Introduction
Motivation
Let k k k k k k k k k [30 Theorem 1.6] ; as such, they possess good homological properties (with minimal dimension). While the structure of separable algebras over a field is well-understood, separable algebras over a general commutative base ring are also extensively studied due to their foundational role in several areas such as algebra, number theory, and algebraic geometry. A special class of separable algebras are Azumaya algebras, which serve as key ingredients in defining the Brauer group of a commutative ring.
In this paper, we study separable simplicial algebras and Azumaya simplicial algebras and introduce the Brauer group of a simplicial commutative ring at the end. One of our ultimate goals is to find a useful simplicial version of the Brauer group [17 ] .
Review of previous results
The classical theory of separable algebras can be found in [2 , 10 , 13 ] . Pareigis initiates the study of separable algebras in symmetric monoidal categories [28 ] , which is extended to the setting of a braided monoidal category by Van Oystaeyen and Y. Zhang [24 ] . Later Borceux and Vitale generalize the notion to separable enriched categories [5 ] . Balmer considers separability in triangulated categories [3 ] , and recently Naumann, Pol, and Ramzi push the notion of separable commutative algebras to symmetric monoidal stable ∞ \infty ∞ [23 ] . We also note there are closely related studies of Azumaya algebras and Brauer groups in the monoidal categorical setting [1 , 6 , 8 , 12 , 37 , 35 ] along with the notion of separable functors [9 , 22 ] .
Although the separability of an algebra is almost unanimously defined by the splitting of the multiplication map (as a bimodule map), an Azumaya algebra can be defined in two natural ways: either as a central separable algebra or as a certain “generator” that is Morita equivalent to the base ring. These two notions are equivalent in the classical setting by the celebrated Auslander-Goldman theorem ([2 Theorem 2.1] and [5 Theorem 1.2] ). In the more general monoidal categorical setting, the Morita-theoretical definition of an Azumaya algebra (for instance, see [12 p. 63] , [28 p. 114] , and [24 Proposition 3.1] ) is the most-often studied.
Why the simplicial setting?
Commutative simplicial algebras/rings have been used in algebraic topology, homological algebra, and derived algebraic geometry. We choose to further investigate separable algebras in the simplicial setting because we are searching for useful and computable invariants of simplicial algebras. Our hope is to understand the correct definition of the Brauer group in the simplicial setting, which is also the main goal of our forthcoming paper [17 ] .
Passing from ordinary algebras to simplicial algebras is often a first step toward a homotopical generalization of classical theory [32 , 33 , 35 , 36 ] ; for instance, Toën defines derived Azumaya algebras over a commutative simplicial ring [35 ] . In general, simplicial algebras carry homotopical information through their homotopy groups while retaining the familiar algebraic language of rings, modules, and homological algebra. In this paper we generalize several classical results concerning separable algebras to the simplicial setting. This becomes possible because we have more tools in the simplicial setting than in the general setting of monoidal categories.
For comparison, let us consider again the notion of an Azumaya algebra in a general setting of monoidal categories as well as in the setting of this paper. Let R R R R R R R R R [12 Theorem 4] , it remains unknown if such a notion agrees with the central separable version. This deficiency leads Pareigis to define the 1-Azumaya algebra ([28 p. 115] , Morita-theoretically) and the 2-Azumaya algebra ([28 p. 121] , central-separability and some extra requirement). Pareigis notes in [28 p. 121] that it is generally unclear if 1-Azumaya algebras and 2-Azumaya algebras coincide. If the unit is projective, then they agree [28 p. 124] . But 2-Azumaya algebras are still not the central-separable ones.
Following Pareigis’ work, Van Oystaeyen and Y. Zhang define two versions of Brauer groups using central separable algebras ([24 p. 123] ) and Azumaya algebras ([24 p. 108] ). Their unifying theorem [24 Theorem 4.9] equates central-separable algebras with Azumaya algebras but relies on the assumption that the base symmetric monoidal category C \mathcal{C} C C \mathcal{C} C [5 Theorem 3.4] is another unifying theorem equating central-separable “strong generators” with Azumaya algebras, which is more akin to [28 Theorem 12] in spirit. When R R R R R R R R R [24 p. 112] ) when R R R [24 ] .
Results of this paper
Let R R R R s M o d R\mathrm{sMod} R sMod R R R R s M o d R\mathrm{sMod} R sMod ⊗ R \otimes_R ⊗ R R R R A A A R R R R s M o d R\mathrm{sMod} R sMod A o A^{\mathrm{o}} A o A e = A ⊗ R A o A^{\mathrm{e}} = A \otimes_R A^{\mathrm{o}} A e = A ⊗ R A o A A A R R R μ : A → A \mu:A \to A μ : A → A
Basics
We first formulate the simplicial centralizers (Definition 3.7) and enriched separable functors (Definition 3.13) to characterize simplicial separability.
Proposition 0.1 (Propositions 3.10 and 3.16). Let A A A R R R
(1) A A A R R R
(2) The centralizer functor ( − ) A : A e s M o d → R s M o d (-)^{A}: A^{e}\mathrm{sMod} \to R\mathrm{sMod} ( − ) A : A e sMod → R sMod
(3) The forgetful functor U : A s M o d → R s M o d U: A\mathrm{sMod} \to R\mathrm{sMod} U : A sMod → R sMod R s M o d R\mathrm{sMod} R sMod
Auslander-Goldman Theorem
Following the classical theory [13 ] , a simplicial algebra A A A R R R Azumaya if A A A R R R [24 , 27 ] , we characterize the simplicial R R R R R R M M M M M M R ⊕ t R^{\oplus t} R ⊕ t R R R M ⊕ t ′ M^{\oplus t'} M ⊕ t ′ R R R [5 , 24 , 28 ] for simplicial algebras.
Theorem 0.2 (Theorem 6.4). (Unifying Theorem) Let A A A R R R
(1) A A A R R R
(2) A A A A e A^{e} A e A A A R R R
(3) A A A R R R A e → Hom R s ( A , A ) A^{e} \to\operatorname{Hom}_{R}^{s}(A,A) A e → Hom R s ( A , A ) A e A^{e} A e R R R
We are wondering if the above unifying theorem holds for other non-simplicial setting. Along this line, we also have
Theorem 0.3 (Theorem 6.6). Let A A A R R R A A A A A A Z ( A ) Z(A) Z ( A ) Z ( A ) Z(A) Z ( A ) R R R
Classifications
We classify certain families of separable simplicial algebras which indicate some rigidity of separability for simplicial algebras. In this subsection, k k k
Theorem 0.4 (Theorem 2.11). Suppose R = k R = k R = k A A A R R R A A A k k k A ≅ A 0 Δ 0 : = k Δ 0 ⊗ k A 0 A \cong A_0\Delta^0 := k\Delta^0 \otimes_k A_0 A ≅ A 0 Δ 0 := k Δ 0 ⊗ k A 0 A 0 A_0 A 0 k k k
Suppose k k k k k k k k k
Proposition 0.5 (Proposition 6.7). Let A A A R R R A A A R R R A ≅ R ⊗ R 0 s A 0 A \cong R \otimes^{s}_{R_0} A_0 A ≅ R ⊗ R 0 s A 0 R R R A 0 A_0 A 0 R 0 R_0 R 0
Proposition 0.6 (Corollary 6.8). Let A A A R R R
(1) A A A R R R
(2) Z ( A ) Z(A) Z ( A ) R R R A 0 A_0 A 0 Z ( A ) 0 Z(A)_0 Z ( A ) 0
A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 A \cong Z(A) \mathbin{\mathop{\otimes}\limits_{Z(A)_0}^{s}} A_0 A ≅ Z ( A ) Z ( A ) 0 ⊗ s A 0 as simplicial Z ( A ) Z(A) Z ( A )
Proposition 0.6 shows that a separable simplicial R R R
(a) a commutative separable simplicial R R R Z ( A ) Z(A) Z ( A )
(b) an ordinary Azumaya Z ( A ) 0 Z(A)_0 Z ( A ) 0 A 0 A_0 A 0
The second datum is controlled, up to Morita equivalence, by the Brauer group B r ( Z ( A ) 0 ) Br(Z(A)_0) B r ( Z ( A ) 0 ) R R R
Lemma 0.7 (Lemma 6.10). Let C C C R R R C ≅ Z ( A ) C \cong Z(A) C ≅ Z ( A ) R R R A A A C C C R R R
Commutative simplicial separable algebras
We characterize commutative simplicial separable algebras as follows.
Theorem 0.8 (Theorem 6.11). Let C C C R R R
(1) C C C R R R
(2) C 0 C_0 C 0 R 0 R_0 R 0
ϕ : R ⊗ R 0 s C 0 → C r ⊗ c ↦ r s 0 n ( c ) , ∀ r ⊗ c ∈ ( R ⊗ R 0 s C 0 ) n \phi: R \mathbin{\mathop{\otimes}\limits_{R_0}^{s}} C_0 \to C
\qquad r \otimes c \mapsto r s_0^n(c), \qquad\forall r \otimes c \in(R \mathbin{\mathop{\otimes}\limits_{R_0}^{s}} C_0)_n ϕ : R R 0 ⊗ s C 0 → C r ⊗ c ↦ r s 0 n ( c ) , ∀ r ⊗ c ∈ ( R R 0 ⊗ s C 0 ) n is a ring epimorphism levelwise.
Brauer group
The definition of the Brauer group of a commutative simplicial ring R R R R 0 R_0 R 0
Theorem 0.9 (Theorem 7.4). (Triviality Theorem) Let R R R B r ( R ) ≅ B r ( R 0 ) Br(R) \cong Br(R_0) B r ( R ) ≅ B r ( R 0 )
The above triviality theorem leads us to consider a simplicial version of the Brauer group [17 ] , denoted by
B r s ( R ) = ( B r s ( R ) n ) n ≥ 0 , Br^{s}(R) = \left(Br^{s}(R)_n\right)_{n \geq0}, B r s ( R ) = ( B r s ( R ) n ) n ≥ 0 , such that B r s ( R ) 0 = B r ( R ) Br^{s}(R)_0 = Br(R) B r s ( R ) 0 = B r ( R )
Currently, the simplicial group B r s ( R ) Br^{s}(R) B r s ( R )
Organization of the paper
In Section 1, we review basic definitions of simplicial algebras and simplicial modules. We define simplicial separability and illustrate the notion by examples over a constant base simplicial ring and simplicial matrix rings in Section 2. Centralizer and separable enriched functors are introduced in Section 3, while more properties of simplicial separability are derived in Section 4. We review and adapt the enriched Morita theory for simplicial algebras in Section 5. Azumaya simplicial algebras are studied in Section 6, and the simplicial Auslander-Goldman Theorem is proved. We conclude the paper by a short discussion on the Brauer groups of commutative simplicial rings in Section 7.
Preliminaries
In this paper we try to provide a comprehensive introduction of separable simplicial algebras with many details.
We start with reviewing basic definitions on simplicial algebras and simplicial modules. A ring is always an associative unital ring, and a ring homomorphism preserves the unit of a ring. The category of rings is denoted by Ring \operatorname{Ring} Ring
Simplicial objects
Let Δ \Delta Δ { [ n ] ∣ n = 0 , 1 , 2 , … } \{[n]\mid n=0,1,2,\ldots\} {[ n ] ∣ n = 0 , 1 , 2 , … } [ n ] : = { 0 , 1 , … , n } [n]:=\{0,1,\ldots,n\} [ n ] := { 0 , 1 , … , n } [ k ] ⟶ [ n ] [k]\longrightarrow[n] [ k ] ⟶ [ n ] k k k n n n C \mathcal{C} C simplicial object X X X C \mathcal{C} C Δ \Delta Δ C \mathcal{C} C n n n X n : = X ( [ n ] ) X_n:=X([n]) X n := X ([ n ]) C \mathcal{C} C θ : [ k ] → [ n ] \theta:[k]\to[n] θ : [ k ] → [ n ] Δ \Delta Δ θ X ∗ : X n → X k \theta_X^*:X_n\to X_k θ X ∗ : X n → X k C \mathcal{C} C { θ X ∗ } \{\theta_X^*\} { θ X ∗ } simplicial structure maps of X X X n n n X n X_n X n level n n n of X X X X X X { X n } n ≥ 0 \{X_n\}_{n\geq0} { X n } n ≥ 0 [38 Proposition 8.1.3] . For n ≥ 1 n\geq1 n ≥ 1 s 0 n s_0^n s 0 n [ n ] → [ 0 ] [n]\to[0] [ n ] → [ 0 ] Δ \Delta Δ d n 0 d_n^0 d n 0 [ 0 ] → [ n ] [0]\to[n] [ 0 ] → [ n ] 0 0 0 n n n Δ \Delta Δ n ≥ 0 n\geq0 n ≥ 0 Δ n \Delta^n Δ n Hom Δ ( − , [ n ] ) \operatorname{Hom}_{\Delta}(-,[n]) Hom Δ ( − , [ n ])
Simplicial rings
A simplicial ring , or an sring , is a simplicial object in Ring \operatorname{Ring} Ring k Δ 0 k^{\Delta^0} k Δ 0 k k k Z Δ 0 \mathbb{Z}^{\Delta^0} Z Δ 0 Z \mathbb{Z} Z
The category of simplicial abelian groups sAb \operatorname{sAb} sAb ⊗ Z \otimes_{\mathbb{Z}} ⊗ Z Z \mathbb{Z} Z Z Δ 0 \mathbb{Z}^{\Delta^0} Z Δ 0 R R R ( sAb , ⊗ Z , Z Δ 0 ) (\operatorname{sAb},\otimes_{\mathbb{Z}},\mathbb{Z}^{\Delta^0}) ( sAb , ⊗ Z , Z Δ 0 ) [32 p. 503] . Namely, there is a simplicial abelian group morphism (the multiplication map) μ R / Z Δ 0 : R ⊗ Z R → R \mu_{R/\mathbb{Z}^{\Delta^0}}:R\otimes_{\mathbb{Z}}R\to R μ R / Z Δ 0 : R ⊗ Z R → R η R / Z Δ 0 : Z Δ 0 → R \eta_{R/\mathbb{Z}^{\Delta^0}}:\mathbb{Z}^{\Delta^0}\to R η R / Z Δ 0 : Z Δ 0 → R ( R , μ R / Z Δ 0 , η R / Z Δ 0 ) (R,\mu_{R/\mathbb{Z}^{\Delta^0}},\eta_{R/\mathbb{Z}^{\Delta^0}}) ( R , μ R / Z Δ 0 , η R / Z Δ 0 )
The tensor product of srings is the tensor product of the underlying simplicial abelian groups. Let sRing \operatorname{sRing} sRing ( sAb , ⊗ Z , Z Δ 0 ) (\operatorname{sAb},\otimes_{\mathbb{Z}},\mathbb{Z}^{\Delta^0}) ( sAb , ⊗ Z , Z Δ 0 ) sRing \operatorname{sRing} sRing [34 Appendix A.1] .
Definition 1.1. Let R R R
(1) A two-sided/left/right simplicial ideal, or two-sided/left/right sideal , I I I R R R I ⊆ R I\subseteq R I ⊆ R I n I_n I n R n R_n R n n ≥ 0 n\geq0 n ≥ 0
(2) A simplicial subring, or sub sring , R ′ R' R ′ R ′ ⊆ R R'\subseteq R R ′ ⊆ R R ′ R' R ′
Unless otherwise specified, an sideal will be a two-sided sideal.
From now on, we always assume that R R R [25 ] , [29 II.6] , and [32 p. 498] .
Definition 1.2. A left simplicial module over R R R R R R R R R ( s A b , ⊗ Z , Z Δ 0 ) (\mathsf{sAb}, \otimes_{\mathbb{Z}}, \mathbb{Z}\Delta^0) ( sAb , ⊗ Z , Z Δ 0 ) R R R M M M μ M : R ⊗ Z M → M \mu_M : R \otimes_{\mathbb{Z}} M \to M μ M : R ⊗ Z M → M ( M , μ M ) (M,\mu_M) ( M , μ M )
Definition 1.3. Let M M M N N N R R R left R R R f : M → N f : M \to N f : M → N f : M → N f : M \to N f : M → N r ∈ R n r \in R_n r ∈ R n m ∈ M n m \in M_n m ∈ M n f ( r m ) = r f ( m ) f(rm)=rf(m) f ( r m ) = r f ( m )
Remark 1.4. Let R R R
Let M M M R R R μ M ( r ⊗ m ) : = r ⋅ M m \mu_M(r\otimes m):=r\cdot_M m μ M ( r ⊗ m ) := r ⋅ M m r ⊗ m ∈ ( R ⊗ Z M ) n r\otimes m\in(R\otimes_{\mathbb{Z}}M)_n r ⊗ m ∈ ( R ⊗ Z M ) n n ≥ 0 n\geq0 n ≥ 0
Right R R R R R R R s M o d R\mathsf{sMod} R sMod R R R s M o d R \mathsf{sMod}R sMod R R R R
Since an sring is a functor Δ op → R i n g \Delta^{\operatorname{op}}\to\mathsf{Ring} Δ op → Ring ( − ) ∘ : R i n g → R i n g (-)^\circ:\mathsf{Ring}\to\mathsf{Ring} ( − ) ∘ : Ring → Ring opposite sring of R R R R ∘ : = { R n ∘ } R^\circ:=\{R_n^\circ\} R ∘ := { R n ∘ } θ R ∘ ∗ : = ( θ R ∗ ) ∘ \theta_{R^\circ}^\ast:=(\theta_R^\ast)^\circ θ R ∘ ∗ := ( θ R ∗ ) ∘ θ : [ m ] → [ n ] \theta:[m]\to[n] θ : [ m ] → [ n ] Δ \Delta Δ
Note that s M o d R \mathsf{sMod}R sMod R R ∘ s M o d R^\circ\mathsf{sMod} R ∘ sMod R R R s M o d R \mathsf{sMod}R sMod R R s M o d R\mathsf{sMod} R sMod R R R R R R R s M o d R\mathsf{sMod} R sMod
Simplicial algebras
The following are mostly standard definitions; see, for instance, [11 §VII] and [32 p. 499] .
Definition 1.5. Let R R R center of R R R Z ( R ) : = { Z ( R ) n } Z(R):=\{Z(R)_n\} Z ( R ) := { Z ( R ) n }
Z ( R ) n : = { x ∈ R n ∣ θ R ∗ ( x ) ∈ Z ( R k ) , ∀ θ : [ k ] → [ n ] in Δ } . Z(R)_n:=\{x\in R_n\mid\theta_R^\ast(x)\in Z(R_k),\ \forall\theta:[k]\to[n]\text{ in }\Delta\}. Z ( R ) n := { x ∈ R n ∣ θ R ∗ ( x ) ∈ Z ( R k ) , ∀ θ : [ k ] → [ n ] in Δ } . It is easy to construct a nonseparable salgebra A A A Z ( A ) n ≠ Z ( A n ) Z(A)_n\ne Z(A_n) Z ( A ) n = Z ( A n ) n n n
Lemma 1.6. The center Z ( R ) Z(R) Z ( R ) R R R
Proof. Let n ≥ 0 n\geq0 n ≥ 0 x ∈ Z ( R ) n x\in Z(R)_n x ∈ Z ( R ) n σ : [ m ] → [ n ] \sigma:[m]\to[n] σ : [ m ] → [ n ] Δ \Delta Δ σ R ∗ ( x ) ∈ Z ( R ) m \sigma_R^\ast(x)\in Z(R)_m σ R ∗ ( x ) ∈ Z ( R ) m θ : [ k ] → [ m ] \theta:[k]\to[m] θ : [ k ] → [ m ]
θ R ∗ ( σ R ∗ ( x ) ) = ( σ θ ) R ∗ ( x ) ∈ Z ( R k ) \theta_R^\ast(\sigma_R^\ast(x))=(\sigma\theta)_R^\ast(x)\in Z(R_k) θ R ∗ ( σ R ∗ ( x )) = ( σ θ ) R ∗ ( x ) ∈ Z ( R k ) which implies σ R ∗ ( x ) ∈ Z ( R ) m \sigma_R^\ast(x)\in Z(R)_m σ R ∗ ( x ) ∈ Z ( R ) m Z ( R ) Z(R) Z ( R ) R R R
Z ( R ) n = ⋂ θ : [ k ] → [ n ] ( θ R ∗ ) − 1 ( Z ( R k ) ) Z(R)_n=\bigcap_{\theta:[k]\to[n]}(\theta_R^\ast)^{-1}(Z(R_k)) Z ( R ) n = θ : [ k ] → [ n ] ⋂ ( θ R ∗ ) − 1 ( Z ( R k )) is a subring of R n R_n R n Z ( R ) Z(R) Z ( R ) □ \square □
So Z ( R ) Z(R) Z ( R ) Z ( R n ) Z(R_n) Z ( R n ) ( R , μ R / Z Δ 0 , η R / Z Δ 0 ) (R,\mu_{R/\mathbb{Z}\Delta^0},\eta_{R/\mathbb{Z}\Delta^0}) ( R , μ R / Z Δ 0 , η R / Z Δ 0 ) commutative if μ R / Z Δ 0 \mu_{R/\mathbb{Z}\Delta^0} μ R / Z Δ 0 R R R Z ( R ) = R Z(R)=R Z ( R ) = R
Definition 1.7. Let R R R tensor product of a right R R R M M M R R R N N N M ⊗ R N M\otimes_RN M ⊗ R N ( M ⊗ R N ) n : = M n ⊗ R n N n (M\otimes_RN)_n:=M_n\otimes_{R_n}N_n ( M ⊗ R N ) n := M n ⊗ R n N n θ M ⊗ R N ∗ \theta_{M\otimes_RN}^\ast θ M ⊗ R N ∗
M n ⊗ R n N n ⟶ M m ⊗ R m N m , x ⊗ y ⟼ θ M ∗ ( x ) ⊗ θ N ∗ ( y ) M_n\otimes_{R_n}N_n\longrightarrow M_m\otimes_{R_m}N_m,\qquad x\otimes y\longmapsto\theta_M^\ast(x)\otimes\theta_N^\ast(y) M n ⊗ R n N n ⟶ M m ⊗ R m N m , x ⊗ y ⟼ θ M ∗ ( x ) ⊗ θ N ∗ ( y ) for every x ⊗ y ∈ M n ⊗ R n N n x\otimes y\in M_n\otimes_{R_n}N_n x ⊗ y ∈ M n ⊗ R n N n θ : [ m ] → [ n ] \theta:[m]\to[n] θ : [ m ] → [ n ] Δ \Delta Δ
Remark 1.8. Let R R R
When R = Z Δ 0 R = \mathbb{Z}\Delta^0 R = Z Δ 0 ⊗ Z Δ 0 s \otimes_{\mathbb{Z}\Delta^0}^{s} ⊗ Z Δ 0 s ⊗ Z s \otimes_{\mathbb{Z}}^{s} ⊗ Z s Z Δ 0 \mathbb{Z}\Delta^0 Z Δ 0 R = k Δ 0 R = k\Delta^0 R = k Δ 0 k k k ⊗ k s : = ⊗ k Δ 0 s \otimes_k^{s} := \otimes_{k\Delta^0}^{s} ⊗ k s := ⊗ k Δ 0 s
As an example, R Δ n : = R ⊗ Z s Z Δ n R\Delta^n := R\otimes_{\mathbb{Z}}^{s}\mathbb{Z}\Delta^n R Δ n := R ⊗ Z s Z Δ n R R R Z Δ n \mathbb{Z}\Delta^n Z Δ n Δ n \Delta^n Δ n
We note R R R [29 §II.6] . The kernels and cokernels of an R R R R R R
When R R R ( R sMod , ⊗ R s , R ) (R\text{ sMod}, \otimes_R^{s}, R) ( R sMod , ⊗ R s , R ) [32 p. 499] .
Definition 1.9. Let R R R A A A R R R η A / R : R → A \eta_{A/R}: R \to A η A / R : R → A η A / R ( R ) ⊆ Z ( A ) \eta_{A/R}(R) \subseteq Z(A) η A / R ( R ) ⊆ Z ( A ) R R R f : A → B f : A \to B f : A → B f ∘ η A / R = η B / R f \circ\eta_{A/R} = \eta_{B/R} f ∘ η A / R = η B / R
Remark 1.10. Let R R R
Equivalently, an R R R A A A R R R [32 p. 499] . In this way, the tensor product of R R R R R R
When k k k k k k k k k k k k [33 ] . A simplicial k k k k Δ 0 k\Delta^0 k Δ 0 k k k
If A A A R R R A A A R R R M M M R R R A ⊗ R s M → M A\otimes_R^{s} M \to M A ⊗ R s M → M [11 §VII, Lemma 1.8] .
Separability of Simplicial Algebras
We define separable simplicial algebras over a commutative sring, characterize separable simplicial algebras over a constant commutative ring, and investigate the separability of the simplicial matrix rings as an example of simplicial algebras over a non-constant commutative sring.
A few projectivity lemmas
Lemma 2.1. Let R R R I I I R R R x ∈ R 0 x \in R_0 x ∈ R 0 ( I x ) n = I n s 0 n ( x ) (Ix)_n = I_n s_0^n(x) ( I x ) n = I n s 0 n ( x ) I x : = { ( I x ) n } n ≥ 0 Ix := \{(Ix)_n\}_{n\geq0} I x := {( I x ) n } n ≥ 0 R R R
Proof. Let θ : [ m ] → [ n ] \theta: [m] \to[n] θ : [ m ] → [ n ] Δ \Delta Δ
θ R ∗ ( ( I x ) n ) = θ R ∗ ( I n s 0 n ( x ) ) = θ R ∗ ( I n ) s 0 m ( x ) ⊆ I m s 0 m ( x ) = ( I x ) m . \theta_R^*((Ix)_n) = \theta_R^*(I_n s_0^n(x)) = \theta_R^*(I_n)s_0^m(x) \subseteq I_m s_0^m(x) = (Ix)_m. θ R ∗ (( I x ) n ) = θ R ∗ ( I n s 0 n ( x )) = θ R ∗ ( I n ) s 0 m ( x ) ⊆ I m s 0 m ( x ) = ( I x ) m . This implies I x Ix I x R R R I n s 0 n ( x ) I_n s_0^n(x) I n s 0 n ( x ) R n R_n R n I x Ix I x R R R
Lemma 2.2. Let R R R e ∈ R 0 e \in R_0 e ∈ R 0
s 0 n ( e ) s_0^n(e) s 0 n ( e ) R n R_n R n n ≥ 0 n \geq0 n ≥ 0
View R R R R R R R R R R R R
As left R R R R = R e ⊕ R ( 1 R 0 − e ) R = R_e \oplus R(1_{R_0} - e) R = R e ⊕ R ( 1 R 0 − e )
Proof. (1): Immediate since s 0 n : R 0 → R n s^n_0:R_0\to R_n s 0 n : R 0 → R n
(2): By Lemma 2.1.
