Notes on tensor products [expository notes]

9 Pages·2016·0.218 MB·English

by  Robert Harron

Notes on tensor products
Robert Harron
Department of Mathematics, Keller Hall, University of Hawai‘i at Ma¯noa, Hon-
olulu, HI 96822, USA
E-mail address: [e-mail]
Abstract. Graduatealgebra
Contents
Chapter 1. Tensor products 5
1. Basic definition 5
Bibliography 9
3

CHAPTER 1
Tensor products
1. Basic definition
Last semester, we quickly defined and constructed the tensor product of modules two modules
M and N over a commutative ring R, defining it as the universal target M ⊗ N of the universal
R
R-bilinear form. Here, we’ll begin by considering the more general setup where R is allowed to be
non-commutative. We first introduce the generalization of a bilinear map to this setting: a ‘balanced’
map.
Let R be a (possibly non-commutative, but always unital) ring. In this section, M will be a right
R-module and N will be a left R-module, unless otherwise indicated.
Definition 1.1. Let A be an abelian group (written additively). A function ψ : M ×N → A is
called R-balanced (or middle R-linear) if for all m ,m ,m∈M,n ,n ,n∈N, and r ∈R,
1 2 1 2
(i) ψ(m +m ,n)=ψ(m ,n)+ψ(m ,n),
1 2 1 2
(ii) ψ(m,n +n )=ψ(m,n )+ψ(m,n ),
1 2 1 2
(iii) ψ(mr,n)=ψ(m,rn).
Let R-Balan (A) denote the set of R-balanced A-valued functions ψ :M ×N →A.
M,N
We may now define the tensor product M ⊗ N by a universal property as we did in the case of
R
commutative R. For simplicity, we will eschew the language of natural transformations.
Definition 1.2. ThetensorproductofM andN overRistheabeliangroupM⊗ N (ifitexists)
R
equippedwithanR-balancedmapψ :M×N →M⊗ N definedbythefollowinguniversalproperty.
univ R
For every abelian group A and every ψ ∈ R-Balan (A), there is a unique group homomorphism
M,N
ϕ:M ⊗ N →A such that
R
M ×N(cid:77)(cid:77)(cid:77)ψ(cid:77)ψ(cid:77)un(cid:77)i(cid:77)v(cid:77)(cid:47)(cid:47)(cid:77)M(cid:77)(cid:77)(cid:38)(cid:38)⊗(cid:15)(cid:15)R∃!ϕN
A
commutes.
Theorem 1.3. The tensor product M ⊗ N exists.
R
Proof. The proof is basically the same as the commutative case, but let’s go through it again.
First, we’ll define the abelian group M ⊗ N and ψ , then we’ll show it satisfies the universal
R univ
property. Let F =Free(M ×N) be the free abelian group (i.e. the free Z-module) on M ×N, i.e. the
abelian group whose elements are formal finite linear combinations
(cid:88)
a ·(m,n)
(m,n)
(m,n)∈M×N
with a ∈ Z. There is a natural group homomorphism M ×N → F sending (m,n) to 1·(m,n).
(m,n)
LetJ ⊂F bethesubgroupgeneratedbyallelementsofoneofthefollowingforms,asm ,m ,mvaries
1 2
over all elements of M, n ,n ,n varies over all elements of N, r varies over all elements of R:
1 2
(i) (m +m ,n)−(m ,n)−(m ,n),
1 2 1 2
(ii) (m,n +n )−(m,n )−(m,n ),
1 2 1 2
(iii) (mr,n)−(m,rn).
5
6 1. TENSOR PRODUCTS
Let M ⊗ N := F/J. Let ψ : M × N → M ⊗ N be the composition of the quotient map
R univ R
F →M ⊗ N with the natural injection M ×N →F. By the definition of J, ψ is R-balanced.
R univ
Let’s now show that this constructed ψ : M ×N → M ⊗ N satisfies the universal property.
univ R
Let A be an abelian group and let ψ :M ×N →A be an R-balanced map. By the universal property
of free Z-modules, there is a unique homomorphism ϕ : F → A such that ϕ((m,n)) = ψ(m,n) for
(cid:101) (cid:101)
all (m,n) ∈ M ×N. Since ψ is R-balanced, ϕ(J) = 0, i.e. J ⊆ ker(ϕ). By the universal property of
(cid:101) (cid:101)
quotients, thereisauniquehomomorphismϕ:M⊗ N →Asuchthatϕ((m,n) mod J)=ϕ((m,n)).
R (cid:101)
We thus have the commutative diagram
ψuniv (cid:42)(cid:42)
(cid:47)(cid:47) (cid:47)(cid:47)
M ×N(cid:71)(cid:71)(cid:71)ψ(cid:71)(cid:71)(cid:71)(cid:71)(cid:71)(cid:71)(cid:35)(cid:35)F(cid:15)(cid:15)∃(cid:122)(cid:122)!ϕ(cid:101) ∃!ϕM ⊗RN
A
yielding the desired universal property. (cid:3)
The image of (m,n) in M ⊗ N is denoted m⊗n as is called a pure tensor or a simple tensor.
R
Note that it denotes an equivalence class and hence may be equal to some other expression m(cid:48)⊗n(cid:48).
A general element of M ⊗ N is a linear combination (with Z coefficients) of pure tensors.
R
Asyoumayhavenoticed, unlikethecasewhereR iscommutative, thetensorproductmaynotbe
an R-module. In order to get that extra structure, we can proceed as follows.
Definition 1.4. Let R and S be two rings. An abelian group M is called an (R,S)-bimodule if
