Non-commutative Algebra

Transcription

1 Non-commutative Algebra Patrick Da Silva Freie Universität Berlin June 2, 2017

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3 Table of Contents 1 Introduction to unital rings Generalities Centralizers and bicentralizers Tensor products and Hom Isotypical modules Finiteness conditions on rings and modules Artinian, noetherian and finite length modules Artinian and noetherian rings Simple modules Completely reducible modules Isotypical components of completely reducible modules Centralizer and bicentralizer of a completely reducible module Centralizer of a simple module Simple and semisimple rings Left & right vector spaces over a skew field Centralizer of a completely reducible module Semisimple rings Radicals Radical of a module Radical of a ring Radical of artinian rings/modules Modules over an artinian ring Tensor products of K-algebras Modules over tensor products Tensor product of K-fields Tensor product of completely reducible modules Separable algebras and modules over a field Page 3

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5 Chapter 1 Introduction to unital rings We assume the reader is familiar with a bit of algebra (at least elementary group theory). All our rings are assumed unital, namely multiplication admits a unique neutral element which will always be denoted by 1. This set of notes will serve as a reference to other sets of notes when proofs are necessary, and thus is not meant to be introductory. The set of non-negative integers N is equal to {0, 1,, n, } (this has the advantage of turning it into a monoid). 1.1 Generalities Definition 1.1. Let A be an abelian group denoted additively, i.e. via the symbol + and with neutral element 0. We say that A is a (unital) ring when it is equipped with a multiplication (usually denoted by juxtaposition) : A A A satisfying the following axioms : Distributivity : For all a 1, a 2, a 3 A, Associativity : For all a 1, a 2, a 3 A, a 1 (a 2 + a 3 ) = a 1 a 2 + a 1 a 3, (a 1 + a 2 )a 3 = a 1 a 3 + a 2 a 3. a 1 (a 2 a 3 ) = (a 1 a 2 )a 3. Neutral element : There exists a unique element, call the unit element (or simply 1) and denoted by 1 A, such that for all a A, 1a = a = a1. A subring of A is a subset of A which contains 1 and is closed under addition and multiplication, and thus becomes a ring of its own. Remark 1.2. If M is an abelian group, the collection of all abelian group endomorphisms of M, denoted by End Z (M), is a ring ; the identity endomorphism of M serves as unit element, addition is performed pointwise and multiplication is given by composition. Definition 1.3. Let A, B be rings. (i) A morphism of rings is a map ϕ : A B such that for all a 1, a 2 A, ϕ(a 1 + a 2 ) = ϕ(a 1 ) + ϕ(a 2 ), ϕ(a 1 a 2 ) = ϕ(a 1 )ϕ(a 2 ), ϕ(1) = 1. It is clear under this definition that the image im ϕ def = ϕ(a) is a subring of B. 5

6 Chapter 1 (ii) A left A-module is an abelian group M together with a morphism of rings ϕ : A End Z (M). This morphism defines a left action of A on M, namely a map A M M, whose output is usually denoted by juxtaposition, and is given by am def def = ϕ a (m), where ϕ a = ϕ(a). The assumptions on a morphism of rings shows that ϕ 1 = id M, ϕ a is an endomorphism of abelian groups and ϕ a1 a 2 = ϕ a1 ϕ a2. (iii) Given a ring A, the opposite ring A opp is a ring whose underlying abelian group is equal to A but if we denote multiplication in A opp by and that of A by juxtaposition, for a 1, a 2 A opp, we set a 1 a 2 def = a 2 a 1. Reading the axioms of a ring in the multiplicatively reversed order shows that A opp is a ring and has the same unit element as A. (iv) A right A-module is an abelian group N together with a morphism of rings ψ : A End Z (M) opp. To make notation more natural, an element ϕ End Z (M) opp acts on m M by setting the endomorphism on the right, not on the left ; in other words, for all m M, ϕ End Z (M) ϕ(m), ϕ End Z (M) opp (m)ϕ. This morphism defines a right action of A on M, namely a map M A M, whose output is also denoted by juxtaposition and is given by ma def = ϕ a (m). This is indeed a right action of A on M : m(a 1 a 2 ) def = (m)ϕ a1 a 2 = ((m)ϕ a1 )ϕ a2. The order has indeed been reversed, since in End Z (M), the composition ϕ a1 ϕ a2 applied to m M is equal to ϕ a1 (ϕ a2 (m)), where as in End Z (M) opp, the composition ϕ a1 ϕ a2 applied to m equals ϕ a2 (ϕ a1 (m)) ; to emphasize the right action and hopefully minimize confusion, we chose the above notation. (v) An (A, B)-bimodule is an abelian group M together with two morphisms of rings, namely ϕ L : A End Z (M) and ϕ R : B End Z (M) opp, corresponding to left and right multiplication. The maps ϕ L and ϕ R are required to turn M into a left A-module and a right B-module respectively, together with the additional property that for all a A, b B and m M, (ϕ L a (m))ϕ R b = ϕl a ((m)ϕ R b ). Note that for left and right A-modules M, we usually denote the action of a A on m M by juxtaposition, namely am def = ϕ a (m). The axioms of an (A, B)-bimodule becomes essentially associativity, namely a(mb) = (am)b. When A = B, we call an (A, A)-bimodule simply an A- bimodule. (vi) If M is an abelian group which is a left A-module and a left B-module, we can canonically turn M into a right B opp -module. We say that the two module structures on M are compatible if they make M a (A, B opp )-module. This is equivalent to asking that m M, a A, b B, a(bm) = b(am). Similarly, if M is a right A-module and a right B-module, we say that these are compatible if it makes M a (A opp, B)-bimodule, which is equivalent to asking that m M, a A, b B, (ma)b = (mb)a. 6