(3): It is clear that 1 R 0 − e ∈ R 0 1_{R_0}-e\in R_0 1 R 0 − e ∈ R 0 R e Re R e R ( 1 R 0 − e ) R(1_{R_0}-e) R ( 1 R 0 − e ) R R R R n = R n s 0 n ( e ) ⊕ R n ( 1 R n − s 0 n ( e ) ) R_n=R_ns^n_0(e)\oplus R_n(1_{R_n}-s^n_0(e)) R n = R n s 0 n ( e ) ⊕ R n ( 1 R n − s 0 n ( e )) n ≥ 0 n\geq0 n ≥ 0
Definition 2.3
Let P P P R R R P P P Hom R -sMod ( P , − ) : R -sMod → Ab \operatorname{Hom}_{R\text{-sMod}}(P,-):R\text{-sMod}\to\operatorname{Ab} Hom R -sMod ( P , − ) : R -sMod → Ab
The projectivity of the unit in a monoidal category is useful in the study of separability of monoids; see [8 Proposition 4.5] and [24 p. 98 and Theorem 4.14] .
Lemma 2.4. Let R R R R R R R R R
Proof. Any left R R R f : R → M f:R\to M f : R → M f ( 1 R 0 ) ∈ M 0 f(1_{R_0})\in M_0 f ( 1 R 0 ) ∈ M 0 f ( r ) = r f ( 1 R n ) = r s 0 n ( f ( 1 R 0 ) ) f(r)=rf(1_{R_n})=rs^n_0(f(1_{R_0})) f ( r ) = r f ( 1 R n ) = r s 0 n ( f ( 1 R 0 )) r ∈ R n r\in R_n r ∈ R n R R R M → R M\to R M → R 1 R 0 1_{R_0} 1 R 0
The following is a generalization of [13 Exercise 1.1.22] in the simplicial setting.
Proposition 2.5. Let R R R I I I R R R
(1) R / I R/I R / I R R R
(2) The sequence of left R R R 0 → I → R → R / I → 0 0\to I\to R\to R/I\to0 0 → I → R → R / I → 0
(3) As a left R R R I I I R R R
(4) There is an element e ∈ R 0 e\in R_0 e ∈ R 0 1 R 0 − e ∈ I 0 1_{R_0}-e\in I_0 1 R 0 − e ∈ I 0 I e = 0 Ie=0 I e = 0
(5) There is an idempotent e ∈ R 0 e\in R_0 e ∈ R 0 I = R ( 1 R 0 − e ) I=R(1_{R_0}-e) I = R ( 1 R 0 − e )
Proof. (1) ⟹ \Longrightarrow ⟹ ⟹ \Longrightarrow ⟹
(3) ⟹ \Longrightarrow ⟹ R = I ⊕ N R=I\oplus N R = I ⊕ N R R R 1 R 0 = ( 1 R 0 − e ) + e 1_{R_0}=(1_{R_0}-e)+e 1 R 0 = ( 1 R 0 − e ) + e e ∈ N 0 e\in N_0 e ∈ N 0 I e = I ( 1 R 0 − ( 1 R 0 − e ) ) ⊆ I + I ( 1 R 0 − e ) ⊆ I Ie=I(1_{R_0}-(1_{R_0}-e))\subseteq I+I(1_{R_0}-e)\subseteq I I e = I ( 1 R 0 − ( 1 R 0 − e )) ⊆ I + I ( 1 R 0 − e ) ⊆ I I e ⊆ I N ⊆ N Ie\subseteq IN\subseteq N I e ⊆ I N ⊆ N N N N R R R I e ⊆ I ∩ N = 0 Ie\subseteq I\cap N=0 I e ⊆ I ∩ N = 0
(4) ⟹ \Longrightarrow ⟹ I e = 0 Ie=0 I e = 0 ( 1 R 0 − e ) e = 0 (1_{R_0}-e)e=0 ( 1 R 0 − e ) e = 0 e e e R = R e ⊕ R ( 1 R 0 − e ) R=Re\oplus R(1_{R_0}-e) R = R e ⊕ R ( 1 R 0 − e ) R R R I I I R ( 1 R 0 − e ) ⊆ I R(1_{R_0}-e)\subseteq I R ( 1 R 0 − e ) ⊆ I
I = I ( 1 R 0 − e + e ) ⊆ I ( 1 R 0 − e ) + I e ⊆ I ( 1 R 0 − e ) ⊆ R ( 1 R 0 − e ) . I=I(1_{R_0}-e+e)\subseteq I(1_{R_0}-e)+Ie\subseteq I(1_{R_0}-e)\subseteq R(1_{R_0}-e). I = I ( 1 R 0 − e + e ) ⊆ I ( 1 R 0 − e ) + I e ⊆ I ( 1 R 0 − e ) ⊆ R ( 1 R 0 − e ) . This completes the proof of I = R ( 1 R 0 − e ) I=R(1_{R_0}-e) I = R ( 1 R 0 − e )
(5) ⟹ \Longrightarrow ⟹ e ∈ R 0 e\in R_0 e ∈ R 0 R = R e ⊕ R ( 1 R 0 − e ) = R e ⊕ I R=Re\oplus R(1_{R_0}-e)=Re\oplus I R = R e ⊕ R ( 1 R 0 − e ) = R e ⊕ I R R R R / I ≅ R e R/I\cong Re R / I ≅ R e R R R R R R
From now on, we always assume that R R R
Definition 2.6 Let A A A R R R A A A R R R
A e : = A ⊗ R A o . A^{e}:=A\mathbin{\otimes_R} A^{o}. A e := A ⊗ R A o . Then A A A A e A^e A e
( a ⊗ b ) ⋅ A x = a x b (a\otimes b)\mathbin{\cdot_A}x=axb ( a ⊗ b ) ⋅ A x = a x b for every a , b , x ∈ A n a,b,x\in A_n a , b , x ∈ A n A e A^e A e
μ A / R : = A e ⟶ A , u ⟼ u ⋅ A 1 A n \mu_{A/R}:=A^e\longrightarrow A,\qquad u\longmapsto u\mathbin{\cdot_A}1_{A_n} μ A / R := A e ⟶ A , u ⟼ u ⋅ A 1 A n for any u ∈ ( A e ) n u\in(A^e)_n u ∈ ( A e ) n J A / R : = ker μ A / R J_{A/R}:=\ker\mu_{A/R} J A / R := ker μ A / R A e A^e A e k ≥ 0 k\geq0 k ≥ 0 ( J A / R ) k = ∑ a ∈ A k ( A e ) k ( a ⊗ 1 A k − 1 A k ⊗ a ) (J_{A/R})_k=\sum_{a\in A_k}(A^e)_k(a\otimes1_{A_k}-1_{A_k}\otimes a) ( J A / R ) k = ∑ a ∈ A k ( A e ) k ( a ⊗ 1 A k − 1 A k ⊗ a )
The following is a generalization of [13 Proposition 4.1.2] . A more general version for braided monoidal categories appears in [24 Proposition 4.1] .
Proposition 2.7. Let A A A R R R
A A A A e A^{e} A e
The exact sequence of left A e A^{e} A e 0 → J A / R → A e → μ A / R A → 0 0 \to J_{A/R} \to A^{e} \xrightarrow{\mu_{A/R}} A \to0 0 → J A / R → A e μ A / R A → 0
There is an element e ∈ ( A e ) 0 e \in(A^{e})_0 e ∈ ( A e ) 0 μ A / R ( e ) = 1 A 0 \mu_{A/R}(e)=1_{A_0} μ A / R ( e ) = 1 A 0 J A / R e = 0 J_{A/R}e=0 J A / R e = 0
There is an idempotent e ∈ ( A e ) 0 e \in(A^{e})_0 e ∈ ( A e ) 0 J A / R = A e ( 1 ( A e ) 0 − e ) J_{A/R}=A^{e}(1_{(A^{e})_0}-e) J A / R = A e ( 1 ( A e ) 0 − e )
Proof. Apply Proposition 2.5 to ( A e , J A / R ) (A^{e},J_{A/R}) ( A e , J A / R ) A ≅ A e / J A / R A \cong A^{e}/J_{A/R} A ≅ A e / J A / R μ ( e ) = 1 A 0 \mu(e)=1_{A_0} μ ( e ) = 1 A 0 1 ( A e ) 0 − e ∈ ( J A / R ) 0 1_{(A^{e})_0}-e \in(J_{A/R})_0 1 ( A e ) 0 − e ∈ ( J A / R ) 0
The following notion of separability of simplicial algebras agrees with the well-studied notion of separability of monoid objects in monoidal categories [8 , 24 , 28 ] .
Definition 2.8. Let A A A R R R A A A separable over R R R , or a separable R R R , if any one of the equivalent conditions in Proposition 2.7 is satisfied. An element e ∈ ( A e ) 0 e \in(A^{e})_0 e ∈ ( A e ) 0 μ A / R ( e ) = 1 A 0 \mu_{A/R}(e)=1_{A_0} μ A / R ( e ) = 1 A 0
( 1 A n ⊗ a ) s 0 n ( e ) = ( a ⊗ 1 A n ) s 0 n ( e ) (1_{A_n}\otimes a)s_0^n(e)=(a\otimes1_{A_n})s_0^n(e) ( 1 A n ⊗ a ) s 0 n ( e ) = ( a ⊗ 1 A n ) s 0 n ( e ) for every a ∈ A n a \in A_n a ∈ A n separability idempotent .
Separable simplicial algebras over a constant simplicial ring
We characterize the separability of simplicial algebras over a constant commutative simplicial ring by a rigidity result.
Lemma 2.10. Let k k k A A A k k k e ∈ ( A e ) 0 e \in(A^{e})_0 e ∈ ( A e ) 0 n ≥ 0 n\geq0 n ≥ 0 f : A n → A n f:A_n\to A_n f : A n → A n k k k ( f ⊗ f o ) ( s 0 n ( e ) ) = s 0 n ( e ) (f\otimes f^o)(s_0^n(e))=s_0^n(e) ( f ⊗ f o ) ( s 0 n ( e )) = s 0 n ( e ) f = I d A n f=Id_{A_n} f = I d A n
Proof. Since f ⊗ f o f\otimes f^o f ⊗ f o s 0 n ( e ) s_0^n(e) s 0 n ( e ) f f f
(E2.10.1)
( f ⊗ I d A 0 o ) ( s 0 n ( e ) ) = ( f ⊗ I d A 0 o ) ( f ⊗ f o ) ( s 0 n ( e ) ) = ( f ⊗ f o ) ( s 0 n ( e ) ) = s 0 n ( e ) . (f\otimes Id_{A_0^o})(s_0^n(e))=(f\otimes Id_{A_0^o})(f\otimes f^o)(s_0^n(e))=(f\otimes f^o)(s_0^n(e))=s_0^n(e). ( f ⊗ I d A 0 o ) ( s 0 n ( e )) = ( f ⊗ I d A 0 o ) ( f ⊗ f o ) ( s 0 n ( e )) = ( f ⊗ f o ) ( s 0 n ( e )) = s 0 n ( e ) . For any a ∈ A n a\in A_n a ∈ A n e e e
(E2.10.2)
( a ⊗ 1 A n ) s 0 n ( e ) = ( 1 A n ⊗ a ) s 0 n ( e ) . (a\otimes1_{A_n})s_0^n(e)=(1_{A_n}\otimes a)s_0^n(e). ( a ⊗ 1 A n ) s 0 n ( e ) = ( 1 A n ⊗ a ) s 0 n ( e ) . Applying the k k k f ⊗ I d A 0 o f\otimes Id_{A_0^o} f ⊗ I d A 0 o
(E2.10.3)
( f ( a ) ⊗ 1 A n ) s 0 n ( e ) = ( 1 A n ⊗ a ) s 0 n ( e ) = ( a ⊗ 1 A n ) s 0 n ( e ) . (f(a)\otimes1_{A_n})s_0^n(e)=(1_{A_n}\otimes a)s_0^n(e)=(a\otimes1_{A_n})s_0^n(e). ( f ( a ) ⊗ 1 A n ) s 0 n ( e ) = ( 1 A n ⊗ a ) s 0 n ( e ) = ( a ⊗ 1 A n ) s 0 n ( e ) . Then apply the left A e A^e A e μ A / k Δ 0 \mu_{A/k\Delta^0} μ A / k Δ 0
f ( a ) = f ( a ) μ A / k Δ 0 ( s 0 n ( e ) ) = a μ A / k Δ 0 ( s 0 n ( e ) ) = a f(a)=f(a)\mu_{A/k\Delta^0}(s_0^n(e))=a\mu_{A/k\Delta^0}(s_0^n(e))=a f ( a ) = f ( a ) μ A / k Δ 0 ( s 0 n ( e )) = a μ A / k Δ 0 ( s 0 n ( e )) = a since μ A / k Δ 0 ( e ) = 1 A 0 \mu_{A/k\Delta^0}(e)=1_{A_0} μ A / k Δ 0 ( e ) = 1 A 0 f = I d A n f=Id_{A_n} f = I d A n
Theorem 2.11. Let k k k A A A k k k A A A k k k A ≅ A 0 Δ 0 : = k Δ 0 ⊗ k A 0 A\cong A_0\Delta^0:=k\Delta^0\otimes_k A_0 A ≅ A 0 Δ 0 := k Δ 0 ⊗ k A 0 A 0 A_0 A 0 A 0 A_0 A 0 k k k
Proof. ( ⟹ ) (\Longrightarrow) ( ⟹ ) A A A k \mathbb{k} k e ∈ ( A e ) 0 e \in(A^{e})_{0} e ∈ ( A e ) 0 k \mathbb{k} k A 0 Δ 0 ↪ A A_{0}\Delta^{0} \hookrightarrow A A 0 Δ 0 ↪ A n ≥ 0 n \ge0 n ≥ 0 k \mathbb{k} k s 0 n : A 0 → A n s_{0}^{n}: A_{0} \to A_{n} s 0 n : A 0 → A n d 0 n s 0 n = Id A 0 d_{0}^{n}s_{0}^{n} = \operatorname{Id}_{A_{0}} d 0 n s 0 n = Id A 0 s 0 n d 0 n : A n → A n s_{0}^{n}d_{0}^{n}: A_{n} \to A_{n} s 0 n d 0 n : A n → A n
( ( s 0 n d 0 n ) ⊗ ( s 0 n d 0 n ) ) s 0 n ( e ) = ( ( s 0 n d 0 n ) ⊗ ( s 0 n d 0 n ) ) ( s 0 n ⊗ s 0 n ) ( e ) = s 0 n ( e ) . \left((s_{0}^{n}d_{0}^{n}) \otimes(s_{0}^{n}d_{0}^{n})\right)s_{0}^{n}(e) = \left((s_{0}^{n}d_{0}^{n}) \otimes(s_{0}^{n}d_{0}^{n})\right)(s_{0}^{n} \otimes s_{0}^{n})(e) = s_{0}^{n}(e). ( ( s 0 n d 0 n ) ⊗ ( s 0 n d 0 n ) ) s 0 n ( e ) = ( ( s 0 n d 0 n ) ⊗ ( s 0 n d 0 n ) ) ( s 0 n ⊗ s 0 n ) ( e ) = s 0 n ( e ) . By Lemma 2.10, s 0 n d 0 n = Id A n s_{0}^{n}d_{0}^{n} = \operatorname{Id}_{A_{n}} s 0 n d 0 n = Id A n s 0 n s_{0}^{n} s 0 n A 0 Δ 0 ↪ A A_{0}\Delta^{0} \hookrightarrow A A 0 Δ 0 ↪ A k \mathbb{k} k e e e A 0 A_{0} A 0 k \mathbb{k} k
( ⟸ ) (\Longleftarrow) ( ⟸ ) A 0 A_{0} A 0 k \mathbb{k} k e ∈ ( A 0 ) e e \in(A_{0})^{e} e ∈ ( A 0 ) e k \mathbb{k} k A 0 Δ 0 A_{0}\Delta^{0} A 0 Δ 0
Corollary 2.12. Let k \mathbb{k} k A A A k \mathbb{k} k A sMod A\operatorname{sMod} A sMod
Proof. By Theorem 2.11, A ≅ k Δ 0 ⊗ k s A 0 A \cong\mathbb{k}\Delta^{0} \otimes_{\mathbb{k}}^{s} A_{0} A ≅ k Δ 0 ⊗ k s A 0 A 0 A_{0} A 0 k \mathbb{k} k [16 Theorem 3.1] , A sMod ≅ Ch ≥ 0 ( A 0 Mod ) A\operatorname{sMod} \cong\operatorname{Ch}_{\ge0}(A_{0}\operatorname{Mod}) A sMod ≅ Ch ≥ 0 ( A 0 Mod ) Ch ≥ 0 ( A 0 Mod ) \operatorname{Ch}_{\ge0}(A_{0}\operatorname{Mod}) Ch ≥ 0 ( A 0 Mod ) M M M A 0 A_{0} A 0 M M M Ch ≥ 0 ( A 0 Mod ) \operatorname{Ch}_{\ge0}(A_{0}\operatorname{Mod}) Ch ≥ 0 ( A 0 Mod )
0 → M → I 0 → I 1 → ⋯ → I n → ⋯ 0 \to M \to I^{0} \to I^{1} \to\cdots\to I^{n} \to\cdots 0 → M → I 0 → I 1 → ⋯ → I n → ⋯ where I n = Σ n ( k → k ) ⊗ M I^{n} = \Sigma^{n}(\mathbb{k} \to\mathbb{k}) \otimes M I n = Σ n ( k → k ) ⊗ M n ≥ 0 n \ge0 n ≥ 0 Ch ≥ 0 ( A 0 Mod ) \operatorname{Ch}_{\ge0}(A_{0}\operatorname{Mod}) Ch ≥ 0 ( A 0 Mod ) [16 Lemma A.3(5)] . Hence M M M A sMod A\operatorname{sMod} A sMod
Separable simplicial algebras over a non-constant simplicial ring
Separability of a simplicial algebra depends on the base sring. When the base sring is not constant, the rigidity in Theorem 2.11 fails.
Definition 2.13. Let k \mathbb{k} k A : Δ o p → k -Alg A : \Delta^{\mathrm{op}} \to\mathbb{k}\text{-Alg} A : Δ op → k -Alg k \mathbb{k} k k ≥ 1 k \ge1 k ≥ 1 M k ( A ) M_{k}(A) M k ( A )
M k ( A ) : = Δ o p → A k -Alg → M k ( − ) k -Alg , M_{k}(A) := \Delta^{\mathrm{op}} \xrightarrow{A} \mathbb{k}\text{-Alg} \xrightarrow{M_{k}(-)} \mathbb{k}\text{-Alg}, M k ( A ) := Δ op A k -Alg M k ( − ) k -Alg , where M k ( − ) M_{k}(-) M k ( − ) k × k k \times k k × k k \mathbb{k} k M k ( A ) m : = M k ( A m ) M_{k}(A)_{m} := M_{k}(A_{m}) M k ( A ) m := M k ( A m ) θ : [ p ] → [ m ] \theta: [p] \to[m] θ : [ p ] → [ m ] Δ \Delta Δ θ M k ( A ) : M k ( A ) m → M k ( A ) p \theta^{M_{k}(A)} : M_{k}(A)_{m} \to M_{k}(A)_{p} θ M k ( A ) : M k ( A ) m → M k ( A ) p ( a i j ) (a_{ij}) ( a ij ) ( θ A ( a i j ) ) (\theta^{A}(a_{ij})) ( θ A ( a ij ))
For every k ≥ 1 k \ge1 k ≥ 1 M k ( A ) M_{k}(A) M k ( A ) k \mathbb{k} k
δ : A → M k ( A ) a ↦ a I k . \delta: A \to M_{k}(A) \qquad a \mapsto aI_{k}. δ : A → M k ( A ) a ↦ a I k . Note that δ ( A ) ⊆ Z ( M k ( A ) ) \delta(A) \subseteq Z(M_{k}(A)) δ ( A ) ⊆ Z ( M k ( A )) A A A A A A M k ( A ) M_{k}(A) M k ( A ) A A A
(a) View A A A k \mathbb{k} k M k ( A ) M_{k}(A) M k ( A ) k \mathbb{k} k
(b) Suppose A A A M k ( A ) M_{k}(A) M k ( A ) A A A
The following lemma is well-known.
Lemma 2.14. Let k \mathbb{k} k B B B k \mathbb{k} k k ≥ 1 k \ge1 k ≥ 1 M k ( B ) M_{k}(B) M k ( B ) k \mathbb{k} k B B B k \mathbb{k} k
Question (a) can be answered by a simplicial version of Lemma 2.14.
Proposition 2.15. Let k k k A A A k k k k ≥ 1 k \ge1 k ≥ 1
(1) M k ( A ) M_k(A) M k ( A ) k k k
(2) M k ( A ) ≅ M k ( A 0 ) Δ 0 M_k(A) \cong M_k(A_0)^{\Delta0} M k ( A ) ≅ M k ( A 0 ) Δ0 M k ( A 0 ) M_k(A_0) M k ( A 0 ) k k k
(3) A ≅ A 0 Δ 0 A \cong A_0^{\Delta0} A ≅ A 0 Δ0 A 0 A_0 A 0 k k k
(4) A A A k k k
Proof. (1) ⟺ \Longleftrightarrow ⟺ Theorem 2.11 .
(2) ⟺ \Longleftrightarrow ⟺ M k ( A ) M_k(A) M k ( A ) A A A Lemma 2.14 , M k ( A 0 ) M_k(A_0) M k ( A 0 ) k k k A 0 A_0 A 0 k k k
(3) ⟺ \Longleftrightarrow ⟺ Theorem 2.11 . □ \square □
If k k k M k ( k ) M_k(k) M k ( k ) k k k [13 Example 4.2.1] . Question (b) can be answered by a simplicial generalization of this fact.
Proposition 2.16. Suppose A A A k ≥ 1 k \ge1 k ≥ 1 M k ( A ) M_k(A) M k ( A ) A A A
Proof. Let 1 ≤ j ≤ k 1 \le j \le k 1 ≤ j ≤ k E p q E_{pq} E pq k × k k \times k k × k 1 A n 1_{A_n} 1 A n ( p , q ) (p,q) ( p , q ) 0 0 0 I k I_k I k n n n M k ( A ) M_k(A) M k ( A ) M k ( A ) e = M k ( A ) ⊗ A M k ( A ) o M_k(A)^e = M_k(A) \mathop{\otimes}_{A} M_k(A)^{\mathrm{o}} M k ( A ) e = M k ( A ) ⊗ A M k ( A ) o e = ∑ i = 1 k E i j ⊗ A 0 E j i ∈ M k ( A 0 ) e e = \sum_{i=1}^{k} E_{ij} \mathop{\otimes}_{A_0} E_{ji} \in M_k(A_0)^e e = ∑ i = 1 k E ij ⊗ A 0 E j i ∈ M k ( A 0 ) e s 0 m ( e ) = ∑ i = 1 k E i j ⊗ A m E j i s_0^m(e) = \sum_{i=1}^{k} E_{ij} \mathop{\otimes}_{A_m} E_{ji} s 0 m ( e ) = ∑ i = 1 k E ij ⊗ A m E j i
It is clear that μ M k ( A ) / A ( e ) = ∑ i = 1 k E i i = I k = 1 M k ( A 0 ) \mu_{M_k(A)/A}(e) = \sum_{i=1}^{k} E_{ii} = I_k = 1_{M_k(A_0)} μ M k ( A ) / A ( e ) = ∑ i = 1 k E ii = I k = 1 M k ( A 0 ) Proposition 2.7 , it remains to show that J M k ( A ) / A e = 0 J_{M_k(A)/A}e = 0 J M k ( A ) / A e = 0 m ≥ 0 m \ge0 m ≥ 0
( J M k ( A ) / A ) m = ∑ E ∈ M k ( A m ) M k ( A m ) e ( E ⊗ A m I k − I k ⊗ A m E ) \left(J_{M_k(A)/A}\right)_m = \sum_{E \in M_k(A_m)} M_k(A_m)^e\left(E \mathop{\otimes}_{A_m} I_k - I_k \mathop{\otimes}_{A_m} E\right) ( J M k ( A ) / A ) m = E ∈ M k ( A m ) ∑ M k ( A m ) e ( E ⊗ A m I k − I k ⊗ A m E ) for every E = ( a p q ) ∈ M k ( A m ) E = (a_{pq}) \in M_k(A_m) E = ( a pq ) ∈ M k ( A m ) E = ∑ a p q E p q = ∑ E p q a p q E = \sum a_{pq}E_{pq} = \sum E_{pq}a_{pq} E = ∑ a pq E pq = ∑ E pq a pq A m A_m A m
a p q E p q ⊗ A m I k − I k ⊗ A m a p q E p q = ( a p q I k ⊗ A m I k ) ( E p q ⊗ A m I k − I k ⊗ A m E p q ) . a_{pq}E_{pq} \mathop{\otimes}_{A_m} I_k - I_k \mathop{\otimes}_{A_m} a_{pq}E_{pq} = (a_{pq}I_k \mathop{\otimes}_{A_m} I_k)(E_{pq} \mathop{\otimes}_{A_m} I_k - I_k \mathop{\otimes}_{A_m} E_{pq}). a pq E pq ⊗ A m I k − I k ⊗ A m a pq E pq = ( a pq I k ⊗ A m I k ) ( E pq ⊗ A m I k − I k ⊗ A m E pq ) . It suffices to note that, for every m ≥ 0 m \ge0 m ≥ 0
( E p q ⊗ A m I k ) s 0 m ( e ) = ∑ i = 1 k E p q E i j ⊗ A m E j i = E p j ⊗ A m E j q = ∑ i = 1 k E i j ⊗ A m E j i E p q = ( I k ⊗ A m E p q ) s 0 m ( e ) \begin{aligned}
(E_{pq} \mathop{\otimes}_{A_m} I_k)s_0^m(e) = \sum_{i=1}^{k} E_{pq}E_{ij} \mathop{\otimes}_{A_m} E_{ji} = E_{pj} \mathop{\otimes}_{A_m} E_{jq} \\
&= \sum_{i=1}^{k} E_{ij} \mathop{\otimes}_{A_m} E_{ji}E_{pq} = (I_k \mathop{\otimes}_{A_m} E_{pq})s_0^m(e)
\end{aligned} ( E pq ⊗ A m I k ) s 0 m ( e ) = i = 1 ∑ k E pq E ij ⊗ A m E j i = E p j ⊗ A m E j q = i = 1 ∑ k E ij ⊗ A m E j i E pq = ( I k ⊗ A m E pq ) s 0 m ( e ) which shows ( J M k ( A ) / A ) m s 0 m ( e ) = 0 \left(J_{M_k(A)/A}\right)_m s_0^m(e) = 0 ( J M k ( A ) / A ) m s 0 m ( e ) = 0 e e e M k ( A ) M_k(A) M k ( A ) M k ( A ) M_k(A) M k ( A ) A A A □ \square □
Separability in Simplicial Enrichment
We follow [18 ] for the general theory of enriched categories. Let R R R R s M o d R\mathrm{sMod} R sMod A s M o d A\mathrm{sMod} A sMod R s M o d R\mathrm{sMod} R sMod R R R A A A U A / R : A s M o d → R s M o d U_{A/R}:A\mathrm{sMod}\to R\mathrm{sMod} U A / R : A sMod → R sMod
Basics of simplicial enrichment
Let R R R R R R R Δ n = R ⊗ Z Z Δ n R\Delta^n=R\mathop{\otimes}_{\mathbb{Z}}\mathbb{Z}\Delta^n R Δ n = R ⊗ Z Z Δ n R s M o d R\mathrm{sMod} R sMod [29 II.6] .
