over a PID
24.1
Modules over a Commutative Ring
In this chapter, we introduce modules over a commutative ring (with unity). After a quick
overview of fundamental concepts such as free modules, torsion modules, and some basic
results about them, we focus on finitely generated modules over a PID and we prove the
structure theorems for this class of modules (invariant factors and elementary divisors). Our
main goal is not to give a comprehensive exposition of modules, but instead to apply the
structure theorem to the K[X]-module Ef defined by a linear map f acting on a finite-
dimensional vector space E, and to obtain several normal forms for f , including the rational
canonical form.
A module is the generalization of a vector space E over a field K obtained replacing
the field K by a commutative ring A (with unity 1). Although formally, the definition is
the same, the fact that some nonzero elements of A are not invertible has some serious
conequences. For example, it is possible that λ · u = 0 for some nonzero λ ∈ A and some
nonzero u ∈ E, and a module may no longer have a basis.
For the sake of completeness, we give the definition of a module, although it is the same
as Definition 2.9 with the field K replaced by a ring A. In this chapter, all rings under
consideration are assumed to be commutative and to have an identity element 1.
Definition 24.1. Given a ring A, a (left) module over A (or A-module) is a set M (of vectors)
together with two operations + : M × M → M (called vector addition),1 and ·: A × M → M
(called scalar multiplication) satisfying the following conditions for all α, β ∈ A and all
u, v ∈ M;
(M0) M is an abelian group w.r.t. +, with identity element 0;
1The symbol + is overloaded, since it denotes both addition in the ring A and addition of vectors in M .
It is usually clear from the context which + is intended.
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
(M1) α · (u + v) = (α · u) + (α · v);
(M2) (α + β) · u = (α · u) + (β · u);
(M3) (α ∗ β) · u = α · (β · u);
(M4) 1 · u = u.
Given α ∈ A and v ∈ M, the element α · v is also denoted by αv. The ring A is often
called the ring of scalars.
Unless specified otherwise or unless we are dealing with several different rings, in the rest
of this chapter, we assume that all A-modules are defined with respect to a fixed ring A.
Thus, we will refer to a A-module simply as a module.
From (M0), a module always contains the null vector 0, and thus is nonempty. From
(M1), we get α · 0 = 0, and α · (−v) = −(α · v). From (M2), we get 0 · v = 0, and
(−α) · v = −(α · v). The ring A itself can be viewed as a module over itself, addition of
vectors being addition in the ring, and multiplication by a scalar being multiplication in the
ring.
When the ring A is a field, an A-module is a vector space. When A = Z, a Z-module is
just an abelian group, with the action given by
0 · u = 0,
n · u = u + · · · + u,
n > 0
n
n · u = −(−n) · u,
n < 0.
All definitions from Section 2.3, linear combinations, linear independence and linear
dependence, subspaces renamed as submodules, apply unchanged to modules. Proposition
2.5 also holds for the module spanned by a set of vectors. The definition of a basis (Definition
2.12) also applies to modules, but the only result from Section 2.4 that holds for modules
is Proposition 2.11. Unfortunately, it is longer true that every module has a basis. For
example, for any nonzero integer m ∈ Z, the Z-module Z/mZ has no basis. Similarly, Q,
as a Z-module, has no basis. In fact, any two distinct nonzero elements p1/q1 and p2/q2 are
linearly dependent, since
p
p
(p
1
2
2q1)
− (p
= 0.
q
1q2)
1
q2
Definition 2.13 can be generalized to rings and yields free modules.
Definition 24.2. Given a commutative ring A and any (nonempty) set I, let A(I) be the
subset of the cartesian product AI consisting of all families (λi)i∈I with finite support of
scalars in A.2 We define addition and multiplication by a scalar as follows:
(λi)i∈I + (µi)i∈I = (λi + µi)i∈I,
2Where AI denotes the set of all functions from I to A.