7 More generally, if {A i } i I and {B j } j J are two families of rings and M is an (A i, B j )-bimodule for each i I, j J, we say that these structures are compatible if the module structures given by A i, A i are compatible for all i, i I and the module structures iven by B j, B j are compatible for all j, j J. (vii) A morphism of left A-modules/right A-modules/(A, B)-bimodules is a morphism of abelian groups between two left A-modules/right A-modules/(A, B)-bimodules which commutes with the action of A on M. For instance, if ϕ : M N is a morphism of abelian groups between two left A-modules, it is a morphism of left A-modules precisely when a A, m M, ϕ(am) = aϕ(m). Remark 1.4. The collection of (small) left A-modules together with their morphisms form a category, which we denote by A-Mod. Similarly, right A-modules and (A, B)-bimodules form categories ; we denote them by Mod-A and A-Mod-B, respectively. Therefore, left and right A-modules inherit notions of endomorphism, automorphism, monomorphism, epimorphism, isomorphism, etc. The composition of two maps (which are morphisms of some type, most of the time) is denoted by (for example, ϕ ψ), except when the maps composed belong to some ring, in which case we might adopt juxtaposition as the symbol for composition. Definition 1.5. Let A be a ring and M be a left A-module. The endomorphism ring of M is the ring End A (M) consisting of all endomorphisms of M, i.e. morphisms of left A-modules ϕ : M M. Addition is done element-wise and multiplication is given by composition. If M is a right A-module, we also denote its endomorphism ring by End A (M). When M is an (A, B)-bimodule, we denote the ring of all (A, B)-module endomorphisms of M by End (A,B) (M). Remark 1.6. A morphism of rings ϕ : A B is equivalent to a morphism of rings ϕ opp : A opp B opp ; the underlying set maps ϕ and ϕ opp are equal, but the requirement that ϕ opp respects multiplication is the multiplicative axiom of ϕ read in reverse. It follows that a right A-module is nothing more but a left A opp - module, so anything proven about all left A-modules for any ring A is also proven for right A-modules, when the statement is read appropriately. To prevent ourselves from dealing with left and right A-modules all the time, unless we specifically need both notions, we will stick to left A-modules. Given a left A-module M, the right A opp -module structure on M corresponding to its left A-module structure will be denoted by M opp (i.e. the morphism of rings ϕ : A End Z (M) corresponds to ϕ : A opp End Z (M) opp, so it is given by the same data but in a different notation). Definition 1.7. Let A be a ring and M a left A-module. A subset N M is called a left A-submodule of M (or simply an A-submodule) if it is an abelian subgroup of M which is stable under the action of A, i.e. for all a A and n N, an N. We also say that N is stable under A to say that it is an A-submodule of M. In this case, we write N M to indicate that N is a submodule of M. Given a family of submodules {N i } i I, we can define their sum as i I N i def = { } n i n i N i i I where the superscript indicates that only finitely many of the n i s are non-zero. A family of submodules {N i } i I of M is said to be in a direct sum if they are in a direct sum as abelian groups, namely every m i I N i can be written uniquely as a sum m = i I n i, n i N i. When i I N i = M, we write M = i I N i and we say that M is the direct sum of the N i s. Remark 1.8. To check if M is a direct sum of two of its submodules N 1, N 2, it suffices to check if N 1 +N 2 = M and N 1 N 2 = 0, where 0 denotes the trivial submodule consisting only of the zero element of M. One 7

8 Chapter 1 needs to be careful when trying to check that M is the direct sum of k submodules, k N ; in this case, one must verify that k k N i = M, 1 i k, N i N j = 0. i=1 Proposition 1.9. (Isomorphism theorems) Let A be a ring and ϕ : M N be a morphism of left A- modules. j=1 j i (i) If N M, the quotient module M/N is defined as the quotient abelian group. structure is given by a(m + N) def = am + N. It is indeed a left A-module. Its A-module (ii) The kernel and the image of ϕ is defined as its kernel and image as abelian groups. We denote them by ker ϕ M and im ϕ N. It induces a natural morphism ϕ : M/ ker ϕ N which is onto im ϕ, thus giving the first isomorphism theorem : M/ ker ϕ im ϕ. (iii) If N 1, N 2 M are two left A-submodules, then N 1 + N 2, N 1 N 2 M and we have the second isomorphism theorem (N 1 + N 2 )/N 1 N 1 /(N 1 N 2 ). (iv) If N 1 N 2 M, then N 1 is a submodule of M and (M/N 1 )/(N 2 /N 1 ) M/N 2. Proof. This is the same proof as in the commutative case, hence is omitted. Definition Let A be a ring. One can see A as an A-bimodule over itself by using multiplication, i.e. ϕ L a 1 (a 2 ) = a 1 a 2 = (a 1 )ϕ R a 2. A left ideal (resp. right ideal, two-sided ideal) of A is a left A-submodule (resp. right A-submodule, A-bisubmodule) of A when seen as an A-bimodule over itself. Given a left A-module M and a subset S M, we can define the left A-submodule of M generated by S, written A S, as the smallest A-submodule of M containing S ; it can be constructed as the intersection of all A-submodules of M which contain S, since this collection is non-empty because M is in it. Similarly, one defines the right A-submodule of M generated by S as S A when M is a right A-module. If B is a ring and M is an (A, B)-bimodule, then the (A, B)-bisubmodule of M generated by S is defined the same way and is written A S B. When A is seen as an A-bimodule over itself, we use a different notation. For S A, the left ideal, right ideal and two-sided ideal generated by S are denoted respectively by A (S), (S) A and A (S) A. If S = {m 1,, m n } is a finite subset of M, we write A m 1,, m n instead (similar notations when M is a right A-module or an (A, B)-bimodule, or when A = M is an A-bimodule over itself and we speak of ideals). If S = {m} is a singleton, we may write Am def = A m, ma def = m A and AmB def = A m B, so that n n n A m 1,, m n = Am i, m 1,, m n A = m i A, A m 1,, m n B = Am i B. 8 i=1 i=1 i=1

