottom
sets are commutative. Furthermore, Rad Dsg (R) = Tame Dsg (R) if and only if the maps s : Esp Dsg (R) ⇄Sing R : S are mutually inverse bijections.
Proof. The only statement essentially diﬀerent from Theorem 3.14 is the new commutativity relations
Spp ·inc = s−1 · Supp and inc · Supp−1 = Spp−1 ·s−1 , but the former follows from the latter and the twohorizontal one-to-one correspondences in the diagram. Take any W ∈ Spcl(Sing R). Then we have(inc · Supp−1 )(W ) = Supp−1 W = (Spp−1 · Spp)(Supp−1 W ) = (Spp−1 ·s−1 )(W ),where the last equality follows from Proposition 4.9(3). We conclude that inc · Supp−1 = Spp−1 ·s−1 .

!

The following result corresponds to Corollary 3.15, which also gives a partial answer to Question 2.10.
Corollary 4.11. Let R be a local hypersurface with closed singular locus. One then has the following
equalities and one-to-one correspondence, and the maps in the diagram in Theorem 4.10 are all bijections.
Th Dsg (R) = Tame Dsg (R) = Rad Dsg (R) "

Spp
Spp

−1

! Param D (R) = Thom(Esp D (R)).
sgsg

CONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES

13

Proof. It follows from [20, Main Theorem] that every thick subcategory of Dsg (R) is tame. It is easy to
see that a subset of Sing R is Thomason if and only if it is specialization-closed. Now essentially the same
argument as in the proof of Corollary 3.15 works.
!
Next, we further study prime thick subcategories of Dsg (R). We introduce the following notion.
Definition 4.12. For a thick subcategory X of Dsg (R), we define the tame interior Xtame of X as thelargest tame thick subcategory of Dsg (R) contained in X .
In the case where R is a Gorenstein local ring, one can describe tame interiors explicitly as follows.
Lemma 4.13. Let R be a Gorenstein local ring. For a thick subcategory X of Dsg (R) one has
Xtame = thick{R/p | p ∈ Sing R and V(p) ∩ max I(X ) = ∅}.

Proof. Let Y be the right-hand side. The set W := {p ∈ Sing R | V(p) ∩ max I(X ) = ∅} is a specializationclosed subset of Sing R, and Y is a tame thick subcategory with singular support W by [20, Corollary 4.11].
If p ∈ Sing R is such that R/p ∈/ X , then p ∈ I(X ) and V(p) ∩ max I(X ) ∕= ∅. Therefore, Supp R/p = V(p)is not contained in W , which shows that R/p is not in Supp−1 W = Y. Thus, Y is contained in X .
It remains to show the maximality of Y, and for this, it suﬃces to verify that if p ∈ Sing R is suchthat Supp−1 V(p) is contained in X , then Supp−1 V(p) is contained in Y. Assume that Supp−1 V(p) is
not contained in Y. Then V(p) is not contained in W , and we find an element q ∈ V(p) such that
V(q) ∩ max I(X ) ∕= ∅. Hence p is contained in some r ∈ max I(X ). Then R/r ∈ Supp−1 V(p) but R/r ∕∈ X ,which is a contradiction. Consequently, Supp−1 V(p) is contained in Y, and we are done.
!
We obtain a characterization of the prime thick subcategories of Dsg (R) in terms of tame interiors.
Proposition 4.14. Let R be a Gorenstein local ring. A thick subcategory X of Dsg (R) is prime if andonly if Xtame = S(p) for some p ∈ Sing R. When this is the case, one has s(X ) = p.
Proof. Suppose that X is a prime thick subcategory of Dsg (R). Set p = s(X ). Proposition 4.5(3) implies
that S(p) is contained in X , while S(p) is tame by (4.4.1). Hence S(p) is contained in Xtame . Letq ∈ Sing R be such that V(q) ∩ max I(X ) = ∅. Then q is not contained in p, and R/q is in S(p). Using
Lemma 4.13, we obtain Xtame = S(p).
Conversely, assume Xtame = S(p). Then the equality (Supp(Xtame ))∁ = (Supp S(p))∁ holds, which gives{q ∈ Sing R | V(q) ∩ max I(X ) ∕= ∅} = {q ∈ Sing R | q ⊆ p}