Definition 3.1. For any R R R M , N M,N M , N simplicial Hom-set Hom R ( M , N ) \operatorname{Hom}_{R}(M,N) Hom R ( M , N )
Hom R ( M , N ) n : = Hom R s M o d ( R Δ n ⊗ R M , N ) . \operatorname{Hom}_{R}(M,N)_n:=\operatorname{Hom}_{R\mathrm{sMod}}(R\Delta^n\mathop{\otimes}_{R}M,N). Hom R ( M , N ) n := Hom R sMod ( R Δ n ⊗ R M , N ) . where each θ : [ k ] → [ n ] \theta:[k]\to[n] θ : [ k ] → [ n ] Δ \Delta Δ θ ∗ : Hom R ( M , N ) n → Hom R ( M , N ) k \theta^*:\operatorname{Hom}_{R}(M,N)_n\to\operatorname{Hom}_{R}(M,N)_k θ ∗ : Hom R ( M , N ) n → Hom R ( M , N ) k f : R Δ n ⊗ R M → N f:R\Delta^n\mathop{\otimes}_{R}M\to N f : R Δ n ⊗ R M → N
θ ∗ ( f ) : R Δ k ⊗ R M → ( θ ∘ − ) ∗ ⊗ R Id M R Δ n ⊗ R M → f N . (1) \theta^*(f):R\Delta^k\mathop{\otimes}_{R}M\xrightarrow{(\theta\circ-)^*\mathop{\otimes}_{R}\operatorname{Id}_{M}}R\Delta^n\mathop{\otimes}_{R}M\xrightarrow{f}N. \tag*{(1)} θ ∗ ( f ) : R Δ k ⊗ R M ( θ ∘− ) ∗ ⊗ R Id M R Δ n ⊗ R M f N . (1) Lemma 3.2. For any R R R M M M N N N Hom R ( M , N ) \operatorname{Hom}_{R}(M,N) Hom R ( M , N ) R R R
Proof. It is clear that Hom R ( M , N ) \operatorname{Hom}_{R}(M,N) Hom R ( M , N ) R R R Hom R ( M , N ) \operatorname{Hom}_{R}(M,N) Hom R ( M , N )
μ : R ⊗ Z Hom R ( M , N ) → Hom R ( M , N ) r ⊗ f ↦ r ⋅ f \mu:R\mathop{\otimes}_{\mathbb{Z}}\operatorname{Hom}_{R}(M,N)\to\operatorname{Hom}_{R}(M,N)\qquad r\mathop{\otimes} f\mapsto r\cdot f μ : R ⊗ Z Hom R ( M , N ) → Hom R ( M , N ) r ⊗ f ↦ r ⋅ f for any r ⊗ f ∈ R n ⊗ Z Hom R ( M , N ) n r\mathop{\otimes}f\in R_n\mathop{\otimes}_{\mathbb{Z}}\operatorname{Hom}_{R}(M,N)_n r ⊗ f ∈ R n ⊗ Z Hom R ( M , N ) n σ ∈ Δ k n \sigma\in\Delta^n_k σ ∈ Δ k n x ∈ M k x\in M_k x ∈ M k
( r ⋅ f ) ( σ ⊗ x ) : = σ R ∗ ( r ) ⋅ N f ( σ ⊗ x ) . (2) (r\cdot f)(\sigma\mathop{\otimes}x):=\sigma_R^*(r)\cdot_N f(\sigma\mathop{\otimes}x). \tag*{(2)} ( r ⋅ f ) ( σ ⊗ x ) := σ R ∗ ( r ) ⋅ N f ( σ ⊗ x ) . (2) By (E3.2.1) and (E3.1.1), one can verify that θ ∗ ( r ⋅ f ) = θ R ∗ ( r ) ⋅ θ ∗ ( f ) \theta^*(r\cdot f)=\theta_R^*(r)\cdot\theta^*(f) θ ∗ ( r ⋅ f ) = θ R ∗ ( r ) ⋅ θ ∗ ( f ) θ : [ k ] → [ n ] \theta:[k]\to[n] θ : [ k ] → [ n ] Δ \Delta Δ μ \mu μ μ \mu μ N N N R R R □ \square □
Remark 3.3. Retain the notation in Definition 3.1.
We call a morphism in Hom R ( M , N ) n \operatorname{Hom}_{R}(M,N)_n Hom R ( M , N ) n n n n Hom R ( M , N ) 0 = Hom R s M o d ( M , N ) \operatorname{Hom}_{R}(M,N)_0=\operatorname{Hom}_{R\mathrm{sMod}}(M,N) Hom R ( M , N ) 0 = Hom R sMod ( M , N ) ( 0 (0 ( 0
Recall that R s M o d R\mathrm{sMod} R sMod R s M o d R\mathrm{sMod} R sMod 0 0 0
Definition 3.4. Let M , N M,N M , N L L L R R R
∘ : Hom R ( N , L ) ⊗ R Hom R ( M , N ) → Hom R ( M , L ) . (3) \circ:\operatorname{Hom}_{R}(N,L)\mathop{\otimes}_{R}\operatorname{Hom}_{R}(M,N)\to\operatorname{Hom}_{R}(M,L). \tag*{(3)} ∘ : Hom R ( N , L ) ⊗ R Hom R ( M , N ) → Hom R ( M , L ) . (3) where for every f ∈ Hom R ( N , L ) n f\in\operatorname{Hom}_{R}(N,L)_n f ∈ Hom R ( N , L ) n g ∈ Hom R ( M , N ) n g\in\operatorname{Hom}_{R}(M,N)_n g ∈ Hom R ( M , N ) n
f ∘ g : = R Δ n ⊗ R M → diag ⊗ R Id M R Δ n ⊗ R R Δ n ⊗ R M → Id R Δ n ⊗ g R Δ n ⊗ R N → f L . (4) f\circ g:=R\Delta^n\mathop{\otimes}_{R}M\xrightarrow{\operatorname{diag}\mathop{\otimes}_{R}\operatorname{Id}_{M}}R\Delta^n\mathop{\otimes}_{R}R\Delta^n\mathop{\otimes}_{R}M\xrightarrow{\operatorname{Id}_{R\Delta^n}\mathop{\otimes}g}R\Delta^n\mathop{\otimes}_{R}N\xrightarrow{f}L. \tag*{(4)} f ∘ g := R Δ n ⊗ R M diag ⊗ R Id M R Δ n ⊗ R R Δ n ⊗ R M Id R Δ n ⊗ g R Δ n ⊗ R N f L . (4) as a composition of R R R σ ∈ Δ k n \sigma\in\Delta_k^n σ ∈ Δ k n x ∈ M k x \in M_k x ∈ M k
( f ∘ g ) ( σ ⊗ x ) = f ( σ ⊗ g ( σ ⊗ x ) ) (5) (f \circ g)(\sigma\otimes x)=f\left(\sigma\otimes g(\sigma\otimes x)\right) \tag*{(5)} ( f ∘ g ) ( σ ⊗ x ) = f ( σ ⊗ g ( σ ⊗ x ) ) (5) By (E3.2.1) and (E3.4.3), the composition ∘ \circ ∘ R R R
( r ⋅ f ) ∘ g = f ∘ ( r ⋅ g ) = r ⋅ ( f ∘ g ) (6) (r \cdot f) \circ g=f \circ(r \cdot g)=r \cdot(f \circ g) \tag*{(6)} ( r ⋅ f ) ∘ g = f ∘ ( r ⋅ g ) = r ⋅ ( f ∘ g ) (6) for any r ∈ R n r \in R_n r ∈ R n θ ∗ ( f ∘ g ) = θ ∗ ( f ) ∘ θ ∗ ( g ) \theta^*(f \circ g)=\theta^*(f)\circ\theta^*(g) θ ∗ ( f ∘ g ) = θ ∗ ( f ) ∘ θ ∗ ( g ) ∘ \circ ∘ ∘ \circ ∘
Recall the notion of a tensored and cotensored enriched category [31 p. 38] .
Theorem 3.6. Let R R R commutative sring. Suppose A A A R R R A A A R R R R R R
s Hom A ( V ⊗ R M , N ) ≅ s Hom R ( V , s Hom A ( M , N ) ) ≅ s Hom A ( M , s Hom R ( V , N ) ) {}^{s}\operatorname{Hom}_A(V\otimes_R M,N)
\cong{}^{s}\operatorname{Hom}_R\left(V,{}^{s}\operatorname{Hom}_A(M,N)\right)
\cong{}^{s}\operatorname{Hom}_A\left(M,{}^{s}\operatorname{Hom}_R(V,N)\right) s Hom A ( V ⊗ R M , N ) ≅ s Hom R ( V , s Hom A ( M , N ) ) ≅ s Hom A ( M , s Hom R ( V , N ) ) that are enriched natural for every R R R V V V A A A M , N M,N M , N
Proof. This follows from [21 p. 395-396] and is implicit in [14 Section 4.1] . □
Centralizers
We extend the classical notion of centralizers [13 (4.5)] to the R R R
Definition 3.7. Let A A A R R R M M M A e A^e A e centralizer of A A A M M M M A M^A M A
( M A ) n : = { x ∈ M n ∣ ( J A / R ) k θ M ∗ ( x ) = 0 , ∀ θ : [ k ] → [ n ] in Δ } (8) (M^A)_n:=\left\{x\in M_n\mid(J_{A/R})_k\theta_M^*(x)=0,\ \forall\theta:[k]\to[n]\text{ in }\Delta\right\} \tag*{(8)} ( M A ) n := { x ∈ M n ∣ ( J A / R ) k θ M ∗ ( x ) = 0 , ∀ θ : [ k ] → [ n ] in Δ } (8) for every n ≥ 0 n\ge0 n ≥ 0 J A / R = ker ( μ A / R : A e → A ) J_{A/R}=\ker(\mu_{A/R}:A^e\to A) J A / R = ker ( μ A / R : A e → A )
Any left A e A^{e} A e ( A , A ) (A,A) ( A , A ) A A A M M M
( M A ) n : = { x ∈ M n ∣ a θ M ∗ ( x ) = θ M ∗ ( x ) a , ∀ θ : [ k ] → [ n ] in Δ and a ∈ A k } . (M^{A})_{n}:=\{x\in M_{n}\mid a\theta^{*}_{M}(x)=\theta^{*}_{M}(x)a,\ \forall\theta:[k]\to[n]\text{ in }\Delta\text{ and }a\in A_{k}\}. ( M A ) n := { x ∈ M n ∣ a θ M ∗ ( x ) = θ M ∗ ( x ) a , ∀ θ : [ k ] → [ n ] in Δ and a ∈ A k } . Therefore, A A = Z ( A ) A^{A}=Z(A) A A = Z ( A ) A A A A e A^{e} A e I I I A A A ( A / I ) A = Z ( A / I ) (A/I)^{A}=Z(A/I) ( A / I ) A = Z ( A / I )
The following generalizes [13 Lemma 4.1.4] in R R R
Lemma 3.9. Let A A A R R R M M M A e A^{e} A e
(1) M A M^{A} M A R R R
(2) ( − ) A : A e (-)^{A}:A^{e} ( − ) A : A e → R \to R → R R R R
(3) ( − ) A ≅ Hom A e ( A , − ) (-)^{A}\cong\operatorname{Hom}_{A^{e}}(A,-) ( − ) A ≅ Hom A e ( A , − ) R R R
Proof. (1): Note that any r ∈ R n r\in R_{n} r ∈ R n M M M r ⊗ 1 A n r\otimes1_{A_{n}} r ⊗ 1 A n 1 A n ⊗ r 1_{A_{n}}\otimes r 1 A n ⊗ r R R R A e A^{e} A e R R R ( J A / R ) n (J_{A/R})_{n} ( J A / R ) n ( M A ) n (M^{A})_{n} ( M A ) n R n R_{n} R n
So it suffices to show that M A M^{A} M A M M M x ∈ ( M A ) n x\in(M^{A})_{n} x ∈ ( M A ) n σ : [ m ] → [ n ] \sigma:[m]\to[n] σ : [ m ] → [ n ] Δ \Delta Δ θ : [ k ] → [ m ] \theta:[k]\to[m] θ : [ k ] → [ m ] Δ \Delta Δ
( J A / R ) k θ M ∗ ( σ M ∗ ( x ) ) = ( J A / R ) k ( σ θ ) M ∗ ( x ) = 0 (J_{A/R})_{k}\theta^{*}_{M}(\sigma^{*}_{M}(x))=(J_{A/R})_{k}(\sigma\theta)^{*}_{M}(x)=0 ( J A / R ) k θ M ∗ ( σ M ∗ ( x )) = ( J A / R ) k ( σ θ ) M ∗ ( x ) = 0 which implies σ M ∗ ( x ) ∈ ( M A ) m \sigma^{*}_{M}(x)\in(M^{A})_{m} σ M ∗ ( x ) ∈ ( M A ) m M A M^{A} M A
(2): Let f ∈ Hom A e ( M , N ) n f\in\operatorname{Hom}_{A^{e}}(M,N)_{n} f ∈ Hom A e ( M , N ) n ( R Δ n ⊗ R M ) A = R Δ n ⊗ R M A (R\Delta^{n}\otimes_{R}M)^{A}=R\Delta^{n}\otimes_{R}M^{A} ( R Δ n ⊗ R M ) A = R Δ n ⊗ R M A A e A^{e} A e M M M f f f Hom R ( M A , N A ) n \operatorname{Hom}_{R}(M^{A},N^{A})_{n} Hom R ( M A , N A ) n σ ∈ Δ m n \sigma\in\Delta^{n}_{m} σ ∈ Δ m n x ∈ ( M A ) m x\in(M^{A})_{m} x ∈ ( M A ) m θ : [ k ] → [ m ] \theta:[k]\to[m] θ : [ k ] → [ m ] Δ \Delta Δ
( J A / R ) k θ N ∗ ( f ( σ ⊗ x ) ) = ( J A / R ) k f ( ( σ θ ) ⊗ θ M ∗ ( x ) ) = f ( ( σ θ ) ⊗ ( J A / R ) k θ M ∗ ( x ) ) = 0 \begin{aligned}
(J_{A/R})_{k}\theta^{*}_{N}(f(\sigma\otimes x))=(J_{A/R})_{k}f((\sigma\theta)\otimes\theta^{*}_{M}(x)) \\
&=f((\sigma\theta)\otimes(J_{A/R})_{k}\theta^{*}_{M}(x))=0
\end{aligned} ( J A / R ) k θ N ∗ ( f ( σ ⊗ x )) = ( J A / R ) k f (( σ θ ) ⊗ θ M ∗ ( x )) = f (( σ θ ) ⊗ ( J A / R ) k θ M ∗ ( x )) = 0 which implies f ( σ ⊗ x ) ∈ ( N A ) m f(\sigma\otimes x)\in(N^{A})_{m} f ( σ ⊗ x ) ∈ ( N A ) m Id M \operatorname{Id}_{M} Id M Id M A \operatorname{Id}_{M^{A}} Id M A ∘ \circ ∘ ( − ) A (-)^{A} ( − ) A A e A^{e} A e → R \to R → R
(3): Define K : Hom A e ( A , M ) → M K:\operatorname{Hom}_{A^{e}}(A,M)\to M K : Hom A e ( A , M ) → M f ↦ K ( f ) = f ( Id [ n ] ⊗ 1 A n ) f\mapsto K(f)=f(\operatorname{Id}_{[n]}\otimes1_{A_{n}}) f ↦ K ( f ) = f ( Id [ n ] ⊗ 1 A n )
K ( θ ∗ ( f ) ) = θ ∗ ( f ) ( Id [ k ] ⊗ 1 A k ) = f ( θ ⊗ 1 A k ) = θ ∗ ( f ( Id [ n ] ⊗ 1 A n ) ) = θ ∗ ( K ( f ) ) \begin{aligned}
K(\theta^{*}(f))=\theta^{*}(f)(\operatorname{Id}_{[k]}\otimes1_{A_{k}})=f(\theta\otimes1_{A_{k}}) \\
&=\theta^{*}(f(\operatorname{Id}_{[n]}\otimes1_{A_{n}}))=\theta^{*}(K(f))
\end{aligned} K ( θ ∗ ( f )) = θ ∗ ( f ) ( Id [ k ] ⊗ 1 A k ) = f ( θ ⊗ 1 A k ) = θ ∗ ( f ( Id [ n ] ⊗ 1 A n )) = θ ∗ ( K ( f )) for any θ : [ k ] → [ n ] \theta:[k]\to[n] θ : [ k ] → [ n ] Δ \Delta Δ K K K R R R
K ( r ⋅ f ) = r f ( Id [ n ] ⊗ 1 A n ) = r K ( f ) K(r\cdot f)=rf(\operatorname{Id}_{[n]}\otimes1_{A_{n}})=rK(f) K ( r ⋅ f ) = r f ( Id [ n ] ⊗ 1 A n ) = r K ( f ) for every r ∈ R n r\in R_{n} r ∈ R n f ∈ Hom A e ( A , M ) n f\in\operatorname{Hom}_{A^{e}}(A,M)_{n} f ∈ Hom A e ( A , M ) n θ : [ k ] → [ n ] \theta:[k]\to[n] θ : [ k ] → [ n ] Δ \Delta Δ
( J A / R ) k θ ∗ ( K ( f ) ) = ( J A / R ) k f ( θ ⊗ 1 A k ) = f ( θ ⊗ ( J A / R ) k 1 A k ) = 0 (J_{A/R})_{k}\theta^{*}(K(f))=(J_{A/R})_{k}f(\theta\otimes1_{A_{k}})=f(\theta\otimes(J_{A/R})_{k}1_{A_{k}})=0 ( J A / R ) k θ ∗ ( K ( f )) = ( J A / R ) k f ( θ ⊗ 1 A k ) = f ( θ ⊗ ( J A / R ) k 1 A k ) = 0 which implies that K ( f ) ∈ ( M A ) n K(f)\in(M^{A})_{n} K ( f ) ∈ ( M A ) n K K K R R R Hom A e ( A , M ) → M A \operatorname{Hom}_{A^{e}}(A,M)\to M^{A} Hom A e ( A , M ) → M A
Conversely, define Ω : M A → Hom A e ( A , M ) \Omega:M^{A}\to\operatorname{Hom}_{A^{e}}(A,M) Ω : M A → Hom A e ( A , M ) x ∈ ( M A ) n x\in(M^{A})_{n} x ∈ ( M A ) n A e A^{e} A e Ω n ( x ) \Omega_{n}(x) Ω n ( x )
Ω ( x ) ( σ ⊗ a ) = ( a ⊗ 1 A k ) ⋅ M σ M ∗ ( x ) = ( 1 A k ⊗ a ) ⋅ M σ M ∗ ( x ) \Omega(x)(\sigma\otimes a)=(a\otimes1_{A_{k}})\cdot_{M}\sigma^{*}_{M}(x)=(1_{A_{k}}\otimes a)\cdot_{M}\sigma^{*}_{M}(x) Ω ( x ) ( σ ⊗ a ) = ( a ⊗ 1 A k ) ⋅ M σ M ∗ ( x ) = ( 1 A k ⊗ a ) ⋅ M σ M ∗ ( x ) for every σ ∈ Δ k n \sigma\in\Delta_k^n σ ∈ Δ k n a ∈ A k a \in A_k a ∈ A k A e A^e A e Ω ( x ) \Omega(x) Ω ( x ) x ∈ ( M A ) n x \in(M^A)_n x ∈ ( M A ) n Ω \Omega Ω R R R K K K K K K Ω \Omega Ω R R R
It remains to show that K K K R R R ϕ ∈ Hom A e ( M , N ) n \phi\in\operatorname{Hom}_{A^e}(M,N)_n ϕ ∈ Hom A e ( M , N ) n
Hom A e ( M , N ) n → ( − ) A Hom R ( M A , N A ) n Hom A e ( A , − ) ↓ ↓ ∗ K Hom R ( Hom A e ( A , M ) , Hom A e ( A , N ) ) n → K ∗ Hom R ( Hom A e ( A , M ) , N A ) n \begin{CD}
\operatorname{Hom}_{A^e}(M,N)_n @>{(-)^A}>> \operatorname{Hom}_{R}(M^A,N^A)_n \\
@V{\operatorname{Hom}_{A^e}(A,-)}VV @VV{\,^*K}V \\
\operatorname{Hom}_{R}\left(\operatorname{Hom}_{A^e}(A,M),\operatorname{Hom}_{A^e}(A,N)\right)_n @>{K^*}>> \operatorname{Hom}_{R}\left(\operatorname{Hom}_{A^e}(A,M),N^A\right)_n
\end{CD} Hom A e ( M , N ) n Hom A e ( A , − ) ↓ ⏐ Hom R ( Hom A e ( A , M ) , Hom A e ( A , N ) ) n ( − ) A K ∗ Hom R ( M A , N A ) n ↓ ⏐ ∗ K Hom R ( Hom A e ( A , M ) , N A ) n where K ∗ = K ∘ − K^* = K \circ- K ∗ = K ∘ − ∗ K = − ∘ ( 1 R Δ n ⊗ R K ) ^*K = - \circ(1_{R^{\Delta^n}} \otimes_R K) ∗ K = − ∘ ( 1 R Δ n ⊗ R K ) 0 0 0 σ ∈ Δ k n \sigma\in\Delta_k^n σ ∈ Δ k n f ∈ Hom A e ( A , M ) k f \in\operatorname{Hom}_{A^e}(A,M)_k f ∈ Hom A e ( A , M ) k
( ϕ ∘ ( 1 R Δ n ⊗ R K ) ) ( σ ⊗ f ) = ϕ ( σ ⊗ K ( f ) ) = ϕ ( σ ⊗ f ( 1 [ k ] ⊗ 1 A k ) ) = σ ∗ ( ϕ ) ( 1 [ k ] ⊗ f ( 1 [ k ] ⊗ 1 A k ) ) = K ( σ ∗ ( ϕ ) ∘ f ) . \begin{aligned}
(\phi\circ(1_{R\Delta_n} \otimes_R K))(\sigma\otimes f)
&= \phi(\sigma\otimes K(f)) \\
&= \phi(\sigma\otimes f(1_{[k]} \otimes1_{A_k})) \\
&= \sigma^*(\phi)(1_{[k]} \otimes f(1_{[k]} \otimes1_{A_k})) \\
&= K(\sigma^*(\phi) \circ f).