9 Remark When ϕ : A B is a morphism of rings, im ϕ is a subring of B and ker ϕ is a two-sided ideal of A, which allows to turn the quotient A/ ker ϕ into an A-bimodule. This A-bimodule structure turns A/ ker ϕ into a ring of its own and A/ ker ϕ im ϕ. Proposition Let A be a ring and M a left A-module. An element π End A (M) is called a projection if π 2 = π. (i) If π is a projection, then we have M = ker π im π. (ii) If π is a projection, then so is 1 π (recall that 1 End A (M) is the identity map of M). (iii) Whenever m M is written as m = m 1 +m 2 where m 1 im π and m 2 ker π, we have m 1 = π(m) and m 2 = (1 π)(m). (iv) We have ker π = im (1 π) and im π = ker(1 π). (v) Let π End A (M) be a projection. We have the identities End A (M)π = {ϕ End A (M) ker ϕ ker π}, πend A (M) = {ϕ End A (M) im ϕ im π}. (vi) If M = N 1 N 2 is the direct sum of two left A-submodules, let π End A (M) be the endomorphism defined by π(n 1 +n 2 ) def = n 1. Then π is the unique projection in End A (M) which satisfies ker π = N 2 and im π = N 1 ; we call it the projection onto N 1 and denote it by π N1 or simply π 1 if the context allows it. More generally, if M = i I N i, there is a unique projection π End A (M) satisfying im π = N i and ker π = j I\{i} N j. Proof. For m M, write m = (m π(m)) + π(m), so that M = ker π + im π. If m ker π im π, then m = π(m ) = π 2 (m ) = π(m) = 0, which proves (i). Note that π = (1 (1 π)), which proves (ii), (iii) and (iv) because m = (m π(m)) + π(m). To prove (v), we proceed with the first equality. If ϕ = ψπ, then π(m) = 0 implies ψ(π(m)) = 0, hence the inclusion ( ) holds. For the reverse, if ϕ : M M satisfies ker ϕ ker π, then ϕ is entirely determined by the values it takes on im π. But for m im π, we have ϕ(m) = ϕ(π(m)), hence ϕ = ϕπ End A (M)π. As for the second equality, it is clear that im πϕ im π, which gives ( ). For the reverse inclusion, suppose ϕ End A (M) satisfies im ϕ im π. Then ϕ(m) im π for all m M, hence π(ϕ(m)) = ϕ(m), which means ϕ = πϕ πend A (M), as desired. Finally, the endomorphism defined in (vi) is clearly a projection. Its unicity comes from part (iii). Definition Let A be a ring, M a left A-module and N M an A-submodule. We say that N is a direct summand if there exists a projection on M onto N. In other words, N is a direct summand if there exists an A-submodule N M such that M = N N. Definition Let A be a ring and M, N be two left A-modules. The set Hom A (M, N) is defined as the set of all morphisms of left A-modules ϕ : M N. It is obviously an abelian group, and it is turned into a left A-module by (aϕ)(m) def = aϕ(m) = ϕ(am). This construction is functorial in the following sense. Given morphisms ϕ : M 2 M 1 and ψ : N 1 N 2, we obtain a morphism Hom A (ϕ, ψ) : Hom A (M 1, N 1 ) Hom A (M 2, N 2 ) by sending α to ψ α ϕ. The 9

10 Chapter 1 following diagram commutes : Hom A (ϕ,id N1 ) Hom A (M 1, N 1 ) Hom A (M 2, N 1 ) Hom A (id M1,ψ) Hom A (ϕ,ψ) Hom A (id M2,ψ) Hom A (ϕ,id N2 ) Hom A (M 1, N 2 ) Hom A (M 2, N 2 ) We sometimes write Hom A (M, ψ) instead of Hom A (id M, ψ) : Hom A (M, N 1 ) Hom A (M, N 2 ) when ψ : N 1 N 2 is a morphism ; similarly for Hom A (ϕ, N) : Hom A (M 1, N) Hom A (M 2, N). Remark As in the case of abelian groups, we have a notion of short exact sequences ϕ ψ 0 M 1 M 2 M 3 0. This sequence is said to be exact at M 2 when ker ψ = im ϕ, and more generally, it is called a short exact sequence when ϕ is injective, ψ is surjective and ker ψ = im ϕ, in which case M 2 /ϕ(m 1 ) M 3 (this is equivalent to asking the sequence to be exact at M 1, M 2 and M 3 ). Applying Hom A (N, ) on this sequence, we obtain the left-exact sequence Hom A (N,ϕ) Hom A (N,ψ) 0 Hom A (N, M 1 ) Hom A (N, M 2 ) Hom A (N, M 3 ) Left-exactness means that the sequence is exact at Hom A (N, M 1 ) and Hom A (N, M 2 ) but not necessarily at Hom A (N, M 3 ). Applying Hom A (, N) on this sequence, we obtain the left-exact sequence Hom A (ψ,n) Hom A (ϕ,n) 0 Hom A (M 3, N) Hom A (M 2, N) Hom A (M 1, N) (note the order inversion ; this is because Hom A (, N) is a contravariant functor, i.e. it reverses the order in which morphisms are composed). See Proposition 1.16 for proofs. When ψ is a projection, we say that the above exact sequence splits. When applying Hom A (N, ) or Hom A (, N) to split exact sequences, the resulting exact sequence is also split. One easily sees that when N M, N is a direct summand of M if and only if the identity map of M extends to a endomorphism π End A (M) with im π = N, in which case M = N ker π. Proposition Let ϕ ψ 0 M 1 M 2 M 3 0 be a sequence of left A-modules (not necessarily exact). The following are equivalent : (i) The sequence above is right-exact (ii) For all left A-modules N, the following sequence is exact : Hom A (ϕ,n) Hom A (ψ,n) 0 Hom A (M 3, N) Hom A (M 2, N) Hom A (M 1, N) Also, the following are equivalent : 10 (i) The above sequence is left-exact (ii) For all left A-modules N, the following sequence is exact : Hom A (N,ϕ) Hom A (N,ψ) 0 Hom A (N, M 1 ) Hom A (N, M 2 ) Hom A (N, M 3 )

11 Proof. The fact that (i) implies (ii) is a consequence of the left-exactness of Hom and is proved in the same way as in the case of abelian groups, so is left as an exercise. Conversely, in the second case, we can take N = A and notice that we have the natural isomorphism Hom A (A, M i ) M i, so we recover the exactness required in (i). In the first case, we have to work a bit more. Let N def = M 3 /im ψ. The projection map π : M 3 M 3 /im ψ is such that π ψ = 0 by definition, but since π ψ = Hom A (ψ, N)(π) = 0 and Hom A (ψ, N) is injective, we get π = 0, i.e. im ψ = M 3, which means ψ is surjective. Let N = M 3 and consider ψ Hom A (M 2, N). Since Hom A (ψ ϕ, N) = Hom A (ψ, N) Hom A (ϕ, N) = 0 for all left A-modules N, we have ψ ϕ = 0 (take N = M 3 and apply Hom A (ψ ϕ, N) to id M3 ), which implies im ϕ ker ψ. Next, take N def = M 2 /im ϕ. Since π : M 2 M 2 /im ϕ lies in ker Hom A (ϕ, N) = im Hom A (ψ, N), there exists ν : M 3 M 2 /im ϕ such that π = Hom A (ψ, N)(ν) = ν ψ. Therefore, ker ψ ker(ν ψ) = ker π = im ϕ, completing the proof. Definition Let A be a ring. (i) An element e A is called an idempotent if e 2 = e. (ii) A family of idempotents {e i } i I of A is called orthogonal if for any i, j I distinct, we have e i e j = 0. Corollary Let A be a ring and M a left A-module. Consider a family of submodules {M i } i I such that i I M i = M. Then the following are equivalent : (i) We have M = i I M i (ii) Each M i is a direct summand and the family of idempotents {π i } i I End A (M) is orthogonal. Proof. Obvious. Remark The assumption that i I M i = M is not vital ; we can always add an element to the set I such that M = i I ker π i and it follows that M = i I { } M i. If M was already the direct sum of the M i s, this does not change anything to the statement ; if it isn t, we have to replace (i) by the equality M def = i I M i = i I M i and the π i by π i M, which shows that the π i M form a family of orthogonal idempotents. 1.2 Centralizers and bicentralizers Definition Let A be a ring and S A be a subset. The centralizer of S in A is denoted by C A (S) and is defined as C A (S) def = {a A s S, as = sa}. Note that by definition, C A (S) is a subring of A, so we can consider its bicentralizer CA 2 The center of A is equal to its own centralizer and is denoted by Z(A) def = C A (A) = {z A a A, az = za}. (S) def = C A (C A (S)). We say that B is commutative if B = Z(B) (in which case we resolve to the theory of commutative rings). Remark It might be tempting to define recursively the k th centralizer by CA k = C A (C k 1 A (S)), but this notion is pointless, as we can see right away. It is clear that S CA 2 (S) by definition (the elements of the centralizer commute with all elements of S, so the elements of S commute with the elements of (S) def 11

