k−1
k
k−j−1
Xk1 − f = (X1 − f)
(X1)jf
,
j=0
24.5. THE TORSION MODULE ASSOCIATED WITH AN ENDOMORPHISM
689
and
k−1
k−j−1
z = (X1 − f)
(X1)jf
(1 ⊗ uk) ,
k
j=0
which shows that z = ψ(y) for some y ∈ E[X].
Finally, we prove that ψ is injective as follows. We have
ψ(z) = ψ
Xk ⊗ uk
k
= (X1 − f)
Xk ⊗ uk
k
=
Xk+1 ⊗ (uk − f(uk+1)),
k
where (uk) is a family of finite support of uk ∈ E. If ψ(z) = 0, then
Xk+1 ⊗ (uk − f(uk+1)) = 0,
k
and because the Xk form a basis of A[X], we must have
uk − f(uk+1) = 0, for all k.
Since (uk) has finite support, there is a largest k, say m + 1 so that um+1 = 0, and then from
uk = f (uk+1),
we deduce that uk = 0 for all k. Therefore, z = 0, and ψ is injective.
Remark: The exact sequence of Theorem 24.21 yields a presentation of Mf .
Since A[X] is a free A-module, A[X]⊗AM is a free A-module, but A[X]⊗AM is generally
not a free A[X]-module. However, if M is a free module, then M [X] is a free A[X]-module,
since if (ui)i∈I is a basis for M, then (1 ⊗ ui)i∈I is a basis for M[X]. This allows us to define
the characterisctic polynomial χf (X) of an endomorphism of a free module M as
χf (X) = det(X1 − f).
Note that to have a correct definition, we need to define the determinant of a linear map
allowing the indeterminate X as a scalar, and this is what the definition of M [X] achieves
(among other things). Theorem 24.21 can be used to quick a short proof of the Cayley-
Hamilton Theorem, see Bourbaki [12] (Chapter III, Section 8, Proposition 20). Proposition
5.10 is still the crucial ingredient of the proof.
We now develop the theory necessary to understand the structure of finitely generated
modules over a PID.
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24.6
Torsion Modules over a PID; The Primary
Decomposition
We begin by considering modules over a product ring obtained from a direct decomposition,
as in Definition 21.3. In this section and the next, we closely follow Bourbaki [13] (Chapter
VII). Let A be a commutative ring and let (b1, . . . , bn) be ideals in A such that there is
an isomorphism A ≈ A/b1 × · · · × A/bn. From Theorem 21.16 part (b), there exist some
elements e1, . . . , en of A such that
e2i = ei
eiej = 0,
i = j
e1 + · · · + en = 1A,
and bi = (1A − ei)A, for i, j = 1, . . . , n.
Given an A-module M with A ≈ A/b1×· · ·×A/bn, let Mi be the subset of M annihilated
by bi; that is,
Mi = {x ∈ M | bx = 0, for all b ∈ bi}.
Because bi is an ideal, each Mi is a submodule of M. Observe that if λ, µ ∈ A, b ∈ bi, and
if λ − µ = b, then for any x ∈ Mi, since bx = 0,
λx = (µ + b)x = µx + bx = µx,
so Mi can be viewed as a A/bi- module.
Proposition 24.22. Given a ring A ≈ A/b1 × · · · × A/bn as above, the A-module M is the
direct sum
M = M1 ⊕ · · · ⊕ Mn,
where Mi is the submodule of M annihilated by bi.
Proof. For i = 1, . . . , n, let pi : M → M be the map given by
pi(x) = eix,
x ∈ M.
The map pi is clearly linear, and because of the properties satisfied by the eis, we have
p2i = pi
pipj = 0,
i = j
p1 + · · · + pn = id.
This shows that the pi are projections, and by Proposition 4.6 (which also holds for modules),
we have a direct sum
M = p1(M) ⊕ · · · ⊕ pn(M) = e1M ⊕ · · · ⊕ enM.