24.1. MODULES OVER A COMMUTATIVE RING
665
and
λ · (µi)i∈I = (λµi)i∈I.
It is immediately verified that addition and multiplication by a scalar are well defined.
Thus, A(I) is a module. Furthermore, because families with finite support are considered, the
family (ei)i∈I of vectors ei, defined such that (ei)j = 0 if j = i and (ei)i = 1, is clearly a basis
of the module A(I). When I = {1, . . . , n}, we denote A(I) by An. The function ι: I → A(I),
such that ι(i) = ei for every i ∈ I, is clearly an injection.
Definition 24.3. An A-module M is free iff it has a basis.
The module A(I) is a free module.
All definitions from Section 2.5 apply to modules, linear maps, kernel, image, except the
definition of rank, which has to be defined differently. Propositions 2.12, 2.13, 2.14, and
2.15 hold for modules. However, the other propositions do not generalize to modules. The
definition of an isomorphism generalizes to modules. As a consequence, a module is free iff
it is isomorphic to a module of the form A(I).
Section 2.6 generalizes to modules. Given a submodule N of a module M , we can define
the quotient module M/N .
If a is an ideal in A and if M is an A-module, we define aM as the set of finite sums of
the form
a1m1 + · · · + akmk, ai ∈ a, mi ∈ M.
It is immediately verified that aM is a submodule of M .
Interestingly, the part of Theorem 2.10 that asserts that any two bases of a vector space
have the same cardinality holds for modules. One way to prove this fact is to “pass” to a
vector space by a quotient process.
Theorem 24.1. For any free module M , any two bases of M have the same cardinality.
Proof sketch. We give the argument for finite bases, but it also holds for infinite bases. The
trick is to pick any maximal ideal m in A (whose existence is guaranteed by Theorem 31.3).
Then, A/m is a field, and M/mM can be made into a vector space over A/m; we leave the
details as an exercise. If (u1, . . . , un) is a basis of M, then it is easy to see that the image of
this basis is a basis of the vector space M/mM . By Theorem 2.10, the number n of elements
in any basis of M/mM is an invariant, so any two bases of M must have the same number
of elements.
The common number of elements in any basis of a free module is called the dimension
(or rank ) of the free module.
One should realize that the notion of linear independence in a module is a little tricky.
According to the definition, the one-element sequence (u) consisting of a single nonzero
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
vector is linearly independent if for all λ ∈ A, if λu = 0 then λ = 0. However, there are free
modules that contain nonzero vectors that are not linearly independent! For example, the
ring A = Z/6Z viewed as a module over itself has the basis (1), but the zero-divisors, such
as 2 or 4, are not linearly independent. Using language introduced in Definition 24.4, a free
module may have torsion elements. There are also nonfree modules such that every nonzero
vector is linearly independent, such as Q over Z.
All definitions from Section 3.1 about matrices apply to free modules, and so do all the
proposition. Similarly, all definitions from Section 4.1 about direct sums and direct products
apply to modules. All propositions that do not involve extending bases still hold. The
important proposition 4.10 survives in the following form.
Proposition 24.2. Let f : E → F be a surjective linear between two A-modules with F a
free module. Given any basis (v1, . . . , vr) of F , for any r vectors u1, . . . , ur ∈ E such that
f (ui) = vi for i = 1, . . . , r, the vectors (u1, . . . , ur) are linearly independent and the module
E is the direct sum
E = Ker (f ) ⊕ U,
where U is the free submodule of E spanned by the basis (u1, . . . , ur).
Proof. Pick any w ∈ E, write f(w) over the basis (v1, . . . , vr) as f(w) = a1v1 + · · · + arvr,
and let u = a1u1 + · · · + arur. Observe that
f (w − u) = f(w) − f(u)
= a1v1 + · · · + arvr − (a1f(u1) + · · · + arf(ur))
= a1v1 + · · · + arvr − (a1v1 + · · · + arvr)
= 0.
