7.1 Tensor Product as an Abelian Group

Back

Throught this chapter R will denote a ring with identity. All modules will

be unital.

Let M be a right R-module, let N be a left R-module, and let A be an

abelian group. By a balanced map, we mean a map

f : M × N −! A,

such that

(i) f(m1 + m2, n) = f(m1, n) + f(m2, n),

(ii) f(m, n1 + n2) = f(m, n1) + f(m, n2),

(iii) f(mr, n) = f(m, rn)

where m,m1,m2 2 M, n, n1, n2 2 N, r 2 R.

By a tensor product of M and N we mean an abelian group T, together

with a balanced map t : M × N ! T such that given any abelian group A,

and any balanced map f : M × N ! A there exists a unique abelian group

homomorphism _ : T ! A, making the diagram below commute

154

7.1. TENSOR PRODUCT AS AN ABELIAN GROUP 155

M × N A

T

-

􀀀

􀀀

􀀀

􀀀

􀀀_ @

@

@

@

@R

f

t _

The following is the usual application of “abstract nonsense.”

Proposition 7.1.1 The tensor product of the right R-module M and the

left R-module N is unique up to abelian group isomorphism.

This leaves the question of existence, which is also not very difficult.

Indeed, given M and N as above, and let F be the free abelian group on

the set M × N. Let B be the subgroup of F generated by elements of the

form

(m1 + m2, n) − (m1, n) − (m2, n),

(m, n1 + n2) − (m, n1) − (m, n2),

(mr, n) − (m, rn),

where m,m1,m2 2 M, n, n1, n2 2 N, r 2 R. Write MRN = F/B and set

mn = (m, n)+B 2 M RN. Therefore, in M RN we have the relations

(m1 + m2)  n = m1  n + m2  n,

m  (n1 + n2) = m  n1 + m  n2,

mr  n = m  rn,

m,m1,m2 2 M, n, n1, n2 2 N, r 2 R. Furthermore, M R N is generated

by all “simple tensors” m  n, m 2 M, n 2 N.

Define the map t : M ×N ! M R N by setting t(m, n) = m n, m 2

M, n 2 N. Then, by construction, t is a balanced map. In fact

Proposition 7.1.2 The abelian group MRN, together with the balanced

map t : M × N ! M R N is a tensor product of M and N.

A couple of simple examples are in order here.

156 CHAPTER 7. TENSOR PRODUCTS

1. If N is a left R-module, then R R N _= N as abelian groups. The

proof simply amounts to showing that the map t : R × N ! N given

by t(r, n) = rn is balanced and is universal with respect to balanced

maps into abelian groups. Invoke Proposition 1.

2. If A is any torsion abelian group and if D is any divisible abelian

group, then D Z A = 0. If a 2 A, let 0 6= n 2 Z be such that na = 0.

Then for any d 2 D there exists d0 2 D such that d0n = d. Therefore

d  a = d0n  a = d0  na = d0  0 = 0. Therefore every simple tensor

in D Z A is zero; thus D Z A = 0.

Next we wish to discuss the mapping or “functorial” properties of the

tensor product.

Proposition 7.1.3 Let f : M ! M0 be a right R-module homomorphism

and let g : N ! N0 be a left R-module homomorphism. Then there exists

a unique abelian group homomorphism f  g : M R N :! M0 R N0 such

that for all m 2 M, n 2 N, (f  g)(m  n) = f(m)  g(n).

In particular, the following observation is the basis of all so-called “homological”

properties of .

Proposition 7.1.4

(i) Let M0 μ

! M _! M00 ! 0 be an exact sequence of right R-modules, and

let N be a left R-module. Then the sequence

M0 R N

μ1N −! M R N

_1N −! M00 R N ! 0

is exact.

(ii) Let N0 μ

! N _! N00 ! 0 be an exact sequence of left R-modules, and

let M be a right R-module. Then

M R N0 1Mμ

−! M R N

1M_

−! M R N00 ! 0

is exact.

We hasten to warn the reader that in Proposition 4 (i) above, even if

M0 μ

! M is been injective, it need not follow that M0RN

μ1N −! MRN is

7.1. TENSOR PRODUCT AS AN ABELIAN GROUP 157

injective. (A similar comment holds for part (ii).) Put succinctly, the tensor

product does not take short exact sequences to short exact sequences. In

fact a large portion of “homological algebra” is devoted to the study of

functors that do not preserve exactness. As an easy example, consider the

short exact sequence of abelian groups (i.e. Z-modules):

Z μ2! Z ! Z/(2) ! 0,

where μ2(a) = 2a. If we tensor the above short exact sequence on the right

by Z/(2), we get the sequence

Z/(2) 0! Z/(2) _=! Z/(2).

Thus the exactness breaks down.

Exercises 7.1

1. Let M1,M2 be right R-modules and let N be a left R-module. Prove

that

(M1 _M2) R N _= M1 R N _M2 R N.

2. Let N be a left R-module. Say that N is flat if for any injective

homomorphism of right R-modules M0 ! M, then the abelian group

homomorphism M0  N ! M  N is also injective. (The obvious

analogous definition also applies to right R-modules.) Prove that if N

is projective then N is flat.

3. Let A be an abelian group. If n is a positive integer, prove that

Z/nZ Z A _= A/nA.

4. Let m, n be positive integers and let k = g.c.d(m, n). Prove that

Z/mZ Z Z/nZ _= Z/kZ.

158 CHAPTER 7. TENSOR PRODUCTS

Навигация