7.2 Tensor Product as a Left S-Module

Back

In the last section we started with a right R-module M and a left Rmodule

N and constructed the abelian group M R N. In this section,

we shall discussion conditions that will enable M R N to carry a module

structure. To this end let S,R be rings, and let M be an abelian group.

We say that M is an (S,R)-bimodule if M is a left S-module and a right

R-module and that for all s 2 S,m 2 M, r 2 R we have

(sm)r = s(mr).

Next assume that M is an (S,R)-bimodule and that N is a left Rmodule.

As in the last section we have the abelian group M R N. In order

to give MRN the structure of a left S-module we need to construct a ring

homomorphism

_ : S ! EndZ(M R N);

this allows for the definition of an S-scalar multiplication: s·a = _(s)(a), a 2

MRN. For each s 2 S define fs : M×N ! MRN by setting fs(m, n) =

sm  n, s 2 S,m 2 M, n 2 N. Then fs is easily checked to be a balanced

map; by the universality of tensor product, there exists a unique abelian

group homomorphism _s : MRN ! MRN satisfying _s(mn) = smn.

Note that the above uniqueness implies that _s1+s2 = _s1 + _s2 and that

_s1s2 = (_s1) · (_s2). In turn, this immediately implies that the mapping

_ : S ! EndZ(M R N), _(s) = _s is the desired ring homomorphism. In

other words, we have succeeded in giving M R N the structure of a left

S-module.

The relevant universal property giving rise to a module homomorphism

is the following:

Proposition 7.2.1 let M be an (S,R)-bimodule, and let M be a left Rmodule.

If K is a left S-module and if f : M × N ! K is a balanced map

which also satisfies f(sm, n) = s · f(m, n), s 2 S, m 2 M, n 2 N, then the

induced abelian group homomorphism _ : M R N ! K is a left S-module

homomorphism.

Of particular importence is the following. Assume that R is a commutative

ring. If M is a left R-module, then M can be regarded also as a right

7.2. TENSOR PRODUCT AS A LEFT S-MODULE 159

R-module simply by declaring that m · r = r · m, r 2 R,m 2 M. (How

does the commutativity of R comes into play?) Therefore, in this situation,

if M,N are both left R-modules, we can form the left R-module M R N.

Probably the most canonical example in this situation is the construction

of the vector space V F W, where V,W are both F-vector spaces. Also, in

this specific situation, we can say more:

Proposition 7.2.2 Let V andW be F-vector spaces with bases {v1, . . . , vn},

{w1, . . . ,wm}, respectively. Then V FW has basis {vi wj | 1 _ i _ n, 1 _

j _ m}. In particular,

dim V F W = dim V · dim W.

The obvious analogue of the above is also true in the infinite-dimensional

case; see Exercise 2, below.

Exercises 7.2

1. Let R be a ring and letM be a left R-module. Prove that RRM _= M

as left R-modules.

2. V and W be F-vector spaces with bases {v_| _ 2 A}, {w_| _ 2 B}.

Then V F W has basis {v_  w_| _ 2 A, _ 2 B}.

3. Let W be an F-vector space and let T : V1 ! V2 be an injective linear

transformation of F-vector spaces. Prove that the sequence 1  T :

W  V1 ! W  V2 is injective.

4. Let T : V ! V be a linear transformation of the finite-dimensional

F-vector space V . If K _ F is a field extension, prove that mT,F(x) =

m1T,K(x). (Hint: Apply Exercise 3, above.)

5. Let F be a field and let A 2 Mn, B 2 Mm be square matrices. Define

the Kronecker (or tensor) product AB as follows. If A = [aij ], B =

[bkl], then AB is the block matrix [Dpq], where each Dpq is the m×m

matrix Dpq = apqB. Thus, for instance, if

A =

_

a11 a12

a21 a22

_

, B =

_

b11 b12

b21 b22

_

.