Therefore, h = w − u ∈ Ker (f), and since w = h + u with h ∈ Ker (f) and u ∈ U, we have
E = Ker (f ) + U .
If u = a1u1 + · · · + arur ∈ U also belongs to Ker (f), then
0 = f (u) = f (a1u1 + · · · + arur) = a1v1 + · · · + arvr,
and since (v1, . . . , vr) is a basis, ai = 0 for i = 1, . . . , r, which shows that Ker (f ) ∩ U = (0).
Therefore, we have a direct sum
E = Ker (f ) ⊕ U.
Finally, if
a1u1 + · · · + arur = 0,
the above reasoning shows that ai = 0 for i = 1, . . . , r, so (u1, . . . , ur) are linearly indepen-
dent. Therefore, the module U is a free module.
24.1. MODULES OVER A COMMUTATIVE RING
667
One should be aware that if we have a direct sum of modules
U = U1 ⊕ · · · ⊕ Um,
every vector u ∈ U can be written is a unique way as
u = u1 + · · · + um,
with ui ∈ Ui but, unlike the case of vector spaces, this does not imply that any m nonzero
vectors (u1, . . . , um) are linearly independent. For example,
Z = Z/2Z ⊕ Z/2Z
where Z and Z/2Z are view as Z-modules, but (1, 0) and (0, 1) are not linearly independent,
since
2(1, 0) + 2(0, 1) = (0, 0).
A useful fact is that every module is a quotient of some free module. Indeed, if M is
an A-module, pick any spanning set I for M (such a set exists, for example, I = M ), and
consider the unique homomorphism ϕ : A(I) → M extending the identity function from I to
itself. Then we have an isomorphism A(I)/Ker (ϕ) ≈ M.
In particular, if M is finitely generated, we can pick I to be a finite set of generators, in
which case we get an isomorphism An/Ker (ϕ) ≈ M, for some natural number n. A finitely
generated module is sometimes called a module of finite type.
The case n = 1 is of particular interest. A module M is said to be cyclic if it is generated
by a single element. In this case M = Ax, for some x ∈ M. We have the linear map
mx : A → M given by a → ax for every a ∈ A, and it is obviously surjective since M = Ax.
Since the kernel a = Ker (mx) of mx is an ideal in A, we get an isomorphism A/a ≈ Ax.
Conversely, for any ideal a of A, if M = A/a, we see that M is generated by the image x of
1 in M , so M is a cyclic module.
The ideal a = Ker (mx) is the set of all a ∈ A such that ax = 0. This is called the
annihilator of x, and it is the special case of the following more general situation.
Definition 24.4. If M is any A-module, for any subset S of M , the set of all a ∈ A such
that ax = 0 for all x ∈ S is called the annihilator of S, and it is denoted by Ann(S). If
S = {x}, we write Ann(x) instead of Ann({x}). A nonzero element x ∈ M is called a torsion
element iff Ann(x) = (0). The set consisting of all torsion elements in M and 0 is denoted
by Mtor.
It is immediately verified that Ann(S) is an ideal of A, and by definition,
Mtor = {x ∈ M | (∃a ∈ A, a = 0)(ax = 0)}.
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
If a ring has zero divisors, then the set of all torsion elements in an A-module M may not
be a submodule of M . For example, if M = A = Z/6Z, then Mtor = {2, 3, 4}, but 3 + 4 = 1
is not a torsion element. Also, a free module may not be torsion-free because there may be
torsion elements, as the example of Z/6Z as a free module over itself shows.
However, if A is an integral domain, then a free module is torsion-free and Mtor is a
submodule of M . (Recall that an integral domain is commutative).
Proposition 24.3. If A is an integral domain, then for any A-module M , the set Mtor of
torsion elements in M is a submodule of M .
Proof. If x, y ∈ M are torsion elements (x, y = 0), then there exist some nonzero elements
a, b ∈ A such that ax = 0 and by = 0. Since A is an integral domain, ab = 0, and then for
all λ, µ ∈ A, we have
ab(λx + µy) = bλax + aµby = 0.