24.6. TORSION MODULES OVER A PID; PRIMARY DECOMPOSITION
691
It remains to show that Mi = eiM. Since (1 − ei)ei = ei − e2i = ei − ei = 0, we see that
eiM is annihilated by bi = (1 − ei)A. Furthermore, for i = j, for any x ∈ M, we have
(1 − ei)ejx = (ej − eiej)x = ejx, so no nonzero element of ejM is annihilated by 1 − ei, and
thus not annihilated by bi. It follows that eiM = Mi, as claimed.
Given an A-module M , for any nonzero α ∈ A, let
M (α) = {x ∈ M | αx = 0},
the submodule of M annihilated by α. If α divides β, then M (α) ⊆ M(β), so we can define
Mα =
M (αn) = {x ∈ M | (∃n ≥ 1)(αnx = 0)},
n≥1
the submodule of M consisting of all elements of M annihilated by some power of α. If N
is any submodule of M , it is clear that
Nα = M ∩ Mα.
Recall that in a PID, an irreducible element is also called a prime element.
Definition 24.8. If A is a PID and p is a prime element in A, we say that a module M is
p-primary if M = Mp.
Proposition 24.23. Let M be module over a PID A. For every nonzero α ∈ A, if
α = upn1
1 · · · pnr
r
is a factorization of α into prime factors (where u is a unit), then the module M (α) anni-
hilated by α is the direct sum
M (α) = M (pn1
1 ) ⊕ · · · ⊕ M (pnr
r ).
Furthermore, the projection from M (α) onto M (pni) is of the form x
i
→ γix, for some γi ∈ A,
and
M (pni) = M (α)
.
i
∩ Mpi
Proof. First, observe that since M (α) is annihilated by α, we can view M (α) as a A/(α)-
module. By the Chinese Remainder Theorem (Theorem 21.15) applied to the ideals (upn1
1 ) =
(pn1
1 ), (pn2
2 ), . . . , (pnr
r ), we have an isomorphism
A/(α) ≈ A/(pn1
1 ) × · · · × A/(pnr
r ).
Since we also have isomorphisms
A/(pni)
)/(α)),
i
≈ (A/(α))/((pni
i
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
we can apply Proposition 24.22, and we get a direct sum
M (α) = N1 ⊕ · · · ⊕ Nr,
where Ni is the A/(α)-submodule of M(α) annihilated by (pni)/(α), and the projections
i
onto the Ni are of the form stated in the proposition. However, Ni is just the A-module
M (pni) annihilated by pni, because every nonzero element of (pni)/(α) is an equivalence class
i
i
i
modulo (α) of the form apni for some nonzero a
i
∈ A, and by definition, x ∈ Ni iff
0 = apni x = apnix,
for all a
i
i
∈ A − {0},
in particular for a = 1, which implies that x ∈ M(pni).
i
The inclusion M (pni)
is clear. Conversely, pick x
, which
i
⊆ M(α) ∩ Mp
∈ M(α) ∩ M
i
pi
means that αx = 0 and psix = 0 for some s ≥ 1. If s < ni, we are done, so assume s ≥ ni.
Since pni is a gcd of α and ps
i
i , by Bezout, we can write
pni = λps
i
i + µα
for some λ, µ ∈ A, and then pnix = λps
), as
i
i x + µαx = 0, which shows that x ∈ M (pni
i
desired.
Recall that if M is a torsion module over a ring A which is an integral domain, then
every finite set of elements x1, . . . , xn in M is annihilated by a = a1 · · · an, where each ai
annihilates xi.
Since A is a PID, we can pick a set P of irreducible elements of A such that every nonzero
nonunit of A has a unique factorization up to a unit. Then, we have the following structure
theorem for torsion modules which holds even for modules that are not finitely generated.
Theorem 24.24. (Primary Decomposition Theorem) Let M be a torsion-module over a
PID. For every irreducible element p ∈ P , let Mp be the submodule of M annihilated by
some power of p. Then, M is the (possibly infinite) direct sum
M =
Mp.
p∈P
Proof. Since M is a torsion-module, for every x ∈ M, there is some α ∈ A such that
x ∈ M(α). By Proposition 24.23, if α = upn1
1 · · · pnr
r
is a factorization of α into prime factors
(where u is a unit), then the module M (α) is the direct sum
M (α) = M (pn1
1 ) ⊕ · · · ⊕ M (pnr
r ).
