Table Of ContentHOCHSCHILD COHOMOLOGY, SEPARABLE AND FORMALLY
SMOOTH ALGEBRAS
C. MENINI
Let R be a ring. The Jacobson radical of R will be denoted by J(R).
1. Separable algebras
We start by recalling the celebrated
Theorem 1.1. (Wedderburn-Artin Theorem) Let R be a ring and assume that R is
R
semisimple. Then R is a direct product of rings, each isomorphic to a finite matrix
ring M (D) over a division ring D.
n
By Wedderburn Artin Theorem it is clear that for a given ring R we have
R is semisimple ⇐⇒ R is semisimple
R R
Definition 1.2. Let R be a ring. R is called semisimple if R is semisimple.
R
Proposition 1.3. Let R be a ring and assume that R is artinian. Then the
R
following statements are equivalent
(a) R is semisimple.
(b) J(R) = {0}
(c) R has no non-zero two-sided nilpotent ideal.
Proof. (a) ⇒ (b) is trivial in view of Wedderburn Artin Theorem.
(b) ⇒ (c) is trivial since every nilpotent two-sided ideal is contained in J(R) = {0}
(c) ⇒ (b) Since R is artinian, there exists an n ∈ N such that
R
J (R)n = J (R)n+1
Since R is noetherian, J (R)n is finitely generated and hence, by Nakayama’s
R R
Lemma, we get that J (R)n = {0}so that we get J (R) = {0}.
(b) ⇒ (a) Since R is artinian, there exists a finite number of maximal left ideals
R
of R say L ,...,L such that
1 n
L ∩···∩L = {0}.
1 n
Thus R embeds in the direct sum of a finite number of simple lef R-modules and
R
hence it is semisimple. (cid:164)
Corollary 1.4. Let A be a finite dimensional algebra over a field K. Then
A is semisimple ⇐⇒ J (A) = {0} ⇐⇒ A contains no non-zero two-sided nilpotent ideal.
Key words and phrases. Hochschild cohomology, monoidal categories, bialgebras.
This is a report on some results that can be found in [AMS] and [AMS2].
1
2 C. MENINI
Proof. Since A is artinian, just apply Proposition 1.3. (cid:164)
A
Proposition 1.5. Let A be a finite dimensional algebra over a field K. Then the
following are equivalent:
(a) A is classically separable.
(b) For every field extension L of K, the L-algebra A is semisimple.
(L)
(c) For every field extension L of K, the L-algebra A contains no non-zero
(L)
two-sided nilpotent ideal.
Proof. For every field extension L of K, we have that
(cid:161) (cid:162)
dim A = dim (A) < ∞
L (L) K
Apply now Corollary 1.4. (cid:164)
Proposition 1.6. Let F be a finite field extension of a field K. Then
F is a classically separable K-algebra ⇐⇒ every u ∈ F is separable over K.
Proof. (⇒) Let u ∈ F , let f be the minimal polynomial of u over K and let L be
u
a splitting field of f over K. Then
u
L[X]
∼
= K[u]⊗ L ⊆ F ⊗ L.
K K
(f )
u
Let
f = (X −α )t1···(X −α )tn
u 1 n
where α ,...,α are the distinct root of f in L. Then, by the Chinese Remainder’s
1 n u
Theorem, we have a ring isomorphism
L[X] L[X] L[X]
∼
= (cid:161) (cid:162) ×...× (cid:161) (cid:162)
(f ) (X −α )t1 (X −α )tn
u 1 n
Thus, any t > 1 gives rise to a nilpotent ideal of K[u]⊗ L and hence of F ⊗ L.
i K K
Since dim F ⊗ L = dim F < ∞, the conclusion follows in view of Proposition
L K K
1.5.
