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HomR(Rn, M) toMn = M ×M ×··×M and this bijection is a group isomorphism. We
will see later that the product Mn is an R-module with scalar multiplication defined
by (m1, m2, .., mn)r =(m1r, m2r, .., mnr). If R is commutative so that HomR(Rn, M)
is an R-module, this theorem gives an R-module isomorphism from HomR(Rn, M) to
Mn.
This theorem reveals some of the great simplicity of linear algebra. It does not
matter how complicated the ring R is, or which R-module M is selected. Any R-
module homomorphism from Rn to M is determined by its values on the basis, and
any function from that basis to M extends uniquely to a homomorphism from Rn to
M.
Exercise
Suppose R is a field and f : RR → M is a non-zero homomorphism.
Show f is injective.
Chapter 5
Linear Algebra
73
Now let’s examine the special case M = Rm and show HomR(Rn, Rm) ≈ Rm,n.
Theorem
Suppose A =(ai,j) ∈ Rm,n. Then f : Rn → Rm defined by f(B) =AB
is a homomorphism with f(ei) = column i of A. Conversely, if m1, . . . , mn ∈ Rm,
define A ∈ Rm,n to be the matrix with column i = mi. Then f defined by f(B) =AB
is the unique homomorphism from Rn to Rm with f(ei) =mi.
Even though this follows easily from the previous theorem and properties of ma-
trices, it is one of the great classical facts of linear algebra. Matrices over R give
R-module homomorphisms! Furthermore, addition of matrices corresponds to addi-
tion of homomorphisms, and multiplication of matrices corresponds to composition
of homomorphisms. These properties are made explicit in the next two theorems.
Theorem
If f, g : Rn → Rm are given by matrices A, C ∈ Rm,n, then f + g is
given by the matrix A+C. Thus HomR(Rn, Rm) and Rm,n are isomorphic as additive
groups. If R is commutative, they are isomorphic as R-modules.
Theorem
If f : Rn → Rm is the homomorphism given by A ∈ Rm,n and g :
Rm → Rp is the homomorphism given by C ∈ Rp,m, then g ◦ f : Rn → Rp is given by
CA ∈ Rp,n.
That is, composition of homomorphisms corresponds to multiplication
of matrices.
Proof
This is just the associative law of matrix multiplication, C(AB) =(CA)B.
The previous theorem reveals where matrix multiplication comes from. It is the
matrix which represents the composition of the functions. In the case where the
domain and range are the same, we have the following elegant corollary.
Corollary
HomR(Rn, Rn) and Rn are isomorphic as rings. The automorphisms
correspond to the invertible matrices.
This corollary shows one way non-commutative rings arise, namely as endomor-
phism rings. Even if R is commutative, Rn is never commutative unless n =1.
We now return to the general theory of modules (over some given ring R).
74
Linear Algebra
Chapter 5
Cosets and Quotient Modules
After seeing quotient groups and quotient rings, quotient modules go through
without a hitch.
As before, R is a ring and module means R-module.
Theorem
Suppose M is a module and N ⊂ M is a submodule. Since N is a
normal subgroup of M, the additive abelian quotient group M/N is defined. Scalar
multiplication defined by (a + N)r = (ar + N) is well defined and gives M/N the
structure of an R-module. The natural projection π : M → M/N is a surjective
homomorphism with kernel N. Furthermore, if f : M → M is a surjective homomor-
phism with ker(f) =N, then M/N ≈ M (see below).
Proof
On the group level, this is all known from Chapter 2. It is only necessary
to check the scalar multiplication, which is obvious.
The relationship between quotients and homomorphisms for modules is the same
as for groups and rings, as shown by the next theorem.
Theorem
Suppose f : M → M is a homomorphism and N is a submodule of M.
If N ⊂ ker(f), then f : (M/N) → M defined by f(a + N) =f(a) is a well defined
homomorphism making the following diagram commute.
f
M
π
f
M/N
¯
¯
¯
¯
¯
¯
¯
M
¯
Thus defining a homomorphism on a quotient module is the same as defining a homo-
morphism on the numerator that sends the denominator to 0. The image of f is the
image of f, and the kernel of f is ker(f)/N. Thus if N = ker(f), f is injective, and
thus (M/N) ≈ image(f). Therefore for any homomorphism f, (domain(f)/ker(f)) ≈
image(f).
Proof
On the group level this is all known from Chapter 2 (see page 29). It is
only necessary to check that f is a module homomorphism, and this is immediate.
¯
¯
¯
¯
¯
Chapter 5
Linear Algebra
75
Theorem
Suppose M is an R-module and K and L are submodules of M.
K ∩ L. Thus (K/K ∩ L) → (K + L)/L is an isomorphism.
with kernel L/K. Thus (M/K)/(L/K) → M/L is an isomorphism.
Examples
These two examples are for the case R = Z.
i)
The natural homomorphism K → (K + L)/L is surjective with kernel
≈
ii)
Suppose K ⊂ L. The natural homomorphism M/K → M/L is surjective
≈
1)
2)
M = Z
K =3Z
L =5Z
K ∩ L =15Z
K/K ∩ L =3Z/15Z ≈ Z/5Z =(K + L)/L
K + L = Z
M = Z
K =6Z
L =3Z
(K ⊂ L)
(M/K)/(L/K) = (Z/6Z)/(3Z/6Z) ≈ Z/3Z = M/L
Products and Coproducts
Infinite products work fine for modules, just as they do for groups and rings.
This is stated below in full generality, although the student should think of the finite
case. In the finite case something important holds for modules that does not hold
for non-abelian groups or rings – namely, the finite product is also a coproduct. This
makes the structure of module homomorphisms much more simple. For the finite
case we may use either the product or sum notation, i.e., M1 × M2 ×· · ×Mn =
M1 ⊕ M2 ⊕· · ⊕Mn.
Theorem
Suppose T is an index set and for each t ∈ T , Mt is an R-module. On
the additive abelian group
Mt = [ Pobierz całość w formacie PDF ]