\end{aligned} ( ϕ ∘ ( 1 R Δ n ⊗ R K )) ( σ ⊗ f ) = ϕ ( σ ⊗ K ( f )) = ϕ ( σ ⊗ f ( 1 [ k ] ⊗ 1 A k )) = σ ∗ ( ϕ ) ( 1 [ k ] ⊗ f ( 1 [ k ] ⊗ 1 A k )) = K ( σ ∗ ( ϕ ) ∘ f ) . So the diagram commutes for every n ≥ 0 n \ge0 n ≥ 0 R R R K K K R R R □ \square □
The following proposition is the simplicial analogue of [13 Corollary 4.1.7] . The exactness in the proposition below refers to the functor between the (underlying) abelian categories A e A^e A e R R R
Proposition 3.10. Let A A A R R R A A A R R R ( − ) A : A e (-)^A : A^e ( − ) A : A e → R \to R → R
Proof. (⟹ \Longrightarrow ⟹ A A A R R R e ∈ ( A e ) 0 e \in(A^e)_0 e ∈ ( A e ) 0 μ ( e ) = 1 A 0 \mu(e)=1_{A_0} μ ( e ) = 1 A 0 J A / R e = 0 J_{A/R}e=0 J A / R e = 0 A e A^e A e M M M
M A = e M as R -smodules . (9) M^A=eM\text{ as }R\text{-smodules}. \tag*{(9)} M A = e M as R -smodules . (9) It is clear that e M ⊆ M A eM \subseteq M^A e M ⊆ M A J A / R ( e M ) = ( J A / R e ) M = 0 J_{A/R}(eM)=(J_{A/R}e)M=0 J A / R ( e M ) = ( J A / R e ) M = 0 x ∈ ( M A ) k x \in(M^A)_k x ∈ ( M A ) k ( J A / R ) k x = 0 (J_{A/R})_k x=0 ( J A / R ) k x = 0 ( 1 ( A e ) k − s 0 k ( e ) ) x = 0 (1_{(A^e)_k}-s^k_0(e))x=0 ( 1 ( A e ) k − s 0 k ( e )) x = 0 x = s 0 k ( e ) x x=s^k_0(e)x x = s 0 k ( e ) x M A ⊆ e M M^A \subseteq eM M A ⊆ e M e e e M M M R R R
M = e M ⊕ ( 1 ( A e ) 0 − e ) M = M A ⊕ ( 1 ( A e ) 0 − e ) M . M=eM\oplus(1_{(A^e)_0}-e)M=M^A\oplus(1_{(A^e)_0}-e)M. M = e M ⊕ ( 1 ( A e ) 0 − e ) M = M A ⊕ ( 1 ( A e ) 0 − e ) M . Thus, ( − ) A (-)^A ( − ) A e M eM e M
(⟸ \Longleftarrow ⟸ ( − ) A (-)^A ( − ) A ( − ) A (-)^A ( − ) A A e A^e A e 0 → J A / R → A e → μ A / R A → 0 0\to J_{A/R}\to A^e \xrightarrow{\mu_{A/R}} A\to0 0 → J A / R → A e μ A / R A → 0 R R R
μ ‾ A / R : ( A e ) A → A A . \overline{\mu}_{A/R}:(A^e)^A\to A^A. μ A / R : ( A e ) A → A A . Let e ∈ ( ( A e ) A ) 0 e\in((A^e)^A)_0 e ∈ (( A e ) A ) 0 μ ‾ A / R ( e ) = 1 A 0 \overline{\mu}_{A/R}(e)=1_{A_0} μ A / R ( e ) = 1 A 0 e ∈ ( ( A e ) A ) 0 e\in((A^e)^A)_0 e ∈ (( A e ) A ) 0 ( J A / R ) k s 0 k ( e ) = 0 (J_{A/R})_k s^k_0(e)=0 ( J A / R ) k s 0 k ( e ) = 0 k ≥ 0 k\ge0 k ≥ 0 ( J A / R ) e = 0 (J_{A/R})e=0 ( J A / R ) e = 0 e e e A A A □ \square □
:::
Corollary 3.11. Let R R R A A A R R R Hom A e ( A , − ) \operatorname{Hom}_{A^e}(A,-) Hom A e ( A , − ) Hom A e sMod ( A , − ) \operatorname{Hom}_{A^e\text{ sMod}}(A,-) Hom A e sMod ( A , − )
Proof. By Lemma 3.9(3), Proposition 3.10, and Proposition 2.7,
Hom A e ( A , − ) is exact ⟺ ( − ) A is exact ⟺ A is separable over R ⟺ A is a projective left A e -smodule ⟺ Hom A e sMod ( A , − ) is exact . \operatorname{Hom}_{A^e}(A,-)\text{ is exact}\Longleftrightarrow(-)^A\text{ is exact}
\Longleftrightarrow A\text{ is separable over }R
\Longleftrightarrow A\text{ is a projective left }A^e\text{-smodule}
\Longleftrightarrow\operatorname{Hom}_{A^e\text{ sMod}}(A,-)\text{ is exact}. Hom A e ( A , − ) is exact ⟺ ( − ) A is exact ⟺ A is separable over R ⟺ A is a projective left A e -smodule ⟺ Hom A e sMod ( A , − ) is exact . Corollary 3.12. Let R R R A A A R R R R R R
A ⊗ R s − : R s M o d ⇄ A e s M o d : ( − ) A . (10) A\mathbin{\otimes_R^{\mathrm{s}}}- : R\mathrm{sMod}\rightleftarrows A^e\mathrm{sMod} : (-)^A. \tag*{(10)} A ⊗ R s − : R sMod ⇄ A e sMod : ( − ) A . (10) Proof. By Theorem 3.6 and Lemma 3.9(3), there are R R R
Hom A e s ( A ⊗ R s V , M ) ≅ Hom R s ( V , Hom A e s ( A , M ) ) ≅ Hom R s ( V , M A ) \operatorname{Hom}_{A^e}^{\mathrm{s}}(A\mathbin{\otimes_R^{\mathrm{s}}}V,M)\cong\operatorname{Hom}_{R}^{\mathrm{s}}(V,\operatorname{Hom}_{A^e}^{\mathrm{s}}(A,M))\cong\operatorname{Hom}_{R}^{\mathrm{s}}(V,M^A) Hom A e s ( A ⊗ R s V , M ) ≅ Hom R s ( V , Hom A e s ( A , M )) ≅ Hom R s ( V , M A ) natural for any R R R V V V A e A^e A e M M M
Separable enriched functors
It is well-known that separability of a monoid object or a monad is related to the separability of some functors ([6 2.9(1)] , [7 Proposition 6.3] , and [22 Proposition 1.3(1)] ). We now formulate an enriched version of [22 Sect. 1] .
Definition 3.13. Let C , D \mathcal{C},\mathcal{D} C , D R R R R R R F : C → D F:\mathcal{C}\to\mathcal{D} F : C → D separable if for every pair of objects X , Y X,Y X , Y C \mathcal{C} C R R R
F = F X , Y : Hom C s ( X , Y ) ⟶ Hom D s ( F ( X ) , F ( Y ) ) F=F_{X,Y}:\operatorname{Hom}_{\mathcal{C}}^{\mathrm{s}}(X,Y)\longrightarrow\operatorname{Hom}_{\mathcal{D}}^{\mathrm{s}}(F(X),F(Y)) F = F X , Y : Hom C s ( X , Y ) ⟶ Hom D s ( F ( X ) , F ( Y )) there is an R R R H = H X , Y : Hom D s ( F ( X ) , F ( Y ) ) ⟶ Hom C s ( X , Y ) H=H_{X,Y}:\operatorname{Hom}_{\mathcal{D}}^{\mathrm{s}}(F(X),F(Y))\longrightarrow\operatorname{Hom}_{\mathcal{C}}^{\mathrm{s}}(X,Y) H = H X , Y : Hom D s ( F ( X ) , F ( Y )) ⟶ Hom C s ( X , Y )
(1) For every X , Y ∈ Obj C X,Y\in\operatorname{Obj}\mathcal{C} X , Y ∈ Obj C
H ( F ( f ) ) = f , ∀ f ∈ Hom C s ( X , Y ) . H(F(f))=f,\qquad\forall f\in\operatorname{Hom}_{\mathcal{C}}^{\mathrm{s}}(X,Y). H ( F ( f )) = f , ∀ f ∈ Hom C s ( X , Y ) . (2) For any objects M , N , M ′ , N ′ M,N,M',N' M , N , M ′ , N ′ C \mathcal{C} C f ∈ Hom C s ( N , N ′ ) n f\in\operatorname{Hom}_{\mathcal{C}}^{\mathrm{s}}(N,N')_n f ∈ Hom C s ( N , N ′ ) n f ′ ∈ Hom C s ( M , M ′ ) n f'\in\operatorname{Hom}_{\mathcal{C}}^{\mathrm{s}}(M,M')_n f ′ ∈ Hom C s ( M , M ′ ) n g ∈ Hom D s ( F ( M ) , F ( N ) ) n g\in\operatorname{Hom}_{\mathcal{D}}^{\mathrm{s}}(F(M),F(N))_n g ∈ Hom D s ( F ( M ) , F ( N ) ) n g ′ ∈ Hom D s ( F ( M ′ ) , F ( N ′ ) ) n g'\in\operatorname{Hom}_{\mathcal{D}}^{\mathrm{s}}(F(M'),F(N'))_n g ′ ∈ Hom D s ( F ( M ′ ) , F ( N ′ ) ) n
F ( f ) ∘ g = g ′ ∘ F ( f ′ ) ⟹ f ∘ H ( g ) = H ( g ′ ) ∘ f ′ . F(f)\circ g=g'\circ F(f')\quad\Longrightarrow\quad f\circ H(g)=H(g')\circ f'. F ( f ) ∘ g = g ′ ∘ F ( f ′ ) ⟹ f ∘ H ( g ) = H ( g ′ ) ∘ f ′ . Remark 3.14. Retain the notation in Definition 3.13. We note
(1) For any object X X X C \mathcal{C} C H ( Id F ( X ) ) = H ( F ( Id X ) ) = Id X H(\operatorname{Id}_{F(X)})=H(F(\operatorname{Id}_X))=\operatorname{Id}_X H ( Id F ( X ) ) = H ( F ( Id X )) = Id X
(2) If f ′ = Id M f'=\operatorname{Id}_M f ′ = Id M g ′ = F ( f ) ∘ g g'=F(f)\circ g g ′ = F ( f ) ∘ g f ∘ H ( g ) = H ( F ( f ) ∘ g ) f\circ H(g)=H(F(f)\circ g) f ∘ H ( g ) = H ( F ( f ) ∘ g ) H H H
Lemma 3.15. Let A A A R R R M , N ∈ A s M o d M,N\in A\mathrm{sMod} M , N ∈ A sMod Hom R s ( M , N ) A = Hom A s ( M , N ) \operatorname{Hom}_{R}^{\mathrm{s}}(M,N)^A=\operatorname{Hom}_{A}^{\mathrm{s}}(M,N) Hom R s ( M , N ) A = Hom A s ( M , N ) R R R
Proof. Note that Hom R s ( M , N ) \operatorname{Hom}_{R}^{\mathrm{s}}(M,N) Hom R s ( M , N ) ( A , A ) (A,A) ( A , A ) M M M N N N ( A , R ) (A,R) ( A , R ) Hom R s ( M , N ) \operatorname{Hom}_{R}^{\mathrm{s}}(M,N) Hom R s ( M , N ) A e A^e A e
( ( a ⊗ b ) ⋅ f ) ( σ ⊗ x ) = σ ∗ ( a ) f ( σ ⊗ σ ∗ ( b ) x ) (11) ((a\otimes b)\cdot f)(\sigma\otimes x)=\sigma^*(a)f(\sigma\otimes\sigma^*(b)x) \tag*{(11)} (( a ⊗ b ) ⋅ f ) ( σ ⊗ x ) = σ ∗ ( a ) f ( σ ⊗ σ ∗ ( b ) x ) (11) for any a , b ∈ A m a,b \in A_m a , b ∈ A m f ∈ Hom R ( M , N ) m f \in\operatorname{Hom}_{R}(M,N)_m f ∈ Hom R ( M , N ) m σ ∈ Δ k m \sigma\in\Delta^m_k σ ∈ Δ k m x ∈ M k x \in M_k x ∈ M k
Given any f ∈ ( Hom R ( M , N ) A ) n s f \in\left(\operatorname{Hom}_{R}(M,N)^A\right)^s_n f ∈ ( Hom R ( M , N ) A ) n s σ ∈ Δ k n \sigma\in\Delta^n_k σ ∈ Δ k n x ∈ M k x \in M_k x ∈ M k a ∈ A k a \in A_k a ∈ A k
f ( σ ⊗ a x ) = σ ∗ ( f ) ( [ k ] ⊗ a x ) = ( ( 1 A k ⊗ a ) ⋅ σ ∗ ( f ) ) ( [ k ] ⊗ x ) f(\sigma\otimes ax)=\sigma^*(f)([k] \otimes ax)=((1_{A_k}\otimes a)\cdot\sigma^*(f))([k]\otimes x) f ( σ ⊗ a x ) = σ ∗ ( f ) ([ k ] ⊗ a x ) = (( 1 A k ⊗ a ) ⋅ σ ∗ ( f )) ([ k ] ⊗ x ) by (E3.15.1)
= ( ( a ⊗ 1 A k ) ⋅ σ ∗ ( f ) ) ( [ k ] ⊗ x ) =((a\otimes1_{A_k})\cdot\sigma^*(f))([k]\otimes x) = (( a ⊗ 1 A k ) ⋅ σ ∗ ( f )) ([ k ] ⊗ x ) as $f \in \left(\operatorname{Hom}_{R}(M,N)^A\right)^s_n
= a σ ∗ ( f ) ( [ k ] ⊗ x ) = a f ( σ ⊗ x ) =a\sigma^*(f)([k]\otimes x)=af(\sigma\otimes x) = a σ ∗ ( f ) ([ k ] ⊗ x ) = a f ( σ ⊗ x ) which implies f f f A A A Hom A ( M , N ) n \operatorname{Hom}_A(M,N)_n Hom A ( M , N ) n
Conversely, given any g ∈ Hom A ( M , N ) n s g \in\operatorname{Hom}_A(M,N)^s_n g ∈ Hom A ( M , N ) n s θ : [ m ] → [ n ] \theta:[m]\to[n] θ : [ m ] → [ n ] Δ \Delta Δ a ∈ A m a\in A_m a ∈ A m
( ( a ⊗ 1 A m ) ⋅ θ ∗ ( g ) ) ( σ ⊗ x ) = σ ∗ ( a ) θ ∗ ( g ) ( σ ⊗ x ) ((a\otimes1_{A_m})\cdot\theta^*(g))(\sigma\otimes x)=\sigma^*(a)\theta^*(g)(\sigma\otimes x) (( a ⊗ 1 A m ) ⋅ θ ∗ ( g )) ( σ ⊗ x ) = σ ∗ ( a ) θ ∗ ( g ) ( σ ⊗ x ) by (E3.15.1)
= σ ∗ ( a ) g ( θ σ ⊗ x ) =\sigma^*(a)g(\theta\sigma\otimes x) = σ ∗ ( a ) g ( θ σ ⊗ x ) by (E3.1.1)
= g ( θ σ ⊗ σ ∗ ( a ) x ) = θ ∗ ( g ) ( σ ⊗ σ ∗ ( a ) x ) = ( ( 1 A m ⊗ a ) ⋅ θ ∗ ( g ) ) ( σ ⊗ x ) \begin{aligned}
=g(\theta\sigma\otimes\sigma^*(a)x)=\theta^*(g)(\sigma\otimes\sigma^*(a)x) \\
&=((1_{A_m}\otimes a)\cdot\theta^*(g))(\sigma\otimes x)
\end{aligned} = g ( θ σ ⊗ σ ∗ ( a ) x ) = θ ∗ ( g ) ( σ ⊗ σ ∗ ( a ) x ) = (( 1 A m ⊗ a ) ⋅ θ ∗ ( g )) ( σ ⊗ x ) for σ ∈ Δ k m \sigma\in\Delta^m_k σ ∈ Δ k m x ∈ M k x\in M_k x ∈ M k ( a ⊗ 1 A m ) ⋅ θ ∗ ( g ) = ( 1 A m ⊗ a ) ⋅ θ ∗ ( g ) (a\otimes1_{A_m})\cdot\theta^*(g)=(1_{A_m}\otimes a)\cdot\theta^*(g) ( a ⊗ 1 A m ) ⋅ θ ∗ ( g ) = ( 1 A m ⊗ a ) ⋅ θ ∗ ( g ) g ∈ ( Hom R ( M , N ) A ) n s g\in\left(\operatorname{Hom}_{R}(M,N)^A\right)^s_n g ∈ ( Hom R ( M , N ) A ) n s
The following extends [22 Proposition 1.3(1)] to R R R
Proposition 3.16. Let A A A R R R A A A R R R U = U A / R : A sMod → R sMod U=U_{A/R}:A\text{ sMod}\to R\text{ sMod} U = U A / R : A sMod → R sMod R R R
Proof. ( ⟹ ) (\Longrightarrow) ( ⟹ ) A A A R R R A e = A ⊗ R A o A^e=A\otimes_R A^o A e = A ⊗ R A o e = ∑ i = 1 t x i ⊗ y i ∈ ( A e ) 0 e=\sum_{i=1}^{t}x_i\otimes y_i\in(A^e)_0 e = ∑ i = 1 t x i ⊗ y i ∈ ( A e ) 0 R R R
Hom A ( M , N ) = Hom R ( M , N ) A = e Hom R ( M , N ) . (12) \operatorname{Hom}_A(M,N)=\operatorname{Hom}_{R}(M,N)^A=e\operatorname{Hom}_{R}(M,N). \tag*{(12)} Hom A ( M , N ) = Hom R ( M , N ) A = e Hom R ( M , N ) . (12) Therefore, we may define a simplicial map
H : Hom R ( M , N ) → Hom A ( M , N ) f ↦ s 0 m ( e ) ⋅ f H:\operatorname{Hom}_{R}(M,N)\to\operatorname{Hom}_A(M,N)\qquad f\mapsto s^m_0(e)\cdot f H : Hom R ( M , N ) → Hom A ( M , N ) f ↦ s 0 m ( e ) ⋅ f for every f ∈ Hom R ( M , N ) m f\in\operatorname{Hom}_{R}(M,N)_m f ∈ Hom R ( M , N ) m Hom R ( M , N ) \operatorname{Hom}_{R}(M,N) Hom R ( M , N ) Hom A ( M , N ) \operatorname{Hom}_A(M,N) Hom A ( M , N ) R R R R R R η A / R ( R ) ⊗ 1 \eta_{A/R}(R)\otimes1 η A / R ( R ) ⊗ 1 e e e H H H R R R
H ( r ⋅ f ) = s 0 m ( e ) ⋅ ( ( η A / R ( r ) ⊗ 1 A m ) ⋅ f ) = ( η A / R ( r ) ⊗ 1 A m ) ⋅ ( s 0 m ( e ) ⋅ f ) = r ⋅ H ( f ) . \begin{aligned}
H(r\cdot f)=s^m_0(e)\cdot((\eta_{A/R}(r)\otimes1_{A_m})\cdot f) \\
&=(\eta_{A/R}(r)\otimes1_{A_m})\cdot(s^m_0(e)\cdot f)=r\cdot H(f).
\end{aligned} H ( r ⋅ f ) = s 0 m ( e ) ⋅ (( η A / R ( r ) ⊗ 1 A m ) ⋅ f ) = ( η A / R ( r ) ⊗ 1 A m ) ⋅ ( s 0 m ( e ) ⋅ f ) = r ⋅ H ( f ) . for every r ∈ R m r\in R_m r ∈ R m f ∈ Hom R ( M , N ) m f\in\operatorname{Hom}_{R}(M,N)_m f ∈ Hom R ( M , N ) m
It remains to show H H H
We first show H H H h ∈ Hom A ( M , N ) m h\in\operatorname{Hom}_A(M,N)_m h ∈ Hom A ( M , N ) m h = s 0 m ( e ) ⋅ h ˉ h=s^m_0(e)\cdot\bar{h} h = s 0 m ( e ) ⋅ h ˉ h ˉ ∈ Hom R ( M , N ) m \bar{h}\in\operatorname{Hom}_{R}(M,N)_m h ˉ ∈ Hom R ( M , N ) m
H ( U ( h ) ) = s 0 m ( e ) ⋅ ( s 0 m ( e ) ⋅ h ˉ ) = s 0 m ( e ) ⋅ h ˉ = h H(U(h))=s^m_0(e)\cdot(s^m_0(e)\cdot\bar{h})=s^m_0(e)\cdot\bar{h}=h H ( U ( h )) = s 0 m ( e ) ⋅ ( s 0 m ( e ) ⋅ h ˉ ) = s 0 m ( e ) ⋅ h ˉ = h as s 0 m ( e ) s^m_0(e) s 0 m ( e ) H H H H H H U ( f ) ∘ g = g ′ ∘ U ( f ′ ) U(f) \circ g = g' \circ U(f') U ( f ) ∘ g = g ′ ∘ U ( f ′ ) A A A m m m f , f ′ f, f' f , f ′ R R R m m m g , g ′ g,g' g , g ′
( f ∘ H ( g ) ) ( σ ⊗ x ) = f ( σ ⊗ ( s 0 m ( e ) ⋅ g ) ( σ ⊗ x ) ) (f \circ H(g))(\sigma\otimes x)=f(\sigma\otimes(s_0^m(e)\cdot g)(\sigma\otimes x)) ( f ∘ H ( g )) ( σ ⊗ x ) = f ( σ ⊗ ( s 0 m ( e ) ⋅ g ) ( σ ⊗ x )) by (E3.4.3)
= ∑ i = 1 t s 0 k ( x i ) f ( σ ⊗ g ( σ ⊗ s 0 k ( y i ) x ) ) =\sum_{i=1}^{t}s_0^k(x_i)f(\sigma\otimes g(\sigma\otimes s_0^k(y_i)x)) = i = 1 ∑ t s 0 k ( x i ) f ( σ ⊗ g ( σ ⊗ s 0 k ( y i ) x )) by (E3.15.1)
= ∑ i = 1 t s 0 k ( x i ) f ( σ ⊗ g ( σ ⊗ s 0 k ( y i ) x ) ) =\sum_{i=1}^{t}s_0^k(x_i)f(\sigma\otimes g(\sigma\otimes s_0^k(y_i)x)) = i = 1 ∑ t s 0 k ( x i ) f ( σ ⊗ g ( σ ⊗ s 0 k ( y i ) x )) f f f A A A
= ∑ i = 1 t s 0 k ( x i ) ( U ( f ) ∘ g ) ( σ ⊗ s 0 k ( y i ) x ) =\sum_{i=1}^{t}s_0^k(x_i)(U(f)\circ g)(\sigma\otimes s_0^k(y_i)x) = i = 1 ∑ t s 0 k ( x i ) ( U ( f ) ∘ g ) ( σ ⊗ s 0 k ( y i ) x ) by (E3.4.3)
= ( s 0 m ( e ) ⋅ ( U ( f ) ∘ g ) ) ( σ ⊗ x ) =(s_0^m(e)\cdot(U(f)\circ g))(\sigma\otimes x) = ( s 0 m ( e ) ⋅ ( U ( f ) ∘ g )) ( σ ⊗ x ) by (E3.15.1)
for every σ ∈ Δ m k \sigma\in\Delta_m^k σ ∈ Δ m k x ∈ M k x \in M_k x ∈ M k f ∘ H ( g ) = s 0 m ( e ) ⋅ ( U ( f ) ∘ g ) f\circ H(g)=s_0^m(e)\cdot(U(f)\circ g) f ∘ H ( g ) = s 0 m ( e ) ⋅ ( U ( f ) ∘ g ) H ( g ′ ) ∘ f ′ = s 0 m ( e ) ⋅ ( g ′ ∘ U ( f ′ ) ) H(g')\circ f'=s_0^m(e)\cdot(g'\circ U(f')) H ( g ′ ) ∘ f ′ = s 0 m ( e ) ⋅ ( g ′ ∘ U ( f ′ )) U ( f ) ∘ g = g ′ ∘ U ( f ′ ) U(f)\circ g=g'\circ U(f') U ( f ) ∘ g = g ′ ∘ U ( f ′ )
f ∘ H ( g ) = s 0 m ( e ) ⋅ ( U ( f ) ∘ g ) = s 0 m ( e ) ⋅ ( g ′ ∘ U ( f ′ ) ) = H ( g ′ ) ∘ f ′ . f\circ H(g)=s_0^m(e)\cdot(U(f)\circ g)=s_0^m(e)\cdot(g'\circ U(f'))=H(g')\circ f'. f ∘ H ( g ) = s 0 m ( e ) ⋅ ( U ( f ) ∘ g ) = s 0 m ( e ) ⋅ ( g ′ ∘ U ( f ′ )) = H ( g ′ ) ∘ f ′ . Therefore, U U U R R R
( ⟸ ) (\Longleftarrow) ( ⟸ ) U U U A e = A ⊗ R A o A^e=A\otimes_R A^o A e = A ⊗ R A o R R R A A A A e A^e A e R R R
j : A ⟶ A e x ↦ 1 A n ⊗ x j:A\longrightarrow A^e \qquad x\mapsto1_{A_n}\otimes x j : A ⟶ A e x ↦ 1 A n ⊗ x for every x ∈ A n x\in A_n x ∈ A n A A A H ( j ) : A → A e H(j):A\to A^e H ( j ) : A → A e
We claim H ( j ) H(j) H ( j ) ( A , A ) (A,A) ( A , A ) H ( j ) H(j) H ( j ) A A A a ∈ A n a\in A_n a ∈ A n A A A
r a ∈ Hom A ( A , A ) n σ ⊗ x ↦ x ⋅ σ ∗ ( a ) r_a\in\operatorname{Hom}_A(A,A)_n \qquad\sigma\otimes x\mapsto x\cdot\sigma^*(a) r a ∈ Hom A ( A , A ) n σ ⊗ x ↦ x ⋅ σ ∗ ( a ) ρ a ∈ Hom A ( A e , A e ) n σ ⊗ z ↦ ( 1 A k ⊗ σ ∗ ( a ) ) ⋅ z \rho_a\in\operatorname{Hom}_A(A^e,A^e)_n \qquad\sigma\otimes z\mapsto(1_{A_k}\otimes\sigma^*(a))\cdot z ρ a ∈ Hom A ( A e , A e ) n σ ⊗ z ↦ ( 1 A k ⊗ σ ∗ ( a )) ⋅ z for every σ ∈ Δ k n \sigma\in\Delta_k^n σ ∈ Δ k n x ∈ A k x\in A_k x ∈ A k z ∈ ( A e ) k z\in(A^e)_k z ∈ ( A e ) k r a r_a r a a a a A A A ρ a \rho_a ρ a a a a A e A^e A e
For any a ∈ A n a\in A_n a ∈ A n
U ( ρ a ) ∘ s 0 n ( j ) = s 0 n ( j ) ∘ U ( r a ) . (13) U(\rho_a)\circ s_0^n(j)=s_0^n(j)\circ U(r_a). \tag*{(13)} U ( ρ a ) ∘ s 0 n ( j ) = s 0 n ( j ) ∘ U ( r a ) . (13) Indeed, for every σ ∈ Δ k n \sigma\in\Delta_k^n σ ∈ Δ k n x ∈ A k x\in A_k x ∈ A k
( U ( ρ a ) ∘ s 0 n ( j ) ) ( σ ⊗ x ) = ρ a ( σ ⊗ 1 A k ⊗ x ) = ( 1 A k ⊗ σ ∗ ( a ) ) ( 1 A k ⊗ x ) = 1 A k ⊗ x σ ∗ ( a ) = 1 A k ⊗ r a ( σ ⊗ x ) = ( s 0 n ( j ) ∘ U ( r a ) ) ( σ ⊗ x ) . \begin{aligned}
(U(\rho_a)\circ s_0^n(j))(\sigma\otimes x)=\rho_a(\sigma\otimes1_{A_k}\otimes x)=(1_{A_k}\otimes\sigma^*(a))(1_{A_k}\otimes x) \\
&=1_{A_k}\otimes x\sigma^*(a)=1_{A_k}\otimes r_a(\sigma\otimes x)=(s_0^n(j)\circ U(r_a))(\sigma\otimes x).