(⇐) Assume that every element u ∈ F is separable over K. Then, by the Theorem
of the Primitive Element, there exists an u ∈ F such that F = K(u) and the
minimal polynomial f of u over K is separable over K. Let L be a field extension
u
of K and let
f = h ···h
u 1 t
be the factorization of f as a product of irreducible factors in L[X]. Let M be a
u
spliting field of f over K. Then in M [X] we can write
u
f = (X −α )···(X −α )
u 1 n
where α ,...,α are all distinct. Considering the field extension L(α ,...,α ),
1 n 1 n
we deduce that h ,...,h are two by two not associated. Then, by the Chinese
1 t
Remainder’s Theorem, we get
K[X] L[X] L[X] L[X]
∼ ∼ ∼
F ⊗ L = K(u)⊗ L = ⊗ L = = ×...× .
K K K
(f ) (f ) (h ) (h )
u u 1 t
SinceeachL[X]/(h )isafield, itfollowsthatF⊗ Lcontainsnonon-zeronilpotent
i K
ideal. (cid:164)
Hochschild cohomology, formally smooth algebras and bosonization. 3
Definition 1.7. Let R be a commutative ring. An R-algebra A is called separable
if the multiplication map
m : A⊗ A → A
A R
has a section σ (i.e. m σ = Id ) which is an A-bimodule homomorphism.
A A
Proposition 1.8. Let R be a commutative ring and let A be a separable R-algebra.
Given a section σ of m which is an A-bimodule homomorphism, set
A
(cid:88)
n
e = σ(1 ) and write e = x ⊗ y for suitable n ∈ N and x ,y ∈ A for every i = 1,...,n.
A i R i i i
i=1
Then we have
(cid:88)
n
(1) m (e) = 1 i.e. x y = 1
A A i i A
i=1
and
(cid:88) (cid:88)
n n
(2) ae = ea i.e. ax ⊗ y = x ⊗ y a for every a ∈ A.
i R i i R i
i=1 i=1
Proof. Equalities (1) and (2) follows directly from being σ an A-bimodule section of
m .
A
(cid:164)
Definition 1.9. Let A be an algebra over a commutative ring R. An element
e ∈ A⊗ A is called a separability element (or also an idempotent) for A (over R)
R
if e fulfills (1) and (2).
Proposition 1.10. Let A be an algebra over a commutative ring R. Then
A is a separable R-algebra ⇔ A⊗ A contains a separability element for A over R.
R
Moreover any separability element of A is an idempotent element of the ring A⊗
R
Aop.
Proof. Let e be a separability element for A and define a map
σ : A → A⊗ A
R
by setting
σ(a) = ae.
Then σ is an A-bimodule homomorphism and a section of m . Write
A
(cid:88)
n
e = x ⊗ y .
i R i
i=1
Then we have:
(cid:179)(cid:88) (cid:180) (cid:179)(cid:88) (cid:180) (cid:88)
n n n
e = σ(1 ) = σ x y = σ x 1 y = x σ(1 )y =
A i i i A i i A i
i=1 i=1 i=1
= σ(1 )· σ(1 ) = e2.
A A⊗RAop A
The other implication is Proposition 1.8. (cid:164)
Proposition 1.11. Let R be a commutative ring and let n ∈ N, n ≥ 1. Then the
matrix ring M (R) is a separable R-algebra.
n
4 C. MENINI
Proof. Let e ∈ M (R) = A be the matrix defined by
i,j n
(e ) = 1 and (e ) = 0 for every (h,k) (cid:54)= (i,j)
i,j (i,j) R i,j (h,k)
and set
(cid:88)
n
e = e ⊗ e
i,1 R 1,i
i=1
Then
(cid:88)
n
m (e) = e = 1
R i,i A
i=1
and, for every h,k = 1,...n, we have
(cid:88)
n
e ·e = e ·e ⊗ e = e ⊗ e
h,k h,k i,1 R 1,i h,1 R 1,k
(cid:88)i=1
n
e·e = e ⊗ e ·e = e ⊗ e
h,k i,1 R 1,i h,k h,1 R 1.k
i=1
Therefore e is a separability element for M (R) over R. (cid:164)
n
Proposition 1.12. Let R be a commutative ring and let G be a finite group whose
order n is a invertible in R. Then the group algebra A = RG is a separable R-
algebra.