12 Chapter 1 the centralizer). Letting C A (S) play the role of S in the latter inclusion gives C A (S) CA 3 (S). When S T A, we have C A (S) C A (T ) (elements which commute with every element of T also commute with every element of S), therefore S CA 2 (S) implies C A(S) CA 3 (S), which gives C A(S) = CA 3 (S). Therefore, the centralizer of the bicentralizer is the centralizer itself. If B A is a subring, then Z(B) = C A (B) B. In particular, if B is commutative, then B C A (B), thus C 2 A (B) C A(B). It follows that Z(C A (B)) = C CA (B)(C A (B)) = C A (C A (B)) C A (B) = C 2 A(B) C A (B). One also sees that C 2 A (A) = C A(Z(A)) = A follows from the definition. Definition Let A be a commutative ring. An associative A-algebra is a ring B equipped with a morphism ϕ : A B, called the structure map of the algebra, such that im A Z(B). More generally, an A-algebra is defined as an A-bimodule B equipped with an A-bilinear map (, ) : B B B ; when this bilinear map satisfies (a, 1) = (1, a) = a and is associative (meaning that (b 1, (b 2, b 3 )) = ((b 1, b 2 ), b 3 ) for all b 1, b 2, b 3 B), we recover the notion of an associative A-algebra. In this document, all our A-algebras are associative unless otherwise mentioned, so we will call them A-algebras. If B 1, B 2 are two A-algebras, since A is commutative, we can consider the tensor product B 1 A B 2 which becomes a ring via the definition (b 1 b 2 )(b 1 b 2) def = (b 1 b 1) (b 2 b 2) which is extended by A-bilinearity to all of B 1 A B 2. It canonically becomes an A-algebra via the morphism A B 1 A B 2 defined by a a 1 = 1 a. An A-subalgebra of a A-algebra B is a subring C B such that the structure map ϕ : A B of A satisfies ϕ(a) C, in which case C becomes an A-algebra using the same structure map with codomain restricted to C since ϕ(a) Z(B) C Z(C). Definition A ring K in which for every a A, there exists a unique element a 1 A satisfying aa 1 = a 1 a = 1, is called a skew field or a division ring. If K is commutative, we call K a field. When K is a field and A is a K-algebra, A is in particular a K-vector space and thus we can apply the techniques of linear algebra. One remark : since K has no nonzero ideals, a K-algebra always contains an isomorphic copy of K as a K-subalgebra, so we usually write K A with no regards to the structure map. Proposition Let K be a field and A, B be two K-algebras. If A A and B B are K-subalgebras, then C A (A ) is a K-subalgebra of A, C B (B ) is a K-subalgebra of B and As a corollary, Z(A K B) = Z(A) K Z(B). C A K B(A K B ) = C A (A ) K C B (B ). Proof. Since A A, we have K Z(A) = C A (A) C A (A ). Since K A C 2 A (A ), we have K C A (A ) C 2 A (A ) = Z(C A (A )) by Remark 1.21, hence C A (A ) (and similarly C B (B )) are K-subalgebras. 12

13 As for the equality, note that the inclusion ( ) is clear. Consider the following direct sum decompositions = A K B = A = C A (A A, B = C B (B ) B ) ( C A (A ) K C B (B ) ( (A K C B (B )) (C A (A ) K B ) (A K B ) which shows that C A (A ) K C B (B ) is a direct summand of the K-vector space A K B and ), C A (A ) K C B (B ) = ( (A K C B (B ) ) ( C A (A ) K B )). By the symmetry of our argument, it will suffice to show that C A K B(A K B ) C A (A ) K B. Let z = i I a i b i C A K B(A K B ) where a i A, b i B. Without loss of generality, we can suppose that the b i are linearly independent over K. For any a A, we have z(a 1) = (a 1)z = i I (a i a aa i ) b i = 0. Since the b i are linearly independent, this means that a i a aa i = 0 for all i, which means a i C A (A ), as desired. Definition Let A be a ring and M an A-module. The structure map ϕ : A End Z (M) is a morphism of rings, so let A M = ϕ(a), which is a subring of End Z (M). For each a A, we use the notation def def a M = ϕ a A M. The map ϕ : A A M defined by a A M is onto and A M is called the ring of homotheties of M (its elements are called homotheties of M). The centralizer and bicentralizer of M are defined as the centralizer/bicentralizer of A M in End Z (M) : C A (M) def = C EndZ (M)(A M ), C 2 A(M) def = C 2 End Z (M) (A M). By definition, the centralizer of M is the endomorphism ring of the A-module M since C A (M) = {ϕ End Z (M) a A, ϕa M = a M ϕ} = End A (M). The countermodule 1 of M, denoted by M is the abelian group M seen as a module over its centralizer, namely, the endomorphism ring End A (M). Multiplication is defined the obvious way, namely if ϕ End A (M) and m M, we define ϕm def = ϕ(m). Note that the A-module and End A (M)-module structures of M are compatible since ϕ(am) = aϕ(m) for all a A, ϕ End A (M) and m M. We define the double countermodule of M, denoted by M, as the countermodule of its countermodule, namely M def = (M ). Remark Even when A is commutative, C A (M) = End A (M) is, in general, a noncommutative ring. Therefore, results that we build in this document here can also be useful for the theory of commutative rings, since considering only commutative rings prevents the study of noncommutative rings such as End A (M). Note that compatibility of module structures is not a transitive notion, i.e. if M is a left module over the three rings A, B, C, the module structures of A and B are compatible and the module structures of B and C are compatible, this does not mean that the structures of A and C are compatible. A particularly relevant example is that of the A-module, C A (M)-module and CA 2 (M)-module structures on the A-module M. The first two and last two are compatible, but A and CA 2 (M) give compatible structures on M if and only if CA 2 (M) C A(M), which is not always the case (see Example 1.34). 1 The French term for this is contre-module, but we have not found an English translation which is commonly used. 13

