Free Modules
Definition:
An \(R\)-module \(M\) is a free \(R\)-module if \(M\) has a basis. A subset \(A\) of an \(R\)-module \(M\) is a basis of \(M\) if \(A\) is linearly independent and generates \(M\).
Examples:
- \(R=R\{1_R\}\) is a free \(R\)-module
- \(R^2=R\oplus R\) has basis \(\{(1_R,0_R),(0_R,1_R)\}\)
- \(R[x]\) is a free \(R\)-module with (infinite) basis \(\{1,x,x^2,...,x^i,...\}\)
- \(R[x,y]\) has basis \(\{x^n,y^m\ |\ n,m\geq 0 \}\) as free \(R\)-module
- \(R[x,y]\) has basis \(\{1,y,y^2, ...,y^i,...\}\) as free \(R[x]\)-module
All free \(R\)-modules \(M \) have some basis \(B=\{b_1,...,b_n\} \) so have rank \(n \), and can be written as \(R^n\cong M \). Additionally every element \(m\in M \) can be uniquely written \( m=\sum_{i=1}^n r_ib_i\).
Theorems:
Theorem: (UMP for free modules) Let R be a ring, M be a free R-module with basis B, N be any R-module, and let j : B → N be any function. Then there is a unique R-module homomorphism h : M → N such that h(b) = j(b) for all b ∈ B.
Corollary: If A and B are sets of the same cardinality, and fix a bijection j : A → B. If M and N are free R-modules with bases A and B respectively, then there is an isomorphism of R-modules M ∼= N.
Theorem:(Uniqueness of rank over commutative rings). Let R be a commutative ring with 1 6= 0 and let M be a free R-module. If A and B are both bases for M, then A and B have the same cardinality, meaning that there exists a bijection A → B.
Theorem: Let R be a commutative ring with 1 6= 0. Let V and W be finitely generated free R-modules of ranks n and m respectively. Fixing ordered bases B for V and C for W gives an isomorphism of R-modules
\(\text{Hom}_R(V,W)\cong \text{M}_{m\times n}(R) \quad f \mapsto [f]_B^C \)
If V = W, so that in particular m = n, and B = C, then the above map is an R-algebra isomorphism EndR(V ) ∼= Mn(R).
Lemma. Given any ring \(R\) with \(1 \ne 0\), any direct sum of copies of R is always a free
R-module.
Theorem. Every R-module is a quotient of a free R-module