Therefore, Mtor is a submodule of M.
The module Mtor is called the torsion submodule of M. If Mtor = (0), then we say that
M is torsion-free, and if M = Mtor, then we say that M is a torsion module.
If M is not finitely generated, then it is possible that Mtor = 0, yet the annihilator of
Mtor is reduced to 0 (find an example). However, if M is finitely generated, this cannot
happen, since if x1, . . . , xn generate M and if a1, . . . , an annihilate x1, . . . , xn, then a1 · · · an
annihilates every element of M .
Proposition 24.4. If A is an integral domain, then for any A-module M , the quotient
module M/Mtor is torsion free.
Proof. Let x be an element of M/Mtor and assume that ax = 0 for some a = 0 in A. This
means that ax ∈ Mtor, so there is some b = 0 in A such that bax = 0. Since a, b = 0 and A
is an integral domain, ba = 0, so x ∈ Mtor, which means that x = 0.
If A is an integral domain and if F is a free A-module with basis (u1, . . . , un), then F
can be embedded in a K-vector space FK isomorphic to Kn, where K = Frac(A) is the
fraction field of A. Similarly, any submodule M of F is embedded into a subspace MK of
FK. Note that any linearly independent vectors (u1, . . . , um) in the A-module M remain
linearly independent in the vector space MK, because any linear dependence over K is of
the form
a1
a
u
m u
b
1 + · · · +
m = 0
1
bm
for some ai, bi ∈ A, with b1 · · · bm = 0, so if we multiply by b1 · · · bm = 0, we get a lin-
ear dependence in the A-module M . Then, we see that the maximum number of linearly
independent vectors in the A-module M is at most n. The maximum number of linearly
independent vectors in a finitely generated submodule of a free module (over an integral
domain) is called the rank of the module M . If (u1, . . . , um) are linearly independent where
24.1. MODULES OVER A COMMUTATIVE RING
669
m is the rank of m, then for every nonzero v ∈ M, there are some a, a1, . . . , am ∈ A, not all
zero, such that
av = a1u1 + · · · + amum.
We must have a = 0, since otherwise, linear independence of the ui would imply that
a1 = · · · = am = 0, contradicting the fact that a, a1, . . . , am ∈ A are not all zero.
Unfortunately, in general, a torsion-free module is not free. For example, Q as a Z-module
is torsion-free but not free. If we restrict ourselves to finitely generated modules over PID’s,
then such modules split as the direct sum of their torsion module with a free module, and a
torsion module has a nice decomposition in terms of cyclic modules.
The following proposition shows that over a PID, submodules of a free module are free.
There are various ways of proving this result. We give a proof due to Lang [65] (see Chapter
III, Section 7).
Proposition 24.5. If A is a PID and if F is a free A-module of dimension n, then every
submodule M of F is a free module of dimension at most n.
Proof. Let (u1, . . . , un) be a basis of F , and let Mr = M ∩ (Au1 ⊕ · · · ⊕ Aur), the intersection
of M with the free module generated by (u1, . . . , ur), for r = 1, . . . , n. We prove by induction
on r that each Mr is free and of dimension at most r. Since M = Mr for some r, this will
prove our result.
Consider M1 = M ∩ Au1. If M1 = (0), we are done. Otherwise let
a = {a ∈ A | au1 ∈ M}.
It is immediately verified that a is an ideal, and since A is a PID, a = a1A, for some a1 ∈ A.
Since we are assuming that M1 = (0), we have a1 = 0, and a1u1 ∈ M. If x ∈ M1, then
x = au1 for some a ∈ A, so a ∈ a1A, and thus a = ba1 for some b ∈ A. It follows that
M1 = Aa1u1, which is free.
