That is, eventually p^{k}m = 0. Right now, it might be possible that k could change for different
m ∈ M. Note that if p is invertible, then M_{p} = 0 so nothing is lost from considering only
non-invertible primes. It is obvious that M_{p} is a subgroup of the module M and is itself a
module.
Proposition 9.3.1Let p_{1},
⋅⋅⋅
,p_{n}be distinct primes (not associates), none of which is invertible. Let Mbe a module (torsion or not) over a p.i.d. D. Then
( )
Mp1 + ⋅⋅⋅+ Mpj −1 + Mpj+1 + ⋅⋅⋅+ Mpn ∩Mpj = 0
Proof:This follows from the observation that ∏_{i≠j}p_{i}^{ki} and p_{j}^{kj} are relatively prime. If q divides the
second, then by Proposition 9.1.12 it is of the form ξp_{j}^{mj},m_{j}≤ k_{j} where ξ is an invertible. If q divides
the first, then by the same proposition, it is ζ∏_{i≠j}p_{i}^{mi},m_{i}≤ k_{i} for an invertible ζ. Since
the primes are distinct, we must have all m_{j},m_{i} equal to 0 and q is some invertible ξ and so
these two, ∏_{i≠j}p_{i}^{ki} and p_{j}^{kj} are relatively prime as claimed. Therefore, there exist σ,τ such
that
∏ k k
1 = σ pii + τpjj (*)
i⁄=j
(*)
If m ∈
( )
Mp1 + ⋅⋅⋅+ Mpj−1 + Mpj+1 + ⋅⋅⋅+ Mpn
∩M_{pj}, then m = ∑_{i≠j}m_{i} and so there exist k_{i},k_{j} such
that p_{i}^{ki}m_{i} = 0 and p_{j}^{kj}m = 0. Then do both sides of ∗ to m.
( ) ( =m )
m = ( σ∏ pki) ( ∑ m ) +τ pkjm = 0
i⁄=j i i⁄=j i j
This yields m = 0 and verifies the conclusion of the proposition. ■
It follows from Proposition 9.3.1 that if m_{i}∈ M_{pi}, and if ∑_{i}m_{i} = 0, then each m_{i} = 0 because
m_{j}∈ M_{pj}∩
( )
Mp1 + ⋅⋅⋅+Mpj −1 + Mpj+1 + ⋅⋅⋅+ Mpn
. The converse is also true. If whenever
∑_{i}m_{i} = 0,m_{i}∈ M_{pi} then each m_{i} = 0, then if m ∈ M_{pj}∩
( )
Mp1 + ⋅⋅⋅+Mpj −1 + Mpj+1 + ⋅⋅⋅+ Mpn
,
then m = ∑_{j≠i}m_{i},m_{i}∈ M_{pi} and so 0 = −m + ∑_{j≠i}m_{i} which implies m = 0 and incidentally, each
m_{i} = 0.
Now suppose that M is a finitely generated torsion module for a p.i.d. Then the following lemma will be
the basis for a decomposition theorem in terms of a direct sum of some M_{pi} for p_{i} a non-invertible
prime.
Lemma 9.3.2Let M be a non zero finitely generated (M = Dz_{1} +
⋅⋅⋅
+Dz_{s}) torsion module overa p.i.d. D. Then there exist primes
{p ,⋅⋅⋅,p }
1 n
and corresponding positive integers k_{i}such thatann
(M )
=
(∏n ki)
i=1pi
.
Proof:By assumption, M = Dz_{1} +
⋅⋅⋅
+ Dz_{s}. Since M is a torsion module, ann
(zi)
is
nonempty, equal to
(αi)
. Say α_{i} = ∏_{j=1}^{mi}p_{ij}^{kj} where the p_{ij} are non-invertible primes. Then
define
∏n ki
σ ≡ pi
i=1
as the least common multiple of the α_{i}. Let I ≡
( )
∏n pki
i=1 i
. Then I is an ideal and if α ∈ I, then αz_{i} = 0
for each i. Hence αm = 0 for every m ∈ M. Thus I ⊆ann
(M )
, the set of all β such that βm = 0 for all m.
Conversely, suppose β ∈ann
(M )
. Then in particular βz_{i} = 0 and so α_{i}∕β because β ∈ann
(zi)
. Thus β is
a multiple of all of the α_{i} and it follows that the least common multiple σ must divide β. Hence β ∈
(σ)
.
■
Next is a lemma about direct sums.
Lemma 9.3.3Let m ∈ M a module over a p.i.d. D, and let αm = 0, for α = ∏_{i=1}^{n}p_{i}^{ki}where the p_{i}arenon-invertible distinct primes. Let
{ }
Mˆpi ≡ m ∈ M : pkii m = 0 ⊆ Mpi,
(Note that here k_{i}is fixed. ) Then
Dm ⊆ ˆMp ⊕ ⋅⋅⋅⊕ Mˆp ⊆ Mp ⊕ ⋅⋅⋅⊕ Mp
1 n 1 n
Proof: If n = 1, then this is clearly true. Suppose it is true for n − 1 ≥ 1. Say αm = 0
where
∏n ki
α = pi
i=1
Let M^{′}≡
{ ∏n −1 ki }
k ∈ M : i=1 pi k = 0
. Then this is a module contained in M. Also p_{n}^{kn}m ∈ M^{′}. Therefore,
by induction,
kn ˆ ′ ˆ′
Dpn m ⊆ Mp1 ⊕ ⋅⋅⋅⊕ M pn−1 (*)
(*)
where
{ }
Mˆ ′≡ a ∈ M ′ : pkia = 0
pi i
Now there are σ,τ such that
kn n∏−1 ki
1 = σpn +τ pi (**)
i=1
(**)
Therefore, if k ∈ M^{′}, do both sides of the above to k to conclude that k = σp_{n}^{kn}k. Then from
∗,
k from ∗n∑−1 ′
pnnm = ki, ki ∈ Mˆpi ⊆ ˆMpi
i=1
Now also each k_{i}∈ M^{′} and so doing both sides of ∗∗ to k_{i} gives k_{i} = σp_{n}^{kn}k_{i}. It follows that the sum on
the right equals
Theorem 9.3.4Let M be a non zero torsion module for a p.i.d. D and suppose that
M = Dz1 + ⋅⋅⋅+ Dzp
so that it is finitely generated. Then there exist primes
{pi}
_{i=1}^{n}distinct in the sense that none is aninvertible multiple of another such that
M = Mp1 ⊕ ⋅⋅⋅⊕ Mpn
For
(β)
= ann
(M )
,β = ∏_{i=1}^{n}p_{i}^{ki}, a product of non-invertible primes, it follows
{ } { }
Mˆpi ≡ m ∈ M : pkiim = 0 = m ∈ M : pkim = 0 for some k = Mpi
which is not dependent on the spanning set.
Proof: Let
(β)
= ann
(M )
. It exists by Lemma 9.3.2 above. Thus β is not invertible. Also if
(α )
= ann
(M )
, one needs α∕β and β∕α so the two are associates meaning that one is an invertible times
the other. Let β = ∏_{i=1}^{n}p_{i}^{ki} where this is the prime factorization of β in terms of non-invertible primes.
Note this is well defined up to associates. Let