by Lemma 4.13 and (4.4.1). The right-hand side has a unique maximal element, which is p. Note thatevery maximal element of I(X ) is also maximal in the left-hand side. Thus p is the only maximal elementof I(X ), which means that X is a prime thick subcategory with s(X ) = p.
!
Except trivial examples S(p), it is diﬃcult in general to find prime thick subcategories of Dsg (R). Our
next aim is to provide methods to construct new prime thick subcategories from given or trivial ones.
Proposition 4.15. Let R be a Gorenstein local ring. Let P, Q be prime thick subcategories of Dsg (R).
(1) Assume P ⊆ Q and s(P) = s(Q). Then any thick subcategory X with P ⊆ X ⊆ Q is a prime thicksubcategory of Dsg (R) satisfying s(X ) = s(Q).
(2) If s(P) ⊆ s(Q), then P ∩ Q is a prime thick subcategory of Dsg (R) satisfying s(P ∩ Q) = s(Q).
Proof. (1) There are inclusions Ptame ⊆ Xtame ⊆ X ⊆ Q. By Proposition 4.14 we have Ptame = S(s(P)) =S(s(Q)) = Qtame , which is the largest tame thick subcategory contained in Q. Therefore, Xtame = Ptame =S(s(P)). Applying Proposition 4.14 again, we see that X is prime and s(X ) = s(Q).
(2) Proposition 4.14 implies Qtame = S(s(Q)) ⊆ S(s(P)) = Ptame . As Qtame ⊆ Q and Qtame ⊆ Ptame ⊆P, we get Qtame ⊆ P ∩ Q ⊆ Q. It follows from (1) that P ∩ Q is prime and s(P ∩ Q) = s(Q).
!
We introduce the notion of covers to state our next results.
Definition 4.16. Let T be a triangulated category. Let X , Y be thick subcategories of T . Then we
say that Y is a cover of X if Y properly contains X and there are no thick subcategories Z of T withX ⊊ Z ⊊ Y.

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It is unclear in general whether a cover of a given thick subcategory exists or not. The following
proposition gives us suﬃcient conditions for the existence of covers.
Proposition 4.17. Let R be either• a complete intersection which is a quotient of a regular local ring, or• a Cohen–Macaulay local ring with quasi-decomposable maximal ideal which is locally a hypersurface on the punctured spectrum.
Then the zero subcategory 0 of Dsg (R) admits a cover. If in addition Sing R is finite, then S(p) admits
a cover as well for each p ∈ Sing R.
Proof. In either case, there is an order isomorphism between Th(Dsg (R)) and the set of specializationclosed subsets of a noetherian topological space X; see [15, 19]. By [13, Lemma 2.9], it restricts to anorder isomorphism between the set T := {thick M | M ∈ Dsg (R)} and the set of closed subsets of X.
Thus T satisfies the descending chain condition (with respect to the inclusion relation), and we can takea minimal element X := thick M of T \ {0}. We can easily check that X is minimal among all nonzero
thick subcategories, i.e., X is a cover of 0. The last assertion of the proposition is shown similarly.
!
We obtain a suﬃcient condition for a given thick subcategory to be prime, using the notion of covers.
Proposition 4.18. Let R be a Gorenstein local ring. Let X be a non-tame thick subcategory of Dsg (R)and p ∈ Sing R. If X is a cover of S(p), then X is a prime thick subcategory of Dsg (R) with s(X ) = p.
Proof. By (4.4.1), the thick subcategory S(p) is tame. As S(p) is contained in X and X is not tame, wehave S(p) ⊆ Xtame ⊊ X . The assumption that X is a cover of S(p) implies S(p) = Xtame . It follows from
Proposition 4.14 that X is a prime thick subcategory of Dsg (R) such that s(X ) = p.
!
As another application of covers, we get criteria for a Cohen–Macaulay local ring to be a hypersurface.
Theorem 4.19. Let R be a singular Cohen–Macaulay local ring possessing a complex of finite positive
complexity. Assume that 0 admits at least one cover (e.g., R is a complete intersection which is a quotientof a regular local ring). Then the following are equivalent.
(1) R is a hypersurface.
(2) Every cover of 0 is tame.
(3) There is a tame cover of 0.
Proof. The implication (1) ⇒ (2) follows from [20, Theorem 6.8], while the implication (2) ⇒ (3) istrivial. Let us show the implication (3) ⇒ (1). We denote by m the maximal ideal of R, by k = R/m
the residue field of R, and put d = dim R. Let X be a tame cover of 0. Then X = Supp−1 {m} since it isminimal among the nonzero tame thick subcategories of Dsg (R). Again the minimality of X shows thatfor each nonzero object X ∈ X one has thick X = X = Supp−1 {m}, which contains k.
By assumption, there exists a maximal Cohen–Macaulay R-module M with finite positive complexity.
Take a maximal regular sequence x = x1 , . . . , xd on R, and set N = M/xM . Then N is an R-module offinite length and with finite positive complexity. As N belongs to Supp−1 (m), the above argument saysthat thick N contains k. Since the full subcategory of objects of Dsg (R) with finite complexity is thick,k has finite complexity. Therefore, R is complete intersection by [10, Theorem 2.3] and cxR C = cxR k =codim R for any 0 ∕= C ∈ Dsg (R). It follows from [6, Proposition 2.2] that there is a finitely generated
R-module L with complexity 1. Then codim R = cxR L = 1, which means that R is a hypersurface. !
Remark 4.20. Let (R, m) be a Gorenstein singular local ring. If s−1 (m) consists only of S(m) = 0, thenSupp−1 {m} is a unique cover of 0. Indeed, let X ∕= 0 be a thick subcategory of Dsg (R). If Xtame = 0, thenX ∈ s−1 (m) by Proposition 4.14, and X = 0. This contradiction shows Xtame ∕= 0. As m ∈ Supp(Xtame ),0 ⊊ Supp−1 {m} ⊆ Supp−1 Supp(Xtame ) = Xtame ⊆ X .
This proves that Supp−1 {m} is a unique cover of 0. In particular, every cover of 0 is tame. By Theorem
4.19, we obtain another proof of Theorem 4.6 in the Gorenstein case.
Finally, we prove a result corresponding to Theorem 3.17, which is an application of Theorem 4.19.
Theorem 4.21. (1) One has the inequality dim Dsg (R) ≥ dim Sing R.