This means that x can be written as
x =
xp,
xp ∈ Mp,
p∈P
24.6. TORSION MODULES OVER A PID; PRIMARY DECOMPOSITION
693
with only finitely many xp nonzero. If
xp =
yp
p∈P
p∈P
for all p ∈ P , with only finitely many xp and yp nonzero, then xp and yp are annihilated by
some common nonzero element a ∈ A, so xp, yp ∈ M(a). By Proposition 24.23, we must
have xp = yp for all p, which proves that we have a direct sum.
It is clear that if p and p are two irreducible elements such that p = up for some unit u,
then Mp = Mp . Therefore, Mp only depends on the ideal (p).
Definition 24.9. Given a torsion-module M over a PID, the modules Mp associated with
irreducible elements in P are called the p-primary components of M .
The p-primary components of a torsion module uniquely determine the module, as shown
by the next proposition.
Proposition 24.25. Two torsion modules M and N over a PID are isomorphic iff for
every every irreducible element p ∈ P , the p-primary components Mp and Np of M and N
are isomorphic.
Proof. Let f : M → N be an isomorphism. For any p ∈ P , we have x ∈ Mp iff pkx = 0 for
some k ≥ 1, so
0 = f (pkx) = pkf (x),
which shows that f (x) ∈ Np. Therefore, f restricts to a linear map f | Mp from Mp to
Np. Since f is an isomorphism, we also have a linear map f −1 : M → N, and our previous
reasoning shows that f −1 restricts to a linear map f −1 | Np from Np to Mp. But, f | Mp and
f −1 | Np are mutual inverses, so Mp and Np are isomorphic.
Conversely, if Mp ≈ Np for all p ∈ P , by Theorem 24.24, we get an isomorphism between
M =
M
N
p∈P
p and N =
p∈P
p.
In view of Proposition 24.25, the direct sum of Theorem 24.24 in terms of its p-primary
components is called the canonical primary decomposition of M .
If M is a finitely generated torsion-module, then Theorem 24.24 takes the following form.
Theorem 24.26. (Primary Decomposition Theorem for finitely generated torsion modules)
Let M be a finitely generated torsion-module over a PID A. If Ann(M ) = (a) and if a =
upn1
1 · · · pnr
r
is a factorization of a into prime factors, then M is the finite direct sum
r
M =
M (pni).
i
i=1
Furthermore, the projection of M over M (pni) is of the form x
i
→ γix, for some γi ∈ A.
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
Proof. This is an immediate consequence of Proposition 24.23.
In particular, Theorem 24.26 applies when A = Z. In this case, M is a finitely generated
torsion abelian group, and the theorem says that such a group is the direct sum of a finite
number of groups whose elements have order some power of a prime number p.
Theorem 24.24 has several useful corollaries.
Proposition 24.27. If M is a torsion module over a PID, for every submodule N of M ,
we have a direct sum
N =
N ∩ Mp.
p∈P
Proof. It is easily verified that N ∩ Mp is the p-primary component of N.
Proposition 24.28. If M is a torsion module over a PID, a submodule N of M is a direct
factor of M iff Np is a direct factor of Mp for every irreducible element p ∈ A.
Proof. This is because if N and N are two submodules of M , we have M = N ⊕ N iff, by
Proposition 24.27, Mp = Np ⊕ Np for every irreducible elements p ∈ A.
An A-module M is said to be semi-simple iff for every submodule N of M , there is some
submodule N of M such that M = N ⊕ N .
Proposition 24.29. Let A be a PID which is not a field, and let M be any A-module. Then,
M is semi-simple iff it is a torsion module and if Mp = M(p) for every irreducible element
p ∈ A (in other words, if x ∈ M is annihilated by a power of p, then it is already annihilated
by p).
Proof. Assume that M is semi-simple. Let x ∈ M and pick any irreducible element p ∈ A.
Then, the submodule pAx has a supplement N such that
M = pAx ⊕ N,
so we can write x = pax + y, for some y ∈ N and some a ∈ A. But then,
y = (1 − pa)x,
and since p is irreducible, p is not a unit, so 1 − pa = 0. Observe that
p(1 − ap)x = py ∈ pAx ∩ N = (0).
