Difference between revisions of "Free Modules"

From Queer Beagle Wiki
 
(3 intermediate revisions by the same user not shown)
Line 14: Line 14:
  
 
===Theorems:===
 
===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,
+
[[UMP for Free Modules| Theorem:]] (UMP for free modules) Let <math>R</math> be a ring, <math>M</math> a free <math>R</math>-module with basis <math>B</math>, <math>N</math> an <math>R</math>-module, <math>j:B\to N</math> a function. Then there exists a unique <math>R</math>-module homomorphism <math>h:M\to N</math> such that <math>h(b)=j(b)\ \forall b\in B</math>.  
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.
 
Corollary: If A and B are sets of the same cardinality, and fix a bijection j : A → B.
Line 22: Line 20:
 
of R-modules M ∼= N.
 
of R-modules M ∼= N.
  
Theorem:(Uniqueness of rank over commutative rings). Let R be a commutative ring
+
[[Uniqueness of Rank of Free Modules Over Commutative Rings|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
 
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.
 
have the same cardinality, meaning that there exists a bijection A → B.
Line 39: Line 37:
 
R-module.
 
R-module.
  
Theorem. Every R-module is a quotient of a free R-module
+
Theorem. Every R-module is a quotient of a free <math>R</math>-module
  
 
[[Category:Modules]] [[Category:Free Modules]]
 
[[Category:Modules]] [[Category:Free Modules]]

Latest revision as of 21:38, 9 March 2023

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:

  1. \(R=R\{1_R\}\) is a free \(R\)-module
  2. \(R^2=R\oplus R\) has basis \(\{(1_R,0_R),(0_R,1_R)\}\)
  3. \(R[x]\) is a free \(R\)-module with (infinite) basis \(\{1,x,x^2,...,x^i,...\}\)
  4. \(R[x,y]\) has basis \(\{x^n,y^m\ |\ n,m\geq 0 \}\) as free \(R\)-module
  5. \(R[x,y]\) 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, \(M\) a free \(R\)-module with basis \(B\), \(N\) an \(R\)-module, \(j:B\to N\) a function. Then there exists a unique \(R\)-module homomorphism \(h:M\to N\) such that \(h(b)=j(b)\ \forall b\in B\).

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