\end{aligned} ( U ( ρ a ) ∘ s 0 n ( j )) ( σ ⊗ x ) = ρ a ( σ ⊗ 1 A k ⊗ x ) = ( 1 A k ⊗ σ ∗ ( a )) ( 1 A k ⊗ x ) = 1 A k ⊗ x σ ∗ ( a ) = 1 A k ⊗ r a ( σ ⊗ x ) = ( s 0 n ( j ) ∘ U ( r a )) ( σ ⊗ x ) . This proves (E3.16.2). By condition (2) of H H H
ρ a ∘ H ( s 0 n ( j ) ) = H ( s 0 n ( j ) ) ∘ r a . (14) \rho_a\circ H(s_0^n(j))=H(s_0^n(j))\circ r_a. \tag*{(14)} ρ a ∘ H ( s 0 n ( j )) = H ( s 0 n ( j )) ∘ r a . (14) Evaluating (E3.16.3) at I d [ n ] ⊗ x Id_{[n]}\otimes x I d [ n ] ⊗ x x ∈ A n x\in A_n x ∈ A n H ( s 0 n ( j ) ) = s 0 n ( H ( j ) ) H(s_0^n(j))=s_0^n(H(j)) H ( s 0 n ( j )) = s 0 n ( H ( j ))
( 1 A n ⊗ a ) ⋅ H ( j ) ( x ) = H ( j ) ( x a ) . (1_{A_n}\otimes a)\cdot H(j)(x)=H(j)(xa). ( 1 A n ⊗ a ) ⋅ H ( j ) ( x ) = H ( j ) ( x a ) . Therefore, H ( j ) H(j) H ( j ) A A A H ( j ) H(j) H ( j ) ( A , A ) (A,A) ( A , A ) A e A^e A e
Last, we note H ( ȷ ) H(\jmath) H ( ) μ A / R : A e → A \mu_{A/R}: A^{e} \to A μ A / R : A e → A μ A / R ∘ H ( ȷ ) = H ( U ( μ ) ∘ ȷ ) = H ( I d A ) = I d A \mu_{A/R} \circ H(\jmath) = H(U(\mu) \circ\jmath) = H(Id_A) = Id_A μ A / R ∘ H ( ) = H ( U ( μ ) ∘ ) = H ( I d A ) = I d A A A A R R R □ \square □
More Properties on Separability
We collect basic properties of separable simplicial algebras, many of which have apparent analogues in the classical theory of separable algebras [13 Chapter 4] .
Invariance under base change, tensor product, and quotient
Lemma 4.1. Let R R R S S S R R R A A A R R R S ⊗ R A S \otimes_R A S ⊗ R A S S S
Proof. Let T = S ⊗ R A T = S \otimes_R A T = S ⊗ R A
T e = ( S ⊗ R A ) ⊗ S ( S ⊗ R A ) o ≅ S ⊗ R A e T^{e} = (S \mathop{\otimes}\limits_R A) \mathop{\otimes}\limits_S (S \mathop{\otimes}\limits_R A)^{o} \cong S \mathop{\otimes}\limits_R A^{e} T e = ( S R ⊗ A ) S ⊗ ( S R ⊗ A ) o ≅ S R ⊗ A e where for every s 0 , s 1 ∈ S n s_0, s_1 \in S_n s 0 , s 1 ∈ S n a 0 , a 1 ∈ A n a_0, a_1 \in A_n a 0 , a 1 ∈ A n
s 0 ⊗ R a 0 ⊗ S s 1 ⊗ R a 1 ↦ s 0 s 1 ⊗ R a 0 ⊗ R a 1 . s_0 \otimes_R a_0 \otimes_S s_1 \otimes_R a_1 \mapsto s_0s_1 \otimes_R a_0 \otimes_R a_1. s 0 ⊗ R a 0 ⊗ S s 1 ⊗ R a 1 ↦ s 0 s 1 ⊗ R a 0 ⊗ R a 1 . Therefore, the left T e T^{e} T e T T T S ⊗ R A e S \otimes_R A^{e} S ⊗ R A e T T T
( s ⊗ R a 0 ⊗ R a 1 ) ⋅ ( x ⊗ R a ) = s x ⊗ R a 0 a a 1 (s \otimes_R a_0 \otimes_R a_1) \cdot(x \otimes_R a) = sx \otimes_R a_0aa_1 ( s ⊗ R a 0 ⊗ R a 1 ) ⋅ ( x ⊗ R a ) = s x ⊗ R a 0 a a 1 for s , x ∈ S n s, x \in S_n s , x ∈ S n a , a 0 , a 1 ∈ A n a, a_0, a_1 \in A_n a , a 0 , a 1 ∈ A n T e T^{e} T e μ T / S : T e → T \mu_{T/S}: T^{e} \to T μ T / S : T e → T S ⊗ R A e S \otimes_R A^{e} S ⊗ R A e μ ~ : S ⊗ R A e → T \widetilde{\mu}: S \otimes_R A^{e} \to T μ : S ⊗ R A e → T
s ⊗ a 0 ⊗ a 1 ↦ s ⊗ a 0 a 1 . s \otimes a_0 \otimes a_1 \mapsto s \otimes a_0a_1. s ⊗ a 0 ⊗ a 1 ↦ s ⊗ a 0 a 1 . It follows that μ ~ = I d S ⊗ R μ A / R \widetilde{\mu} = Id_S \otimes_R \mu_{A/R} μ = I d S ⊗ R μ A / R A A A R R R μ A / R \mu_{A/R} μ A / R A e A^{e} A e h h h S ⊗ R A e S \otimes_R A^{e} S ⊗ R A e I d S ⊗ R h Id_S \otimes_R h I d S ⊗ R h μ ~ \widetilde{\mu} μ μ T / S \mu_{T/S} μ T / S T T T S S S □ \square □
Lemma 4.2. Let R R R S , T S, T S , T R R R A A A S S S B B B T T T A ⊗ R B A \otimes_R B A ⊗ R B S ⊗ R T S \otimes_R T S ⊗ R T
Proof. Let A e = A ⊗ S A o A^{e} = A \otimes_S A^{o} A e = A ⊗ S A o B e = B ⊗ T B o B^{e} = B \otimes_T B^{o} B e = B ⊗ T B o D = A ⊗ R B D = A \otimes_R B D = A ⊗ R B K = S ⊗ R T K = S \otimes_R T K = S ⊗ R T D e = D ⊗ K D o D^{e} = D \otimes_K D^{o} D e = D ⊗ K D o
D e = ( A ⊗ R B ) ⊗ K ( A ⊗ R B ) o ≅ A e ⊗ R B e D^{e} = (A \otimes_R B) \otimes_K (A \otimes_R B)^{o} \cong A^{e} \otimes_R B^{e} D e = ( A ⊗ R B ) ⊗ K ( A ⊗ R B ) o ≅ A e ⊗ R B e where for every a 0 , a 1 ∈ A n a_0, a_1 \in A_n a 0 , a 1 ∈ A n b 0 , b 1 ∈ B n b_0, b_1 \in B_n b 0 , b 1 ∈ B n
a 0 ⊗ R b 0 ⊗ K a 1 ⊗ R b 1 ↦ a 0 ⊗ S a 1 ⊗ R b 0 ⊗ T b 1 . a_0 \otimes_R b_0 \otimes_K a_1 \otimes_R b_1 \mapsto a_0 \otimes_S a_1 \otimes_R b_0 \otimes_T b_1. a 0 ⊗ R b 0 ⊗ K a 1 ⊗ R b 1 ↦ a 0 ⊗ S a 1 ⊗ R b 0 ⊗ T b 1 . The left D e D^{e} D e D D D A e ⊗ R B e A^{e} \otimes_R B^{e} A e ⊗ R B e
( a 0 ⊗ S a 1 ⊗ R b 0 ⊗ T b 1 ) ⋅ ( a ⊗ R b ) = a 0 a a 1 ⊗ R b 0 b b 1 (a_0 \otimes_S a_1 \otimes_R b_0 \otimes_T b_1) \cdot(a \otimes_R b) = a_0aa_1 \otimes_R b_0bb_1 ( a 0 ⊗ S a 1 ⊗ R b 0 ⊗ T b 1 ) ⋅ ( a ⊗ R b ) = a 0 a a 1 ⊗ R b 0 b b 1 for any a , a 0 , a 1 ∈ A n a, a_0, a_1 \in A_n a , a 0 , a 1 ∈ A n b , b 0 , b 1 ∈ B n b, b_0, b_1 \in B_n b , b 0 , b 1 ∈ B n D e D^{e} D e μ D / K : D e → D \mu_{D/K}: D^{e} \to D μ D / K : D e → D μ ~ : A e ⊗ R B e → A ⊗ R B \widetilde{\mu}: A^{e} \otimes_R B^{e} \to A \otimes_R B μ : A e ⊗ R B e → A ⊗ R B
a 0 ⊗ S a 1 ⊗ R b 0 ⊗ T b 1 ↦ a 0 a 1 ⊗ R b 0 b 1 . a_0 \otimes_S a_1 \otimes_R b_0 \otimes_T b_1 \mapsto a_0a_1 \otimes_R b_0b_1. a 0 ⊗ S a 1 ⊗ R b 0 ⊗ T b 1 ↦ a 0 a 1 ⊗ R b 0 b 1 . Clearly μ ~ = μ A / S ⊗ R μ B / T \tilde{\mu}=\mu_{A/S}\otimes_R\mu_{B/T} μ ~ = μ A / S ⊗ R μ B / T A A A S S S B B B T T T μ A / S \mu_{A/S} μ A / S A e A^e A e h A h_A h A μ B / T \mu_{B/T} μ B / T B e B^e B e h B h_B h B h A ⊗ R h B h_A\otimes_R h_B h A ⊗ R h B A e ⊗ R B e A^e\otimes_R B^e A e ⊗ R B e μ ~ \tilde{\mu} μ ~ μ D / K \mu_{D/K} μ D / K D D D K K K
Remark 4.3. Let A A A R R R e = ∑ i = 1 t x i ⊗ y i e=\sum_{i=1}^{t}x_i\otimes y_i e = ∑ i = 1 t x i ⊗ y i
A o A^o A o R R R e = ∑ i = 1 t y i ⊗ x i e=\sum_{i=1}^{t}y_i\otimes x_i e = ∑ i = 1 t y i ⊗ x i A A A R R R A o A^o A o R R R
Then by Lemma 4.2, A e = A ⊗ R s A o A^e=A\otimes_R^s A^o A e = A ⊗ R s A o R R R
Lemma 4.4. Let R R R A A A R R R I I I A A A A / I A/I A / I R R R
Proof. Let M M M ( A / I ) ⊗ R ( A / I ) o (A/I)\otimes_R(A/I)^o ( A / I ) ⊗ R ( A / I ) o A ↠ A / I A\twoheadrightarrow A/I A ↠ A / I M M M A ⊗ R A o A\otimes_R A^o A ⊗ R A o A / I A/I A / I M M M A A A M M M M A = M A / I M^A=M^{A/I} M A = M A / I R R R A A A R R R ( − ) A (-)^A ( − ) A ( − ) A / I (-)^{A/I} ( − ) A / I A / I A/I A / I R R R
Simplicial center is computed levelwise
Let R R R
Lemma 4.5. Let A A A R R R R R R P : A → A P:A\to A P : A → A n ≥ 0 n\ge0 n ≥ 0
im ( P n ) = Z ( A n ) and P n ( a ) = a \operatorname{im}(P_n)=Z(A_n)\quad\text{and}\quad P_n(a)=a im ( P n ) = Z ( A n ) and P n ( a ) = a for all a ∈ Z ( A n ) a\in Z(A_n) a ∈ Z ( A n )
Proof. Let e = ∑ i = 1 t x i ⊗ y i ∈ ( A e ) 0 e=\sum_{i=1}^{t}x_i\otimes y_i\in(A^e)_0 e = ∑ i = 1 t x i ⊗ y i ∈ ( A e ) 0 μ A / R ( e ) = 1 A 0 \mu_{A/R}(e)=1_{A_0} μ A / R ( e ) = 1 A 0 ( b ⊗ 1 A k ) s 0 k ( e ) = ( 1 A k ⊗ b ) s 0 k ( e ) (b\otimes1_{A_k})s_0^k(e)=(1_{A_k}\otimes b)s_0^k(e) ( b ⊗ 1 A k ) s 0 k ( e ) = ( 1 A k ⊗ b ) s 0 k ( e ) b ∈ A k b\in A_k b ∈ A k A A A A e A^e A e R R R
φ : A e → Hom R ( A , A ) \varphi:A^e\to\operatorname{Hom}_R(A,A) φ : A e → Hom R ( A , A ) given by the left A e A^e A e P : = φ ( e ) P:=\varphi(e) P := φ ( e ) a ↦ s 0 n ( e ) ⋅ a a\mapsto s_0^n(e)\cdot a a ↦ s 0 n ( e ) ⋅ a a ∈ A n a\in A_n a ∈ A n
im ( P ) = e A = A A = Z ( A ) . \operatorname{im}(P)=eA=A^A=Z(A). im ( P ) = e A = A A = Z ( A ) . Moreover, if a ∈ Z ( A n ) a\in Z(A_n) a ∈ Z ( A n )
a = μ A / R ( s 0 n ( e ) ) a = ∑ i = 1 t s 0 n ( x i ) a s 0 n ( y i ) = s 0 n ( e ) ⋅ a = P ( a ) . a=\mu_{A/R}(s_0^n(e))a=\sum_{i=1}^{t}s_0^n(x_i)as_0^n(y_i)=s_0^n(e)\cdot a=P(a). a = μ A / R ( s 0 n ( e )) a = i = 1 ∑ t s 0 n ( x i ) a s 0 n ( y i ) = s 0 n ( e ) ⋅ a = P ( a ) . This completes the proof.
Proposition 4.6 (Level-wise center lemma). Let A A A R R R n ≥ 0 n\ge0 n ≥ 0
Z ( A ) n = Z ( A n ) . Z(A)_n=Z(A_n). Z ( A ) n = Z ( A n ) . Proof. By definition, Z ( A ) n ⊆ Z ( A n ) Z(A)_n\subseteq Z(A_n) Z ( A ) n ⊆ Z ( A n ) a ∈ Z ( A n ) a\in Z(A_n) a ∈ Z ( A n ) a = P ( a ) ∈ im ( P ) = Z ( A ) n a=P(a)\in\operatorname{im}(P)=Z(A)_n a = P ( a ) ∈ im ( P ) = Z ( A ) n
Corollary 4.7. Let A A A B B B R R R
Z ( A ⊗ R s B ) = Z ( A ) ⊗ R s Z ( B ) . Z(A\otimes_R^s B)=Z(A)\otimes_R^s Z(B). Z ( A ⊗ R s B ) = Z ( A ) ⊗ R s Z ( B ) . Proof. By Lemma 4.2 , A ⊗ R B A \otimes_R B A ⊗ R B R R R n ≥ 0 n \ge0 n ≥ 0 A n ⊗ R n B n A_n \otimes_{R_n} B_n A n ⊗ R n B n R n R_n R n [4 Chapter 3, Corollary 2.7] ,
Z ( A n ⊗ R n B n ) = Z ( A n ) ⊗ R n Z ( B n ) . Z(A_n \otimes_{R_n} B_n) = Z(A_n) \otimes_{R_n} Z(B_n). Z ( A n ⊗ R n B n ) = Z ( A n ) ⊗ R n Z ( B n ) . It follows from Proposition 4.6 that for every n ≥ 0 n \ge0 n ≥ 0
Z ( A ⊗ R B ) n = Z ( A n ⊗ R n B n ) = Z ( A n ) ⊗ R n Z ( B n ) = Z ( A ) n ⊗ R n Z ( B ) n . Z(A \otimes_R B)_n = Z(A_n \otimes_{R_n} B_n) = Z(A_n) \otimes_{R_n} Z(B_n) = Z(A)_n \otimes_{R_n} Z(B)_n. Z ( A ⊗ R B ) n = Z ( A n ⊗ R n B n ) = Z ( A n ) ⊗ R n Z ( B n ) = Z ( A ) n ⊗ R n Z ( B ) n . This completes the proof.
Contrasts with the classical theory
Section 4.1 demonstrates a large portion of the classical theory of separable algebras is retained in the simplicial setting. There are, nevertheless, important differences. In particular,
(a) Classically, simple commutative rings are necessarily fields. This observation is used in Van Oystaeyen and Y. Zhang’s proof of [24 Proposition 4.6] . But simple commutative rings may not be simplicial fields.
(b) Simplicial separability implies levelwise separability but not conversely. This shows simplicial separability is strictly stronger than levelwise separability.
Let k k k W ⟨ 0 ⟩ : = W ⟨ 0 ⟩ / k W^{\langle0\rangle} := W^{\langle0\rangle}/k W ⟨ 0 ⟩ := W ⟨ 0 ⟩ / k k k k [38 Theorem 8.4.1] and [16 Theorem 3.1] ), to the chain complex k ≅ k k \cong k k ≅ k k k k ( W ⟨ 0 ⟩ ) n (W^{\langle0\rangle})_n ( W ⟨ 0 ⟩ ) n { ξ j n } j = 0 n \{\xi_j^n\}_{j=0}^n { ξ j n } j = 0 n
d i ( ξ j n ) = { ξ j − 1 n − 1 i ≥ j + 1 , ξ j n − 1 i ≤ j d_i(\xi_j^n) =
\begin{cases}
\xi_{j-1}^{n-1} & i \ge j+1,\\
\xi_j^{n-1} & i \le j
\end{cases} d i ( ξ j n ) = { ξ j − 1 n − 1 ξ j n − 1 i ≥ j + 1 , i ≤ j with the convention that ξ − 1 n = 0 \xi_{-1}^n = 0 ξ − 1 n = 0
s i ( ξ j n ) = { ξ j n + 1 i ≥ j + 1 , ξ j + 1 n + 1 i ≤ j . s_i(\xi_j^n) =
\begin{cases}
\xi_{j}^{n+1} & i \ge j+1,\\
\xi_{j+1}^{n+1} & i \le j.
\end{cases} s i ( ξ j n ) = { ξ j n + 1 ξ j + 1 n + 1 i ≥ j + 1 , i ≤ j . Moreover, there is a unital commutative k k k W ⟨ 0 ⟩ W^{\langle0\rangle} W ⟨ 0 ⟩
ξ i n ξ j n = ξ j n ξ i n = ξ min { i , j } n \xi_i^n \xi_j^n = \xi_j^n \xi_i^n = \xi_{\min\{i,j\}}^n ξ i n ξ j n = ξ j n ξ i n = ξ m i n { i , j } n where ξ n n = 1 ( W ⟨ 0 ⟩ ) n \xi_n^n = 1_{(W^{\langle0\rangle})_n} ξ n n = 1 ( W ⟨ 0 ⟩ ) n ( W ⟨ 0 ⟩ ) n (W^{\langle0\rangle})_n ( W ⟨ 0 ⟩ ) n
p j n = ξ j n − ξ j − 1 n p 0 n = ξ 0 n p_j^n = \xi_j^n - \xi_{j-1}^n \qquad p_0^n = \xi_0^n p j n = ξ j n − ξ j − 1 n p 0 n = ξ 0 n form a complete set of orthogonal idempotents of ( W ⟨ 0 ⟩ ) n (W^{\langle0\rangle})_n ( W ⟨ 0 ⟩ ) n ( W ⟨ 0 ⟩ ) n ≅ k ⊕ ( n + 1 ) (W^{\langle0\rangle})_n \cong k^{\oplus(n+1)} ( W ⟨ 0 ⟩ ) n ≅ k ⊕ ( n + 1 ) k k k n > 0 n > 0 n > 0 W ⟨ 0 ⟩ / k W^{\langle0\rangle}/k W ⟨ 0 ⟩ / k simplicial winding algebra (with base ring k k k [16 Sect. 6.2] .
Lemma 4.8. Let k k k W ⟨ 0 ⟩ / k W^{\langle0\rangle}/k W ⟨ 0 ⟩ / k
Proof. A simplicial field is necessarily constant because the face maps are surjective by the simplicial identities. But any nonzero surjection between fields is an isomorphism. Since W ⟨ 0 ⟩ W^{\langle0\rangle} W ⟨ 0 ⟩
Suppose I I I W ⟨ 0 ⟩ / k W^{\langle0\rangle}/k W ⟨ 0 ⟩ / k N ( I ) N(I) N ( I ) I I I N ( W ⟨ 0 ⟩ / k ) = ( k ≅ k ) N(W^{\langle0\rangle}/k) = (k \cong k) N ( W ⟨ 0 ⟩ / k ) = ( k ≅ k ) I 0 = N ( I ) 0 ≠ 0 I_0 = N(I)_0 \ne0 I 0 = N ( I ) 0 = 0 N ( I ) 1 = 0 N(I)_1 = 0 N ( I ) 1 = 0 N ( I ) 1 → N ( I ) 0 N(I)_1 \to N(I)_0 N ( I ) 1 → N ( I ) 0 I 0 I_0 I 0 ( W ⟨ 0 ⟩ / k ) 0 = k (W^{\langle0\rangle}/k)_0 = k ( W ⟨ 0 ⟩ / k ) 0 = k I 0 = k I_0 = k I 0 = k 1 ( W ⟨ 0 ⟩ / k ) 0 ∈ I 0 1_{(W^{\langle0\rangle}/k)_0} \in I_0 1 ( W ⟨ 0 ⟩ / k ) 0 ∈ I 0 1 ( W ⟨ 0 ⟩ / k ) n ∈ I n 1_{(W^{\langle0\rangle}/k)_n} \in I_n 1 ( W ⟨ 0 ⟩ / k ) n ∈ I n n ≥ 0 n \ge0 n ≥ 0 I = W ⟨ 0 ⟩ / k I = W^{\langle0\rangle}/k I = W ⟨ 0 ⟩ / k I I I
Example 4.9. Let K / k K/k K / k
( W ⟨ 0 ⟩ / k ) ⊗ k s K ≅ W ⟨ 0 ⟩ / K (W^{\langle0\rangle}/k) \mathop{\otimes}\limits_{k}^{s} K \cong W^{\langle0\rangle}/K ( W ⟨ 0 ⟩ / k ) k ⊗ s K ≅ W ⟨ 0 ⟩ / K is a simple commutative K K K Theorem 2.11 that any separable K K K A A A K ⊗ K s A 0 K\mathop{\otimes}\limits_{K}^{s} A_{0} K K ⊗ s A 0
(1) As W ⟨ 0 ⟩ / K W^{\langle0\rangle}/K W ⟨ 0 ⟩ / K W ⟨ 0 ⟩ / K W^{\langle0\rangle}/K W ⟨ 0 ⟩ / K K K K ( W ⟨ 0 ⟩ / K ) n = K ⊕ ( n + 1 ) (W^{\langle0\rangle}/K)_n=K^{\oplus(n+1)} ( W ⟨ 0 ⟩ / K ) n = K ⊕ ( n + 1 ) K K K Remark 2.9 (3).
(2) Note that K K K k k k [13 Theorem 4.5.6] . So W ⟨ 0 ⟩ / K W^{\langle0\rangle}/K W ⟨ 0 ⟩ / K W ⟨ 0 ⟩ / k W^{\langle0\rangle}/k W ⟨ 0 ⟩ / k Lemma 4.1 . This gives yet another example, showing that over a non-constant commutative sring, a separable simplicial algebra need not be constant, in contrast to Theorem 2.11 .
Enriched Morita Theory
The enriched Morita theory of monoids in a monoidal category is well-developed from a categorical perspective [5 , 20 , 27 ] . We record some of the general theory of enriched Morita equivalence and characterize the relevant notions of finitely generated projective modules and generators for simplicial modules as finite retracts.
