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Dummit & Foote Chapter 12 - Modules over Principal Ideal Domains

Section 12.1 The Basic Theory

Definition. The left $R$-module M is said to be a Noetherian $R$-module if there are no infinitely increasing chains of submodules. That is, given


M_1 \subseteq M_2 \subseteq \ldots

there exists some k \in \mathbb{N} such that given any n \in \mathbb{N} with n \geq k, then M_n = M_k.

Definition. A ring R is Noetherian if it is Noetherian when viewed as a left $R$-module over itself.

Theorem. Let R be a ring and M a left $R$-module. Then, the following are equivalent:

  1. M is Noetherian
  2. Every nonempty set of submodules of M contains a maximal element under inclusion
  3. Every submodule of M is finitely-generated

Corollary. If R is a principal ideal domain (PID), then all nonempty set of ideals of R has a maximal element. Additionally, R is as Noetherian ring.

Proposition. Let R be an integral domain, and M be a free $R$-module of rank n < \infty. Then, given S is subset M with |S| > n, the elements of S are $R$-linearly dependent.

Definition. Given R an integral domain and M an $R$-module,


\Tor(M) = \{ x \in M | rx = 0 \text{ for any } r \neq 0 \}

This is the torsion submodule of M. If \Tor(M) is empty, then M is torsion-free.

Definition Let R be an integral domain and M be an $R$-module. Then, given a submodule N,


\Ann_R(N) = \{r \in R | rn = 0 \text{ for all } n \in N \}

This ideal of R is the annihilator of $N$. That is, \Ann(N) is the set of elements of R such that (r)N = \{ 0 \}.

Note that if N is not a torsion submodule of M, then \Ann(N) = (0)R. Additionally, given N, L are submodules of M with N \subseteq L, then \Ann(N) \subseteq \Ann(L).

Additionally, if R is a PID, as \Ann_R(N) is an ideal, \Ann(N) = (n)R and \Ann(L) = (l)R for some n, l \in R such that n | l.

Definition. Given any integral domain R, the rank of an $R$-module M is the maximum number of $R$-linearly independent elements of M.

Corollary. The rank of a free module is the number of generating elements.

Theorem. Let R be a principal ideal domain, and M be a free $R$-module of finite rank m, and N be a submodule of M. Then,

  1. N is a free submodule with rank n \leq m.
  2. There exists a basis y_1, y_2, \ldots, y_m of M so that r_1 y_1, r_2 y_2, \ldots, r_m y_n is a basis of N for some r_i \in R and r_1 | r_2 | \ldots | r_n