160 CHAPTER 7. TENSOR PRODUCTS

then

A  B =

2

664

a11b11 a11b12 a12b11 a12b12

a11b21 a11b22 a12b21 a12b22

a21b11 a21b12 a22b11 a22b12

a21b21 a21b22 a22b21 a22b22

3

775

.

Now Let V,W be F-vector spaces with ordered bases A = (v1, v2, . . . , vn),

B = (w1,w2, . . . ,wm), respectively. Let T : V ! V, S : W ! W be

linear transformations with matrix representations TA = A, SB = B.

Assume that A  B is the ordered basis of V F W given by A  B =

(v1w1, v1w2, . . . , v1wm; v2w1, . . . , v2wm; . . . , vnwm). Show

that the matrix representation of T  S relative to A  B is given by

(T  S)AB = A  B.

6. Let V be a two-dimensional vector space over the field F, and let

T, S : V ! V be linear transformations. Assume that mT (x) =

(x − a)2, mS(x) = (x − b)2. (Therefore T and S can be represented

by Jordan blocks, J2(a), J2(b), respectively.) Compute the invariant

factors of T  S : V  V ! V  V . (See Exercise 5 of Section 5.4).

7. Let M be a left R-module, and let I _ R be a 2-sided ideal in R.

Prove that, as left R-modules,

R/I R M _= M/IM.

8. Let R be a commutative ring and let M1,M2,M3 be R-modules. Prove

that there is an isomorphism of R-modules:

(M1 R M2) R M3

_= M1 R (M2 R M3).

9. Let R be a commutative ring and let M1,M2, . . . ,Mk be R-modules.

Assume that there is an R-multilinear map

f : M1 ×M2 × . . . ×Mk −! N

into the R-module N. Prove that there is a unique R-module homomorphism

_ : M1 R M2 R . . . R Mk −! N

satisfying _(m1m2. . .mk) = f(m1,m2, . . . ,mk), where all mi 2

Mi, i = 1, . . . k.

7.2. TENSOR PRODUCT AS A LEFT S-MODULE 161

10. Let G be a finite group and let H be a subgroup of G. Let F be a

field, and let FG, FH be the F-group algebras, as in Section 5.10. If V

is a finite-dimensional FH-module, prove that

dim FG FH V = [G : H] · dim V.

11. Let A be an abelian group. Prove that a ring structure on A is equivalent

to an abelian group homomorphism μ : A Z A ! A, together

with an element e 2 A such that μ(ea) = μ(ae) = a, for all a 2 A,

and such that

A Z A Z A

A Z A

A Z A

A

-

-

? ?

μ

1  μ

μ  1 μ

commutes. (The above diagram, of course, stipulates that multiplication

is associative.)

12. Let R be a Dedekind domain with fraction field E, and let I, J _ E

be fractional ideals. If [I] = [J] 2 CR (the ideal class group of R), then

I _=R J. (The converse is easier, see Exercise 4, of Section 5.1. Hint:

Consider the commutative diagram below:

I J

E R I E R J E

-

-

6 6

_

_

_

_>

-

_

1  _

iI iJ

_

i

where iI , iJ are injections, given by iI (a) = 1  a, iJ (b) = 1  b, a 2

I, b 2 J, _ : EJ ! E is given by _(_b) = _b, and where i : J ,! E.

Note also that 1  _ : E R I ! E  J is a E-linear transformation.

Next, if 0 6= a0 2 I, note that for all a 2 I, we have 1a = a(a−1

0 a0).

Indeed, a(a−1

0  a0) = aa−1

0 (1  a) = a−1

0 (a  a0) = a−1

0 (1  aa0) =

162 CHAPTER 7. TENSOR PRODUCTS

a−1

0 (a0a) = a−1

0 a0(1a) = 1a. Thus, if we set _0 = _(1_)(a−1

0

a0) we have _(a) = _) · a 2 E. In other words, _ : I ! J is given by

left multiplication by _0, i.e., J = _0I and the result follows.)

7.3. TENSOR PRODUCT AS AN ALGEBRA 163

Навигация