Difference between revisions of "Free Modules"

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