Simplicial Morita context
Let R R R Morita context is a sextuple ( A , B , A P B , B Q A , f , g ) (A,B,{}_A P_B,{}_B Q_A,f,g) ( A , B , A P B , B Q A , f , g ) R R R A A A B B B ( A , B ) (A,B) ( A , B ) P P P ( B , A ) (B,A) ( B , A ) Q Q Q
f : Q ⊗ A s P ⟶ B a ( B , B ) -sbimodule morphism f:Q\mathop{\otimes}\limits_{A}^{s}P\longrightarrow B\quad\text{a }(B,B)\text{-sbimodule morphism} f : Q A ⊗ s P ⟶ B a ( B , B ) -sbimodule morphism g : P ⊗ B s Q ⟶ A an ( A , A ) -sbimodule morphism g:P\mathop{\otimes}\limits_{B}^{s}Q\longrightarrow A\quad\text{an }(A,A)\text{-sbimodule morphism} g : P B ⊗ s Q ⟶ A an ( A , A ) -sbimodule morphism making the following diagrams commutative:
Q ⊗ A s P ⊗ B s Q → f ⊗ B s Id B ⊗ B s Q Id ⊗ A s g ↓ ↓ ≅ Q ⊗ A s A → ≅ Q P ⊗ B s Q ⊗ A s P → g ⊗ A s Id A ⊗ A s P Id ⊗ B s f ↓ ↓ ≅ P ⊗ B s B → ≅ P \begin{CD}
Q\mathop{\otimes}\limits_{A}^{s}P\mathop{\otimes}\limits_{B}^{s}Q @>{f\mathop{\otimes}_{B}^{s}\operatorname{Id}}>> B\mathop{\otimes}\limits_{B}^{s}Q \\
@V{\operatorname{Id}\mathop{\otimes}_{A}^{s}g}VV @VV{\cong}V \\
Q\mathop{\otimes}\limits_{A}^{s}A @>{\cong}>> Q
\end{CD}
\qquad
\begin{CD}
P\mathop{\otimes}\limits_{B}^{s}Q\mathop{\otimes}\limits_{A}^{s}P @>{g\mathop{\otimes}_{A}^{s}\operatorname{Id}}>> A\mathop{\otimes}\limits_{A}^{s}P \\
@V{\operatorname{Id}\mathop{\otimes}_{B}^{s}f}VV @VV{\cong}V \\
P\mathop{\otimes}\limits_{B}^{s}B @>{\cong}>> P
\end{CD} Q A ⊗ s P B ⊗ s Q Id ⊗ A s g ↓ ⏐ Q A ⊗ s A f ⊗ B s Id ≅ B B ⊗ s Q ↓ ⏐ ≅ Q P B ⊗ s Q A ⊗ s P Id ⊗ B s f ↓ ⏐ P B ⊗ s B g ⊗ A s Id ≅ A A ⊗ s P ↓ ⏐ ≅ P where ≅ \cong ≅
Definition 5.1. Let P P P B B B S P : = Hom B ( P , P ) S_P:=\operatorname{Hom}_{B}(P,P) S P := Hom B ( P , P ) P P P ( S P , B ) (S_P,B) ( S P , B )
P ⊗ R s B ⟶ P p ⊗ b ↦ p b P\mathop{\otimes}\limits_{R}^{s}B\longrightarrow P\qquad p\otimes b\mapsto pb P R ⊗ s B ⟶ P p ⊗ b ↦ p b S P ⊗ R s P ⟶ P s ⊗ p ↦ s p : = s ( Id [ n ] ⊗ p ) . S_P\mathop{\otimes}\limits_{R}^{s}P\longrightarrow P\qquad s\otimes p\mapsto sp:=s(\operatorname{Id}_{[n]}\otimes p). S P R ⊗ s P ⟶ P s ⊗ p ↦ s p := s ( Id [ n ] ⊗ p ) . Consequently, Q P : = Hom B ( P , B ) Q_P:=\operatorname{Hom}_{B}(P,B) Q P := Hom B ( P , B ) ( B , S P ) (B,S_P) ( B , S P )
Q P ⊗ R s S P ⟶ Q P q ⊗ s ↦ q s : σ ⊗ p ↦ q ( σ ⊗ s ( σ ⊗ p ) ) Q_P\mathop{\otimes}\limits_{R}^{s}S_P\longrightarrow Q_P\qquad q\otimes s\mapsto qs:\sigma\otimes p\mapsto q(\sigma\otimes s(\sigma\otimes p)) Q P R ⊗ s S P ⟶ Q P q ⊗ s ↦ q s : σ ⊗ p ↦ q ( σ ⊗ s ( σ ⊗ p )) B ⊗ R s Q P ⟶ Q P b ⊗ q ↦ b q : σ ⊗ p ↦ σ ∗ ( b ) q ( σ ⊗ p ) . B\mathop{\otimes}\limits_{R}^{s}Q_P\longrightarrow Q_P\qquad b\otimes q\mapsto bq:\sigma\otimes p\mapsto\sigma^{*}(b)q(\sigma\otimes p). B R ⊗ s Q P ⟶ Q P b ⊗ q ↦ b q : σ ⊗ p ↦ σ ∗ ( b ) q ( σ ⊗ p ) . Define a ( B , B ) (B,B) ( B , B ) ( S P , S P ) (S_P,S_P) ( S P , S P )
a : Q P ⊗ S P P ⟶ B q ⊗ S P p ⟼ ( q p = q ( Id [ n ] ⊗ p ) ) \mathbf{a}: Q_P \otimes_{S_P} P \longrightarrow B \qquad q \otimes_{S_P} p \longmapsto(qp=q(\operatorname{Id}_{[n]}\otimes p)) a : Q P ⊗ S P P ⟶ B q ⊗ S P p ⟼ ( q p = q ( Id [ n ] ⊗ p )) b : P ⊗ B Q P ⟶ S P p ⊗ B q ⟼ ( p q : σ ′ ⊗ p ′ ⟼ ( σ ′ ) ∗ ( p ) q ( σ ′ ⊗ p ′ ) ) . \mathbf{b}: P \otimes_B Q_P \longrightarrow S_P \qquad p \otimes_B q \longmapsto(pq:\sigma' \otimes p' \longmapsto(\sigma')^*(p)q(\sigma' \otimes p')). b : P ⊗ B Q P ⟶ S P p ⊗ B q ⟼ ( pq : σ ′ ⊗ p ′ ⟼ ( σ ′ ) ∗ ( p ) q ( σ ′ ⊗ p ′ )) . We say ( S P , B , S P P B , B Q S P , a , b ) (S_P,B,{}_{S_P}P_B,{}_{B}Q_{S_P},\mathbf{a},\mathbf{b}) ( S P , B , S P P B , B Q S P , a , b ) canonical Morita context associated with P B P_B P B R s M o d \mathsf{RsMod} RsMod
(a) P B P_B P B B B B b \mathbf{b} b ( S P , S P ) (S_P,S_P) ( S P , S P )
(b) P B P_B P B B B B a \mathbf{a} a ( B , B ) (B,B) ( B , B )
(c) P B P_B P B B B B P B P_B P B B B B
Remark 5.2. (1) For a left A A A A P {}_AP A P A P {}_AP A P A A A A A A A A A
(2) Definition 5.1 agrees with [24 §2] and [27 ] specialized to simplicial modules.
(3) Definition 5.1 is also the direct simplicial analogue of [19 §18] . Moreover, it is not hard to verify that the simplicial analogues of the usual associativity identities in [19 §18] hold. Consequently, if a \mathbf{a} a b \mathbf{b} b
For simplicial modules, basic projectives and sgenerators admit a simple characterization similar to that of finitely generated projectives and generators in the classical ring and module theory.
Lemma 5.3. Let P P P B B B
(1) P B P_B P B P P P B ⊕ t B^{\oplus t} B ⊕ t
(2) P B P_B P B B B B P ⊕ t P^{\oplus t} P ⊕ t
Proof. Set Q : = Q P = Hom B ( P , B ) Q:=Q_P=\operatorname{Hom}_B(P,B) Q := Q P = Hom B ( P , B ) S : = S P = Hom B ( P , P ) S:=S_P=\operatorname{Hom}_B(P,P) S := S P = Hom B ( P , P ) P B P_B P B
( S , B , S P B , B Q S , a , b ) . (S,B,{}_{S}P_B,{}_{B}Q_{S},\mathbf{a},\mathbf{b}). ( S , B , S P B , B Q S , a , b ) . (1) Suppose P B P_B P B b : P ⊗ B Q ⟶ S \mathbf{b}:P\otimes_B Q\longrightarrow S b : P ⊗ B Q ⟶ S p i ∈ P 0 p_i\in P_0 p i ∈ P 0 q i ∈ Q 0 q_i\in Q_0 q i ∈ Q 0 1 S 0 = ∑ i = 1 t p i q i 1_{S_0}=\sum_{i=1}^{t}p_iq_i 1 S 0 = ∑ i = 1 t p i q i p ∈ P k p\in P_k p ∈ P k
p = 1 S k ⋅ p = s 0 k ( ∑ i = 1 t p i q i ) p = ∑ i = 1 t ( s 0 k ( p i ) s 0 k ( q i ) ) p = ∑ i = 1 t s 0 k ( p i ) ( s 0 k ( q i ) ( Id [ k ] ⊗ p ) ) = ∑ i = 1 t s 0 k ( p i ) ( q i ( p ) ) . \begin{aligned}
p=1_{S_k}\cdot p=s_0^k\left(\sum_{i=1}^{t}p_iq_i\right)p=\sum_{i=1}^{t}(s_0^k(p_i)s_0^k(q_i))p \\
&=\sum_{i=1}^{t}s_0^k(p_i)\left(s_0^k(q_i)(\operatorname{Id}_{[k]}\otimes p)\right)=\sum_{i=1}^{t}s_0^k(p_i)(q_i(p)).
\end{aligned} p = 1 S k ⋅ p = s 0 k ( i = 1 ∑ t p i q i ) p = i = 1 ∑ t ( s 0 k ( p i ) s 0 k ( q i )) p = i = 1 ∑ t s 0 k ( p i ) ( s 0 k ( q i ) ( Id [ k ] ⊗ p ) ) = i = 1 ∑ t s 0 k ( p i ) ( q i ( p )) . Define right B B B
f : B ⊕ t ⟶ P ( b 1 , … , b t ) ⟼ ∑ i = 1 t s 0 k ( p i ) b i f:B^{\oplus t}\longrightarrow P \qquad(b_1,\ldots,b_t)\longmapsto\sum_{i=1}^{t}s_0^k(p_i)b_i f : B ⊕ t ⟶ P ( b 1 , … , b t ) ⟼ i = 1 ∑ t s 0 k ( p i ) b i h : P ⟶ B ⊕ t p ⟼ ( q 1 ( p ) , … , q t ( p ) ) . h:P\longrightarrow B^{\oplus t}\qquad p\longmapsto(q_1(p),\ldots,q_t(p)). h : P ⟶ B ⊕ t p ⟼ ( q 1 ( p ) , … , q t ( p )) . It follows that f h = Id P fh=\operatorname{Id}_P f h = Id P P P P B ⊕ t B^{\oplus t} B ⊕ t P P P B ⊕ t B^{\oplus t} B ⊕ t
P → h B ⊕ t → f P f h = Id P P \xrightarrow{h} B^{\oplus t} \xrightarrow{f} P \qquad fh=\operatorname{Id}_P P h B ⊕ t f P f h = Id P Let ȷ i : B → B ⊕ t \jmath_i:B\to B^{\oplus t} i : B → B ⊕ t π i : B ⊕ t → B \pi_i:B^{\oplus t}\to B π i : B ⊕ t → B B ⊕ t B^{\oplus t} B ⊕ t
p i = f ȷ i ( 1 B 0 ) ∈ P 0 q i = π i h ∈ Q 0 . p_i=f\jmath_i(1_{B_0})\in P_0 \qquad q_i=\pi_i h\in Q_0. p i = f i ( 1 B 0 ) ∈ P 0 q i = π i h ∈ Q 0 . Then for every p ∈ P k p\in P_k p ∈ P k f f f B B B
p = f h ( p ) = f ( ∑ i = 1 t ȷ i ( 1 B k ) π i h ( p ) ) = ∑ i = 1 t f ( ȷ i ( s 0 k ( 1 B 0 ) ) π i h ( p ) ) = ∑ i = 1 t s 0 k ( f ȷ i ( 1 B 0 ) ) π i h ( p ) = ∑ i = 1 t s 0 k ( p i ) q i ( p ) . \begin{aligned}
p&=fh(p)=f\left(\sum_{i=1}^{t}\jmath_i(1_{B_k})\pi_i h(p)\right)=\sum_{i=1}^{t}f\left(\jmath_i(s_0^k(1_{B_0}))\pi_i h(p)\right)\\
&=\sum_{i=1}^{t}s_0^k(f\jmath_i(1_{B_0}))\pi_i h(p)=\sum_{i=1}^{t}s_0^k(p_i)q_i(p).
\end{aligned} p = f h ( p ) = f ( i = 1 ∑ t i ( 1 B k ) π i h ( p ) ) = i = 1 ∑ t f ( i ( s 0 k ( 1 B 0 )) π i h ( p ) ) = i = 1 ∑ t s 0 k ( f i ( 1 B 0 )) π i h ( p ) = i = 1 ∑ t s 0 k ( p i ) q i ( p ) . This implies that 1 S 0 = ∑ i = 1 t p i q i 1_{S_0}=\sum_{i=1}^{t}p_iq_i 1 S 0 = ∑ i = 1 t p i q i s ∈ S k s\in S_k s ∈ S k
s = s ⋅ 1 S k = s ∑ i = 1 t s 0 k ( p i ) s 0 k ( q i ) = ∑ i = 1 t b ( s s 0 k ( p i ) ⊗ s 0 k ( q i ) ) s=s\cdot1_{S_k}=s\sum_{i=1}^{t}s_0^k(p_i)s_0^k(q_i)=\sum_{i=1}^{t}b(ss_0^k(p_i)\otimes s_0^k(q_i)) s = s ⋅ 1 S k = s i = 1 ∑ t s 0 k ( p i ) s 0 k ( q i ) = i = 1 ∑ t b ( s s 0 k ( p i ) ⊗ s 0 k ( q i )) as b b b ( S , S ) (S,S) ( S , S ) b b b
(2) Suppose B P {}_BP B P S S S a : Q ⊗ S P → B a:Q\otimes_S P\to B a : Q ⊗ S P → B q i ′ ∈ Q 0 q'_i\in Q_0 q i ′ ∈ Q 0 p i ′ ∈ P 0 p'_i\in P_0 p i ′ ∈ P 0 1 B 0 = ∑ i = 1 t ′ q i ′ p i ′ 1_{B_0}=\sum_{i=1}^{t'}q'_ip'_i 1 B 0 = ∑ i = 1 t ′ q i ′ p i ′ B B B
f ′ : P ⊕ t ′ → B ( x 1 , … , x t ′ ) ↦ ∑ i = 1 t ′ q i ′ ( x i ) f':P^{\oplus t'}\to B\qquad(x_1,\ldots,x_{t'})\mapsto\sum_{i=1}^{t'}q'_i(x_i) f ′ : P ⊕ t ′ → B ( x 1 , … , x t ′ ) ↦ i = 1 ∑ t ′ q i ′ ( x i ) h ′ : B → P ⊕ t ′ y ↦ ( s 0 k ( p 1 ′ ) y , … , s 0 k ( p t ′ ′ ) y ) . h':B\to P^{\oplus t'}\qquad y\mapsto(s_0^k(p'_1)y,\ldots,s_0^k(p'_{t'})y). h ′ : B → P ⊕ t ′ y ↦ ( s 0 k ( p 1 ′ ) y , … , s 0 k ( p t ′ ′ ) y ) . Then by construction, f ′ h ′ ( 1 B 0 ) = 1 B 0 f'h'(1_{B_0})=1_{B_0} f ′ h ′ ( 1 B 0 ) = 1 B 0 B B B f ′ h ′ f'h' f ′ h ′ I d B I d_B I d B B B B P ⊕ t ′ P^{\oplus t'} P ⊕ t ′
Conversely, suppose B B B P ⊕ t ′ P^{\oplus t'} P ⊕ t ′
B → h ′ P ⊕ t ′ → f ′ B f ′ h ′ = I d B B\xrightarrow{h'}P^{\oplus t'}\xrightarrow{f'}B\qquad f'h'=I d_B B h ′ P ⊕ t ′ f ′ B f ′ h ′ = I d B Write h ′ ( 1 B 0 ) = ( p 1 ′ , … , p t ′ ′ ) h'(1_{B_0})=(p'_1,\ldots,p'_{t'}) h ′ ( 1 B 0 ) = ( p 1 ′ , … , p t ′ ′ )
q i ′ = P → ȷ i P ⊕ t ′ → f ′ B . q'_i=P\xrightarrow{\jmath_i}P^{\oplus t'}\xrightarrow{f'}B. q i ′ = P i P ⊕ t ′ f ′ B . Then 1 B 0 = f ′ h ′ ( 1 B 0 ) = ∑ i = 1 t ′ q i ′ p i ′ = a ( ∑ i = 1 t ′ q i ′ ⊗ p i ′ ) 1_{B_0}=f'h'(1_{B_0})=\sum_{i=1}^{t'}q'_ip'_i=a(\sum_{i=1}^{t'}q'_i\otimes p'_i) 1 B 0 = f ′ h ′ ( 1 B 0 ) = ∑ i = 1 t ′ q i ′ p i ′ = a ( ∑ i = 1 t ′ q i ′ ⊗ p i ′ ) a a a
Simplicial Morita equivalence
Theorem 5.4. [27 Theorems 5.1 and 5.3] Let A A A B B B R R R
There exists an R R R F : A s M o d ⇄ B s M o d : G F:A\mathrm{sMod}\rightleftarrows B\mathrm{sMod}:G F : A sMod ⇄ B sMod : G
There exists a Morita context ( A , B , A P B , B Q A , f , g ) (A,B,{}_AP_B,{}_BQ_A,f,g) ( A , B , A P B , B Q A , f , g ) f f f g g g
In particular, there are R R R
F ≅ Q ⊗ A − G ≅ P ⊗ B − F\cong Q\mathop{\otimes}\limits_A- \qquad G\cong P\mathop{\otimes}\limits_B- F ≅ Q A ⊗ − G ≅ P B ⊗ − and A P {}_AP A P A A A B Q {}_BQ B Q B B B
Theorem 5.5. [24 Theorem 2.1] Let ( A , B , A P B , B Q A , f , g ) (A,B,{}_AP_B,{}_BQ_A,f,g) ( A , B , A P B , B Q A , f , g ) R R R f , g f,g f , g
There are canonical R R R P P P Q Q Q
A ≅ Hom B s ( P , P ) A \cong\operatorname{Hom}_{B}^{\mathrm{s}}(P,P) A ≅ Hom B s ( P , P ) A o ≅ Hom B s ( Q , Q ) A^{\mathrm{o}} \cong\operatorname{Hom}_{B}^{\mathrm{s}}(Q,Q) A o ≅ Hom B s ( Q , Q ) B ≅ Hom A s ( Q , Q ) B \cong\operatorname{Hom}_{A}^{\mathrm{s}}(Q,Q) B ≅ Hom A s ( Q , Q ) B o ≅ Hom A s ( P , P ) . B^{\mathrm{o}} \cong\operatorname{Hom}_{A}^{\mathrm{s}}(P,P). B o ≅ Hom A s ( P , P ) . The smodules A P B {}_{A}P_{B} A P B B Q A {}_{B}Q_{A} B Q A
Corollary 5.6. Let A A A B B B R R R A P B {}_{A}P_{B} A P B ( A , B ) (A,B) ( A , B )
P B P_{B} P B R R R λ : A → Hom B s ( P , P ) \lambda:A\to\operatorname{Hom}_{B}^{\mathrm{s}}(P,P) λ : A → Hom B s ( P , P ) A A A
A P {}_{A}P A P R R R B o → Hom A s ( P , P ) B^{\mathrm{o}}\to\operatorname{Hom}_{A}^{\mathrm{s}}(P,P) B o → Hom A s ( P , P ) B B B
Proof. (1) ⟹ \Longrightarrow ⟹ P B P_{B} P B
( S P , B , S P P B , B ( Q P ) S P , a , b ) . (S_{P},B,{}_{S_{P}}P_{B},{}_{B}(Q_{P})_{S_{P}},\mathfrak{a},\mathfrak{b}). ( S P , B , S P P B , B ( Q P ) S P , a , b ) . Since P B P_{B} P B B B B a , b \mathfrak{a},\mathfrak{b} a , b
λ : A → S P = Hom B s ( P , P ) \lambda:A\to S_{P}=\operatorname{Hom}_{B}^{\mathrm{s}}(P,P) λ : A → S P = Hom B s ( P , P ) the S P S_{P} S P P P P Q P Q_{P} Q P A A A λ \lambda λ A A A ( A , B ) (A,B) ( A , B ) P P P ( B , A ) (B,A) ( B , A ) Q P Q_{P} Q P
a ⋅ p = λ ( a ) ⋅ p . a\cdot p=\lambda(a)\cdot p. a ⋅ p = λ ( a ) ⋅ p . Therefore, there is a canonical ( B , B ) (B,B) ( B , B ) λ \lambda λ
λ ˉ : Q P ⊗ A P ≅ Q P ⊗ S P P q ⊗ A p ↦ q ⊗ S P p . \bar{\lambda}:Q_{P}\mathbin{\otimes_{A}}P\cong Q_{P}\mathbin{\otimes_{S_{P}}}P
\qquad q\otimes_{A}p\mapsto q\otimes_{S_{P}}p. λ ˉ : Q P ⊗ A P ≅ Q P ⊗ S P P q ⊗ A p ↦ q ⊗ S P p . This induces a Morita context ( A , B , A P B , B ( Q P ) A , f , g ) (A,B,{}_{A}P_{B},{}_{B}(Q_{P})_{A},f,g) ( A , B , A P B , B ( Q P ) A , f , g )
f : Q P ⊗ A P → λ ˉ Q P ⊗ S P P → a B f:Q_{P}\mathbin{\otimes_{A}}P\xrightarrow{\bar{\lambda}}Q_{P}\mathbin{\otimes_{S_{P}}}P\xrightarrow{\mathfrak{a}}B f : Q P ⊗ A P λ ˉ Q P ⊗ S P P a B g : P ⊗ B Q P → b S P → λ − 1 A . g:P\mathbin{\otimes_{B}}Q_{P}\xrightarrow{\mathfrak{b}}S_{P}\xrightarrow{\lambda^{-1}}A. g : P ⊗ B Q P b S P λ − 1 A . Indeed, f f f g g g a ⋅ p ′ = λ ( a ) ⋅ p ′ a\cdot p'=\lambda(a)\cdot p' a ⋅ p ′ = λ ( a ) ⋅ p ′
g ( p ⊗ B q ) ⋅ p ′ = λ ( g ( p ⊗ B q ) ) ⋅ p ′ = λ ( λ − 1 b ( p ⊗ B q ) ) ⋅ p ′ = b ( p ⊗ B q ) ⋅ p ′ = p ⋅ a ( q ⊗ S P p ′ ) = p ⋅ f ( q ⊗ A p ′ ) g(p\otimes_{B}q)\cdot p'
=\lambda(g(p\otimes_{B}q))\cdot p'
=\lambda(\lambda^{-1}\mathfrak{b}(p\otimes_{B}q))\cdot p'
=\mathfrak{b}(p\otimes_{B}q)\cdot p'
=p\cdot\mathfrak{a}(q\otimes_{S_{P}}p')
=p\cdot f(q\otimes_{A}p') g ( p ⊗ B q ) ⋅ p ′ = λ ( g ( p ⊗ B q )) ⋅ p ′ = λ ( λ − 1 b ( p ⊗ B q )) ⋅ p ′ = b ( p ⊗ B q ) ⋅ p ′ = p ⋅ a ( q ⊗ S P p ′ ) = p ⋅ f ( q ⊗ A p ′ ) and similarly
f ( q ⊗ A p ) ⋅ q ′ = a ( q ⊗ S P p ) ⋅ q ′ = q ⋅ b ( p ⊗ B q ′ ) = q ⋅ λ ( λ − 1 b ( p ⊗ B q ′ ) ) = q ⋅ g ( p ⊗ B q ′ ) . f(q\otimes_{A}p)\cdot q'
=\mathfrak{a}(q\otimes_{S_{P}}p)\cdot q'
=q\cdot\mathfrak{b}(p\otimes_{B}q')
=q\cdot\lambda(\lambda^{-1}\mathfrak{b}(p\otimes_{B}q'))
=q\cdot g(p\otimes_{B}q'). f ( q ⊗ A p ) ⋅ q ′ = a ( q ⊗ S P p ) ⋅ q ′ = q ⋅ b ( p ⊗ B q ′ ) = q ⋅ λ ( λ − 1 b ( p ⊗ B q ′ )) = q ⋅ g ( p ⊗ B q ′ ) . By Theorem 5.5, A P {}_{A}P A P B o → Hom A s ( P , P ) B^{\mathrm{o}}\to\operatorname{Hom}_{A}^{\mathrm{s}}(P,P) B o → Hom A s ( P , P )
(2) ⟹ \Longrightarrow ⟹ □ \square □
Definition 5.7. An ( A , B ) (A,B) ( A , B ) P P P ( B , A ) (B,A) ( B , A ) Q Q Q
P ⊗ B Q ≅ A P\mathbin{\otimes_{B}}Q\cong A P ⊗ B Q ≅ A as ( A , A ) (A,A) ( A , A )
Q ⊗ A P ≅ B Q\mathbin{\otimes_{A}}P\cong B Q ⊗ A P ≅ B as ( B , B ) (B,B) ( B , B )
Theorem 5.8. Let A A A B B B R R R The following are equivalent.
There exists an R R R G : B G : B G : B → A \to A → A
There exists a B B B A P B _A P_B A P B A A A λ : A → Hom B s ( P , P ) \lambda: A \to\operatorname{Hom}^{s}_{B}(P,P) λ : A → Hom B s ( P , P ) P P P A A A B B B
There exists an invertible ( A , B ) (A,B) ( A , B ) A P B _A P_B A P B
In particular, any R R R G : B G : B G : B → A \to A → A A P B ⊗ B s − _A P_B \mathbin{\otimes}^{s}_{B} - A P B ⊗ B s −
Proof. (1) ⟹ \Longrightarrow ⟹
( A , B , A P B , B Q A , f , g ) (A,B,{}_A P_B,{}_B Q_A,f,g) ( A , B , A P B , B Q A , f , g ) where f , g f,g f , g P B P_B P B B B B λ : A → Hom B s ( P , P ) \lambda: A \to\operatorname{Hom}^{s}_{B}(P,P) λ : A → Hom B s ( P , P )
(2) ⟹ \Longrightarrow ⟹ P B P_B P B
( S P , B , S P S P P B , B ( Q P ) S P , a , b ) (S_P,B,{}_{S_P}S_P P_B,{}_B(QP)_{S_P},\mathbf{a},\mathbf{b}) ( S P , B , S P S P P B , B ( QP ) S P , a , b ) where a , b \mathbf{a},\mathbf{b} a , b Q P Q_P Q P P P P P P P ( S P , B ) (S_P,B) ( S P , B ) λ : A → S P \lambda: A \to S_P λ : A → S P P P P ( A , B ) (A,B) ( A , B )
(3) ⟹ \Longrightarrow ⟹ A P B _A P_B A P B Q Q Q F = Q ⊗ A s − F=Q\mathbin{\otimes}^{s}_{A}- F = Q ⊗ A s − G = P ⊗ B s − G=P\mathbin{\otimes}^{s}_{B}- G = P ⊗ B s − R R R □ \square □
Definition 5.9. Let A A A B B B R R R A A A B B B simplicial Morita equivalent if any one of the equivalent conditions in Theorem 5.8 is satisfied.
