Alternative definition on free modules.
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I 'm trying to compare Free Modules and Free Abelian Groups.
We know that,
Definition. An abelian group $G$ is called free abelian group with rank $nin Bbb N^*$, if $G$ is the direct sum of $n$ infinite cyclic groups. That is,
$$G=langle x_1 rangle oplus ... oplus langle x_n rangle$$
where $U_i:=langle x_i rangle, i=1,...,n$ are infinite cyclic groups. Then, $x_1,...,x_n$ are called free generators of $G$, $X:={x_1,...,x_n}$ is called basis (or free generators set) of $G$ and $n$ is called rank of $G$.
We already know the following definition that seems equivalent.
Definition. Let $R$ be a ring with $1_R$. An $R$-module $F$ is called free on a subset $A$ of $F$, if for every nonzero element $xin F$, there exist unique nonzero elements $r_1,...,r_nin R$ and unique
$a_1,...a_n in A$ such that $$x=r_1a_1+...+r_na_n$$ for some $nin
mathbb{Z}^+$ In this situation we say $A$ is a basis or set of
free generators for $F$. If $R$ is commutative ring, the cardinality
of $A$ is called the rank of $F$.
Questions.
Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?
Is it useful to make such comparisons?
Thank you in advance.
Reference: Abstract Algebra, Dummit and Foote, 3rd ed.
abstract-algebra ring-theory modules free-modules free-abelian-group
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up vote
0
down vote
favorite
I 'm trying to compare Free Modules and Free Abelian Groups.
We know that,
Definition. An abelian group $G$ is called free abelian group with rank $nin Bbb N^*$, if $G$ is the direct sum of $n$ infinite cyclic groups. That is,
$$G=langle x_1 rangle oplus ... oplus langle x_n rangle$$
where $U_i:=langle x_i rangle, i=1,...,n$ are infinite cyclic groups. Then, $x_1,...,x_n$ are called free generators of $G$, $X:={x_1,...,x_n}$ is called basis (or free generators set) of $G$ and $n$ is called rank of $G$.
We already know the following definition that seems equivalent.
Definition. Let $R$ be a ring with $1_R$. An $R$-module $F$ is called free on a subset $A$ of $F$, if for every nonzero element $xin F$, there exist unique nonzero elements $r_1,...,r_nin R$ and unique
$a_1,...a_n in A$ such that $$x=r_1a_1+...+r_na_n$$ for some $nin
mathbb{Z}^+$ In this situation we say $A$ is a basis or set of
free generators for $F$. If $R$ is commutative ring, the cardinality
of $A$ is called the rank of $F$.
Questions.
Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?
Is it useful to make such comparisons?
Thank you in advance.
Reference: Abstract Algebra, Dummit and Foote, 3rd ed.
abstract-algebra ring-theory modules free-modules free-abelian-group
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I 'm trying to compare Free Modules and Free Abelian Groups.
We know that,
Definition. An abelian group $G$ is called free abelian group with rank $nin Bbb N^*$, if $G$ is the direct sum of $n$ infinite cyclic groups. That is,
$$G=langle x_1 rangle oplus ... oplus langle x_n rangle$$
where $U_i:=langle x_i rangle, i=1,...,n$ are infinite cyclic groups. Then, $x_1,...,x_n$ are called free generators of $G$, $X:={x_1,...,x_n}$ is called basis (or free generators set) of $G$ and $n$ is called rank of $G$.
We already know the following definition that seems equivalent.
Definition. Let $R$ be a ring with $1_R$. An $R$-module $F$ is called free on a subset $A$ of $F$, if for every nonzero element $xin F$, there exist unique nonzero elements $r_1,...,r_nin R$ and unique
$a_1,...a_n in A$ such that $$x=r_1a_1+...+r_na_n$$ for some $nin
mathbb{Z}^+$ In this situation we say $A$ is a basis or set of
free generators for $F$. If $R$ is commutative ring, the cardinality
of $A$ is called the rank of $F$.
Questions.
Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?
Is it useful to make such comparisons?
Thank you in advance.
Reference: Abstract Algebra, Dummit and Foote, 3rd ed.
abstract-algebra ring-theory modules free-modules free-abelian-group
I 'm trying to compare Free Modules and Free Abelian Groups.
We know that,
Definition. An abelian group $G$ is called free abelian group with rank $nin Bbb N^*$, if $G$ is the direct sum of $n$ infinite cyclic groups. That is,
$$G=langle x_1 rangle oplus ... oplus langle x_n rangle$$
where $U_i:=langle x_i rangle, i=1,...,n$ are infinite cyclic groups. Then, $x_1,...,x_n$ are called free generators of $G$, $X:={x_1,...,x_n}$ is called basis (or free generators set) of $G$ and $n$ is called rank of $G$.
