Definition 9.5.1Let M be a module. It is called a Noetherian moduleif whenever there is an increasingsequence of sub modules
{Nk }
,
⋅⋅⋅
N_{k}⊆ N_{k+1}
⋅⋅⋅
, it follows that for large enough k all thesubmodules are equal. (Ascending chains of submodules are eventually constant.) A mapping θfrom a R module M to another R module N is a morphismif it preserves operations. Thatis,
θ(m1 + m2) = θm1 + θm2, θ(σm ) = σθ (m )
It is just like “linear” except that here the scalars come from a ring.
When is a module M Noetherian? A sufficient condition is for it to be a finitely generated module over
a p.i.d. This is explained next.
Lemma 9.5.2Suppose
0 → A θ→ B →η C → 0
is a short exact sequence of modules for a p.i.d. D. Recall this means that these mappings are morphisms,η ∘ θ = 0,θ is one to one,η is onto, and ker
(η)
= θ
(A)
. Then if A,C are Noetherian, so isB.
Proof: Suppose B_{n} is an ascending chain of sub modules in B. Then η
(Bn)
is an ascending chain
of sub modules in C and so it is eventually constant. Also θ^{−1}B_{n} is an ascending chain of
sub modules in A. These are sub modules because they are clearly additive subgroups and if
α ∈ D,αθ^{−1}
(Bn)
= θ^{−1}
(αBn )
= θ^{−1}
(Bn )
. Note that θ^{−1}B_{n} cannot be empty because it must contain 0.
Thus there exists m such that if n ≥ m,θ^{−1}
(Bn)
= θ^{−1}
(Bm)
and η
(Bn )
= η
(Bm )
. Let b ∈ B_{n}. It must be
shown that b ∈ B_{m}. Since η
(Bn)
= η
(Bm)
, it follows that there is
ˆb
∈ B_{m} such that ηb = η
ˆb
so
( )
b− ˆb
∈ ker
(η)
= θ
(A)
and so θ^{−1}
( )
b− ˆb
∈ θ^{−1}B_{n} = θ^{−1}B_{m}. Hence b −
ˆb
= c ∈ B_{m}. Then
b =
ˆb
+ c ∈ B_{m} showing that B_{n} = B_{m} for n ≥ m. ■
Now observe the following.
Lemma 9.5.3Suppose A,B are modules and τ : A → B is a morphism. Then if C is a sub moduleof B, it follows that τ^{−1}
(C)
is a submodule of A which contains ker
(τ )
.
Proof: If α ∈ D,τ
(ατ−1 (C ))
= ατ
(τ− 1(C))
= αC = C and so ατ^{−1}
(C)
∈ τ^{−1}
(C)
. Similarly, if
a,b ∈ τ^{−1}
(C )
, then τ
(a)
+ τ
(b)
= τ
(a+ b)
∈ C and so a + b ∈ τ^{−1}
(C)
. It is similarly the case that if
a ∈ τ^{−1}
(C)
, then so is −a. Thus τ^{−1}
(C)
is a sub module of A. If b ∈ ker
(τ)
, then τ
(b)
= 0 ∈ C and so,
b ∈ τ^{−1}
(C)
. ■
Now note that a p.i.d. D is a Noetherian module for D. This is because it has the ascending chain
condition on ideals which are just the modules for D. Consider
0 → Dn−1 θ→ Dn →η D → 0
where θ
(b)
≡
(b,0)
,η
(c)
≡ c_{n} where c =
(c1,⋅⋅⋅,cn)
. Then this is a short exact sequence. It is clear that θ is
one to one and that η is onto. Also η
(θ(b))
= η
((b,0))
= 0 and ker
(η)
=
{ }
(b,0) : b ∈ Dn −1
= θ
( )
Dn− 1
.
Now consider the above lemma for n = 2 to conclude D^{2} is Noetherian and then for n = 3 to conclude
that D^{3} is Noetherian and so forth. Thus D^{n} is Noetherian.