14 Chapter 1 Definition Let A be a ring. We have already seen that A can be interpreted as an A-bimodule over itself, but we can relax these assumptions and only consider the left A-module A, which we denote by A l. Equivalently, we can consider the right A-module A, which we denote by A r. Proposition Let A be a ring. (i) The centralizer of the left A-module A l is the opposite ring of homotheties of the right A-module A r. In symbols, C A (A l ) = (A Ar ) opp = (A opp ) (A opp ) l A opp. This means that A l is canonically a left A opp -module. Namely, as left A opp -modules, A l (A opp ) l. (ii) The centralizer of the right A-module A r is the opposite ring of homotheties of the left A-module A l. In symbols, C A (A r ) = (A Al ) opp = (A opp ) (A opp ) r A opp. The same comments as in part (i) apply, i.e. A r is a right A opp -module satisfying A r (A opp ) r. Proof. We begin by proving (i). If ϕ End Z (A l ) = End Z ((A opp ) r ) commutes with a Al for each a A, then ϕ(a) = ϕ(a1) = (ϕ a Al )(1) = (a Al ϕ)(1) = aϕ(1), so that ϕ corresponds to multiplication on the right by the element ϕ(1). Conversely, any such endomorphism (of the form a A opp) belongs to C r A (A l ) since multiplication on the left commutes with multiplication on the right ; this is associativity of multiplication. Therefore, C A (A l ) and (A Ar ) opp are identified as abelian groups. As for multiplication, if ψ, ϕ C A (A l ), then (ψ ϕ)(a) = ψ(ϕ(a)) = ψ(aϕ(1)) = aϕ(1)ψ(1) = a(ψ(1) Ar ϕ(1) Ar ). where stands for multiplication in (A Ar ) opp. Finally, we show that the identity map of A induces the isomorphism A l (A opp ) l. For ϕ, ψ C A (A l ), we have (ψ ϕ)(a) = a(ψ(1) ϕ(1)) = ψ(1) ϕ(1) a. The left-most term is the left C A (A l )-action on A l, and identifying the action of C A (A l ) with that of A opp via the map ϕ ϕ(1) (A opp ) l, this completes the proof. (Note that C A (A l ) and (A Al ) opp are not just isomorphic, they are equal ; this is because they are defined as the same subsets of End Z (A).) The statement of (ii) is proved along the lines of the proof of part (i), everything being written in reverse order. Remark The centralizer of a left A-module M is equal to the ring of homotheties of M (in symbols, C A (M) = C A (M) M ). Since it is clear that C A (M) M C A (M), this is the statement that for ϕ C A (M) = C EndZ (M)(A M ), as endomorphisms of M, we have ϕ M = ϕ. In other words, the map ϕ ϕ M from C A (M) to End Z (M) is an injective morphism of rings. This is clear because the endomorphism of M which multiplies by ϕ M is exactly ϕ. It follows that C A (M) M = C A (M) and thus C 2 A(M) = C 2 End Z (M) (A M) = C EndZ (M)(C EndZ (M)(A M )) = C EndZ (M)(C A (M) M ) = C CA (M)(M ), C 2 A(M) = C CA (M)(M ) = C CA (M)(M ) M = C 2 A(M) M. This means that C 2 A (M) is the ring of homotheties of the double countermodule M and the centralizer of M. The latter can be re-written as C 2 A (M) = End End A (M)(M ), but since for any a A, the endomorphism a M is an End A (M)-linear map on M by definition of End A (M), we obtain A M C 2 A (M) = A M. Also note that since C EndZ (M)(A M ) = C 3 End Z (M) (A M), the centralizer of M is just the centralizer of M, so that we do not speak of M. 14

15 Corollary Let A be a ring. We have C 2 A(A l ) = A Al, C 2 A(A r ) = A Ar. Proof. It suffices to prove that the first equality holds. By using Proposition 1.28 and Remark 1.29, we see that C 2 A(A l ) = C A opp(a l ) = C A opp((a opp ) r ) = A Al. The proof of the second equality is done analogously. Proposition Let A be a ring and M a left A-module. Suppose N M is an A-submodule which is a direct summand. (i) N is stable under C 2 A (M), i.e. is also a submodule of M. Letting N be a complement for N, since N is also a direct summand, it is also stable under C 2 A (M), so N is also a direct summand of M. (ii) If N is another direct summand of M, any A-linear map ϕ : N N is also C 2 A (M)-linear. (iii) For every ψ C 2 A (M), the restriction of ψ to N lies in C2 A (N), i.e. ψ M N = (ψ N ) N. This implies that restriction to N is a morphism of rings ( ) N : C 2 A (M) C2 A (N). Proof. Let π N End A (M) be the projection of M onto N. (i) For ψ C 2 A (M), since elements of C2 A (M) commute with elements of C A(M) = End A (M) by definition, we see that ψ(n) = ψ(π(m)) = π(ψ(m)) π(m) = N. (ii) Let ϕ : N N be an A-linear map. Then ϕ π N End A (M) = C A (M), so for ψ C 2 A (M) and n N, we have ϕ(ψn) = ((ϕ π N ) ψ)(n) = (ψ (ϕ π N ))(n) = ψ(ϕ(n)). (iii) Let ψ CA 2 (M) and ϕ C A(N) = End A (N). By part (ii), if we set N = N, we see that ϕ End C 2 A (M) (N), which means that ψ N ϕ = ϕ ψ N because ψ acts on N via ψ N. This implies ψ N CA 2 (N), so we are done. Definition Let {A i } i I be a family of rings. Their product is the ring A def = i I A i with addition and multiplication defined pointwise, so that its unit element is equal to (1) i I. Given a family {M i } i I where each M i is a left A i -module, the abelian group i I M i is a left i I A i-module via (a i ) i I (m i ) i I def = (a i m i ) i I. Proposition Let A 1, A 2 be rings and M i be a left A i -module, i = 1, 2. Then C A1 A 2 (M 1 M 2 ) = C A1 (M 1 ) C A2 (M 2 ), C 2 A 1 A 2 (M 1 M 2 ) = C 2 A 1 (M 1 ) C 2 A 2 (M 2 ). Proof. The second equality follows from the first since the bicentralizer is the centralizer of the counter- 15