A Proposition
The following generalizes [13 Proposition 4.4.2] .
Proposition 5.10. Let S S S R R R A A A S S S
If A A A S S S S S S R R R A A A R R R
If A A A R R R A A A S S S
If A A A R R R A A A S S S S S S R R R
Proof. (1) By Proposition 3.16, U S / R : S U_{S/R} : S U S / R : S → R \to R → R R R R U A / S : A U_{A/S} : A U A / S : A → S \to S → S S S S U A / S U_{A/S} U A / S R R R [22 Lemma 1.1(2)] ). The assertion follows from Proposition 3.16 since U A / R = U S / R ∘ U A / S : A U_{A/R}=U_{S/R}\circ U_{A/S}: A U A / R = U S / R ∘ U A / S : A → R \to R → R R R R
(2) Let e ∈ ( A ⊗ R A o ) 0 e\in(A\mathbin{\otimes}_{R}A^{o})_{0} e ∈ ( A ⊗ R A o ) 0 π : A ⊗ R A o → A ⊗ S A o \pi:A\mathbin{\otimes}_{R}A^{o}\to A\mathbin{\otimes}_{S}A^{o} π : A ⊗ R A o → A ⊗ S A o e ˉ : = π ( e ) \bar e:=\pi(e) e ˉ := π ( e ) μ A / R = μ A / S ∘ π \mu_{A/R}=\mu_{A/S}\circ\pi μ A / R = μ A / S ∘ π μ A / S ( e ˉ ) = μ A / S ( π ( e ) ) = μ A / R ( e ) = 1 A 0 \mu_{A/S}(\bar e)=\mu_{A/S}(\pi(e))=\mu_{A/R}(e)=1_{A_{0}} μ A / S ( e ˉ ) = μ A / S ( π ( e )) = μ A / R ( e ) = 1 A 0 a ∈ A n a\in A_n a ∈ A n
( a ⊗ S 1 A n ) s 0 n ( e ˉ ) = π ( ( a ⊗ R 1 A n ) s 0 n ( e ) ) = π ( ( 1 A n ⊗ R a ) s 0 n ( e ) ) = ( 1 A n ⊗ S a ) s 0 n ( e ˉ ) . \begin{aligned}
(a\mathbin{\otimes}_{S}1_{A_n})s^{n}_{0}(\bar e)=\pi((a\mathbin{\otimes}_{R}1_{A_n})s^{n}_{0}(e)) \\
&=\pi((1_{A_n}\mathbin{\otimes}_{R}a)s^{n}_{0}(e))=(1_{A_n}\mathbin{\otimes}_{S}a)s^{n}_{0}(\bar e).
\end{aligned} ( a ⊗ S 1 A n ) s 0 n ( e ˉ ) = π (( a ⊗ R 1 A n ) s 0 n ( e )) = π (( 1 A n ⊗ R a ) s 0 n ( e )) = ( 1 A n ⊗ S a ) s 0 n ( e ˉ ) . So e ˉ \bar e e ˉ A A A S S S A A A S S S
::: (3) Let A e = A ⊗ R A o A^{e}=A\otimes_R A^{o} A e = A ⊗ R A o S e = S ⊗ R S S^{e}=S\otimes_R S S e = S ⊗ R S A A A S S S A A A S S S A A A S ⊕ m S^{\oplus m} S ⊕ m S S S A o = A A^{o}=A A o = A S S S A e A^{e} A e ( S e ) ⊕ m 2 (S^{e})^{\oplus m^2} ( S e ) ⊕ m 2 S e S^{e} S e A e A^{e} A e S e S^{e} S e A e A^{e} A e S e S^{e} S e
Since A A A R R R A A A A e A^{e} A e A e A^{e} A e η A / S : S → A \eta_{A/S}:S\to A η A / S : S → A A A A A e A^{e} A e S e S^{e} S e A A A S e S^{e} S e
Finally, since A A A S S S S S S A ⊕ t A^{\oplus t} A ⊕ t S S S S S S S S S A A A S S S S e S^{e} S e S S S A ⊕ t A^{\oplus t} A ⊕ t S e S^{e} S e S S S S e S^{e} S e A A A S e S^{e} S e S S S R R R
Azumaya Simplicial Algebras
We define Azumaya simplicial algebras and prove several structure theorems.
Basics of Azumaya simplicial algebras
Let R R R
Definition 6.1. Let A A A R R R
A A A R R R R ≅ Z ( A ) R\cong Z(A) R ≅ Z ( A ) η A / R : R → A \eta_{A/R}:R\to A η A / R : R → A
A A A R R R R R R A A A R R R
Remark 6.2. Let A A A B B B R R R
Recall from Remark 2.9(3) that A n A_n A n R n R_n R n A n A_n A n R n R_n R n Z ( A n ) = Z ( A ) n ≅ R n Z(A_n)=Z(A)_n\cong R_n Z ( A n ) = Z ( A ) n ≅ R n
By (E3.10.1) and Remark 3.8(2), R ≅ Z ( A ) = A A = e A R\cong Z(A)=A^{A}=eA R ≅ Z ( A ) = A A = e A A A A R R R e e e
The tensor product A ⊗ R B A\otimes_R B A ⊗ R B R R R A ⊗ R B A\otimes_R B A ⊗ R B R ⊗ R R ≅ R R\otimes_R R\cong R R ⊗ R R ≅ R
Z ( A ⊗ R B ) = Z ( A ) ⊗ R Z ( B ) ≅ R ⊗ R R ≅ R . Z(A\otimes_R B)=Z(A)\otimes_R Z(B)\cong R\otimes_R R\cong R. Z ( A ⊗ R B ) = Z ( A ) ⊗ R Z ( B ) ≅ R ⊗ R R ≅ R . Lemma 6.3. Let A A A R R R
There is a one-to-one correspondence between sideals I I I R R R J J J A A A I ↦ I A I\mapsto IA I ↦ I A J ↦ J ∩ R J\mapsto J\cap R J ↦ J ∩ R
If M M M A A A I I I R R R I A = M IA=M I A = M
Proof. (1) It suffices to show that
I A ∩ R = I ( J ∩ R ) A = J IA\cap R=I \qquad(J\cap R)A=J I A ∩ R = I ( J ∩ R ) A = J which is equivalent to showing for every $n\ge 0
( I A ) n ∩ R n = I n ( J ∩ R ) n A n = J n . (IA)_n\cap R_n=I_n \qquad(J\cap R)_nA_n=J_n. ( I A ) n ∩ R n = I n ( J ∩ R ) n A n = J n . Since I n I_n I n R n R_n R n J n J_n J n A n A_n A n [13 Corollary 7.1.7] as A n A_n A n R n R_n R n M M M A A A I = M ∩ R I=M\cap R I = M ∩ R M = ( M ∩ R ) A = I A M=(M\cap R)A=IA M = ( M ∩ R ) A = I A I ⊆ I ′ ⊆ R I\subseteq I'\subseteq R I ⊆ I ′ ⊆ R I ′ I' I ′ R R R
M = I A ⊆ I ′ A ⊆ A . M=IA\subseteq I'A\subseteq A. M = I A ⊆ I ′ A ⊆ A . But M M M I ′ A = M I'A=M I ′ A = M A A A
I ′ = I ′ A ∩ R = M ∩ R or A ∩ R = I or R I'=I'A\cap R=M\cap R\text{ or }A\cap R=I\text{ or }R I ′ = I ′ A ∩ R = M ∩ R or A ∩ R = I or R which shows I I I □ \square □
Auslander-Goldman Theorem
Theorem 6.4. (Unifying Theorem) Let A A A R R R
(1) A A A R R R
(2) A A A A e A^{e} A e A A A R R R
(3) A A A R R R A e → Hom R ( A , A ) A^{e}\to\operatorname{Hom}_{R}(A,A) A e → Hom R ( A , A ) A e A^{e} A e R R R
Proof. (1) ⟹ \Longrightarrow ⟹ A A A R R R A A A R R R A A A A e A^{e} A e Proposition 2.7 (2), A A A A e A^{e} A e A e A^{e} A e A A A A e A^{e} A e Lemma 5.3 (1)). So it remains to show A A A A e A^{e} A e A e A^{e} A e A ⊕ m A^{\oplus m} A ⊕ m
Let e ∈ ( A e ) 0 e\in(A^{e})_{0} e ∈ ( A e ) 0
A e e A e = A e . A^{e}eA^{e}=A^{e}. A e e A e = A e . If not, there is a maximal sideal M M M A e e A e ⊆ M ⊆ A e A^{e}eA^{e}\subseteq M\subseteq A^{e} A e e A e ⊆ M ⊆ A e A e A^{e} A e R R R
for some maximal sideal I ⊆ R I\subseteq R I ⊆ R
1 A 0 = μ A / R ( e ) ∈ μ ( I A e ) ⊆ I A . 1_{A_{0}}=\mu_{A/R}(e)\in\mu(IA^{e})\subseteq IA. 1 A 0 = μ A / R ( e ) ∈ μ ( I A e ) ⊆ I A . Therefore, I A = A IA=A I A = A
A e = A ⊗ R s A o = ( I A ) ⊗ R s A o = I ( A ⊗ R s A o ) = I A e = M . A^{e}=A\mathbin{\otimes_{R}^{s}}A^{o}=(IA)\mathbin{\otimes_{R}^{s}}A^{o}=I(A\mathbin{\otimes_{R}^{s}}A^{o})=IA^{e}=M. A e = A ⊗ R s A o = ( I A ) ⊗ R s A o = I ( A ⊗ R s A o ) = I A e = M . It contradicts that M M M A e e A e = A e A^{e}eA^{e}=A^{e} A e e A e = A e e e e A e e A^{e}e A e e A e A^{e} A e [19 Remark 18.10(D)] ). Specifically, there exist finitely many u i , v i ∈ ( A e ) 0 u_i,v_i\in(A^{e})_{0} u i , v i ∈ ( A e ) 0
1 ( A e ) 0 = ∑ i = 1 m u i v i . 1_{(A^{e})_{0}}=\sum_{i=1}^{m}u_iv_i. 1 ( A e ) 0 = i = 1 ∑ m u i v i . Define left A e A^{e} A e
h : A e ⟶ ( A e e ) ⊕ m x ⟼ ( x s 0 n ( u 1 ) s 0 n ( e ) , … , x s 0 n ( u m ) s 0 n ( e ) ) h:A^{e}\longrightarrow(A^{e}e)^{\oplus m}
\qquad
x\longmapsto\left(xs_{0}^{n}(u_{1})s_{0}^{n}(e),\ldots,xs_{0}^{n}(u_{m})s_{0}^{n}(e)\right) h : A e ⟶ ( A e e ) ⊕ m x ⟼ ( x s 0 n ( u 1 ) s 0 n ( e ) , … , x s 0 n ( u m ) s 0 n ( e ) ) f : ( A e e ) ⊕ m ⟶ A e ( y 1 , … , y m ) ⟼ ∑ i = 1 m y i s 0 n ( v i ) . f:(A^{e}e)^{\oplus m}\longrightarrow A^{e}
\qquad
(y_{1},\ldots,y_{m})\longmapsto\sum_{i=1}^{m}y_{i}s_{0}^{n}(v_{i}). f : ( A e e ) ⊕ m ⟶ A e ( y 1 , … , y m ) ⟼ i = 1 ∑ m y i s 0 n ( v i ) . Then f h = I d A e fh=Id_{A^{e}} f h = I d A e A e A^{e} A e ( A e e ) ⊕ m (A^{e}e)^{\oplus m} ( A e e ) ⊕ m e e e A e ≅ A e e ⊕ A e ( 1 ( A e ) 0 − e ) A^{e}\cong A^{e}_{e}\oplus A^{e}(1_{(A^{e})_{0}}-e) A e ≅ A e e ⊕ A e ( 1 ( A e ) 0 − e ) A e A^{e} A e μ A / R : A e → A \mu_{A/R}:A^{e}\to A μ A / R : A e → A J A / R = A e ( 1 ( A e ) 0 − e ) J_{A/R}=A^{e}(1_{(A^{e})_{0}}-e) J A / R = A e ( 1 ( A e ) 0 − e ) Proposition 2.7 ). Therefore,
A ≅ A e / J A / R ≅ A e / A e ( 1 ( A e ) 0 − e ) ≅ A e e A\cong A^{e}/J_{A/R}\cong A^{e}/A^{e}(1_{(A^{e})_{0}}-e)\cong A^{e}e A ≅ A e / J A / R ≅ A e / A e ( 1 ( A e ) 0 − e ) ≅ A e e as left A e A^{e} A e A e A^{e} A e ( A e e ) ⊕ m ≅ A ⊕ m (A^{e}e)^{\oplus m}\cong A^{\oplus m} ( A e e ) ⊕ m ≅ A ⊕ m A A A A e A^{e} A e
( 2 ) ⟹ ( 1 ) (2)\Longrightarrow(1) ( 2 ) ⟹ ( 1 ) A A A A e A^{e} A e R R R A A A A e A^{e} A e Lemma 5.3 (1) and Lemma 2.4 ) which implies A A A R R R Proposition 2.7 . The assertion follows.
( 2 ) ⟺ ( 3 ) (2)\Longleftrightarrow(3) ( 2 ) ⟺ ( 3 ) A A A R R R
R ≅ Z ( A ) = A A ≅ Hom A e ( A , A ) . R\cong Z(A)=A^{A}\cong\operatorname{Hom}_{A^{e}}(A,A). R ≅ Z ( A ) = A A ≅ Hom A e ( A , A ) . Applying Theorem 5.8 and Corollary 5.6 to A e A R {}_{A^{e}}A_{R} A e A R R R R A e A^{e} A e
Remark 6.5. By Theorem 6.4 , A A A R R R A e = A ⊗ R A s A^{e}=A\otimes_{R}A^{\mathrm{s}} A e = A ⊗ R A s R R R Theorem 6.4 answers the remaining questions of Pareigis [28 p. 121] and Borceux and Vitale [5 p. 458] in the setting of simplicial algebras without assuming the extra conditions as Van Oystaeyen and Y. Zhang did in [24 §4] .
Theorem 6.6. An R R R A A A A A A Z ( A ) Z(A) Z ( A ) Z ( A ) Z(A) Z ( A ) R R R
Proof. ( ⟹ ) (\Longrightarrow) ( ⟹ ) A A A R R R A A A Z ( A ) Z(A) Z ( A ) A A A Z ( A ) Z(A) Z ( A ) Theorem 6.4 , A A A Z ( A ) Z(A) Z ( A ) Z ( A ) Z(A) Z ( A ) R R R
( ⟸ ) (\Longleftarrow) ( ⟸ )
Classification type results
Proposition 6.7. Let A A A R R R A A A R R R A ≅ R ⊗ R 0 A 0 A\cong R\otimes_{R_{0}}A_{0} A ≅ R ⊗ R 0 A 0 R R R A 0 A_{0} A 0 R 0 R_{0} R 0
Proof. ( ⟹ ) (\Longrightarrow) ( ⟹ ) A A A R R R A A A R R R Theorem 6.4 . Therefore, we have a retraction of R R R
A → ι R ⊕ m → π A . A\xrightarrow{\iota}R^{\oplus m}\xrightarrow{\pi}A. A ι R ⊕ m π A . Note R R R A 0 A_{0} A 0 A A A R 0 R_{0} R 0 η : R → A \eta:R\to A η : R → A ι : A 0 → A \iota:A_{0}\to A ι : A 0 → A R 0 R_{0} R 0
f : R ⊗ R 0 A 0 → η ⊗ R 0 ι A ⊗ R 0 A → μ A A . (15) f:R\otimes_{R_{0}}A_{0}\xrightarrow{\eta\otimes_{R_{0}}\iota}A\otimes_{R_{0}}A\xrightarrow{\mu_{A}}A. \tag*{(15)} f : R ⊗ R 0 A 0 η ⊗ R 0 ι A ⊗ R 0 A μ A A . (15) Then f f f R R R
f ( ( r 1 ⊗ R 0 a 1 ) ( r 2 ⊗ R 0 a 2 ) ) = η ( r 1 r 2 ) s 0 n ( a 1 a 2 ) = η ( r 1 ) s 0 n ( a 1 ) η ( r 2 ) s 0 n ( a 2 ) = f ( r 1 ⊗ R 0 a 1 ) f ( r 2 ⊗ R 0 a 2 ) \begin{aligned}
f((r_{1}\otimes_{R_{0}}a_{1})(r_{2}\otimes_{R_{0}}a_{2}))=\eta(r_{1}r_{2})s_{0}^{n}(a_{1}a_{2}) \\
&=\eta(r_{1})s_{0}^{n}(a_{1})\eta(r_{2})s_{0}^{n}(a_{2}) \\
&=f(r_{1}\otimes_{R_{0}}a_{1})f(r_{2}\otimes_{R_{0}}a_{2})
\end{aligned} f (( r 1 ⊗ R 0 a 1 ) ( r 2 ⊗ R 0 a 2 )) = η ( r 1 r 2 ) s 0 n ( a 1 a 2 ) = η ( r 1 ) s 0 n ( a 1 ) η ( r 2 ) s 0 n ( a 2 ) = f ( r 1 ⊗ R 0 a 1 ) f ( r 2 ⊗ R 0 a 2 ) for every r 1 ⊗ a 1 , r 2 ⊗ a 2 ∈ ( R ⊗ R 0 s A 0 ) n r_1\otimes a_1,r_2\otimes a_2\in(R\otimes_{R_0}^{s}A_0)_n r 1 ⊗ a 1 , r 2 ⊗ a 2 ∈ ( R ⊗ R 0 s A 0 ) n
f ( 1 R n ⊗ R 0 1 A 0 ) = 1 R n s 0 n ( 1 A 0 ) = 1 A n . f(1_{R_n}\otimes_{R_0}1_{A_0})=1_{R_n}s_0^n(1_{A_0})=1_{A_n}. f ( 1 R n ⊗ R 0 1 A 0 ) = 1 R n s 0 n ( 1 A 0 ) = 1 A n . In particular, f f f R R R
Let u 1 , … , u m u_1,\ldots,u_m u 1 , … , u m R 0 ⊕ m R_0^{\oplus m} R 0 ⊕ m a j = π ( u j ) ∈ A 0 a_j=\pi(u_j)\in A_0 a j = π ( u j ) ∈ A 0
π ( r 1 , … , r m ) = ∑ j = 1 m η ( r j ) s 0 n ( a j ) \pi(r_1,\ldots,r_m)=\sum_{j=1}^{m}\eta(r_j)s_0^n(a_j) π ( r 1 , … , r m ) = j = 1 ∑ m η ( r j ) s 0 n ( a j ) for any ( r 1 , … , r m ) ∈ ( R ⊕ m ) n (r_1,\ldots,r_m)\in(R^{\oplus m})_n ( r 1 , … , r m ) ∈ ( R ⊕ m ) n π ȷ = I d A \pi\jmath=Id_{A} π = I d A x ∈ A n x\in A_n x ∈ A n
x = π ( ȷ ( x ) 1 , … , ȷ ( x ) m ) = ∑ j = 1 m η ( ȷ ( x ) j ) s 0 n ( a j ) . (16) x=\pi(\jmath(x)_1,\ldots,\jmath(x)_m)=\sum_{j=1}^{m}\eta(\jmath(x)_j)s_0^n(a_j). \tag*{(16)} x = π ( ( x ) 1 , … , ( x ) m ) = j = 1 ∑ m η ( ( x ) j ) s 0 n ( a j ) . (16) Define another morphism
g : A ⟶ R ⊗ R 0 s A 0 x ↦ ∑ j = 1 m ȷ ( x ) j ⊗ R 0 a j (17) g:A\longrightarrow R\otimes_{R_0}^{s}A_0\qquad x\mapsto\sum_{j=1}^{m}\jmath(x)_j\otimes_{R_0}a_j \tag*{(17)} g : A ⟶ R ⊗ R 0 s A 0 x ↦ j = 1 ∑ m ( x ) j ⊗ R 0 a j (17) for every x ∈ A n x\in A_n x ∈ A n g g g R R R
Indeed, for any θ : [ k ] → [ n ] \theta:[k]\to[n] θ : [ k ] → [ n ] Δ \Delta Δ
θ ∗ ( g ( x ) ) = ∑ j = 1 m θ ∗ ( ȷ ( x ) j ) ⊗ R 0 a j = ∑ j = 1 m ȷ ( θ ∗ ( x ) ) j ⊗ R 0 a j = g ( θ ∗ ( x ) ) \theta^*(g(x))=\sum_{j=1}^{m}\theta^*(\jmath(x)_j)\otimes_{R_0}a_j=\sum_{j=1}^{m}\jmath(\theta^*(x))_j\otimes_{R_0}a_j=g(\theta^*(x)) θ ∗ ( g ( x )) = j = 1 ∑ m θ ∗ ( ( x ) j ) ⊗ R 0 a j = j = 1 ∑ m ( θ ∗ ( x ) ) j ⊗ R 0 a j = g ( θ ∗ ( x )) which implies g g g r ∈ R n r\in R_n r ∈ R n
g ( r x ) = ∑ j = 1 m ȷ ( r x ) j ⊗ R 0 a j = r ( ∑ j = 1 m ȷ ( x ) j ⊗ R 0 a j ) = r g ( x ) g(rx)=\sum_{j=1}^{m}\jmath(rx)_j\otimes_{R_0}a_j=r\left(\sum_{j=1}^{m}\jmath(x)_j\otimes_{R_0}a_j\right)=rg(x) g ( r x ) = j = 1 ∑ m ( r x ) j ⊗ R 0 a j = r ( j = 1 ∑ m ( x ) j ⊗ R 0 a j ) = r g ( x ) which shows g g g R R R
We show f f f g g g R R R x ∈ A n x\in A_n x ∈ A n
f g ( x ) = f ( ∑ j = 1 m ȷ ( x ) j ⊗ R 0 a j ) fg(x)=f\left(\sum_{j=1}^{m}\jmath(x)_j\otimes_{R_0}a_j\right) f g ( x ) = f ( j = 1 ∑ m ( x ) j ⊗ R 0 a j ) by E6.7.1
= ∑ j = 1 m η ( ȷ ( x ) j ) s 0 n ( a j ) = x . =\sum_{j=1}^{m}\eta(\jmath(x)_j)s_0^n(a_j)=x. = j = 1 ∑ m η ( ( x ) j ) s 0 n ( a j ) = x . by E6.7.2
Therefore, f g = I d A fg=Id_A f g = I d A
Conversely, for every r ⊗ R 0 a ∈ ( R ⊗ R 0 s A 0 ) n r\otimes_{R_0}a\in(R\otimes_{R_0}^{s}A_0)_n r ⊗ R 0 a ∈ ( R ⊗ R 0 s A 0 ) n
g f ( r ⊗ R 0 a ) = g ( r s 0 n ( a ) ) = r g ( s 0 n ( a ) ) gf(r\otimes_{R_0}a)=g(rs_0^n(a))=rg(s_0^n(a)) g f ( r ⊗ R 0 a ) = g ( r s 0 n ( a )) = r g ( s 0 n ( a )) as g g g R R R
= r ∑ j = 1 m s 0 n ( ȷ ( a ) j ) ⊗ R 0 a j =r\sum_{j=1}^{m}s_0^n(\jmath(a)_j)\otimes_{R_0}a_j = r j = 1 ∑ m s 0 n ( ( a ) j ) ⊗ R 0 a j by E6.7.3
= r ⊗ R 0 ∑ j = 1 m ȷ ( a ) j a j =r\otimes_{R_0}\sum_{j=1}^{m}\jmath(a)_ja_j = r ⊗ R 0 j = 1 ∑ m ( a ) j a j since R n R_n R n R 0 R_0 R 0
= r ⊗ R 0 π ȷ ( a ) = r ⊗ a . =r\otimes_{R_0}\pi\jmath(a)=r\otimes a. = r ⊗ R 0 π ( a ) = r ⊗ a . Then g f = I d R ⊗ R 0 s A 0 gf=Id_{R\otimes_{R_0}^{s}A_0} g f = I d R ⊗ R 0 s A 0 f f f R R R
Lastly, we note that A 0 A_0 A 0 R 0 R_0 R 0
( ⟸ ) (\Longleftarrow) ( ⟸ ) A 0 A_0 A 0 R 0 R_0 R 0 A ≅ R ⊗ R 0 s A 0 A \cong R \mathop{\otimes}_{R_0}^{s} A_0 A ≅ R ⊗ R 0 s A 0 R R R Lemma 4.1 . Moreover, as A 0 A_0 A 0 R 0 R_0 R 0 A n ≅ R n ⊗ R 0 s A 0 A_n \cong R_n \mathop{\otimes}_{R_0}^{s} A_0 A n ≅ R n ⊗ R 0 s A 0 R n R_n R n [13 Corollary 7.1.9] . Therefore, by Proposition 4.6,
Z ( A ) n = Z ( A n ) ≅ R n . Z(A)_n = Z(A_n) \cong R_n. Z ( A ) n = Z ( A n ) ≅ R n . Then Z ( A ) ≅ R Z(A) \cong R Z ( A ) ≅ R A A A R R R
Corollary 6.8. Let A A A R R R
A A A R R R
Z ( A ) Z(A) Z ( A ) R R R A 0 A_0 A 0 Z ( A ) 0 Z(A)_0 Z ( A ) 0
A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 A \cong Z(A) \mathop{\otimes}_{Z(A)_0}^{s} A_0 A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 as Z ( A ) Z(A) Z ( A )
Proof. ( ⟹ ) (\Longrightarrow) ( ⟹ ) A A A R R R A A A Z ( A ) Z(A) Z ( A ) Z ( A ) Z(A) Z ( A ) R R R Theorem 6.6 . By Proposition 6.7 A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 A \cong Z(A) \mathop{\otimes}_{Z(A)_0}^{s} A_0 A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 Z ( A ) Z(A) Z ( A ) A 0 A_0 A 0 Z ( A 0 ) Z(A_0) Z ( A 0 ) Z ( A ) 0 Z(A)_0 Z ( A ) 0
( ⟸ ) (\Longleftarrow) ( ⟸ ) A 0 A_0 A 0 Z ( A ) 0 Z(A)_0 Z ( A ) 0 Lemma 4.1 , A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 A \cong Z(A) \mathop{\otimes}_{Z(A)_0}^{s} A_0 A ≅ Z ( A ) ⊗ Z ( A ) 0 s A 0 Z ( A ) Z(A) Z ( A ) Z ( A ) Z(A) Z ( A ) R R R A A A R R R
Remark 6.9. By Corollary 6.8 , the data of a separable R R R
a commutative separable R R R Z ( A ) Z(A) Z ( A )
an ordinary Azumaya Z ( A ) 0 Z(A)_0 Z ( A ) 0 A 0 A_0 A 0
So to classify separable R R R R R R Z ( A ) 0 Z(A)_0 Z ( A ) 0 B r ( Z ( A ) 0 ) Br(Z(A)_0) B r ( Z ( A ) 0 )
Which commutative R R R R R R
Lemma 6.10. Let C C C R R R C ≅ Z ( A ) C \cong Z(A) C ≅ Z ( A ) R R R A A A C C C R R R
Proof. ( ⟹ ) (\Longrightarrow) ( ⟹ ) A A A R R R C ≅ Z ( A ) C \cong Z(A) C ≅ Z ( A ) Z ( A ) Z(A) Z ( A ) R R R Theorem 6.6 .