Proposition 9.5.4Let A be a finitely generated module for D a p.i.d. Then A is Noetherian.
Proof:Let A = span
(a ,⋅⋅⋅,a )
1 n
. Let τ : D^{n}→ A be given by τ
(x)
≡∑_{i}x_{i}a_{i}. Then this is clearly a
morphism. Let A_{k} be an ascending chain of sub modules of A. Then τ^{−1}
(A )
k
is an ascending chain of
submodules of D^{n} which contain ker
(τ)
. It was shown above that D^{n} is Noetherian. Hence there is an m
such that if k > m, τ^{−1}
(A )
k
= τ^{−1}
(A )
m
and so A_{k} = A_{m} for all k > m showing that A is Noetherian.
■
Recall the elementary matrices which involved doing a row operation to the identity matrix. The
elementary matrices which involve switching two rows or adding a multiple of one row to another result in
elementary matrices which are invertible even if the entries come from a commutative ring
with unity. See Problem 26. Similarly, these two column operations may be accomplished by
multiplying on the right by an elementary matrix which involves adding a multiple of a column
to another column or switching two columns. See Problem 40 on Page 257. Similarly, such
elementary matrices are invertible even if the entries of the matrix come from a commutative
ring.
Theorem 9.5.5Let A be an m×n matrixwhose entries are from a Euclidean integral domain D. Thus ifα,β ∈ D, there exists κ such that
α = κβ + ρ, δ(ρ) < δ(β) or else ρ = 0
where δ is the nonnegative integer valued function in the definition of the Euclidean ring. Then there areinvertible matrices P,Q of the right size such that
PAQ = B
where B is a diagonal matrix. This can be done in such a way that B_{kk}∕B_{}
(k+1)
(k+1)
as long as theseentries are nonzero.
Proof:If A = 0 there is nothing to show. Just let P,Q be appropriate identity matrices. Assume then
that A≠0. Begin with P and Q appropriate sized identity matrices. Let δ
(Aij)
be the smallest
of all entries of A which are not zero. Now choosing a switch of columns and rows, we can
modify P,Q such that B_{11} = A_{ij}. Consider B_{i1}, the first entry in the i^{th} row. By the Euclidean
algorithm,
Bi1 = B11q + r, δ(r) < δ(B11)
or else r = 0. If it is zero, stop. If not, take −q times the first row of B and add to the i^{th} row to place a r_{i1}
in the i1 position in place of B_{i1}. This involves adjusting P to get this new B. Now out of all entries of the
new matrix B the B_{rs} which has δ
(Brs)
the smallest is in the i^{th} row and δ
(Bis)
≤ δ
(r)
. Switch
rows and columns till B_{is} is in the 11 position. Now repeat the argument just given, replacing
the first entry of the i^{th} row with a remainder r^{′} where it is either zero or δ
′
(r )
< δ
(Bis)
.
Continuing in this way, eventually the remainder r must be zero because the process yields a strictly
decreasing sequence of nonnegative integers. Now do a similar process to the other rows of the
resulting matrix. When this is done, do the same thing using column operations to eventually
obtain
( T )
P AQ = B11 0 (*)
0 Bˆ
(*)
If
Bˆ
= 0 we are done. If not, do the same thing working with the rows of
( )
0 ˆB
and then the columns
of
( )
0T
Bˆ
to obtain
( B 0 )
| 11 |
P AQ = ( B22 )
0 ˆB
where
ˆB
is now
(m − 2)
×
(n − 2)
. Eventually, the result is a diagonal matrix.
