Advanced Algebra II

Advanced Algebra II

modules over principal ideal domain

Definition 0.1. Let M ∈ RM. For x ∈ M, we define the annihilator of x, denoted A(x), to be {r ∈ R|rx = 0}.

We can also define the annihilator of M, denoted A(M), to be {r ∈ R|rx = 0, ∀x ∈ M}.

It’s easy to check that A(x) (resp. A(M) ) is a left ideal (hence a submodule of R). By considering the sujective map R → Rx, one has R/A(x) ∼= Rx.

Let M ∈ RM. For x ∈ M, we say that x is torsion if A(x) 6= 0.

Example 0.2. If R is an integral domain, then the set of all torsion elements M_τ is a submodule, called torsion submodule. However, this is not true in general. For example, take R = Z₆, M = Z₆ as an R-module. One sees that 2, 3 are torsion elements, but 5 = 2 + 3 is not.

Let M be a module over an integral domain. We say that M is a torsion module if M = M_τ. And M is said to be a torsion-free module if M_τ = 0.

Theorem 0.3. A finitely generated torsion free module M ∈ _RM over a principal ideal domain R is free.

Proof. Let X = {x₁, ..., x_n} be a spanning set of M. In X, we consider all independent subsets of X. There exists a maximal independent subset, say S := {x₁, ..., x_k} and let F be the submodule generated by S. F is clearly a free module.

Thus for all x_i, i > k, one has

rixi = Xk

j=1

rijxj ∈ F.

Let r = Q_n

i>kr_i. It’s clear that rM < F . Therefore, rM is a free module.

Lastly, we consider f : M → M by f (x) = rx. Since M is torsion free, Kerf = 0. By the isomorphism theorem, one has

rM = Imf ∼= M/Kerf = M.

In particular, M is free. ¤

Corollary 0.4. Let M be a finitely generated module over a principal ideal domain R. Then M = Mτ⊕ F , where F is a free submodule and F ∼= M/Mτ

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Proof. M/Mτ is a finitely generated module. We claim that it’s a torsion-free module. Then by the Theorem, it’s a free module. Hence the sequence 0 → M_τ → M → M/M_τ → 0 splits. Indeed, M ∼= M_τ ⊕ M/M_τ. Let F be the image of the inclusion ı : M/M_τ → M.

The it’s clear that F is free and F ∼= M/Mτ.

It suffices to prove that M/M_τ is torsion-free. For any x + M_τ such that rx + M_τ = 0. Then rx ∈ M_τ, therefore, there exists s ∈ R such that srx = (sr)x = 0. In particular, x ∈ M_τ, x + M_τ = M_τ. ¤ We have seen that a finitely generated module over principal ideal domain can be decomposed into ”free part” and ”torsion part”. Our next goal is to analyze the torsion part.

Proposition 0.5. Let R be a principal ideal domain and p ∈ R a prime element.

(1) if pⁱx = 0, then A(x) = (p^j) for some j ≤ i.

(2) if A(x) = pⁱ, then p^jx 6= 0 for j < i.

Remark 0.6. Let R be a principal ideal domain. Then p ∈ R a prime element if and only if p is irreducible. Please note that the notion of prime and irreducible are not the same in general.

Let R be a principal ideal domain and M ∈_RM. An element x ∈ M is said to be has order r if A(x) = (r). We have the following further decomposition of torsion part.

Theorem 0.7. Let M be a torsion module over a principal ideal do- main R. For a prime p ∈ R, let M(p) := {x ∈ M|A(x) = pⁿ for some n}.

(1) M(p) < M . (2) M = ⊕M(p).

(3) If M is finitely generated, then it’s a finite direct sum.

Proof. The first statement is easy.

The proof is basically the same as Chinese remainder theorem. For a fixed x ∈ M, with A(x) = (r). Since R is a unique factorization domain, one can write r = Q

i=1,...,np^a_iⁱ. Let ri := r/p^a_iⁱ, then it’s clear that (r₁, ..., r_n) = 1. In particular, there exist s₁, ..., s_n ∈ R such that Ps_ir_i = 1. Therefore,

x =X s_ir_ix.

Let x_i := s_ir_ix. Then p^a_iⁱx_i = rs_ix = 0. Thus x_i ∈ M(p). Hence we have seen that M is generated by M(p).

Let M(p)⁰ be the submodule generated by M(q) with q 6= p. It remains to show that the intersection of M(p) ∩ M(p)⁰ = 0 which is easy.

