Advanced Algebra II

Advanced Algebra II

free modules, modules over principal ideal domain We start the discussion of free module and projective modules. Then we work on modules over principal ideal domain. The key result here is that a submodule of a free module over principal ideal domain is again free. As applications, one have an integrated view of fundamental theorem of finite generated abelian groups and the Jordan canonical form.

Definition 0.1. Let R be a ring, and M be a module. One can de- fine the notion of ”linear independent” and ”span” as we did in linear algebra. That is, a subset X ⊂ M is linearly independent if for any x₁, ..., x_n ∈ X, P

r_ix_i = 0 implies that r_i = 0 for all i. A subset X ⊂ M spans M if every element x ∈ M can be written as P

r_ix_i for some x_i ∈ X, r_i ∈ R. A spanning independent subset is called a basis.

Example 0.2. A vector space has a basis. And two basis have the same cardinality.

Example 0.3. There is no basis in general. For example, let M = Z₄ be an Z-module. Then there is no independent subset.

Theorem 0.4. Let R be a ring with identity and F a unitary RM.

The following are equivalent:

(1) F has a non-empty basis.

(2) there is a subset X ⊂ F such that F ∼= L

x∈XRx with each Rx ∼= R as RM.

(3) there is a set X and a function ı : X → F satisfying the fol- lowing ”universal property”: for all function f : X → M to M ∈ _RM, there is a unique _RM-homomorphism ¯f : F → M such that ¯f ◦ ı = f .

A unitary _RM satisfying the above condition is called a free module.

Proof. If F has a basis X, then one can verify that F ∼=L

x∈XRx with each Rx ∼= R asRM. And vice versa.

Moreover, if there is a subset X ⊂ F such that F ∼=L

x∈XRx with each Rx ∼= R as RM. Let ı : X → F be the inclusion. One can verify that there is the universal property.

Lastly, if there is a set X and a function ı : X → F satisfying the following universal property. We claim that ı(X) ⊂ F such that F ∼=L

x∈ı(X)R. To see this, one note that there are ı_x : R → F and there are px : F → R by the universal property. Hence one sees that F ∼=L

x∈XR. ¤

Proposition 0.5. Let 0 → M → N → F → 0 be a short exact^ϕ sequence such that F is free, then the sequence split.

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Proof. Let X ⊂ F be a basis. For each x ∈ X we pick ax ∈ N maps to x. Hence we have a function f : X → N. By the universal property, we have ¯f : F → N. One notices that ϕ ¯f |_X = 1_F|_X. By the uniqueness,

ϕ ¯f = 1_F. Thus the sequence split. ¤

Let F be a free module over R, we would like to define the rank of F , denoted rank(F ), to be the cardinality of its basis. However, this is not always well-defined. We say that a ring with identity has the invariant dimension property if the cardinality of two basis of a free module is the same. Hence the rank is well-defined. The following theorem shows that this is well-defined if R is a commutative ring with identity.

Theorem 0.6. Let R be a ring with identity, I C (6=)R an ideal. Let F be free module with basis X. Then F/IF is a free R/I-module with basis π(X). And |(π(X)| = |X|. (Where π : F → F/IF is the canonical surjection).

Remark 0.7. Let ICR an ideal an M ∈ _RM. Then IM := {P

a_ix_i|a_i ∈ I, x_i ∈ M} is a submodule.

Moreover, M/IM is a R/I-module by considering R/I × M/IM → M/IM sending (r + I)(x + IM ) 7→ rx + IM .

Proof. For u + IF ∈ F/IF , we can write u =P

r_ix_i with x_i ∈ X, r_i ∈ R. It’s clear that π(X) spans F/IF .

On the other hand, we claim that π(X) is linearly independent. If P(r_i + I)(x_i + IF ) = 0, then P

r_ix_i ∈ IF . We can write P

r_ix_i = Pajyj with aj ∈ I, yj ∈ F . Since X is a basis, each yj = P

bjixi for bji ∈ F . Combining all these, one has

Xr_ix_i =X

j

X

i

a_jb_jix_i =X

i

(X

j

r_ib_ji)x_i.

By the linear independence of X, we have ri = P

ribji ∈ I. Hence r_i+ I = 0 + I for all i, it follows that π(X) is linearly independent.

It suffices to show that X → π(X) is injective. (It’s surjective any- way). This is clear by the independence of π(X). For example, if x₁ + IF = x₂ + IF , then x₁ − x₂ + IF = 0x₁ + 0x₂ + IF which is

impossible. ¤

Corollary 0.8. Let R 6= 0 be a commutative ring with identity. Then R has the ”invariant dimensional property”.

Proof. By Zorn’s Lemma, there exist an maximal ideal m C R. For a free module F ∈ RM with basis X1, X2, then π(X1), π(X2) are basis of a free R/m-module F/IF . Note that R/m is a field, and hence F/IF is a vector space. Hence

|X₁| = |π(X₁)| = |π(X₂)| = |X₂|.

¤

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Therefore, for a free module F over a commutative ring with identity.

One can define the rank of F by |X|, where X is a basis of F .

Theorem 0.9. Let R be a PID. Then every submodule G of a free module F is free with rankG ≤ rankF .

Proof. Let X = {x_i|i ∈ I} be a basis of F . We fix a well-ordering ≤ on I. Let i + 1 denote the immediate successor of i. Let F_i :=P

j≤iRx_j. For convenience, we add an extra element ∞ to I, that is, let J = I ∪ {∞} with ∞ 6∈ I and i < ∞ for all i ∈ I. Then every element in I has an immediate successor in J. We can write F = F_∞.

Let G_i := G ∩ F_i. We would like to prove by transfinite induction.

Consider first the case that rankF = 1. F = Rx ∼= R, G < F is a submodule, hence G ∼= I C R. R is PID, so I = (c) = Rc for some c ∈ R. The homomorphism R → Rc is an isomorphism if c 6= 0. Thus G is free of rank 0 or 1.

We consider now

G_i+1/G_i = G_i+1/(G_i+1∩ F_i) ∼= (Gi+1+ F_i)/F_i < F_i+1/F_i. One checks that F_i+1/F_i ∼= Rxi is free. Thus, G_i+1/G_i is 0 or free.

If G_i+1/G_i = 0, then G_i+1 = G_i is free with rankG_i+1 = rankG_i ≤ rankF_i ≤ rankF_i+1. we are done by transfinite induction.

If G_i+1/G_i is free of rank 1. Note that the sequence 0 → G_i → G_i+1 → G_i+1/G_i → 0 split. In particular, G_i+1 ∼= Gi ⊕ G_i+1/G_i. We have G_i+1 is free with rankG_i+1 = rankG_i + 1 ≤ rankF_i + 1 =

rankF_i+1. ¤

Theorem 0.10. A finitely generated torsion free module M ∈ _RM over a principal ideal domain R is free.