Proof. Let
(cid:88)
e = (n1 )−1 g ⊗ g−1.
A R
g∈G
Then
m (e) = (n1 )−1 ·(n1 ) = 1
R A A A
and, for every h ∈ G, we have
(cid:88) (cid:88)
h·e = (n1 )−1 hg ⊗ g−1 = (n1 )−1 t⊗ t−1h = e·h.
A R A R
g∈G t∈G
Therefore the element e is a separability element for RG over R. (cid:164)
Proposition 1.13. Let A be an algebra over a field K. Then
A separable over K ⇒ dim (A) < ∞.
K
Proof. Let
(cid:88)
n
e = x ⊗y
j j
j=1
be a separability element for A over K. For every a ∈ A we have
(cid:88) (cid:88)
n n
ax ⊗y = x ⊗y a.
j j j j
j=1 j=1
Let (e ) be a basis of A over K and for every i ∈ I let e∗ : A → K be the K-linear
i i∈I i
map defined by
e∗(e ) = δ
i j ij
Then, for every i ∈ I, we have
(cid:88) (cid:88)
n n
(3) ax ⊗e∗(y ) = x ⊗e∗(y a).
j i j j i j
j=1 j=1
Now any element r ∈ A can be uniquely written as
(cid:88)
r = e∗(r)e
i i
i∈F(r)
Hochschild cohomology, formally smooth algebras and bosonization. 5
where F (r) is a suitable finite subset of I. Thus, using (3) we obtain
(cid:88) (cid:88) (cid:88)
n n
ax ⊗y = ax ⊗ e∗(y )e
j j j i j i
j=1 (cid:88)j=1(cid:88) i∈F(yj)
n
= ax ⊗e∗(y )e
j i j i
(cid:88)j=1(cid:88)i∈F(yj)
n
= x ⊗e∗(y a)e
j i j i
j=1 i∈F(yj)
Set
(cid:91)
F = F (y ).
j
j=1,...n
Then we have
(cid:88) (cid:88) (cid:88)
n n
(4) ax ⊗y = x ⊗e∗(y a)e
j j j i j i
j=1 (cid:88)j=1(cid:88)i∈F(yj)
n
= x ⊗e∗(y a)e
j i j i
j=1 i∈F
(cid:88) (cid:88)
n
(5) = x ⊗e∗(y a)e
j i j i
i∈F j=1
so that we obtain
(cid:179)(cid:88) (cid:180)
n
a = mσ(a) = m ax ⊗y
j j
j=1
(cid:179)(cid:88) (cid:88) (cid:180)
n
= m x ⊗e∗(y a)e
j i j i
i∈F j=1
(cid:88) (cid:88) (cid:88) (cid:88)
n n
= x e∗(y a)e = e∗(y a)x e .
j i j i i j j i
i∈F j=1 i∈F j=1
It follows that the set { x e | j = 1,...n, i ∈ F} is a set of generators for A over
j i
K. (cid:164)
Proposition 1.14. Let A be a separable algebra over a field K. Then A is semisim-
ple.
Proof. Let σ be a section of the multiplication map m : A⊗ A → A which is an
A K
A-bimodule homomorphism and let
(cid:88)
n
e = a ⊗ b
i K i
i=1
be a separability element of A over K.