16 Chapter 1 module, hence For the first, note that C 2 A 1 A 2 (M 1 M 2 ) = C CA1 A 2 (M 1 M 2 )(M 1 M 2 ) = C CA1 (M 1 ) C A2 (M 2 )(M 1 M 2 ) = C CA1 (M 1 )(M 1 ) C CA2 (M 2 )(M 2 ) = C 2 A 1 (M 1 ) C 2 A 2 (M 2 ). M 1 M 2 = (M 1 {0}) ({0} M 2 ) }{{}}{{} def = M def 1 = M 2. The element ϕ End Z (M 1 M 2 ) belongs in the set C A1 A 2 (M 1 M 2 ) if and only if for all (a 1, a 2 ) A 1 A 2, we have ϕ(a 1, a 2 ) M = (a 1, a 2 ) M ϕ. Letting π i be the projection onto M i, ϕ C A1 A 2 (M 1 M 2 ) implies that π i ϕ C Ai (M i ), so ϕ = π 1 ϕ+π 2 ϕ C A1 (M 1 ) C A2 (M 2 ). The reverse inclusion is obvious, which proves equality. Example We give an example where the restriction morphism ( ) N : CA 2 (M) C2 A (N) is neither injective or surjective. Consider a field K and the K-algebra End K (K 3 ). After fixing the standard basis {e 1, e 2, e 3 }, this is canonically isomorphic to the ring of 3 3 matrices with coefficients in K, which we denote by Mat 3 3 (K). Consider the K-subalgebra A of matrices of the form a 0 0 b c 0, a, b, c K. 0 0 a Considered as a left A-module, M def = K 3 is the direct sum of the A-submodules N def = K e 1, e 2 and N def = K e 3. Note that elements of C A (M) have to be K-linear maps since they have to commute with elements of the form λ M for λ K. The condition that ϕ End Z (M) commutes with all matrices of the above form gives linear conditions on the coefficients of the matrix form, so solving for it tells us that the matrix form of ϕ over the basis {e 1, e 2, e 3 } is of the form α α 0 β 0 γ One sees that this ring is of the same form as A, where the roles of e 2 and e 3 have been permuted. It follows that A = CA 2 (M). The map ( ) N : CA 2 (M) C2 A (N) is injective since it can be written as a 0 0 [ ] b c 0 a 0 b c 0 0 a However, one sees that C A (N) is the set of 2 2 scalar matrices, so CA 2 (N) is the full ring of 2 2 matrices, which means that ( ) N is not surjective. Similarly, C A (N ) = CA 2 (N ) K, so ( ) N is surjective but not injective. Note that A A is a ring (its unit element is (1, 1)). Consider the left A A-module M M with action given by (a 1, a 2 )(m 1, m 2 ) def = (a 1 m 1, a 2 m 2 ). The subset N N is a direct summand and by Proposition 1.33, we see that CA A 2 (M M) = A A and C A A(N N ) = C A (N) C A (N ). The restriction map ( ) N N : CA 2 (M) C2 A (M) C2 A (N) C2 A (N ) is neither injective or surjective since it is the product of the restriction maps ( ) N ( ) N on each component, the first not being surjective and the second not being injective. 16

17 Lemma Let A be a ring, M a left A-module and N, P two A-submodules such that M = N P. Suppose that there exists two families of A-submodules {P i } i I and {N i } i I of P and N respectively, such that P = i I P i and P i N/N i. (i) We have CA (M) N = M. (ii) The restriction morphism ( ) N : CA 2 (M) C2 A (N) is injective. (iii) If A N = C 2 A (N), then A M = C 2 A (M). Proof. (i) This is clear since CA (M) N contains P i by assumption ; just take the surjective map N N/N i P i and lift it to an endomorphism of M to see that CA (M) N P i, so that C A (M) N P. (ii) Assume ϕ C 2 A (M) satisfies ϕ N = 0. The composition N N/N i P i P is A-linear, thus C 2 A (M)-linear by Proposition 1.31 (ii). If ϕ N = 0, then ϕ(p i ) = ϕ(n/n i ) = 0, which means ϕ(p ) = i I ϕ(p i) = 0. This means ϕ(m) = ϕ(n) + ϕ(p ) = 0, i.e. ϕ = 0, so ( ) N is injective. (iii) Let ϕ C2 A(M). Since A N = CA 2 (N), there exists a A such that ϕ N = a N, i.e. (ϕ a M ) N = 0. Since restriction to N is injective by part (ii), we see that ϕ = a M A M. Since A M C2 A (M), we are done. Proposition Let A be a ring and M a left A-module with an A-submodule such that M A l N (in other words, we assume that M admits a direct summand isomorphic to A l ). Then A M = C 2 A (M). Proof. Consider the restriction map ( ) Al : CA 2 (M) C2 A (A l). By Corollary 1.30, we have CA 2 (A l) = A l, so for any ϕ CA 2 (M), we have ϕ A l = a Al for some a A. Each element n N generates a left A-module A n A l /a n where a n is some left ideal of A (namely, the ideal of all those a A such that an = 0). It follows that the pair (A l, N) satisfies the hypotheses of Lemma 1.35, so ( ) Al is injective, This means ϕ = a M, so we re done. Definition Let A be a ring. A left (resp. right) A-module is said to be free if there exists an isomorphism M i I A l (resp. M i I A r). When I is finite, we write A n def l = i I A l where I = n. Corollary Let A be a ring and M be a left A-module which admits a direct summand isomorphic to A l. Then A M = CA 2 (M) and Z(End A (M)) = Z(A M ), i.e. an element of the center of End A (M) is equal to multiplication by some element of Z(A M ). This holds in particular if M is a free left A-module. Furthermore, if A is commutative, Z(End A (M)) = A M. 17

