Module 3
1
Topic 6
Uniform Modules, Primary Modules, and Nother-Lasker Theorem
Definition 1. A non-zero module M is called uniform if any two non-zero submodules of M have nonzero intersection.
M is said to contain enough uniforms if every non-zero submodule contains a uniform module.
Definition 2. Two uniform modules U and V are said to be subisomorphic, denoted by U ∼ V if U
and V contain non-zero isomorphic submodules.
Definition 3. A module M is called primary if M contains enough uniforms and any two uniform
submodules of M are subisomorphic.
The relation ∼ is easily seen to be an equivalence relation. Clearly, ∼ is reflexive and symmetric. Now
suppose U1 ∼ U2 and U2 ∼ U3 . Then there exist non-zero submodules V1 , V2 of U1 , U2 , respectively, and
W2 , W3 of U2 , U3 respectively, such that V1 ∼
=φ W3 . As U2 is uniform, V2 ∩ W2 6= {0}.
=θ V2 and W2 ∼
Then θ−1 (V2 ∩ W2 ) ∼
= φ(V2 ∩ W2 ). Hence V1 and V3 contain non zero isomorphic submodules,
= V2 ∩ W 2 ∼
which gives V1 ∼ V3 . For a uniform module U , [U ] will denote the equivalence class of U with respect to
the relation ∼.
Examples:
• Every commutative domain is uniform as a module over itself. This is easy to verify.
• A vector space over a field is uniform if and only if it is one-dimensional. Let V be a uniform
vector space over a field K. Suppose dimK V > 1. Let W be a non-zero subspace of V such that
W 6= V . Then there exists a subspace W 0 such that V = W ⊕ W 0 . Then W ∩ W 0 = {0} which
is a contradiction as V is uniform. The converse is trivial. Now every non-zero submodule of V
contains a one-dimensional subspace and hence a uniform space. Also one-dimensional subspaces
of V are isomorphic. Hence it follows that V is primary.
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• Let A be a cyclic group under addition. Then A as a Z-module is uniform if and only if either A
is infinite or |A| = pn for some prime p and positive integer n.
4. Any abelian group contains a cyclic group, which is either infinite or is of prime order. Hence an
abelian group as a Z-module contains enough uniforms.
Nother-Lasker Theorem
Let M be a finitely generated module over a commutative Noetherian ring R. Then there exist finite
family N1 , N2 , . . . , Nm of submodules of M such that
(a) ∩m
i=1 Ni = 0 and ∩i6=j Ni 6= 0 ∀1 ≤ j ≤ m
(b) Each quotient M/Ni is Pi −primary module for some prime ideal Pi .
(c) The Pi ’s are all distinct.
(d) The primary components Ni is unique if and only if Pi does not contain Pj for j 6= i.
In order to prove this Theorem, we require the following results.
Theorem 1. Let M be a Notherian module or a module over a Noetherian ring. Then each non-zero
submodule of M contains a uniform module.
Proof. To prove this, it is enough to prove that every non-zero cyclic submodule of M contains a uniform
submodule. Let 0 6= x ∈ M . Suppose M is Noetherian. Then the submodule Rx is also Noetherian.
On the other hand if R is Noetherian, then Rx ∼
= R/I, where I is the kernel of the onto R-module
homomorphism R → Rx, given by r 7→ rx. And hence again Rx is Noetherian.
Now consider the family
== family of all submodules of Rx which intersect at least one submodule of Rx trivially.
As (0) ∈ =, = =
6 φ. Now Rx is Noetherian implies that = has a maximal element say, K. By definition
of =, there exists a submodule U of Rx such that K ∩ U = (0).
Claim: U is uniform.
Suppose if possible, there exist submodules A and B of U such that A∩B = (0). Now K∩A ⊆ K∩U = (0).
Consider (K ⊕ A) ∩ B. Let x belong to this intersection. Then x = y + a = b, where y ∈ K, a ∈ A and
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b ∈ B. This implies y = b − a ∈ K ∩ U = (0) ⇒ y = 0 ⇒ x = a = b ∈ A ∩ B = (0) ⇒ x = 0. Hence
(K ⊕ A) ∩ B = (0). This implies K ⊕ A ∈ =, a contradiction. Hence U is uniform submodule of Rx.
By the above result every module over a Noetherian ring contains enough Uniforms.
Definition 4. Let R be a commutative Noetherian ring and P be a prime ideal of R. Then P is said to
be associated with an R-module M if R/P embeds in M . Equivalently, P = AnnR (x) for some x ∈ M ,
where AnnR (x) = {r ∈ R |rx = 0} is the annihilator of x in R.