We will prove that any epimorphism
f : M → N
of left A-modules splits in A-Mod. Let s : N → M be a section of f in K-Mod and
let us define a map
σ : N → M
by setting
(cid:88)
n
σ(x) = a s(b x).
i i
i=1
Clearly we have
(cid:104)(cid:88) (cid:105) (cid:88) (cid:88)
n n n
fσ(x) = f a s(b x) = a f [s(b x)] = a (b x) = x
i i i i i i
i=1 i=1 i=1
6 C. MENINI
so that σ is a section of f. Now in A⊗ A we have, for every a ∈ A
K
(cid:88) (cid:88)
n n
aa ⊗ b = a ⊗ b a
i K i i K i
i=1 i=1
so that in A⊗ A⊗ N we have, for every a ∈ A and x ∈ N
K K
(cid:88) (cid:88)
n n
aa ⊗ b ⊗ x = a ⊗ b a⊗ x
i K i K i K i K
i=1 i=1
Let µ (resp. µ ) be the multiplication map on M (resp. on N):
M N
µ : A⊗ M → M.
M K
Then, for every a ∈ A and for every x ∈ N, we have
(cid:88)
n
aσ(x) = aa s(b x)
i i
i=1 (cid:179)(cid:88) (cid:180)
n
= µ (A⊗ s)(A⊗ µ ) aa ⊗ b ⊗ x
N K K M i K i K
(cid:104) (cid:179)(cid:88)i=1 (cid:180)(cid:105)
n
= µ (A⊗ s)(A⊗ µ ) a ⊗ b a⊗ x
N K K M i K i K
i=1
(cid:88)
n
= a s(b ax) = σ(ax)
i i
i=1
and hence σ is a morphism of left A-modules. (cid:164)
Proposition 1.15. Let R be a commutative ring, let A be an R-algebra and let S
be a commutative R-algebra. Then
A is a separable R-algebra ⇒ A = A⊗ S is a separable S-algebra.
(S) R
Moreover if we assume that R is a subring of S and π : S → R is an R-bilinear
retraction of the canonical inclusion ι : R → S, then
A = A⊗ S is a separable S-algebra ⇒ A is a separable R-algebra.
(S) R
Proof. Let us remark that for any R-algebra B, the S-algebra structure (and hence
the S-bimodule structure) of B = B ⊗ S is via the ring homomorphism
(S) R
λ : S → B ⊗ S = B
R (S)
defined by setting
λ(s) = 1⊗ s
R
whose image lies in the center of B . This applies, in particular when B = A and
(S)
B = A⊗ A.
R
The map
φ : (A⊗ S)⊗ (A⊗ S) = A ⊗ A −→ (A⊗ A)⊗ S = (A⊗ A)
R S R (S) S (S) R R R (S)
defined by setting
φ((a⊗ s)⊗ (b⊗ t)) = (a⊗ b)⊗ st
R S R R S
is well defined and an S-algebra isomorphism whose inverse is the map
ψ : A⊗ A⊗ S = (A⊗ A) −→ (A⊗ S)⊗ (A⊗ S) = A ⊗ A
R R R (S) R S R (S) S (S)
defined by setting
ψ(a⊗ b⊗ s) = (a⊗ 1 )⊗ (b⊗ s).
R R R S S R
Hochschild cohomology, formally smooth algebras and bosonization. 7
Let us note that φ is also an A = (A⊗ S)-bimodule homomorphism since the
(S) R
A -bimodule structure on (A⊗ A) is given by
(S) R (S)
(c⊗ w)·[(a⊗ b)⊗ s] = (ca⊗ b)⊗ ws
R R S R S
and
[(a⊗ b)⊗ s]·(c⊗ w) = (a⊗ bc)⊗ sw
R S R R S
so that
φ((c⊗ w)·[(a⊗ s)⊗ (b⊗ t)]) = φ((ca⊗ ws)⊗ (b⊗ t))
R R S R R S R
= (ca⊗ b)⊗ wst = (c⊗ w)·[(a⊗ b)⊗ st]
R S R R S
φ([(a⊗ s)⊗ (b⊗ t)](c⊗ w)) = φ((a⊗ s)⊗ (bc⊗ tw))
R S R R R S R
= (a⊗ bc)⊗ stw = [(a⊗ b)⊗ st]·(c⊗ w).