Assume inductively that Mr is free of dimension at most r < n, and let
a = {a ∈ A | (∃b1 ∈ A) · · · (∃br ∈ A)(b1u1 + · · · + brur + aur+1 ∈ M)}.
It is immediately verified that a is an ideal, and since A is a PID, a = ar+1A, for some
ar+1 ∈ A. If ar+1 = 0, then Mr+1 = Mr, and we are done.
If ar+1 = 0, then there is some v1 ∈ Au1 ⊕ · · · ⊕ Aur such that
w = v1 + ar+1ur+1 ∈ M.
For any x ∈ Mr+1, there is some v ∈ Au1 ⊕· · ·⊕Aur and some a ∈ A such that x = v+aur+1.
Then, a ∈ ar+1A, so there is some b ∈ A such that a = bar+1. As a consequence
x − bw = v − bv1 ∈ Mr,
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
and so x = x − bw + bw with x − bw ∈ Mr, which shows that
Mr+1 = Mr + Aw.
On the other hand, if u ∈ Mr ∩ Aw, then since w = v1 + ar+1ur+1 we have
u = bv1 + bar+1ur+1,
for some b ∈ A, with u, v1 ∈ Au1 ⊕ · · · ⊕ Aur, and if b = 0, this yields the nontrivial linear
combination
bv1 − u + bar+1ur+1 = 0,
contradicting the fact that (u1, . . . , ur+1) are linearly independent. Therefore,
Mr+1 = Mr ⊕ Aw,
which shows that Mr+1 is free of dimension at most r + 1.
Proposition 24.5 implies that if M is a finitely generated module over a PID, then any
submodule N of M is also finitely generated.
Indeed, if (u1, . . . , un) generate M, then we have a surjection ϕ : An → M from the free
module An onto M . The inverse image ϕ−1(N ) of N is a submodule of the free module An,
therefore by Proposition 24.5, ϕ−1(N ) is free and finitely generated. This implies that N is
finitely generated (and that it has a number of generators ≤ n).
We can also prove that a finitely generated torsion-free module over a PID is actually
free. We will give another proof of this fact later, but the following proof is instructive.
Proposition 24.6. If A is a PID and if M is a finitely generated module which is torsion-
free, then M is free.
Proof. Let (y1, . . . , yn) be some generators for M , and let (u1, . . . , um) be a maximal sub-
sequence of (y1, . . . , yn) which is linearly independent. If m = n, we are done. Otherwise,
due to the maximality of m, for i = 1, . . . , n, there is some ai = 0 such that such that
aiyi can be expressed as a linear combination of (u1, . . . , um). If we let a = a1 . . . an, then
a1 . . . anyi ∈ Au1 ⊕ · · · ⊕ Aum for i = 1, . . . , n, which shows that
aM ⊆ Au1 ⊕ · · · ⊕ Aum.
Now, A is an integral domain, and since ai = 0 for i = 1, . . . , n, we have a = a1 . . . an = 0,
and because M is torsion-free, the map x → ax is injective. It follows that M is isomorphic
to a submodule of the free module Au1 ⊕ · · · ⊕ Aum. By Proposition 24.5, this submodule
if free, and thus, M is free.
Although we will obtain this result as a corollary of the structure theorem for finitely
generated modules over a PID, we are in the position to give a quick proof of the following
theorem.
24.2. FINITE PRESENTATIONS OF MODULES
671
Theorem 24.7. Let M be a finitely generated module over a PID. Then M/Mtor is free,
and there exit a free submodule F of M such that M is the direct sum
M = Mtor ⊕ F.
The dimension of F is uniquely determined.
Proof. By Proposition 24.4 M/Mtor is torsion-free, and since M is finitely generated, it is
also finitely generated. By Proposition 24.6, M/Mtor is free. We have the quotient linear
map π : M → M/Mtor, which is surjective, and M/Mtor is free, so by Proposition 24.2, there
is a free module F isomorphic to M/Mtor such that
M = Ker (π) ⊕ F = Mtor ⊕ F.