CONSTRUCTION OF SPECTRA OF TRIANGULATED CATEGORIES

15(2) Consider the following four conditions.
(a) dim Dsg (R) = dim Sing R. (b) dim Esp Dsg (R) = dim Sing R.
(c) Esp Dsg (R) ∼(d) R is a hypersurface.
= Sing R.
Then the implications (c) ⇒ (b) ⇒ (a) holds. The implication (d) ⇒ (c) holds if R is a singular local
ring with closed singular locus. The implication (a) ⇒ (d) holds if R is a Gorenstein local ring withclosed singular locus possessing a complex of finite positive complexity.
Proof. (1) Proposition 4.5(1) shows the assertion.
(2) It is clear that (c) implies (b), while it follows from (1) and Remark 2.6 that (b) implies (a). If R
is local and Sing R ∕= ∅ is closed, then Theorem 4.6 shows that (d) implies (c).
Let us show that (a) implies (d) under the assumption that (R, m) is a Gorenstein local ring with closed
singular locus possessing a complex of finite positive complexity. Set n := dim Dsg (R) = dim Sing R. Takea chain p0 ⊊ · · · ⊊ pn = m in Sing R, and apply S. We get a chain
0 = S(m) = S(pn ) ⊊ S(pn−1 ) ⊊ · · · ⊊ S(p0 )in Esp Dsg (R). As dim Dsg (R) = n, we have ht S(pn−1 ) = 1. Fix a prime thick subcategory P of Dsg (R)with ht P = 1. Then P is a cover of S(m) = 0. If P is not tame, then P ∈ s−1 (m) by Proposition 4.18.
Hence S(pn−1 ) ∩ P ∈ s−1 (m) by Propositions 4.15(2) and 4.5(1). As S(pn−1 ) is tame by (4.4.1), it is notequal to P. Thus S(pn−1 ) and P are distinct prime thick subcategories of height 1, which forces us to haveS(pn−1 ) ∩ P = 0. Similarly to the last paragraph of the proof of Theorem 3.10, for a maximal Cohen–
Macaulay R-module M in P and a maximal regular sequence x on R, we have M/xM ∈ S(pn−1 )∩P = 0,
and M ∼
= 0 in Dsg (R), whence P = 0, a contradiction. Thus any height 1 prime thick subcategory of
Dsg (R) is tame, and so is every cover of 0. It follows from Theorem 4.19 that R is a hypersurface.
!
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HIROKI MATSUI AND RYO TAKAHASHI

Graduate School of Mathematical Sciences, University of Tokyo, 3-8-1 Komaba, Meguro-ku, Tokyo 1538914, Japan
Email address: mhiroki@ms.u-tokyo.ac.jp
Graduate School of Mathematics, Nagoya University, Furocho, Chikusaku, Nagoya, Aichi 464-8602,
Japan/Department of Mathematics, University of Kansas, Lawrence, KS 66045-7523, USA
Email address: takahashi@math.nagoya-u.ac.jp
URL: http://www.math.nagoya-u.ac.jp/~takahashi/