What of the last claim that we can get B_{kk}∕B_{}
(k+1)
(k+1)
? This just involves a little more care. When
you get to ∗, if any of the entries of
Bˆ
are not a multiple of B_{11}, say B_{ij}, then take the column containing
this entry and add to the first column. This yields
( )
B11 0T
∗ ˆB
with the offending B_{ij} the first entry of the i^{th} row. Then do a row operation to obtain r in this position
with 0 < δ
(r)
< δ
(B11)
You can’t have 0 = δ
(r)
because it is given that B_{ij} is not a multiple of B_{11}. Now
switch to obtain
( )
r ∗
∗ ˜B
Now you repeat the first part of the argument till you obtain
( r 0T )
˜ , new B˜
0 B
If any entry of
˜B
is not a multiple of r, then you repeat what was just done. This process must stop
because you have a sequence δ
(r1)
> δ
(r2)
,
⋅⋅⋅
which are each positive integers with r_{i} in the upper left
corner. Thus eventually, every entry of
˜B
must be a multiple of r in the above. Now you do the first part of
the argument to obtain
( )
r1 0
|( r2 |)
0 ˆB
where r_{1}∕r_{2}. The various linear combinations of entries which are multiples of r_{1} will yield multiples of r_{1}
and so it will indeed be as shown above. Then you repeat what was just considered if
ˆB
contains an entry
which is not a multiple of r_{2}. All are multiples of r_{1}. Eventually, you obtain, the above in which every entry
of
Bˆ
is a multiple of r_{2} which is a multiple of r_{1}. Eventually you obtain that B is a diagonal matrix and
B_{11}∕B_{22}∕B_{33}
⋅⋅⋅
. ■
The above is the main result, but in fact you can generalize it to an arbitrary p.i.d. in place of a
Euclidean domain. See Jacobsen [20]. I will sketch the proof below. It is similar to the above but you have
to include a slightly harder argument.
Corollary 9.5.6The above conclusion holds for a general p.i.d. in place of a Euclidean domain.
Proof: If A is 1 × 1, then there is nothing to show. So always it can be assumed that either m or n is
larger than 1. The proof is based on the following two claims.
Claim 1:Suppose a≠0. Then there exists an invertible 2 × 2 matrix R such that
( ) ( )
R a ∗ = d ∗
b ∗ 0 ∗
where d is a greatest common divisor of a,b if both are nonzero. If b = 0, you can take R = I.
Proof of the claim:If a∕b, then b = ka for some k and so you can let
( )
R = 1 0
− k 1
So assume that a does not divide b and let d be a greatest common divisor. This is unique up to associates.
Then there are s,t such that
d = sa+ tb
Now a = rd,b = qd. Then the above yields
d = srd +tqd, 1 = sr+ tq
Consider the following product.
( ) ( ) ( ) ( )
s t r − t = rs + tq − st+ st = 1 0
− q r q s − qr+ rq tq +rs 0 1
Thus both matrices are invertible. Now
( )( ) ( )
s t a ∗ as+ tb ∗
− q r b ∗ = − aq+ rb ∗
( )
= d ∗
− rdq + rqd ∗
( )
= d ∗
0 ∗
Then analogous reasoning yields
Claim 2:Suppose a≠0. Then there exists an invertible 2 × 2 matrix R such that
( ) ( )
a b d 0
∗ ∗ R = ∗ ∗
where d is a greatest common divisor of a and b. This is left as an exercise. See Problem 28 on Page
606.
Now the difference here is that there is no function δ. Let l
(a)
denote the number of non-invertible
primes in a prime factorization of a. Thus, this is an integer. If a is invertible, let l
(a)
≡ 0. As before, begin
with
PAQ = B
where B = A and P,Q are identity matrices which will be adjusted. Let l
(Aij)
be the smallest. Adjust P,Q
so that A_{ij} = B_{11}. Now if any entries of the first column are a multiple of B_{11}, use an elementary matrix
on the left to modify P and zero these out. If B_{i1} is not a multiple of B_{11}, move it two the 2,1 position and
multiply on the left by
( )
R 0
0 I
in such a way as to get a zero in this position. Keep doing this till you obtain
( T )
PAQ = d1 a
0 ˆB
Then use claim 2 as needed to eventually obtain
( )
d1 0T
PAQ = 0 ˆB
If d_{1} fails to divide any entry of
ˆB
, say B_{ij}, add that column to the left column. Then do
the above process to zero it out and replace the top left corner with the greatest common
divisor of B_{ij} and d_{1}. Continue doing this till you obtain a column of zeros below the top left
corner. Then use claim 2 to get a row of zeros to the right of the top left corner. Thus you
obtain
( )
d2 0T
PAQ = ˆ
0 B
where d_{2}∕d_{1}. Hence l
(d2)
< l
(d1)
. Iterate this procedure till
ˆB
has no entries which are not multiples of d_{i}.