Since p(1 − ap) = 0, x is a torsion element, and thus M is a torsion module. The above
argument shows that
p(1 − ap)x = 0,
24.7. FINITELY GENERATED MODULES OVER A PID
695
which implies that px = ap2x, and by induction,
px = anpn+1x,
for all n ≥ 1.
If we pick x in Mp, then there is some m ≥ 1 such that pmx = 0, and we conclude that
px = 0.
Therefore, Mp = M(p), as claimed.
Conversely, assume that M is a torsion-module and that Mp = M(p) for every irreducible
element p ∈ A. By Proposition 24.28, it is sufficient to prove that a module annihilated by a
an irreducible element is semi-simple. This is because such a module is a vector space over
the field A/(p) (recall that in a PID, an ideal (p) is maximal iff p is irreducible), and in a
vector space, every subspace has a supplement.
Theorem 24.26 shows that a finitely generated torsion module is a direct sum of p-primary
modules Mp. We can do better. In the next section, we show that each primary module Mp
is the direct sum of cyclic modules of the form A/(pn).
24.7
Finitely Generated Modules over a PID; Invariant
Factor Decomposition
There are several ways of obtaining the decomposition of a finitely generated module as a
direct sum of cyclic modules. One way to proceed is to first use the Primary Decomposition
Theorem and then to show how each primary module Mp is the direct sum of cyclic modules of
the form A/(pn). This is the approach followed by Lang [65] (Chapter III, section 7), among
others. We prefer to use a proposition that produces a particular basis for a submodule of
a finitely generated free module, because it yields more information. This is the approach
followed in Dummitt and Foote [30] (Chapter 12) and Bourbaki [13] (Chapter VII). The
proof that we present is due to Pierre Samuel.
Proposition 24.30. Let F be a finitely generated free module over a PID A, and let M be
any submodule of F . Then, M is a free module and there is a basis (e1, ..., en) of F , some
q ≤ n, and some nonzero elements a1, . . . , aq ∈ A, such that (a1e1, . . . , aqeq) is a basis of M
and ai divides ai+1 for all i, with 1 ≤ i ≤ q − 1.
Proof. The proposition is trivial when M = {0}, thus assume that M is nontrivial. Pick some
basis (u1, . . . , un) for F . Let L(F, A) be the set of linear forms on F . For any f ∈ L(F, A),
it is immediately verified that f (M ) is an ideal in A. Thus, f (M ) = ahA, for some ah ∈ A,
since every ideal in A is a principal ideal. Since A is a PID, any nonempty family of ideals
in A has a maximal element, so let f be a linear map such that ahA is a maximal ideal in A.
Let πi : F → A be the i-th projection, i.e., πi is defined such that πi(x1u1 + · · · + xnun) = xi.
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
It is clear that πi is a linear map, and since M is nontrivial, one of the πi(M) is nontrivial,
and ah = 0. There is some e ∈ M such that f(e ) = ah.
We claim that, for every g ∈ L(F, A), the element ah ∈ A divides g(e ).
Indeed, if d is the gcd of ah and g(e ), by the Bézout identity, we can write
d = rah + sg(e ),
for some r, s ∈ A, and thus
d = rf (e ) + sg(e ) = (rf + sg)(e ).
However, rf + sg ∈ L(F, A), and thus,
ahA ⊆ dA ⊆ (rf + sg)(M),
since d divides ah, and by maximality of ahA, we must have ahA = dA, which implies that
d = ah, and thus, ah divides g(e ). In particular, ah divides each πi(e ) and let πi(e ) = ahbi,
with bi ∈ A.
Let e = b1u1 + · · · + bnun. Note that
e = π1(e )u1 + · · · + πn(e )un = ahb1u1 + · · · + ahbnun,
and thus, e = ahe. Since ah = f (e ) = f (ahe) = ahf (e), and since ah = 0, we must have
f (e) = 1.
Next, we claim that
F = Ae ⊕ f−1(0)
and
M = Ae ⊕ (M ∩ f−1(0)),
with e = ahe.