R S R S R
Note that
(m ⊗ S)◦φ = m
A R A
(S)
so that
(6) m ◦ψ = m ⊗ S
A A R
(S)
Let
σ : A → A⊗ A
R
be an A-bimodule homomorphism which is a section of m . Then the map
A
σ ⊗ S : A⊗ S = A → (A⊗ A)⊗ S = (A⊗ A)
R R (S) R R R (S)
is clearly a section of m ⊗ S which is an A -bimodule homomorphism. In fact,
A R (S)
for every a ∈ A and s ∈ S, we have
(σ ⊗ S)(a⊗ s) = σ(a)⊗ s = aσ(1 )⊗ s = (a⊗s)(σ(1 )⊗ 1 )
R R R R R R R S
(σ ⊗ S)(a⊗ s) = σ(a)⊗ s = σ(1 )a⊗ s = (σ(1 )⊗ 1 )(a⊗s).
R R R R R R R S
Then the map
ψ ◦(σ ⊗ S) : A⊗ S = A → (A⊗ S)⊗ (A⊗ S) = A ⊗ A
R R (S) R S R (S) S (S)
is an A -bimodule homomorphism which is a section of m . In fact, in view of
(S) A
(S)
(6) we have
m ◦ψ ◦(σ ⊗ S) = (m ⊗S)◦(σ ⊗ S) = A⊗ S.
A R A R R
(S)
Conversely, assume that θ : A → (A⊗ S)⊗ (A⊗ S) = A ⊗ A is A -
(S) R S R (S) S (S) (S)
bimodule homomorphism which is a section of m . Then we have
A
(S)
m ◦r ◦[(A⊗ A)⊗ π]◦φ◦θ◦(A⊗ ι)r−1
A A⊗RA R R R A
= r (m ⊗ R)[(A⊗ A)⊗ π]φ◦θ◦(A⊗ ι)r−1
A A R R R R A
= r [A⊗ π](m ⊗ S)φ◦θ◦(A⊗ ι)r−1
A R A R R A
= r [A⊗ π]◦m ◦θ◦(A⊗ ι)r−1
A R A(S) R A
= r [A⊗ π]◦Id (A⊗ ι)r−1 = Id
A R A(S) R A A
8 C. MENINI
where r : A ⊗ R → A and r : A ⊗ A ⊗ R → A ⊗ A are the usual
A R A⊗RA R R R
isomorphisms. Thus
σ = r ◦[(A⊗ A)⊗ π]◦φ◦θ◦(A⊗ι)r−1
A⊗RA R R A
isasectionofm . Theproofthatσ isanA-bimoduleisomorphismisstraightforward
A
and is left as an exercise to the reader. (cid:164)
Lemma 1.16. Let A and A be separable algebras over a commutative ring R. Then
1 2
A ×A is a separable R-algebra.
1 2
Proof. Let i : A → A ×A and let i : A → A ×A the usual injective R-module
1 1 1 2 2 2 1 2
homomorphisms and let us consider the codiagonal map
θ = ∇((i ⊗ i ),(i ⊗ i )) : (A ⊗ A )×(A ⊗ A ) → (A ×A )⊗ (A ×A ).