18 Chapter 1 Proof. We have Z(End A (M)) = C EndZ (M)(End A (M)) End A (M) = C EndZ (M)(C A (M)) End A (M) = C 2 A(M) End A (M) = A M End A (M) = Z(A M ). The last equality is clear since saying that a M is an A-linear map is exactly saying that it commutes with all other b M for b A, i.e. lies in Z(A M ). Remark We have Z(A) M Z(A M ), but we don t have equality in general. The point is, two endomorphisms a M and b M may commute without a and b commuting in A. Definition Let A be a ring and M a left A-module. We say that M is finitely generated or is a finite A-module if there exists a surjective map A n l M of left A-modules for some n 1. Corollary Let A be a PID (note that principal ideal domains are assumed commutative by definition!) and M be a finitely generated left A-module. Then A M = C 2 A (M) and Z(End A(M)) A/Ann A (M) (recall that Ann A (M) = ker(a a M )). Proof. By the classification of finitely generated modules over a PID, we have an isomorphism M n i=1 A/a i where we can choose the ideals such that a 1 a 2 a n. It follows that Ann A (M) = a 1, so we can interpret M as an A/a 1 -module. Without loss of generality, assume a 1 = 0. We can then apply Proposition 1.36 and Corollary Corollary Let V be a finite-dimensional vector space over a field K and ϕ 1, ϕ 2 End K (V ). The following are equivalent : (i) There exists a polynomial p K[T ] such that p(ϕ 1 ) = ϕ 2 (ii) For all ψ End K (V ) which commutes with ϕ 1, ψ commutes with ϕ 2. Proof. Turn V into a K[T ]-module V ϕ1 by setting pv def = p(ϕ 1 )(v). The ring End K[T ] (V ) equals the set of K-linear endomorphisms of V which commute with polynomials in ϕ 1 ; this is equivalent to commuting with ϕ 1. By Corollary 1.41, we have K[T ] V /Ann K[T ] (V ) Z(End K[T ] (V )), so after translating both sides of this equality into the statements (i) and (ii), we are done. 1.3 Tensor products and Hom For the rest of this chapter, A denotes a ring. Definition Let M be a right A-module and N be a left A-module. If L is an abelian group, a map ϕ : M N L is called A-balanced if for any m 1, m 2 M, n 1, n 2 N and a A, the following holds : ϕ(m 1 + m 2, n 1 ) = ϕ(m 1, n 1 ) + ϕ(m 2, n 1 ) ϕ(m 1, n 1 + n 2 ) = ϕ(m 1, n 1 ) + ϕ(m 1, n 2 ) ϕ(m 1 a, n 1 ) = ϕ(m 1, an 1 ). In other words, it is a Z-bilinear map for which the action of a A can be done either on M or on N without changing the result of the application of ϕ. 18

19 The tensor product of M and N over A is the abelian group M A N defined as the quotient of the free abelian group Z (M N) modulo the subgroup generated by the relations (m 1 + m 2, n 1 ) (m 1, n 1 ) (m 2, n 1 ) (m 1, n 1 + n 2 ) (m 1, n 1 ) (m 1, n 2 ) (m 1 a, n 1 ) (m 1, an 1 ). The coset of the Z-module basis element (m, n) M N in M A N is denoted by m n. It comes with a canonical A-balanced map ι : M N M N given by (m, n) m n. Proposition (Universal property of the tensor product) Let M be a right A-module, N a left A-module and L an abelian group. There is a canonical bijection {ϕ : M N L ϕ is A-balanced} Hom Z (M A N, L) sending ϕ to ϕ in such a way that the following diagram commutes : M N ι ϕ M A N L ϕ Proof. Obvious from our definition of M A N. Details are left to the reader. Remark Since we have shown that M A N satisfies the above universal property, the pair (M A N, ι) is unique up to a unique isomorphism making a commutative triangle with M N as above. More generally, if we assume additionally that we have rings B, C, that M is a left B-module, N a right C-module and L a (B, C)-bimodule, then M A N is a (B, C)-bimodule via b B, c C, m M, n N, b(m n)c def = bm nc and the set of A-balanced maps ϕ : M N L satisfying ϕ(bm, nc) = bϕ(m, n)c are in bijection with Hom (B,C) (M A N, L). Even more generally, the module structures on M and N which are compatible with their respective A-module structures induce module structures on M A N (if the module structure on M is on the right or the structure on N is on the left, just replace it by the corresponding structure on the opposite ring and use the same construction). Furthermore, the construction (M, N) M A N gives a bifunctor : Mod-A A-Mod Ab. If any of M or N admits module structures compatible with their A-module structure, will also be functorial in the category of pairs of A-modules admitting those structures. Note that is an additive functor in both of its arguments, meaning that if f, f End A (M) and g, g End A (N), then (f + f ) g = f g + f g, f (g + g ) = f g + f g. Definition Let A, B be rings and M, N be (A, B)-modules. The abelian group Hom (A,B) (M, N) automatically becomes an (End (A,B) (N), End (A,B) (M))-bimodule via post-/pre-composition : for (ϕ B, ϕ A ) End B (N) End A (M), f Hom (A,B) (M, N) ϕ B fϕ A def = ϕ B f ϕ A. One easily checks that this is still a morphism of (A, B)-bimodules : for (a, b) A B and m M, (ϕ B fϕ A )(amb) = (ϕ B f)(aϕ A (m)b) = ϕ B (a(fϕ A )(m)b) = a(ϕ B fϕ A )(m)b. 19

20 Chapter 1 Remark When M, N are two left A-modules, we can define module structures on Hom A (M, N) using module structures on M and N compatible with their A-module structures. For instance, if M is also a left C-module, N a left B-module and the structures are compatible, we can turn Hom A (M, N) into a (B, C)-bimodule via m M, b B, c C, (bfc)(m) def = bf(cm). This is possible because the compatibility condition is equivalent to asking that the endomorphisms c M and b N are A-linear, giving rise to morphisms of rings C End A (M) and B End A (N). The (End A (N), End A (M))-bimodule structure on Hom A (M, N) can therefore be restricted to a (B, C)-bimodule structure. We can also apply this technique if one of the module structures on M or N is a right module structure instead of a left one, as long as it is compatible with the left A-module structures on M or N ; it suffices to replace these right module structures by left module structures on the opposite ring. This is possible, in particular, when M is an (A, B)-bimodule and N a (A, C)-bimodule, making Hom A (M, N) a (C, B)- bimodule via m M, b B, c C, (cfb)(m) def = f(mb)c. Furthermore, the construction (M, N) Hom A (M, N) gives a bifunctor : A-Mod opp A-Mod Ab. If any of M or N admits module structures compatible with their A-module structure, Hom A (, ) will also be functorial in the category of pairs of A-modules admitting those structures. By linearity of composition, Hom A (, ) is also additive in both of its arguments. Lemma Let M, N, {M i } i I, {N i } i I be left A-modules. We have natural isomorphisms Hom A ( i I M i, N) i I Hom A (M i, N), Hom A (M, i I N i ) i I Hom A (M, N i ). Proof. The first statement is equivalent to saying that a morphism of left A-modules ϕ : M N is determined by its restrictions ϕ Mi : M i N since ϕ = i I ϕ M i. For the second one, letting π i : N N i be the canonical projection, this is the fact that a morphism ϕ : M N is determined by its components π i ϕ : M N i. Corollary Let M, N be two left A-modules and assume N = i I N i for a family {N i } i I submodules. If M is finitely generated, then the isomorphism of Hom A (M, i I N i ) i I Hom A (M, N i ) restricts to the following isomorphism on the two following subsets : Hom A (M, i I N i ) i I Hom A (M, N i ). Proof. The isomorphism of Lemma 1.48 puts in correspondence a morphism ϕ : M i I N i with its projections {π i ϕ} i I. Since M is finitely generated, such a morphism ϕ is determined by the image of a subset {m 1,, m k }. Because ϕ(m i ) = i I n i, only finitely many indices i I contain non-zero components of the elements ϕ(m 1 ),, ϕ(m k ). It follows that only finitely many of the maps π i ϕ are non-zero. Conversely, if only π i1 ϕ,, π is ϕ are non-zero, then the corresponding map ϕ : M i I N i satisfies ϕ(m) s j=1 N i j, so ϕ(m) lands inside N in a well-defined manner. 20