Indeed, every x ∈ F can be written as
x = f (x)e + (x − f(x)e),
and since f (e) = 1, we have f (x − f(x)e) = f(x) − f(x)f(e) = f(x) − f(x) = 0. Thus,
F = Ae + f −1(0). Similarly, for any x ∈ M, we have f(x) = rah, for some r ∈ A, and thus,
x = f (x)e + (x − f(x)e) = rahe + (x − f(x)e) = re + (x − f(x)e),
we still have x − f(x)e ∈ f−1(0), and clearly, x − f(x)e = x − rahe = x − re ∈ M, since
e ∈ M. Thus, M = Ae + (M ∩ f−1(0)).
To prove that we have a direct sum, it is enough to prove that Ae ∩ f−1(0) = {0}. For
any x = re ∈ Ae, if f(x) = 0, then f(re) = rf(e) = r = 0, since f(e) = 1 and, thus, x = 0.
Therefore, the sums are direct sums.
24.7. FINITELY GENERATED MODULES OVER A PID
697
We can now prove that M is a free module by induction on the size, q, of a maximal
linearly independent family for M .
If q = 0, the result is trivial. Otherwise, since
M = Ae ⊕ (M ∩ f−1(0)),
it is clear that M ∩ f−1(0) is a submodule of F and that every maximal linearly independent
family in M ∩ f−1(0) has at most q − 1 elements. By the induction hypothesis, M ∩ f−1(0)
is a free module, and by adding e to a basis of M ∩ f−1(0), we obtain a basis for M, since
the sum is direct.
The second part is shown by induction on the dimension n of F .
The case n = 0 is trivial. Otherwise, since
F = Ae ⊕ f−1(0),
and since, by the previous argument, f −1(0) is also free, f −1(0) has dimension n − 1. By
the induction hypothesis applied to its submodule M ∩ f−1(0), there is a basis (e2, . . . , en)
of f −1(0), some q ≤ n, and some nonzero elements a2, . . . , aq ∈ A, such that, (a2e2, . . . , aqeq)
is a basis of M ∩ f−1(0), and ai divides ai+1 for all i, with 2 ≤ i ≤ q − 1. Let e1 = e, and
a1 = ah, as above. It is clear that (e1, . . . , en) is a basis of F , and that that (a1e1, . . . , aqeq)
is a basis of M , since the sums are direct, and e = a1e1 = ahe. It remains to show that a1
divides a2. Consider the linear map g : F → A such that g(e1) = g(e2) = 1, and g(ei) = 0,
for all i, with 3 ≤ i ≤ n. We have ah = a1 = g(a1e1) = g(e ) ∈ g(M), and thus ahA ⊆ g(M).
Since ahA is maximal, we must have g(M) = ahA = a1A. Since a2 = g(a2e2) ∈ g(M), we
have a2 ∈ a1A, which shows that a1 divides a2.
We need the following basic proposition.
Proposition 24.31. For any commutative ring A, if F is a free A-module and if (e1, . . . , en)
is a basis of F , for any elements a1, . . . , an ∈ A, there is an isomorphism
F/(Aa1e1 ⊕ · · · ⊕ Aanen) ≈ (A/a1A) ⊕ · · · ⊕ (A/anA).
Proof. Let σ : F → A/(a1A) ⊕ · · · ⊕ A/(anA) be the linear map given by
σ(x1e1 + · · · + xnen) = (x1, . . . , xn),
where xi is the equivalence class of xi in A/aiA. The map σ is clearly surjective, and its
kernel consists of all vectors x1e1 + · · · + xnen such that xi ∈ aiA, for i = 1, . . . , n, which
means that
Ker (σ) = Aa1e1 ⊕ · · · ⊕ Aanen.
Since M/Ker (σ) is isomorphic to Im(σ), we get the desired isomorphism.
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CHAPTER 24. INTRODUCTION TO MODULES; MODULES OVER A PID
We can now prove the existence part of the structure theorem for finitely generated
modules over a PID.