1 R 1 2 R 2 1 R 1 2 R 2 1 2 R 1 2
We have
θ((a ⊗ b ),(a ⊗ b )) = [(a 0 )⊗ (b 0 )]+[(0 ,a )⊗ (0 ,b )]
1 R 1 2 R 2 1, A2 R 1, A2 A1 2 R A1 2
(m ◦θ)((a ⊗ b ),(a ⊗ b ))
A1×A2 1 R 1 2 R 2
= m ([(a 0 )⊗ (b 0 )]+[(0 ,a )⊗ (0 ,b )])
A1×A2 1, A2 R 1, A2 A1 2 R A1 2
= (a b ,0 )+(0 ,a b ) = (a b ,a b )
1 1 A2 A1 2 2 1 1 2 2
= (m ×m )((a ⊗ b ),(a ⊗ b ))
A1 A2 1 R 1 2 R 2
so that
m ◦θ = m ×m
A1×A2 A1 A2
Let σ be an A -bimodule sections of m and let σ be an A -module section of
1 1 A1 2 2
m . It follows that
A2
m ◦θ◦(σ ×σ ) = (m ×m )◦(σ ×σ ) = Id
A1×A2 1 2 A1 A2 1 2 A1×A2
and hence θ◦(σ ×σ ) is a section of m . Let us prove that θ◦(σ ×σ ) is an
1 2 A1×A2 1 2
A ×A -bimodule homomophism. From
1 2
θ((α a ⊗ b ,α a ⊗ b )) = (α a ,0 )⊗ (b ,0 )+(0 ,α a )⊗ (0 ,b )
1 1 R 1 2 2 R 2 1 1 A2 R 1 A2 A1 2 2 R A1 2
= (α ,α )([(a 0 )⊗ (b 0 )]+[(0 ,a )⊗ (0 ,b )])
1 2 1, A2 R 1, A2 A1 2 R A1 2
= (α ,α )θ((a ⊗ b ),(a ⊗ b ))
1 2 1 R 1 2 R 2
we deduce that θ is a left A ×A -module homomorphism. An analogous result on
1 2
the right gives us that θ is in fact an A ×A -bimodule homorphism. Since we have
1 2
(σ ×σ )((a b ,a b )) = (σ (a b ),σ (a b )) =
1 2 1 1 2 2 1 1 1 2 2 2
= (a σ (b ),a σ (b )) =
1 1 1 2 2 2
= (a ,a )(σ (b ),σ (b ))
1 2 1 1 2 2
and similarly on the right side, we also conclude that σ ×σ is an A ×A -bimodule
1 2 1 2
homorphism. (cid:164)
Proposition 1.17. Let A be an algebra over a field K. Then
A separable over K ⇔ dim (A) < ∞ and A is classically separable over K.
K
Hochschild cohomology, formally smooth algebras and bosonization. 9
Proof. (⇒) By Proposition 1.13, we already know that dim (A) < ∞. Let now L
K
be a field extension of K. Then, by Proposition 1.15, A is a separable L-algebra
(L)
and hence it is semisimple by Proposition 1.14. Then, in view of Proposition 1.5, A
is classically separable.
(⇐) Let L be an algebraic closure of K. Then A = A⊗ L has finite dimension
(L) K
over L and hence it is left (and right) artinian. Moreover, since A is classically
(cid:161) (cid:162)
separable over K, we know that J A = 0. Hence, by Corollary 1.4, we get that
(L)
A is semisimple. Then, by Wedderburn-Artin Theorem, we obtain that A is
(L) (L)
a direct product of rings, each isomorphic to a finite matrix ring M (D) over a
n
division ring D:
∼
A = M (D )×...×M (D )
(L) n1 (cid:161) 1 (cid:162) nt t
The natural embedding of L in Z A gives rise to the embeddings of L in
(L)
Z(M (D )). It follows that each D is a finite dimensional L-algebra. Since L
ni i i
is algebraically closed we get that each D is isomorphic to L so that
i
∼
A = M (L)×...×M (L).
(L) n1 nt
By Proposition 1.11 each M (L) is separable over L. In view of Lemma 1.16, we
ni
conclude. (cid:164)
Proposition 1.18. Let K be a field and let H be a Hopf algebra over K. Then the
following statements are equivalent:
(a) H is separable.
(b) H is semisimple.
(c) There exists a left integral t in H such that ε (t) = 1.
H
Moreover, if one of these conditions hold, then dim (H) < ∞.
K
Proof. (a) ⇒ (b) is Proposition 1.14.
