Basic Concepts of Module Theory

E. L. Lady

Definition. Essential submodule. Definition. M/N Definition. Linear independence (infinite sets). Basis. Theorem. If K is a field then K[X] is a principal ideal domain. Proposition. If an ideal in a commutative ring is free then it is principal. Euclidean rings and greatest common divisors. Proposition. If M has a basis and N is any R-module, there exists a homomorphism from M to N taking the basis elements to any arbitrary set of elements in N. Rephrased. If M has a basis {m_{i}}_{i∈I} then for any R-module N there is an isomorphism of abelian groups θ: Hom(M, N) → ⨁_{I} N. This is defined by θ(φ) = {φ(m_{i})}_{i∈I}. Exercise? Proposition. {1} is a basis for the R-module R. Proposition. Existence of the free R-module on a set X. Proposition. Every module is a homomorphic image of a free module. Internal and external direct sums. Direct products. Characterization of the internal direct sum. Equivalent conditions for a family of submodules to be independent. Proposition. Every subspace of a vector space is a direct summand. Questions. For what rings R is it true that every submodule of an R-module M is a direct summand? What rings R have the property that every R-module is free? Proposition. Let N ⊆ M and let φ: M → M/N be the quotient map. (In Hungerford p. 172, this is called the canonical epimorphism.) Then N is a direct summand of M if and only if there exists a submodule L ⊆ M such that φ maps L isomorphically onto M/N. In this case, M = L ⊕ N. (Exercise.) Proposition. Let N be a submodule of M. The following conditions are equivalent:

(1) N is a direct summand of M.

(2) There exists π ∈ End M such that N = π(M) and π restricts to the identity map on N.

(3) There exists π ∈ End M such that N = π(M) and π^{2} = π. (Exercise.)

Theorem. If φ is an epimorphism from M onto P and P is a free R-module, then Ker φ is a direct summand of M.

Restatement. If N is a submodule of M such that M/N is free, then N is a direct summand of M.

Proposition. If M is a finitely generated torsion free module over an integral domain R then M is isomorphic to a submodule of a free R-module.

Example. ℚ is not isomorphic to a submodule of a free ℤ-module. (Hom_{ℤ}(ℚ, ℤ) = 0.)

Proposition. If M is a finitely generated torsion free module over an integral domain R then some homomorphic image of M is isomorphic to an ideal in R.

Example. No homomorphic image of ℤ(p) is isomorphic to an ideal of ℤ.

Proposition. If R is a principal ideal domain then a submodule of a finitely generated free R-module is free.

Theorem. A finitely generated torsion free module over a principal ideal domain is free.

Corollary. If M is a finitely generated module over a principal ideal domain R and t(M) is the torsion submodule of M, then t(M) is a direct summand of M.

Example. If R = ℤ and M = ∏_{p} ℤ(p), then t(M) = ⨁_{p} ℤ(p) and t(M) is not a direct summand of M.

Definition. An irreducible element in an integral domain and a prime ideal in a commutative ring. The associated primes for a module over a principal ideal domain and the p-primary components.

Proposition. An ideal (a) in a principal ideal domain R is a prime ideal if and only if a is an irreducible element.

Example. In the ring ℤ[\sqrt{−5}], 2 is irreducible but (2) is not a prime ideal.

Theorem. If M is a torsion module (not necessarily finitely generated) over a principal ideal domain then M = ⨁_{p} M_{p}.

Lemma. Suppose that M is a module over a principal ideal domain such that p^{n}M = 0 for some prime element p and n ≥ 1. If n is the smallest integer such that p^{n}M = 0 then there exists an element m ∈ M such that p^{n−1}m ≠ 0. Furthermore, the cyclic submodule of M generated by m is a direct summand.