Since F is isomorphic to M/Mtor, the dimension of F is uniquely determined.
Theorem 24.7 reduces the study of finitely generated module over a PID to the study
of finitely generated torsion modules. This is the path followed by Lang [65] (Chapter III,
section 7).
24.2
Finite Presentations of Modules
Since modules are generally not free, it is natural to look for techniques for dealing with
nonfree modules. The hint is that if M is an A-module and if (ui)i∈I is any set of generators
for M , then we know that there is a surjective homomorphism ϕ : A(I) → M from the free
module A(I) generated by I onto M . Furthermore M is isomorphic to A(I)/Ker (ϕ). Then,
we can pick a set of generators (vj)j∈J for Ker (ϕ), and again there is a surjective map
ψ : A(J) → Ker (ϕ) from the free module A(J) generated by J onto Ker (ϕ). The map ψ can
be viewed a linear map from A(J) to A(I), we have
Im(ψ) = Ker (ϕ),
and ϕ is surjective. Note that M is isomorphic to A(I)/Im(ψ). In such a situation we say
that we have an exact sequence and this is denoted by the diagram
A(J)
ψ
/ A(I) ϕ / M
/ 0.
Definition 24.5. Given an A-module M , a presentation of M is an exact sequence
A(J)
ψ
/ A(I) ϕ / M
/ 0
which means that
1. Im(ψ) = Ker (ϕ).
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
2. ϕ is surjective.
Consequently, M is isomorphic to A(I)/Im(ψ). If I and J are both finite, we say that this is
a finite presentation of M .
Observe that in the case of a finite presentation, I and J are finite, and if |J| = n and
|I| = m, then ψ is a linear map ψ : An → Am, so it is given by some m × n matrix R with
coefficients in A called the presentation matrix of M . Every column Rj of R may thought
of as a relation
aj1e1 + · · · + ajmem = 0
among the generators e1, . . . , em of Am, so we have n relations among these generators. Also
the images of e1, . . . , em in M are generators of M, so we can think of the above relations
as relations among the generators of M . The submodule of Am spanned by the columns of
R is the set of relations of M , and the columns of R are called a complete set of relations
for M . The vectors e1, . . . , em are called a set of generators for M. We may also say that
the generators e1, . . . , em and the relations R1, . . . , Rn (the columns of R) are a (finite)
presentation of the module M .
For example, the Z-module presented by the 1 × 1 matrix R = (5) is the quotient, Z/5Z,
of Z by the submodule 5Z corresponding to the single relation
5e1 = 0.
But Z/5Z has other presentations. For example, if we consider the matrix of relations
2 −1
R =
,
1
2
presenting the module M , then we have the relations
2e1 + e2 = 0
−e1 + 2e2 = 0.
From the first equation, we get e2 = −2e1, and substituting into the second equation we get
−5e1 = 0.
It follows that the generator e2 can be eliminated and M is generated by the single generator
e1 satisfying the relation
5e1 = 0,
which shows that M ≈ Z/5Z.
The above example shows that many different matrices can present the same module.
Here are some useful rules for manipulating a relation matrix without changing the isomor-
phism class of the module M it presents.
24.2. FINITE PRESENTATIONS OF MODULES
673
Proposition 24.8. If R is an m × n matrix presenting an A-module M, then the matrices
S of the form listed below present the same module (a module isomorphic to M ):
(1) S = QRP −1, where Q is a m × m invertible matrix and P a n × n invertible matrix
(both over A).
(2) S is obtained from R by deleting a column of zeros.
(3) The jth column of R is ei, and S is obtained from R by deleting the ith row and the
jth column.
Proof. (1) By definition, we have an isomorphism M ≈ Am/RAn, where we denote by RAn
the image of An by the linear map defined by R. Going from R to QRP −1 corresponds
to making a change of basis in Am and a change of basis in An, and this yields a quotient
module isomorphic to M .
(2