it is a left R-module, a right S-module, and
r(ms)=(rm)s
for all r ∈R,s∈S,m∈M.
A simple, and important, example is the case where R is commutative and M is a left (or right)
R-module. In this case, you can define a right (or left) R-module structure on M by m·r := r·m.
This makes M into a (R,R)-bimodule called the standard bimodule structure on M.
Proposition1.5. LetRandS berings. LetM bean(R,S)-bimoduleandletN bealeftS-module.
Then, M ⊗ N is a left R-module with scalar multiplication defined by
S
(cid:32) k (cid:33) k
(cid:88) (cid:88)
r· m ⊗n := (r·m )⊗n .
i i i i
i=1 i=1
Proof. We’ll take advantage of the universal property. For each r ∈ R, you can check that the
map ψ :(m,n)(cid:55)→(rm)⊗n is S-balanced; indeed, checking condition (iii),
r
ψ (ms,n)=r(ms)⊗n
r
=(rm)s⊗n
=(rm)⊗(sn)
=ψ (m,sn).
r
By the universal property of tensor products, there is a unique homormophism
ϕ :M ⊗ N →M ⊗ N
r S S
such that
ϕ (m⊗n)=ψ (m,n)=(rm)⊗n.
r r
The existence of ϕ shows that the definition of the scalar multiplication in the statement of the
r
propositioniswell-defined(independentofthewayofrepresentingtheinputasasumofpuretensors)
and shows that it gives a left R-module structure. (cid:3)
1. BASIC DEFINITION 7
In particular, if R is a commutative ring, M and N are two left R-modules, and we view M with
its standard bimodule structure, then M ⊗ N is a left R-module and this left R-module structure is
R
the same we defined near the end of last semester (i.e. the one that satisfies the universal property of
being the target of the universal R-bilinear form).
Lemma 1.6. In any tensor product M⊗ N, the pure tensors m⊗0 and 0⊗n are 0 for all m∈M
R
and all n∈N.
Proof. For m∈M,
m⊗0=m⊗(0+0)
=m⊗0+m⊗0,
so canceling one m⊗0 on each side gives the desired result. Similarly, for 0⊗n=0. (cid:3)
Lemma 1.7. For m,n∈Z ,
≥1
Z/mZ⊗ Z/nZ∼=Z/gcd(m,n)Z.
Z
Proof. First, note that Z/mZ⊗ Z/nZ is a cyclic abelian group generated by 1⊗1; indeed,
Z
a⊗b=a⊗(b·1)=(ab)⊗1=((ab)·1)⊗1=ab(1⊗1),
so every tensor is a multiple of (1⊗1).
Letg :=gcd(m,n). ByB´ezout’sidentity,thereareintegersu,v ∈Zsuchthatg =mu+nv. Thus,
g(1⊗1)=(mu+nv)(1⊗1)
=mu(1⊗1)+nv(1⊗1)
=(mu·1)⊗1+1⊗(nv·1)
=0⊗1+1⊗0
=0,
andsoZ/mZ⊗ Z/nZiscyclicoforderdividingg. Toshowitsorderisatleastg,we’lltakeadvantage
Z
of the universal property of tensor products. Consider the map
ψ : Z/mZ×Z/nZ → Z/gZ
,
(a,b) (cid:55)→ ab (mod g)
whichiswell-definedbecausegdividesbothmandn. ItisstraightforwardtocheckthatψisZ-bilinear
and so, by the universal property, there is a Z-module homomorphism ϕ : Z/mZ⊗ Z/nZ → Z/gZ.
Z
Since ϕ ‘agrees with’ ψ, ϕ(1⊗1) = 1. But 1 ∈ Z/gZ has order g, so 1⊗1 must have order at least
g. (cid:3)
Lemma 1.8. Suppose R is commutative and M and N are two left R-modules with their standard
bimodulestructures. IfM isafreeR-modulewithbasisB ={m :i∈I}andN isafreeR-modulewith
i
basis C ={n :j ∈J}, then M ⊗ N is a free R-module with basis B⊗C :={m ⊗n :i∈I,j ∈J},
j R i j
so that rk (M ⊗ N)=rk (M)·rk (N).
R R R R
Proof. B⊗C spans: we know that M ⊗ N is generated by pure tensors m⊗n. Furthermore,
R
a given m is a finite linear combination
(cid:88)
m= a m
i i
i∈I
and similarly
(cid:88)
n= b n .
j j
j∈J
Thus,
(cid:88)
m⊗n= a b (m ⊗n ),
i j i j
i∈I,j∈J
and so B⊗C is a generating set.
8 1. TENSOR PRODUCTS
B⊗C is linearly independent: suppose
(cid:88)
α (m ⊗n )=0
i,j i j
i∈I,j∈J
(where only finitely many α are non-zero). We want to show that all α are zero. To do this,
i,j i,j
we will ‘extract’ this coefficient. This is done by constructing, for each pair (i,j), a linear functional
f :M ⊗ N →R that takes the value 1 on m ⊗n and 0 on m ⊗n for i(cid:48) (cid:54)=i and j(cid:48) (cid:54)=j. If we
i,j R i j i(cid:48) j(cid:48)
have such linear functionals, then, applying each one independently to the above relation gives
 
(cid:88)
0=fi,j(0)=fi,j αi,j(mi⊗nj)=αi,j.
i∈I,j∈J
So,howdoyouconstructalinearfunctionM⊗ N →R? Bytheuniversalpropertyoftensorproducts,
R
you just have to define a bilinear form ψ : M ×N → R. Given what we want from this bilinear
i,j
form, we define it as follows. For i ∈I, j ∈J, and
0 0
(cid:88) (cid:88)
m= a m and n= b n ,
i i j j
i∈I j∈J
define
ψ (m,n)=a b .
i0,j0 i0 j0
ThisgivesanR-valuedbilinearformandalinearfunctionalonM⊗ N withthedesiredproperty. (cid:3)
R
Bibliography
9