Module Theory From now on, R will denote an arbitrary ring with 1 ...

2
4. Submodules of M/N correspond to submodules of M containing N.
Theorem. In the category of R-modules, direct products and direct sums (coproducts) exist.
For any subset S of an R-module M, the set N = { n
1 risi | ri ∈ R, si ∈ S } is
the submodule of N generated by S. M is finitely generated if it has a finite number of
generators. For m ∈ M, Rm is a principal submodule of M.
Proposition. Let M be an R-module, N1,N2 submodules of M. M ∼ = N1 ⊕ N2 iff M =
N1 + N2 and N1 ∩ N2 = {0}.
Theorem. Let I be any set. The free R-module on the set I is the coproduct R (I) .
Definition. Let M be an R-module. A subset S ⊆ M is a basis for M if every element of
M can be written uniquely as a linear combination of elements of S.
Remarks. The free modules are the modules that have bases. Direct sums of free modules
are free since R (I) ⊕ R (J) ∼ = R I ˙∪J) . The zero module (0) is the free module on the empty
set.
Example. The free abelian groups are the groups of the form Z (I)
. If |I| = n,wehave
n
i=1 Z is the free abelian group on n generators. The Z-module Z/mZ is thus an abelian
group which is not free.
Proposition. Every module can be written as the quotient of a free module.
Characterization of finitely generated modules over a PID.
Theorem. Let R be a PID, F a finitely generated free module over R and M a submodule
of F . Then M is free with the number of generators less than or equal to the number of
generators of F . (This is true without the hypothesis of finitely generated [L, Theorem
III.7.1].)
Invariant Basis Property. Any two bases of F have the same number of elements.
The theorem fails for principal ideal rings (not domains): take R = Z/6Z and M =
{0, 2, 4}. The theorem fails for UFD’s: take R = R[x, y] andM=theideal(x, y).