21 Theorem (Tensor-hom adjunction) Let A, B be rings, M a right A-bimodule, N a (A, B)-bimodule and P a right B-bimodule. (It follows that M A N is a right B-module and Hom B (N, P ) is a right A-module.) There is a natural isomorphism of abelian groups Hom B (M A N, P ) Hom A (M, Hom B (N, P )). Moreover, this is an isomorphism of right End A (M)-modules, End (A,B) (N)-modules and End B (P )- modules. This implies that if C is a ring and any of M, N or P admit a C-module structure which is compatible with the module structures we already gave on that module, then both sides admit that corresponding C-module structure and the natural isomorphism above is an isomorphism of C-modules. Proof. Working out the details is easy but long and pointless. Seeing the main idea is the most useful part. A morphism of right B-modules ϕ : M A N P corresponds to an A-balanced map ϕ : M N P for which the restriction ϕ m : N P given by ϕ m (n) = ϕ(m, n) is B-linear. The fact that it is A-balanced implies that for a A, ϕ am (n) = ϕ m (an) = (ϕ m a)(n) by definition of the A-module structure on Hom B (N, P ), which means that the construction m ϕ m is A-linear. This process can be reversed, thus gives the isomorphism. Naturality is obvious. The fact that this is an isomorphism of End A (M)-modules (resp. End (A,B) (N), End B (P )) follows from the naturality of this isomorphism in all three arguments. Restricting this property to any subring of one of these three rings gives the statement about compatible module structures. Theorem Let M be a right A-module and N be a left A-module. (i) The functor of left A-modules to abelian groups given by N M A N is right-exact. (ii) The functor of right A-modules to abelian groups M M A N is right-exact. Proof. Let P be a left A-module. Recall that Hom A (N, P ) is an (A, A)-bimodule since both N and P are left A-modules. Since M is a right A-module, the functor Hom A (M, Hom A (, P )) = Hom Z (M, ) Hom A (, P ) is a composition of left-exact functors, namely Hom A (, P ) : A-Mod A-Mod-A and Hom A (M, ) A-Mod-A : A-Mod-A A-Mod-A (because both M and Hom A (N, P ) are right A-modules). The first functor sends a right-exact sequence to a left-exact one, and the second functor preserves left-exact sequences. Therefore, the composition is a contravariant functor mapping a right-exact sequence to a left-exact one and it is naturally isomorphic to Hom A (M A, P ) by Theorem Since the latter is exact for all left A-modules P, it follows that the functor M A is right-exact by Proposition 1.16 since the functor Hom A (, P ) is contravariant. This proves part (i). For part (ii), note that the functor Hom A (, Hom A (N, P )) is contravariant left-exact for all left A- modules N,P and it is naturally isomorphic to Hom A ( A N, P ). Since the latter is left-exact for all P, the functor A N is right-exact. Definition Let M 0,, M n be abelian groups and A 1,, A n be rings with the following relations : (i) M 0 is a right A 1 -module (ii) M n is a left A n -module (iii) For each 1 i n 1, M i is a (A i, A i+1 )-bimodule. 21

22 Chapter 1 Then we can form the abelian group M 0 A1 M 1 A2 M 2 A2 An M n as follows. Consider the free abelian group Z n i=0 M i on the set n + 1-tuples (m 0,, m n ) n i=0 M i. Quotient out by the submodule generated by the relations (m 0,, m i,, m n ) + (m 0,, m i,, m n ) (m 0,, m i + m i,, m n ) (m 0,, m i a i, m i+1,, m n ) (m 0,, m i, a i m i+1,, m n ). This gives us the abelian group. It is a bimodule for each ring A i by allowing multiplication on the M i 1 - factor on the right and on the M i -factor on the left. If A 0 (resp. A n+1 ) is a ring and M 0 is a left A 0 -module (resp. if M n is a right A n+1 -module), then the corresponding tensor product also admits this structure. Proposition (Associativity of the tensor product) Let A 0, A 1, A 2, A 3 be rings, M 0 a (A 0, A 1 )-bimodule, M 1 a (A 1, A 2 )-bimodule and M 2 a (A 2, A 3 )-bimodule. We have two isomorphism of abelian groups which is simultaneously an isomorphism of A 0, A 1, A 2 and A 3 -bimodules : M 0 A1 (M 1 A2 M 2 ) M 0 A1 M 1 A2 M 2 (M 0 A1 M 1 ) A2 M 2. Proof. It suffices to see that the correspondences m 0 (m 1 m 2 ) m 0 m 1 m 2 (m 0 m 1 ) m 2 produce well-defined maps in each direction and they are inverse to each other, as can be seen when applied on the generators of the respective modules given above. Proposition (Tensor product with respect to opp ) Given a left (resp. right) A-module M, recall Remark 1.6 for the definition of the opposite module structure M opp on its opposite ring A opp. Let M 0 be an (A 0, A 1 )-bimodule and M 1 be an (A 1, A 2 )-bimodule. We have an isomorphism of (A opp 2, A opp 0 )-bimodules (M 0 A1 M 1 ) opp M opp 1 A opp M opp 1 0. In particular, if A 0, A 1, A 2 are commutative rings, the tensor product is commutative up to the natural isomorphism given in the proof. Proof. Given m M 0, write m opp M opp 0 for the same element but which gets affected by multiplication from the opposite side in the reversed order. It suffices to map (m 0 m 1 ) opp to m opp 1 m opp 0. The (A opp 2, A opp 0 )-linearity becomes obvious since for a 2 A 2, m 0 M 0 and m 1 M 1, denoting the opposite multiplication with the symbol, we have a 2 (m 0 m 1 ) opp = (m 0 m 1 a 2 ) opp (m 1 a 2 ) opp m opp 0 = a 2 m opp 1 m opp 0. The argument is similar for a 0 A Isotypical modules Isotypical modules are to be seen as generalizations of free modules over a ring. We will see their importance in Theorem 1.59, where we establish a correspondence between the free modules over End A (M) = C A (M) and the isotypical modules of type M. Definition Let M, N be two left A-modules. We say that M is isotypical of type N if M i I N i where N i N. When this is the case, we write M N I. 22