Difference between revisions of "Free Modules"
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| − | ''' | + | ===Definition:=== |
| + | 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>. | ||
| + | |||
| + | ===Examples:=== | ||
# <math>R=R\{1_R\}</math> is a free <math>R</math>-module | # <math>R=R\{1_R\}</math> is a free <math>R</math>-module | ||
# <math>R^2=R\oplus R</math> has basis <math>\{(1_R,0_R),(0_R,1_R)\}</math> | # <math>R^2=R\oplus R</math> has basis <math>\{(1_R,0_R),(0_R,1_R)\}</math> | ||
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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>. | 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>. | ||
| + | |||
| + | ===Theorems:=== | ||
| + | [[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>. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | [[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 | ||
| + | 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 | ||
| + | |||
| + | <math>\text{Hom}_R(V,W)\cong \text{M}_{m\times n}(R) \quad f \mapsto [f]_B^C </math> | ||
| + | |||
| + | 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 <math>R</math> with <math>1 \ne 0</math>, any direct sum of copies of R is always a free | ||
| + | R-module. | ||
| + | |||
| + | Theorem. Every R-module is a quotient of a free <math>R</math>-module | ||
| + | |||
| + | [[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:
- \(R=R\{1_R\}\) 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, \(M\) 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