This must occur because the sequence
{l(di)}
where d_{i} is in the upper left corner is strictly
decreasing.
After this, do the same thing to
Bˆ
using row and column operations and multiplication by matrices of
the form
( )
1 0
|( R |)
0 I
Thus, eventually you obtain the same conclusion. ■
Now here is the cyclic decomposition theorem.
Theorem 9.5.7Let M be a finitely generated torsion module for a domain D such that the conclusion ofTheorem 9.5.5holds. Then it is the direct sum of cyclic submodules. That is,
M = Dm1 ⊕ ⋅⋅⋅⊕ Dmp
Proof:Let M = Da_{1} + Da_{2} +
⋅⋅⋅
+ Da_{n}. Since it is finitely generated, this is possible. Let
η : D^{n}→ M
( )
∑n n∑
η xiei ≡ xiai
i=1 i=1
where e_{i} is the usual thing having a 1 in the i^{th} position and zeros elsewhere. Clearly η is a morphism. Let
K be the kernel of η. Let
ˆη
: D^{n}∕K → M
ˆη(x+ K ) ≡ η (x)
Thus
ˆη
is one to one and onto. Now it was shown above that D^{n} is Noetherian and so K is finitely
generated by some
{z1,⋅⋅⋅,zm}
. Let the matrix A be defined by
n∑
zk = Aklel
l=1
So A is an m × n matrix whose entries are from D. Now let Q be an invertible n × n matrix and
define
∑n n∑
e′i ≡ Qijej, Q−l1i e′i = el
j=1 i=1
Also let P be an invertible m × m matrix such that
m
z′= ∑ Pkjzj, k = 1,⋅⋅⋅,m (**)
k j=1
(**)
Thus
′ m∑ ∑m ∑n
zk = Pkjzj = Pkj Ajlel
jm=1 n j=1n l=1
∑ ∑ ∑ − 1′
= j=1 PkjAjli=1 Qli ei
n l=1m n
= ∑ ∑ ∑ P A Q−1e′
i=1 j=1l=1 kj jl li i
By Theorem 9.5.5, we can choose P,Q such that PAQ^{−1} is a diagonal. We do this now. Then the result of
the above reduces to
z′= Bkke′ = dke′, dk ⁄= 0,k ≤ m (*)
k k k
(*)
The
{ek}
are linearly independent in the sense that if
n
∑ αiei = 0,
i=1
then each α_{i} = 0 because the left side is just the column vector
( )
α ⋅⋅⋅ α
1 n
^{T}. It is also the case that
the
′
{ei}
are independent. Indeed, if you have
∑n
0 = αie′i
i=1
then you have
n n n ( n )
0 = ∑ α ∑ Q e = ∑ ∑ α Q e.
i=1 ij=1 ij j j=1 i=1 i ij j
Hence ∑_{i=1}^{n}α_{i}Q_{ij} = 0 for each j. That is, α^{T}Q = 0^{T}. But Q^{−1} exists and so α^{T} = 0^{T}. In
addition,
span (e′⋅⋅⋅e′) = Dn
1 n
To see this,
∑ ∑n ∑n −1 ′ ′ ′
x = xlel = xl Q li ei ∈ span(e1⋅⋅⋅en)
l l=1 i=1
Now it was given that span
(z1,⋅⋅⋅,zm)
= K. Also span
′ ′
(z1,⋅⋅⋅,zm)
= K because by ∗∗,
∑m m∑
z′k = Pkjzj, P−ij1z′j = zi
j=1 j=1
and so anything in the span of the z_{i} is also in the span of the z_{j}^{′}. Each z_{k}^{′}∈ K because