Lastly, if M is finitely generated. Then A(M) = (r) for some r. Let r = Q

i=1,...,np^a_iⁱ, then one can easily prove that M = ⊕_i=1,...,nM(p_i).

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Thus it suffices to work on the submodule M(p). We need the fol- lowing lemma taking care of element of maximal order.

Lemma 0.8. Let M be a module over a principal ideal domain R such that pⁿM = 0 and pⁿ⁻¹M 6= 0. Let x ∈ M be a element of orderpⁿ.

(1) If M 6= Rx, then there exist y ∈ M such that Rx ∩ Ry = 0.

(2) There is a submodule N < M such that M = Rx ⊕ N.

Proof. If M 6= Rx, then there is c ∈ MRx. Consider {c, pc, ...pⁿc = 0 ∈ Rx}. One can pick j such that p^jc ∈ Rx but p^j−1c 6∈ Rx.

Let p^jc = rx. Factorize r = r⁰p^k ∈ R. Since 0 = pⁿc = p^n−jrx = p^n−j+kr⁰x.

One has n − j + k ≥ n, thus k ≥ j ≥ 1.

We consider y = p^j−1c − r⁰p^k−1x ∈ M. y 6= 0 since p^j−1c 6∈ Rx and py = 0. If Rx ∩ Ry 6= 0, then there is s ∈ R such that sy ∈ Rx.

Clearly, (p, s) = 1, thus 1 = pα + sβ for some α, β ∈ R. It follows that y = βsy ∈ Rx, and therefore, p^j−1c ∈ Rx. This is a contradiction.

For the second statement, we consider S = {N < M |N ∩ Ra} = 0}.

If M = Rx then nothing to prove. We may assume that M 6= Rx, hence S 6= ∅. By Zorn’s Lemma, there is a maximal element N. We claim that M = Rx + N then we are done.

To this end, we consider M/N. Clearly, pⁿM/N = 0 and pⁿ(x+N) = 0. It’s not difficult to check that x + N ∈ M/N has order pⁿ. If M/N = R(a + N) then we are happy. Otherwise, apply (1) to M/N, there is y ∈ M such that y + N 6∈ R(x + N). Then one checks that N + Ry ∩ Rx = 0 and thus contradicts to the maximality of N. ¤ Theorem 0.9. Keep the notation as above. Let M = M(p) be a finitely generated module. Then M is a direct sum of cyclic R-modules of order pⁿ¹, ...., p^nk, where n₁ ≥ n₂ ≥ .... ≥ n_k ≥ 1.

Proof. Let x₁, ..., x_r be generators whose order are p_m1, ..., p^mr respec- tively. Let n₁ := max{m_i}. We may assume that x₁ ∈ M is an element of order n₁. Then M = Rx₁⊕ N. By induction, we are done. ¤ These pⁿ_i^j are called the elementary divisors. In order to prove the uniqueness of the decomposition and also another description by in- variant factors. We need to work more.

Lemma 0.10. Let M, N be module over a principal ideal domain, r ∈ R and p ∈ R is prime.

(1) rM and M[r] := {x ∈ M|rx = 0} are submodules.

(2) M[p] is a vector space over R/(p).

(3) (R/(pⁿ))[p] ∼= R/(p) and p^m(R/(pⁿ)) ∼= R/(p^n−m) for m < n.

(4) If M ∼= ⊕Mi, then rM ∼= ⊕rMi and M[r] ∼= ⊕Mi[r].

(5) If f : M → N a RM-isomorphism, then f : Mτ ∼= Nτ, and f : M(p) ∼= N(p).

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Lemma 0.11. Let r =Q

i=1,...,kpⁿ_iⁱ. Then R/(r) ∼= ⊕i=1,...,kR/(pⁿ_iⁱ).

Theorem 0.12. Let M be a finitely generated module over a principal ideal domain R. Then

(1) M is a direct sum of a free module F of finite rank and a finite number of cyclic torsion modules of order r₁, ..., r_m respectively.

Where r₁|r₂|...|r_m. And the rank of F and the list of ideals (r₁), ..., (r_m) is unique.

(2) M is a direct sum of a free module F of finite rank and a fi- nite number of cyclic torsion modules of order p^a₁¹, ..., p^a_k^k respec- tively. And the rank of F and the list of ideals (p^a₁¹), ..., (p^a_k^k) is unique.

The r₁, ..., r_m are called invariant factors. And p^a₁¹, ..., p^a_k^k are called elementary divisors.