(b) ⇒ (c). Since
ε : H → K
H
is an algebra homomorphism, K becomes a left H-module via ε and it results that
H
ε is a morphism of left H-modules. Since H is semisimple, the module K is
H H
projective so that, as ε is surjective, there exists a left H-module homomorphism
H
τ : K → H which is a section of ε .
H
Let
t = τ (1 ).
K
Then we have
ε (t) = ε (τ (1 )) = ε ◦τ (1 ) = 1 .
H H K H K K
Also, for every h ∈ H we have
h·t = h·τ (1 ) = τ (h·1 ) = τ (ε (h)·1 ) = ε (h)·τ (1 ) = ε (h)·t
K K H K H K H
and hence t is a left integral in H.
(c) ⇒ (a). Let t ∈ H be a left integral such that ε (t) = 1. Let us prove that
H
(cid:88) (cid:161) (cid:162)
e = t ⊗S t
(1) (2)
is a separability element for H over K. We have
(cid:88) (cid:161) (cid:162)
t S t = ε (t) = 1
(1) (2) H
10 C. MENINI
so that e fulfills (1). Let h ∈ H. We have
(cid:88) (cid:161) (cid:162)
he = ht ⊗S t =
(1) (2)
(cid:88) (cid:161) (cid:162) (cid:161) (cid:162)
= h t ⊗S t ε h
(1) (1) (2) H (2)
(cid:88) (cid:161) (cid:162) (cid:161) (cid:162)
= h t ⊗S t S h h
(1) (1) (2) (2) (3)
(cid:88) (cid:161) (cid:162)
= h t ⊗S h t h
(1) (1) (2) (2) (3)
(cid:88) (cid:161) (cid:162) (cid:161) (cid:162)
= [(Id ⊗S)◦∆] h t · 1⊗h
H (1) (2)
(cid:179)(cid:88) (cid:161) (cid:161) (cid:162) (cid:162)(cid:180) (cid:161) (cid:162)
= [(Id ⊗S)◦∆] ε h t · 1⊗h
H H (1) (2)
(cid:179) (cid:88) (cid:161) (cid:162) (cid:180)
= ([(Id ⊗S)◦∆](t))· 1⊗ ε h h
H H (1) (2)
(cid:104)(cid:88) (cid:161) (cid:162)(cid:105)
= t ⊗S t (1⊗h)
(1) (2)
(cid:88) (cid:161) (cid:162)
= t ⊗S t h = eh
(1) (2)
so that e also fulfills (2).
The last assertion follows by Proposition 1.13. (cid:164)
2. Hochschild cohomology in monoidal categories
2.1. Let (M,⊗,1) be a monoidal category and suppose that (A,m,u) is an algebra
in (M,⊗,1). Then A⊗X ∈ M, for any X ∈ M, where the left structure is given
A
by µ := m⊗X (exercise). Thus we have a functor
T : M → M
A A
which is defined by
T(X) = A⊗X and T(f) = A⊗f.
A A
Similarly X ⊗ A is a right A-module, so we obtain a functor: T : M → M
A A
given by T (X) = X ⊗A and T (f) = f ⊗A.
A A
Let now M ∈ M. Then M ⊗ A is a right A-module as explained above, and
A
is a left A-module via ν = µ ⊗ A. These two structures are compatible, defining
an A-bimodule on M ⊗ A (exercise). Similarly, if M ∈ M then A ⊗ M is an
A
A-bimodule.
In particular (A⊗X)⊗A is an A-bimodule and
T : M → M
A A A A
T (X) = (A⊗X)⊗A and T (f) = (A⊗f)⊗A
A A A A
is a functor.
Analogously, A⊗(X ⊗A) can be regarded as an A-bimodule, and one can easily
prove that
a : (A⊗X)⊗A → A⊗(X ⊗A)
A,X,A
is a functorial isomorphism of bimodules.