Definition 5. A module M is called P -primary for a prime ideal P if P is the only prime ideal associated
with M .
Theorem 2. Let U be a uniform module over a commutative Noetherian ring R. Then U is subisomorphic to R/P for exactly one prime ideal P of R.
Proof. Let = = {AnnR (x) |0 6= x ∈ U } be the family of annihilators of non-zero x ∈ U in R. As R is
Noetherian = has a maximal element, say, AnnR (x) for 0 6= x ∈ U .
Claim: AnnR (x) = P (say) is prime.
Clearly, P is an ideal of R. Now let ab ∈ P . Then abx = 0. Suppose b 6∈ P , that is, bx 6= 0. Now
P = AnnR (x) ⊆ AnnR (bx) as R is commutative. By maximality of P , we get P = AnnR (bx). As
a ∈ AnnR (bx), a ∈ P . Thus P is prime. Also R/P ∼
= Rx ⊆ U and hence U is subisomorphic to R/P .
Now we prove the uniqueness. Suppose Q is a prime ideal of R such that U is subisomorphic to R/Q.
Then [R/Q] = [U ] = [R/P ] ⇒ R/Q is subisomorphic to R/P . Thus there exist cyclic submodules Rx
and Ry of R/P and R/Q such that Rx ∼
= Ry.
As Rx ∼
= R/P and Ry ∼
= R/Q (Prove!)
we get R/P ∼
= R/Q ⇒ P = Q (Verify!)
Theorem 3. Let M be a non-zero finitely generated module over a commutative Noetherian ring R.
Then there are only finitely many prime associated with M .
Proof. Define S to be the family of direct sums of cyclic uniform submodules of M . By Theorem 1,
S =
6 φ. Define partial order in S by
X
X
⊕
xi R ≤ ⊕
yj R if and only if I ⊆ J and xi R ⊆ yi R ∀i ∈ I. By Zorns Lemma S has a maximal
i∈I
j∈J
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element, say, N = ⊕λ∈ω
X
xλ R. As M is Noetherian, N is finitely generated and hence N = ⊕
m
X
xi R
i=1
for some positive integer m.
As each xi R is uniform, by Theorem 2, there exist xi ai ∈ xi R such that Pi = AnnR (xi ai ) is the prime
m
X
ideal associated with xi R. Let K =
xi ai R.
i=1
Claim: If Q is any associated prime ideal of M , then Q = Pi for some i, 1 ≤ i ≤ m.
Now Q = AnnR (x) for some x ∈ M . As N is a maximal member of S , both N and K intersect every
m
X
non-zero submodule of M non-trivially. Let 0 6= y ∈ xR ∩ K. Then y = xr =
xi ai ri .
i=1
Let xi ai ri s = 0 ⇒ ri s ∈ AnnR (xi ai ) = Pi . Suppose xi ai ri 6= 0. Then ri 6∈ Pi . This gives s ∈ Pi . Hence
if xi ai ri 6= 0, AnnR (xi ai ) = AnnR (xi ai ri )
Now AnnR (y)
=
m
\
AnnR (xi ai ri )
i=1
=
\
AnnR (xi ai )
i∈Λ
=
\
Pi , where i ∈ Λ implies xi ai ri 6= 0
i∈Λ
Now R/Q ∼
=θ xR. As θ−1 (yR) is cyclic submodule of R/Q, θ−1 (yR) ∼
= R/Q.
\
Now Q = AnnR (x) = AnnR (y) =
Pi
i∈Λ
Thus Q ⊆ Pi ∀i ∈ Λ. Now suppose Pi * Q ∀i ∈ Λ. Then there exist xi ∈ Pi such that xi 6∈ Q∀i ∈ Λ. As
Πi∈Λ xi ∈ ∩Pi = Q and Q is prime, this leads to a contradiction. Hence Pj ⊆ Q for some j ∈ Λ. Thus
Q ∈ {P1 , P2 , . . . , Pm }.
Now we come to the proof of the Noether Lasker Theorem.
Proof. (Proof of Noether Lasker Theorem) Consider the uniform modules Rxi as defined in the
earlier result for 1 ≤ i ≤ m. Now choose those Rxi ’s such that [Rxi ] 6= [Rxj ] for i 6= j. After renumbering we call these Ui = Rxi , 1 ≤ i ≤ t. Note that if [Rxi ] = [Rxj ], then [R/Pi ] = [R/Pj ] ⇒
R/Pi ∼
= R/Pj ⇒ Pi = Pj .