( ⟸ ) (\Longleftarrow) ( ⟸ ) C C C R R R C = Z ( C ) C = Z(C) C = Z ( C )
This leads to the problem of classifying commutative separable R R R
Theorem 6.11. Let C C C R R R
C C C R R R
C 0 C_0 C 0 R 0 R_0 R 0
ϕ : R ⊗ R 0 s C 0 ⟶ C r ⊗ c ⟼ r s 0 n ( c ) \phi: R \mathop{\otimes}_{R_0}^{s} C_0 \longrightarrow C \qquad r \otimes c \longmapsto r s_0^n(c) ϕ : R ⊗ R 0 s C 0 ⟶ C r ⊗ c ⟼ r s 0 n ( c ) for every r ⊗ c ∈ ( R ⊗ R 0 s C 0 ) n r \otimes c \in(R \mathop{\otimes}_{R_0}^{s} C_0)_n r ⊗ c ∈ ( R ⊗ R 0 s C 0 ) n
In particular, when C n C_n C n R n R_n R n n n n C C C C ≅ ( R ⊗ R 0 s T ) / J C \cong(R \mathop{\otimes}_{R_0}^{s} T)/J C ≅ ( R ⊗ R 0 s T ) / J T T T R 0 R_0 R 0 R 0 R_0 R 0 J J J s s s R ⊗ R 0 s T R \mathop{\otimes}_{R_0}^{s} T R ⊗ R 0 s T J 0 = 0 J_0 = 0 J 0 = 0
Proof. Let A = R ⊗ R 0 s C 0 A=R\otimes_{R_0}^{s}C_0 A = R ⊗ R 0 s C 0
R → η A / R A → ϕ C . R\xrightarrow{\eta_{A/R}}A\xrightarrow{\phi}C. R η A / R A ϕ C . (1) ⟹ \Longrightarrow ⟹ C C C R R R C 0 C_0 C 0 R 0 R_0 R 0
ϕ n : A n = R n ⊗ R 0 C 0 ⟶ C n , r ⊗ c ⟼ r s 0 n ( c ) (18) \phi_n:A_n=R_n\otimes_{R_0}C_0\longrightarrow C_n,\qquad r\otimes c\longmapsto rs_0^n(c) \tag*{(18)} ϕ n : A n = R n ⊗ R 0 C 0 ⟶ C n , r ⊗ c ⟼ r s 0 n ( c ) (18) for every r ⊗ c ∈ ( R ⊗ R 0 C 0 ) n r\otimes c\in(R\otimes_{R_0}C_0)_n r ⊗ c ∈ ( R ⊗ R 0 C 0 ) n n ≥ 0 n\geq0 n ≥ 0
π n : C n ⊗ R n C n ⟶ C n ⊗ A n C n . \pi_n:C_n\otimes_{R_n}C_n\longrightarrow C_n\otimes_{A_n}C_n. π n : C n ⊗ R n C n ⟶ C n ⊗ A n C n . Let e = ∑ i = 1 t x i ⊗ R 0 y i ∈ ( C e ) 0 e=\sum_{i=1}^{t}x_i\otimes_{R_0}y_i\in(C^e)_0 e = ∑ i = 1 t x i ⊗ R 0 y i ∈ ( C e ) 0 C C C R R R ϕ n ( 1 R n ⊗ R 0 x i ) = s 0 n ( x i ) \phi_n(1_{R_n}\otimes_{R_0}x_i)=s_0^n(x_i) ϕ n ( 1 R n ⊗ R 0 x i ) = s 0 n ( x i )
π n ( s 0 n ( e ) ) = ∑ i = 1 t s 0 n ( x i ) ⊗ A n s 0 n ( y i ) = ∑ i = 1 t ϕ n ( 1 R n ⊗ R 0 x i ) ⊗ A n s 0 n ( y i ) = ∑ i = 1 t 1 C n ⊗ A n ϕ n ( 1 R n ⊗ R 0 x i ) s 0 n ( y i ) C is an A -smodule by ϕ : A ⟶ C = ∑ i = 1 t 1 C n ⊗ A n s 0 n ( x i ) s 0 n ( y i ) = 1 C n ⊗ A n 1 C n , since μ ( e ) = 1 C 0 . \begin{aligned}
\pi_n(s_0^n(e))
&=\sum_{i=1}^{t}s_0^n(x_i)\otimes_{A_n}s_0^n(y_i)\\
&=\sum_{i=1}^{t}\phi_n(1_{R_n}\otimes_{R_0}x_i)\otimes_{A_n}s_0^n(y_i)\\
&=\sum_{i=1}^{t}1_{C_n}\otimes_{A_n}\phi_n(1_{R_n}\otimes_{R_0}x_i)s_0^n(y_i)
\qquad C\text{ is an }A\text{-smodule by }\phi:A\longrightarrow C\\
&=\sum_{i=1}^{t}1_{C_n}\otimes_{A_n}s_0^n(x_i)s_0^n(y_i)
=1_{C_n}\otimes_{A_n}1_{C_n},
\qquad\text{since }\mu(e)=1_{C_0}.
\end{aligned} π n ( s 0 n ( e )) = i = 1 ∑ t s 0 n ( x i ) ⊗ A n s 0 n ( y i ) = i = 1 ∑ t ϕ n ( 1 R n ⊗ R 0 x i ) ⊗ A n s 0 n ( y i ) = i = 1 ∑ t 1 C n ⊗ A n ϕ n ( 1 R n ⊗ R 0 x i ) s 0 n ( y i ) C is an A -smodule by ϕ : A ⟶ C = i = 1 ∑ t 1 C n ⊗ A n s 0 n ( x i ) s 0 n ( y i ) = 1 C n ⊗ A n 1 C n , since μ ( e ) = 1 C 0 . Since s 0 n ( e ) s_0^n(e) s 0 n ( e ) C n C_n C n R n R_n R n
1 C n ⊗ A n 1 C n = π n ( s 0 n ( e ) ) 1_{C_n}\otimes_{A_n}1_{C_n}=\pi_n(s_0^n(e)) 1 C n ⊗ A n 1 C n = π n ( s 0 n ( e )) is a separability idempotent of C n C_n C n
c ⊗ A n 1 C n = c ( 1 C n ⊗ A n 1 C n ) = ( 1 C n ⊗ A n 1 C n ) c = 1 C n ⊗ A n c (19) c\otimes_{A_n}1_{C_n}=c(1_{C_n}\otimes_{A_n}1_{C_n})=(1_{C_n}\otimes_{A_n}1_{C_n})c=1_{C_n}\otimes_{A_n}c \tag*{(19)} c ⊗ A n 1 C n = c ( 1 C n ⊗ A n 1 C n ) = ( 1 C n ⊗ A n 1 C n ) c = 1 C n ⊗ A n c (19) for every c ∈ C n c\in C_n c ∈ C n
μ = μ C n / A n : C n ⊗ A n C n ⟶ C n \mu=\mu_{C_n/A_n}:C_n\otimes_{A_n}C_n\longrightarrow C_n μ = μ C n / A n : C n ⊗ A n C n ⟶ C n is an isomorphism with inverse
h : C n ⟶ C n ⊗ A n C n , c ⟼ c ⊗ A n 1 C n h:C_n\longrightarrow C_n\otimes_{A_n}C_n,\qquad c\longmapsto c\otimes_{A_n}1_{C_n} h : C n ⟶ C n ⊗ A n C n , c ⟼ c ⊗ A n 1 C n for every c ∈ C n c\in C_n c ∈ C n μ h = Id C n \mu h=\operatorname{Id}_{C_n} μ h = Id C n
h μ ( c ⊗ A n c ′ ) = h ( c c ′ ) = c c ′ ⊗ A n 1 C n = ( c ⊗ A n 1 C n ) ( c ′ ⊗ A n 1 C n ) = ( c ⊗ A n 1 C n ) ( 1 C n ⊗ A n c ′ ) = c ⊗ A n c ′ . \begin{aligned}
h\mu(c\otimes_{A_n}c')
&=h(cc')=cc'\otimes_{A_n}1_{C_n}\\
&=(c\otimes_{A_n}1_{C_n})(c'\otimes_{A_n}1_{C_n})\\
&=(c\otimes_{A_n}1_{C_n})(1_{C_n}\otimes_{A_n}c')
=c\otimes_{A_n}c'.
\end{aligned} h μ ( c ⊗ A n c ′ ) = h ( c c ′ ) = c c ′ ⊗ A n 1 C n = ( c ⊗ A n 1 C n ) ( c ′ ⊗ A n 1 C n ) = ( c ⊗ A n 1 C n ) ( 1 C n ⊗ A n c ′ ) = c ⊗ A n c ′ . By [39 Tag 04VN] , ϕ n : A n ⟶ C n \phi_n:A_n\longrightarrow C_n ϕ n : A n ⟶ C n
(2) ⟹ \Longrightarrow ⟹ ϕ n \phi_n ϕ n [39 Tag 04VN] , the multiplication map μ C n / A n : C n ⊗ A n C n ⟶ C n \mu_{C_n/A_n}:C_n\otimes_{A_n}C_n\longrightarrow C_n μ C n / A n : C n ⊗ A n C n ⟶ C n
μ C / A : C ⊗ A s C ⟶ C \mu_{C/A}:C\otimes_A^{s}C\longrightarrow C μ C / A : C ⊗ A s C ⟶ C is an isomorphism of srings. View both as C ⊗ A s C C\otimes_A^{s}C C ⊗ A s C μ C / A \mu_{C/A} μ C / A C ⊗ A s C C\otimes_A^{s}C C ⊗ A s C C C C A A A Proposition 2.7 (2). Since C 0 C_0 C 0 R 0 R_0 R 0 A = R ⊗ R 0 s C 0 A=R\otimes_{R_0}^{s}C_0 A = R ⊗ R 0 s C 0 R R R Lemma 4.1 . Thus, C C C R R R
Suppose, in addition, that every C n C_n C n R n R_n R n C n C_n C n A n A_n A n ϕ n : A n → C n \phi_n:A_n\to C_n ϕ n : A n → C n ϕ n \phi_n ϕ n [40 Tag 04VT] . Then it follows that ϕ : A → C \phi:A\to C ϕ : A → C J = ker ϕ J=\ker\phi J = ker ϕ C ≅ A / J C\cong A/J C ≅ A / J
In level 0, the canonical map ϕ 0 : R 0 ⊗ R 0 C 0 → C 0 \phi_0:R_0\otimes_{R_0} C_0\to C_0 ϕ 0 : R 0 ⊗ R 0 C 0 → C 0 J 0 = 0 J_0=0 J 0 = 0
Conversely, if C ≅ ( R ⊗ R 0 s T ) / J C\cong(R\otimes^s_{R_0}T)/J C ≅ ( R ⊗ R 0 s T ) / J R ⊗ R 0 s T R\otimes^s_{R_0}T R ⊗ R 0 s T R R R Lemma 4.1 . It follows that C ≅ ( R ⊗ R 0 s T ) / J C\cong(R\otimes^s_{R_0}T)/J C ≅ ( R ⊗ R 0 s T ) / J R R R
Remark 6.12. We record two consequences of Theorem 6.11 .
Let K K K Example 4.9 , W ⟨ 0 ⟩ / K W^{\langle0\rangle}/K W ⟨ 0 ⟩ / K K K K C = W ⟨ 0 ⟩ / K C=W^{\langle0\rangle}/K C = W ⟨ 0 ⟩ / K R = K Δ 0 R=K\Delta^0 R = K Δ 0
ϕ : R n ⊗ R 0 s ( W ⟨ 0 ⟩ / K ) 0 → ( W ⟨ 0 ⟩ / K ) n \phi:R_n\otimes^s_{R_0}(W^{\langle0\rangle}/K)_0\to(W^{\langle0\rangle}/K)_n ϕ : R n ⊗ R 0 s ( W ⟨ 0 ⟩ / K ) 0 → ( W ⟨ 0 ⟩ / K ) n is K → K ⊕ ( n + 1 ) K\to K^{\oplus(n+1)} K → K ⊕ ( n + 1 ) n ≥ 1 n\ge1 n ≥ 1 Theorem 6.11 reconfirms the non-separability of W ⟨ 0 ⟩ / K W^{\langle0\rangle}/K W ⟨ 0 ⟩ / K K K K
Let k k k R = k Δ 0 R=k\Delta^0 R = k Δ 0 C C C R R R C n C_n C n R n R_n R n Theorem 6.11 implies C ≅ k Δ 0 ⊗ k s C 0 C\cong k\Delta^0\otimes^s_k C_0 C ≅ k Δ 0 ⊗ k s C 0 J J J k Δ 0 ⊗ k s C 0 k\Delta^0\otimes^s_k C_0 k Δ 0 ⊗ k s C 0 J 0 = 0 J_0=0 J 0 = 0 Theorem 6.11 recovers Theorem 2.11 for the finite commutative k k k
Brauer Groups of Simplicial Commutative Rings
We define the Brauer group of a commutative sring R R R R 0 R_0 R 0
Basics of Brauer groups
Let R R R B ( R ) B(R) B ( R ) R R R ( B ( R ) , ⊗ R s ) (B(R),\otimes_R^s) ( B ( R ) , ⊗ R s ) ⊗ R s \otimes_R^s ⊗ R s R R R A , B ∈ B ( R ) A,B\in B(R) A , B ∈ B ( R ) A A A A A A A ∼ B A\sim B A ∼ B A A A B B B
Lemma 7.1. The relation ∼ \sim ∼ ⊗ R s \otimes_R^s ⊗ R s B ( R ) B(R) B ( R )
Proof. We show if A ∼ A ′ A\sim A' A ∼ A ′ B ∼ B ′ B\sim B' B ∼ B ′ ( A ⊗ R s B ) ∼ ( A ′ ⊗ R s B ′ ) (A\otimes_R^s B)\sim(A'\otimes_R^s B') ( A ⊗ R s B ) ∼ ( A ′ ⊗ R s B ′ ) Theorem 5.8 , it suffices to show there exists an invertible ( A ⊗ R s B , A ′ ⊗ R s B ′ ) (A\otimes_R^s B,A'\otimes_R^s B') ( A ⊗ R s B , A ′ ⊗ R s B ′ ) A ∼ A ′ A\sim A' A ∼ A ′ B ∼ B ′ B\sim B' B ∼ B ′ Theorem 5.8 , there is an invertible ( A , A ′ ) (A,A') ( A , A ′ ) P P P ( B , B ′ ) (B,B') ( B , B ′ ) P ′ P' P ′ P ⊗ R s P ′ P\otimes_R^s P' P ⊗ R s P ′ ( A ⊗ R s B , A ′ ⊗ R s B ′ ) (A\otimes_R^s B,A'\otimes_R^s B') ( A ⊗ R s B , A ′ ⊗ R s B ′ ) ∼ \sim ∼ ⊗ R s \otimes_R^s ⊗ R s
Definition 7.2. Define B r ( R ) : = B ( R ) / ∼ Br(R):=B(R)/\sim B r ( R ) := B ( R ) / ∼ R R R ⊗ R s \otimes_R^s ⊗ R s ( − ) ∘ (-)^\circ ( − ) ∘ B r ( R ) Br(R) B r ( R ) R R R A ∈ B ( R ) A\in B(R) A ∈ B ( R ) [ A ] [A] [ A ]
The following proposition recovers the usual descriptions of Brauer groups; see [24 p. 108] and [13 p. 252] .
Proposition 7.3. Let A A A B B B R R R
A A A B B B
There exist R R R N N N M M M
A ⊗ R s Hom R s ( M , M ) ≅ B ⊗ R s Hom R s ( N , N ) . A \otimes_R^{s} \operatorname{Hom}_R^{s}(M,M) \cong B \otimes_R^{s} \operatorname{Hom}_R^{s}(N,N). A ⊗ R s Hom R s ( M , M ) ≅ B ⊗ R s Hom R s ( N , N ) . Proof. (1) ⟹ \Longrightarrow ⟹ A ∼ B A \sim B A ∼ B
( A ⊗ R s B o ) ∼ ( B ⊗ R s B o ) = B e ≅ Hom R s ( B , B ) ∼ R . (A \otimes_R^{s} B^{o}) \sim(B \otimes_R^{s} B^{o}) = B^{e} \cong\operatorname{Hom}_R^{s}(B,B) \sim R. ( A ⊗ R s B o ) ∼ ( B ⊗ R s B o ) = B e ≅ Hom R s ( B , B ) ∼ R . By Theorem 5.8, there is an R R R N N N A ⊗ R B o ≅ Hom R s ( N , N ) A \otimes_R B^{o} \cong\operatorname{Hom}_R^{s}(N,N) A ⊗ R B o ≅ Hom R s ( N , N )
A ⊗ R s Hom R s ( B , B ) ≅ A ⊗ R s ( B ⊗ R s B o ) ≅ ( A ⊗ R s B o ) ⊗ R s B A \otimes_R^{s} \operatorname{Hom}_R^{s}(B,B) \cong A \otimes_R^{s}(B \otimes_R^{s} B^{o}) \cong(A \otimes_R^{s} B^{o}) \otimes_R^{s} B A ⊗ R s Hom R s ( B , B ) ≅ A ⊗ R s ( B ⊗ R s B o ) ≅ ( A ⊗ R s B o ) ⊗ R s B ≅ Hom R s ( N , N ) ⊗ R s B ≅ B ⊗ R s Hom R s ( N , N ) . \cong\operatorname{Hom}_R^{s}(N,N) \otimes_R^{s} B \cong B \otimes_R^{s} \operatorname{Hom}_R^{s}(N,N). ≅ Hom R s ( N , N ) ⊗ R s B ≅ B ⊗ R s Hom R s ( N , N ) . The assertion follows from Theorem 6.4 that M : = B M := B M := B R R R
(2) ⟹ \Longrightarrow ⟹ A ⊗ R s Hom R s ( M , M ) ≅ B ⊗ R s Hom R s ( N , N ) A \otimes_R^{s} \operatorname{Hom}_R^{s}(M,M) \cong B \otimes_R^{s} \operatorname{Hom}_R^{s}(N,N) A ⊗ R s Hom R s ( M , M ) ≅ B ⊗ R s Hom R s ( N , N ) R R R M M M N N N Hom R s ( M , M ) ∼ R ∼ Hom R s ( N , N ) \operatorname{Hom}_R^{s}(M,M) \sim R \sim\operatorname{Hom}_R^{s}(N,N) Hom R s ( M , M ) ∼ R ∼ Hom R s ( N , N )
( A ⊗ R s Hom R s ( M , M ) ) ∼ ( A ⊗ R s R ) ≅ A (A \otimes_R^{s} \operatorname{Hom}_R^{s}(M,M)) \sim(A \otimes_R^{s} R) \cong A ( A ⊗ R s Hom R s ( M , M )) ∼ ( A ⊗ R s R ) ≅ A since ∼ \sim ∼ ⊗ R s \otimes_R^{s} ⊗ R s
( B ⊗ R s Hom R s ( N , N ) ) ∼ ( B ⊗ R s R ) ≅ B . (B \otimes_R^{s} \operatorname{Hom}_R^{s}(N,N)) \sim(B \otimes_R^{s} R) \cong B. ( B ⊗ R s Hom R s ( N , N )) ∼ ( B ⊗ R s R ) ≅ B . It follows that A ∼ A ⊗ R s Hom R s ( M , M ) ≅ B ⊗ R s Hom R s ( N , N ) ∼ B A \sim A \otimes_R^{s} \operatorname{Hom}_R^{s}(M,M) \cong B \otimes_R^{s} \operatorname{Hom}_R^{s}(N,N) \sim B A ∼ A ⊗ R s Hom R s ( M , M ) ≅ B ⊗ R s Hom R s ( N , N ) ∼ B □ \square □
Reduction to degree 0
Theorem 7.4. (Triviality Theorem) Let R R R
B r ( R ) ≅ B r ( R 0 ) . Br(R) \cong Br(R_0). B r ( R ) ≅ B r ( R 0 ) . Proof. If A A A R R R A 0 A_0 A 0 R 0 R_0 R 0 A ∼ B A \sim B A ∼ B A 0 ∼ B 0 A_0 \sim B_0 A 0 ∼ B 0 ( A , B ) (A,B) ( A , B ) P P P ( A 0 , B 0 ) (A_0,B_0) ( A 0 , B 0 ) P 0 P_0 P 0
f : B r ( R ) ⟶ B r ( R 0 ) [ A ] ⟼ [ A 0 ] . f : Br(R) \longrightarrow Br(R_0) \qquad[A] \longmapsto[A_0]. f : B r ( R ) ⟶ B r ( R 0 ) [ A ] ⟼ [ A 0 ] . Conversely, given any Azumaya R 0 R_0 R 0 T T T R ⊗ R 0 s T R \otimes_{R_0}^{s} T R ⊗ R 0 s T R R R T , T ′ T,T' T , T ′ R 0 R_0 R 0 ( T , T ′ ) (T,T') ( T , T ′ ) Q Q Q ( R ⊗ R 0 s T , R ⊗ R 0 s T ′ ) (R \otimes_{R_0}^{s} T,R \otimes_{R_0}^{s} T') ( R ⊗ R 0 s T , R ⊗ R 0 s T ′ ) R ⊗ R 0 s Q R \otimes_{R_0}^{s} Q R ⊗ R 0 s Q R ⊗ R 0 s T ∼ R ⊗ R 0 s T ′ R \otimes_{R_0}^{s} T \sim R \otimes_{R_0}^{s} T' R ⊗ R 0 s T ∼ R ⊗ R 0 s T ′
( R ⊗ R 0 s T ) ⊗ R s ( R ⊗ R 0 s T ′ ) ≅ ( R ⊗ R s R ) ⊗ R 0 s ( T ⊗ R 0 s T ′ ) ≅ R ⊗ R 0 s ( T ⊗ R 0 s T ′ ) . (R \otimes_{R_0}^{s} T) \otimes_R^{s} (R \otimes_{R_0}^{s} T') \cong(R \otimes_R^{s} R) \otimes_{R_0}^{s} (T \otimes_{R_0}^{s} T') \cong R \otimes_{R_0}^{s} (T \otimes_{R_0}^{s} T'). ( R ⊗ R 0 s T ) ⊗ R s ( R ⊗ R 0 s T ′ ) ≅ ( R ⊗ R s R ) ⊗ R 0 s ( T ⊗ R 0 s T ′ ) ≅ R ⊗ R 0 s ( T ⊗ R 0 s T ′ ) . This implies that we can define a group homomorphism
g : B r ( R 0 ) ⟶ B r ( R ) [ T ] ⟼ [ R ⊗ R 0 s T ] . g : Br(R_0) \longrightarrow Br(R) \qquad[T] \longmapsto[R \otimes_{R_0}^{s} T]. g : B r ( R 0 ) ⟶ B r ( R ) [ T ] ⟼ [ R ⊗ R 0 s T ] . Now f f f g g g f g ( [ T ] ) = f ( [ R ⊗ R 0 T ] ) = [ T ] fg([T]) = f([R \otimes_{R_0} T]) = [T] f g ([ T ]) = f ([ R ⊗ R 0 T ]) = [ T ]
g f ( [ A ] ) = g ( [ A 0 ] ) = [ R ⊗ R 0 A 0 ] = [ A ] gf([A]) = g([A_0]) = [R \otimes_{R_0} A_0] = [A] g f ([ A ]) = g ([ A 0 ]) = [ R ⊗ R 0 A 0 ] = [ A ] where the third equality is by Proposition 6.7. This completes the proof. □
Acknowledgments
We used GPT to assist in verifying some of our arguments and searching typos. We have written and checked all proofs on our own, and we take full responsibility for the mathematical content of this paper. This project was partially supported by the US National Science Foundation (No. DMS-2302087).
Xuan Huang: Department of Mathematics, Box 354350, University of Washington, Seattle, Washington 98195, USA Email address: xhuang9@uw.edu
Garcia Sun: Department of Mathematics, Box 354350, University of Washington, Seattle, Washington 98195, USA Email address: zhiqisun@uw.edu
Eric Zhang: Department of Mathematics, Box 354350, University of Washington, Seattle, Washington 98195, USA Email address: zhangj20@uw.edu
James J. Zhang: Department of Mathematics, Box 354350, University of Washington, Seattle, Washington 98195, USA Email address: zhang@math.washington.edu
