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Let d = ps ps · · · ps be the prime factorization of d.
Then the natural map
R/d -→R/ps ⊕ · · · ⊕ R/ps is an isomorphism of R-modules. (The elements ps
are called elementary divisors of R/d.)
Proof
If i = j, ps and ps are relatively prime. By the Lemma above, they are
¯
⊂
¯
j
i
i
j
1
2
t
1
2
t
1
t
i
1
t
i
≈
120
Appendix
Chapter 6
comaximal and thus by the Chinese Remainder Theorem, the natural map is a ring
isomorphism. Since the natural map is also an R-module homomorphism, it is an
R-module isomorphism.
This theorem carries the splitting as far as it can go, as seen by the next exercise.
Exercise
Suppose R is a PID, p ∈ R is a prime element, and s ≥ 1. Then the
R-module R/ps has no proper submodule which is a summand.
To give perspective to this section, here is a brief discussion of torsion submodules.
Definition
Suppose M is a module over a domain R. An element m ∈ M is said
to be a torsion element if ∃ r ∈ R with r = 0 and mr = 0. This is the same as
saying m is dependent. If R = Z, it is the same¯as saying m has finite order. Denote
by T (M) the set of all torsion elements of M. If T (M) =0, we say that M is torsion
free.
Theorem 7
Suppose M is a module over a domain R. Then T (M) is a submodule
of M and M/T (M) is torsion free.
Proof
This is a simple exercise.
Theorem 8
Suppose R is a Euclidean domain and M is a finitely generated
R-module which is torsion free. Then M is a free R-module, i.e., M ≈ Rm.
Proof
This follows immediately from Theorem 5.
Theorem 9
Suppose R is a Euclidean domain and M is a finitely generated
R-module. Then the following s.e.s. splits.
0 -→ T (M) -→ M -→ M/T (M) -→ 0
Proof
By Theorem 7, M/T (M) is torsion free. By Theorem 8, M/T (M) is a free
R-module, and thus there is a splitting map. Of course this theorem is transparent
anyway, because Theorem 5 gives a splitting of M into a torsion part and a free part.
¯
¯
Chapter 6
Appendix
121
Note
It follows from Theorem 9 that ∃ a free submodule V of M such that T (M)⊕
V = M. The first summand T (M) is unique, but the complementary summand V is
not unique. V depends upon the splitting map and is unique only up to isomorphism.
To complete this section, here are two more theorems that follow from the work
we have done.
Theorem 10
Suppose T is a domain and T is the multiplicative group of units
of T .
If G is a finite subgroup of T , then G is a cyclic group.
Thus if F is a finite
field, the multiplicative group F is cyclic.
Thus if p is a prime, (Zp)∗ is cyclic.
Proof
This is a corollary to Theorem 5 with R = Z. The multiplicative group G
is isomorphic to an additive group Z/d1 ⊕ Z/d2 ⊕· · · ⊕Z/dt where each di 1 and
di|di+1 for 1 ≤ i t. Every g in the additive group has the property that gdt =0. So
every u ∈ G is a solution to xd - 1 =0. If t 1, the equation will have degree less
than the number of roots, which is¯impossible. Thus t = 1 and G is cyclic.
Exercise
For which primes p and q is the group of units (Zp ×Zq)∗ a cyclic group?
We know from Exercise 2) on page 59 that an invertible matrix over a field is the
product of elementary matrices. This result also holds for any invertible matrix over
a Euclidean domain.
Theorem 11
Suppose R is a Euclidean domain and A ∈ Rn is a matrix with
non-zero determinant. Then by elementary row and column operations, A may be
transformed to a diagonal matrix
∗
∗
∗
¯
t
¯










d1
d2
...
dn


where each di =0 and di|di+1 for 1 ≤ i n. Also d1 generates the ideal generated
by the entries of A. Furthermore A is invertible iff each di is a unit. Thus if A is
invertible, A is the product of elementary matrices.
¯
122
Appendix
Chapter 6
Proof
It follows from Theorem 3 that A may be transformed to a diagonal matrix
with di|di+1. Since the determinant of A is not zero, it follows that each di = 0.
Furthermore, the matrix A is invertible iff the diagonal matrix is invertible, which is
true iff each di is a unit. If each di is a unit, then the diagonal matrix is the product
of elementary matrices of type 1. Therefore if A is invertible, it is the product of
elementary matrices.
Exercise
Let R = Z, A =
and D =
. Perform elementary
operations on A and D to obtain diagonal matrices where the first diagonal element
divides the second diagonal element. Write D as the product of elementary matrices.
Find the characteristic polynomials of A and D. Find an elementary matrix B over
Z such that B-1AB is diagonal. Find an invertible matrix C in R2 such that C-1DC
is diagonal. Show C cannot be selected in Q2.
Jordan Blocks
In this section, we define the two special types of square matrices used in the
Rational and Jordan canonical forms. Note that the Jordan block B(q) is the sum
of a scalar matrix and a nilpotent matrix. A Jordan block displays its eigenvalue
on the diagonal, and is more interesting than the companion matrix C(q). But as
we shall see later, the Rational canonical form will always exist, while the Jordan
canonical form will exist iff the characteristic polynomial factors as the product of
linear polynomials.
Suppose R is a commutative ring, q = a0 + a1x + · · · + an-1xn-1 + xn ∈ R[x]
is a monic polynomial of degree n ≥ 1, and V is the R[x]-module V = R[x]/q.
V is a torsion module over the ring R[x], but as an R-module, V has a free basis
{1, x, x2, . . . , xn-1}. (See the division algorithm in the chapter on rings.) Multipli-
cation by x defines an R-module endomorphism on V , and C(q) will be the ma-
trix of this endomorphism with respect to this basis. Let T : V → V be defined [ Pobierz całość w formacie PDF ]