Claim:We can consider p^{k}M∕p^{k+1}M as a vector space over
ˆD
as follows.
( k+1 ) k+1
(α + (p)) m + p M ≡ αm + p M
Proof of claim:This operation needs to be well defined and then it will be obvious that indeed this is a
vector space. Suppose then that α +
(p)
=
ˆα
+
(p)
and m + p^{k+1}M =
mˆ
+ p^{k+1}M. Is it the case that
αm −
αˆ
ˆm
∈ p^{k+1}M? The difference equals
α (m − ˆm) +(α − ˆα) ˆm
The first term is in p^{k+1}M because it is given that m −
ˆm
is in p^{k+1}M. Consider the second term. It is
given that
mˆ
= p^{k}
mˆ
_{1} because it is in p^{k}M. Also, it is given that α−
ˆα
is a multiple of p and so indeed the
second term is also in p^{k+1}M. This proves the claim.
If k ≥ l_{s} = q, then p^{k}M = 0 and so for such k,p^{k}M∕p^{k+1}M = 0. The other case is that k < l_{s}. Say
k ∈ [l_{j},l_{j+1}). Then
k k k k
p M = Dp vj+1 + Dp vj+2 + ⋅⋅⋅+ Dp vs.
This is because if k ≥ l_{j}, then for i ≤ j,p^{k}v_{i} = 0 and so all that survives is what is in the above sum.
Is
{ }
pkvj+1 + pk+1M, ⋅⋅⋅,pkvs + pk+1M
linearly independent over
ˆD
? Suppose
∑s ( k k+1 )
(αr + (p)) p vr + p M = 0
r=j+1
Then
∑s k k+1
αrp vr + p M = 0
r=j+1
This requires
∑s
αrpkvr = pk+1m
r=j+1
for some m ∈ M. However, recall that M is a direct sum and so
s s
∑ α pkv = ∑ pk+1β v
r=j+1 r r r=j+1 r r
Thus, since M is a direct sum again, for each r,
k k+1
p αr = p βr
and so α_{r} is a multiple of p. Hence α_{r} +
(p)
= 0 for each r and this is indeed a basis for p^{k}M∕p^{k+1}M over
Dˆ
. It follows that the dimension of p^{k}M∕p^{k+1}M over
ˆD
is s−j where k ∈ [l_{j},l_{j+1}), the number of v_{j+1} for
l_{j+1}> k.
Now suppose that
M = Dw1 ⊕ ⋅⋅⋅⊕ Dwt
such that ann
(Dw1 )
⊇ann
(Dw2 )
⊇
⋅⋅⋅
⊇ann
(Dwt )
and ann
(wj )
=
(pmj)
. The question is whether
s = t and m_{j} = l_{j}. It was just shown that for k a positive integer, the dimension of p^{k}M∕p^{k+1}M is the
number of v_{j} for l_{j}> k. Similarly it is the number of w_{j} for m_{j}> k and this must be the same number. In
other words, for each k there are the same number of m_{j} larger than k as there are l_{j} larger than k.
Thus the l_{j} coincide with the m_{j} and there are the same number of them. Hence s = t and
m_{j} = l_{j}.
This last assertion deserves a little more explanation. The smallest l_{j} and k_{j} can be is 1. Otherwise, you
would have ann
(w1)
or ann
(v1)
=
( )
p0
=
(1)
= D. Thus for every α ∈ D,αw_{1} = 0. In particular, this
would hold if α = 1 and so w_{1} = 0 contrary to assumption that none of the w_{i},v_{i} equal to 0. (Obviously
you can’t get uniqueness if you allow some v_{j} to equal 0 because you can string together as many D0
as you like.) Therefore, you could consider M∕pM such that k = 0 and there are the same
number of v_{i} and w_{j} since each l_{i},m_{j} is larger than 0. Thus s = t. Then a contradiction is
also obtained if some m_{i}≠l_{i}. You just consider the first such i and let k be the smaller of the
two.
Theorem 9.6.1Suppose M is a finitely generated torsion module for a p.i.d. andann
(M )
=
q
(p )
wherep is a prime. Then
M = Dv1 ⊕ ⋅⋅⋅⊕ Dvs, no vj = 0
Lettingann
(v )
j
=
(ν)
, it follows that ν = p^{lj}for some l_{j}≤ q. If the direct summands are listed in theorder that the l_{i}are increasing (or decreasing), then s is independent of the choice of the v_{j}and any othersuch cyclic direct sum for M will have the same sequence of l_{j}.
These considerations about uniqueness yield the following more refined version of Theorem
9.5.9.
Theorem 9.6.2Let 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, no Mpi equal to 0
For
(β)
= ann
(M )
,β = ∏_{i=1}^{n}p_{i}^{ki}, a product of non-invertible primes, it follows
{ ki }
Mˆpi ≡ m ∈ M : pi m = 0
= {m ∈ M : pkm = 0 for some k} = Mp
i
which is not dependent on the spanning set. Also,
M = Da ⊕ ⋅⋅⋅⊕ Da ,m ≤ k
pj j1 jmj j j
whereann
(ajr)
= p^{lr}, such that l_{r}≤ k_{j}and we can have the order arranged such thatl_{1}≤ l_{2}≤
⋅⋅⋅
≤ l_{mj} = k_{j}. The numbers
l1 ≤ l2 ≤ ⋅⋅⋅ ≤ lmj = kj
are the same for each cyclic decomposition of M_{pj}. That is, they do not depend on the particulardecomposition chosen.
Of course we would also have uniqueness if we adopted the convention that the l_{i} should be arranged in
decreasing order.
All of the above is included in [20] and [6] where, in addition to the above, there are some different
presentations along with much more on modules and rings.