Theorem 24.32. Let M be a finitely generated nontrivial A-module, where A a PID. Then,
M is isomorphic to a direct sum of cyclic modules
M ≈ A/a1 ⊕ · · · ⊕ A/am,
where the ai are proper ideals of A (possibly zero) such that
a1 ⊆ a2 ⊆ · · · ⊆ am = A.
More precisely, if a1 = · · · = ar = (0) and (0) = ar+1 ⊆ · · · ⊆ am = A, then
M ≈ Ar ⊕ (A/ar+1 ⊕ · · · ⊕ A/am),
where A/ar+1 ⊕ · · · ⊕ A/am is the torsion submodule of M. The module M is free iff r = m,
and a torsion-module iff r = 0. In the latter case, the annihilator of M is a1.
Proof. Since M is finitely generated and nontrivial, there is a surjective homomorphism
ϕ : An → M for some n ≥ 1, and M is isomorphic to An/Ker (ϕ). Since Ker (ϕ) is a submod-
ule of the free module An, by Proposition 24.30, Ker (ϕ) is a free module and there is a basis
(e1, . . . , en) of An and some nonzero elements a1, . . . , aq (q ≤ n) such that (a1e1, . . . , aqeq) is
a basis of Ker (ϕ) and a1 | a2 | · · · | aq. Let aq+1 = . . . = an = 0.
By Proposition 24.31, we have an isomorphism
An/Ker (ϕ) ≈ A/a1A ⊕ · · · ⊕ A/anA.
Whenever ai is unit, the factor A/aiA = (0), so we can weed out the units. Let r = n − q,
and let s ∈ N be the smallest index such that as+1 is not a unit. Note that s = 0 means that
there are no units. Also, as M = (0), s < n. Then,
M ≈ An/Ker (ϕ) ≈ A/as+1A ⊕ · · · ⊕ A/anA.
Let m = r + q − s = n − s. Then, we have the sequence
as+1, . . . , aq, aq+1, . . . , an,
q−s
r=n−q
where as+1 | as+2 | · · · | aq are nonzero and nonunits and aq+1 = · · · = an = 0, so we define
the m ideals ai as follows:
(0)
if 1 ≤ i ≤ r
ai =
ar+q+1−iA if r + 1 ≤ i ≤ m.
With these definitions, the ideals ai are proper ideals and we have
ai ⊆ ai+1,
i = 1, . . . , m − 1.
When r = 0, since as+1 | as+2 | · · · | an, it is clear that a1 = anA is the annihilator of M.
The other statements of the theorem are clear.
24.7. FINITELY GENERATED MODULES OVER A PID
699
The natural number r is called the free rank or Betti number of the module M . The
generators α1, . . . , αm of the ideals a1, . . . , am (defined up to a unit) are often called the
invariant factors of M (in the notation of Theorem 24.32, the generators of the ideals
a1, . . . , am are denoted by aq, . . . , as+1, s ≤ q).
As corollaries of Theorem 24.32, we obtain again the following facts established in Section
24.1:
1. A finitely generated module over a PID is the direct sum of its torsion module and a
free module.
2. A finitely generated torsion-free module over a PID is free.
It turns out that the ideals a1 ⊆ a2 ⊆ · · · ⊆ am = A are uniquely determined by the
module M . Uniqueness proofs found in most books tend to be intricate and not very intuitive.
The shortest proof that we are aware of is from Bourbaki [13] (Chapter VII, Section 4), and
uses wedge products.
The following preliminary results are needed.
Proposition 24.33. If A is a commutative ring and if a1, . . . , am are ideals of A, then there
is an isomorphism
A/a1 ⊗ · · · ⊗ A/am ≈ A/(a1 + · · · + am).
Sketch of proof. We proceed by induction on m. For m = 2, we define the map
ϕ : A/a1 × A/a2 → A/(a1 + a2) by
ϕ(a, b) = ab (mod a1 + a2).
It is well-defined because if a = a + a1 and b = b + a2 with a1 ∈ a1 and a2 ∈ a2, then
a b = (a + a1)(b + a2) = ab + ba1 + aa2 + a1a2,
and so
a b ≡ ab (mod a1 + a2).
It is also clear that this map is bilinear, so it induces a linear map ϕ : A/a1 ⊗ A/a2 →