Hence still the only prime ideals associated with M are P1 , P2 , . . . , Pt Also this gives us that all Pi0 s
are distinct. Now define Fi to be the family of submodules of M which do not contain any submodule
subisomorphic to Ui , for 1 ≤ i ≤ t.
Let Ni be a maximal element of Fi .
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(a) Suppose ∩Ni 6= (0). Then ∩Ni contains a uniform module U . Let P be the unique prime associated
with U . Then P is associated prime of M . Hence P = Pj for some 1 = j = t.
⇒ [U ] = [R/Pj ] = [Uj ] a contradiction as U ⊆ Nj .
Observation: Every uniform submodule of M/Ni belongs to [Ui ].
Proof. Let N be any submodule of M containing Ni . If Ni ( N , N 6∈ Si . Hence we can find a
uniform submodule U of N such that there exists a 1-1 map θ : U → Ui
Restricting θ to U ∩ Ni , we get an imbedding of U ∩ Ni in Ui . Hence, we have no choice but
U ∩ Ni = (0).
Thus the map
θ∗ :
U + Ni
→ Ui
Ni
given by θ∗ (x + Ni ) = θ(x), for x 6∈ Ni , still remains an embedding. Hence the result.
Now let U be a uniform submodule of M such that U ∈ [Ui ]. Then there does not exist any
monomorphism from U to M/Nj for i 6= j. Hence U ∩ Nj 6= 0 for i 6= j. As U is uniform,


\
\
\
(U ∩ Nj ) 6= (0) ⇒ U ∩  Nj  6= (0) ⇒
Nj 6= (0)
i6=j
i6=j
i6=j
(b) Let Q be a prime ideal associated with M/Ni . Let U be the submodule of M/Ni such that
h i
R
U∼
= [U ] = [Ui ].
= R/Q. Then U is uniform submodule of M/Ni and hence U ∈ [Ui ] ⇒ Q
⇒ Q = Pi as Pi is the unique prime associated with Ui . Thus M/Ni is Pi - primary.
(c) Done!
(d) Assume Ni is unique. Suppose for some j 6= i, Pj ⊆ Pi .
Let f : R/Pj → R/Pi given by f (r + Pj ) = r + Pi
Then if r ∈ Pj , r ∈ Pi . Hence well defined. Clearly this is a non-zero(as f (1 + Pj ) 6= 0) Rhomomorphism. Let U and V be uniform submodules of Uj and Ui , respectively, such that U ∼
=
R/Pj and V ∼
= R/Pi . Hence we have a non-zero R-homomorphism g : U → R/Pj → R/Pi → V .
Let x ∈ U such that g(x) 6= 0.
Claim: r.x = 0 ⇔ r(x − g(x)) = 0.
Suppose r.x = 0. Then g(rx) = 0 ⇒ rg(x) = 0 ⇒ r(x − g(x)) = 0.
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Conversely, let r(x − g(x)) = 0 ⇒ rx = rg(x) ∈ U ∩ V
As U ∈ [Uj ] and V ∈ [Ui ], U ∩ V = (0) ⇒ rx = 0.
Thus Rx ∼
= R/AnnR (x) ∼
= R(x − g(x))
As U ∈ [Uj ], U ∈ Fi .
⇒ Rx ∈ Fi and hence R(x − g(x)) ∈ Fi . Let N0 and N0∗ be maximal elements of Fi containing
Rx and R(x − g(x)), respectively. As maximal element of Fi is unique, N0 = N0∗ .
⇒ Rx + R(x − g(x)) ∈ Fi ⇒ Rg(x) ∈ Fi .
But g(x) ∈ V and [V ] = [Ui ] a contradiction. Hence Pj * Pi ∀j 6= i.
Conversely, suppose Pj * Pi ∀j 6= i.
⇒ there does not exist any non-zero R-homomorphism from R/Pj to R/Pi . Let N and L be two
maximal elements of Fi . Then N * L and L * N , if N 6= L.
⇒ the map M → M/N gives a non-zero homo θ from L to M/N .
Every uniform submodule of M/N belongs to [Ui ]. Hence Pi is the prime ideal associated with
M/N . Let U be the uniform submodule of M/N isomorphic to R/Pi . Now restricting θ to V ,
the preimage of U under θ, we get a homomorphism from V to R/Pi . As V ∈ Fi , Pj is a prime
associated with V for some j 6= i.
Consequently, we get a non-zero homomorphism from R/Pj to R/Pi a contradiction. Hence Fi
has a unique maximal element.
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