We already know the following definition that seems equivalent.
Definition. Let $R$ be a ring with $1_R$. An $R$-module $F$ is called free on a subset $A$ of $F$, if for every nonzero element $xin F$, there exist unique nonzero elements $r_1,...,r_nin R$ and unique
$a_1,...a_n in A$ such that $$x=r_1a_1+...+r_na_n$$ for some $nin
mathbb{Z}^+$ In this situation we say $A$ is a basis or set of
free generators for $F$. If $R$ is commutative ring, the cardinality
of $A$ is called the rank of $F$.
Questions.
Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?
Is it useful to make such comparisons?
Thank you in advance.
Reference: Abstract Algebra, Dummit and Foote, 3rd ed.
abstract-algebra ring-theory modules free-modules free-abelian-group
abstract-algebra ring-theory modules free-modules free-abelian-group
asked Nov 26 at 1:38
Chris
836411
836411
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Abelian groups are precisely $mathbb{Z}$-modules: the data of a $mathbb{Z}$-module $M$ consists of an abelian group $M$ together with the specification of a ring homomorphism $eta: mathbb{Z} to text{End}(M)$, where the latter denotes the ring of group endomorphisms of $M$. But there is only one possible ring homomorphism from $mathbb{Z}$ to any ring $S$, given by sending $1 in mathbb{Z}$ to the multiplicative unit in $S$, so $eta$ above is uniquely specified hence adds no additional data to the abelian group structure of $M$.
Hence, free abelian groups are the same thing as free $mathbb{Z}$-modules. The translation between the definition you gave for free $R$-modules and a formulation given in terms of direct sums can be obtained by observing that having a free $R$-module $F$ on a set of generators $A subset F$ is equivalent to being able to write down an isomorphism of $R$-modules:
$$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Abelian groups are precisely $mathbb{Z}$-modules: the data of a $mathbb{Z}$-module $M$ consists of an abelian group $M$ together with the specification of a ring homomorphism $eta: mathbb{Z} to text{End}(M)$, where the latter denotes the ring of group endomorphisms of $M$. But there is only one possible ring homomorphism from $mathbb{Z}$ to any ring $S$, given by sending $1 in mathbb{Z}$ to the multiplicative unit in $S$, so $eta$ above is uniquely specified hence adds no additional data to the abelian group structure of $M$.
Hence, free abelian groups are the same thing as free $mathbb{Z}$-modules. The translation between the definition you gave for free $R$-modules and a formulation given in terms of direct sums can be obtained by observing that having a free $R$-module $F$ on a set of generators $A subset F$ is equivalent to being able to write down an isomorphism of $R$-modules:
$$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
add a comment |
up vote
2
down vote
Abelian groups are precisely $mathbb{Z}$-modules: the data of a $mathbb{Z}$-module $M$ consists of an abelian group $M$ together with the specification of a ring homomorphism $eta: mathbb{Z} to text{End}(M)$, where the latter denotes the ring of group endomorphisms of $M$. But there is only one possible ring homomorphism from $mathbb{Z}$ to any ring $S$, given by sending $1 in mathbb{Z}$ to the multiplicative unit in $S$, so $eta$ above is uniquely specified hence adds no additional data to the abelian group structure of $M$.
Hence, free abelian groups are the same thing as free $mathbb{Z}$-modules. The translation between the definition you gave for free $R$-modules and a formulation given in terms of direct sums can be obtained by observing that having a free $R$-module $F$ on a set of generators $A subset F$ is equivalent to being able to write down an isomorphism of $R$-modules:
$$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
add a comment |
up vote
2
down vote
up vote
2
down vote
Abelian groups are precisely $mathbb{Z}$-modules: the data of a $mathbb{Z}$-module $M$ consists of an abelian group $M$ together with the specification of a ring homomorphism $eta: mathbb{Z} to text{End}(M)$, where the latter denotes the ring of group endomorphisms of $M$. But there is only one possible ring homomorphism from $mathbb{Z}$ to any ring $S$, given by sending $1 in mathbb{Z}$ to the multiplicative unit in $S$, so $eta$ above is uniquely specified hence adds no additional data to the abelian group structure of $M$.
Hence, free abelian groups are the same thing as free $mathbb{Z}$-modules. The translation between the definition you gave for free $R$-modules and a formulation given in terms of direct sums can be obtained by observing that having a free $R$-module $F$ on a set of generators $A subset F$ is equivalent to being able to write down an isomorphism of $R$-modules:
$$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$
Abelian groups are precisely $mathbb{Z}$-modules: the data of a $mathbb{Z}$-module $M$ consists of an abelian group $M$ together with the specification of a ring homomorphism $eta: mathbb{Z} to text{End}(M)$, where the latter denotes the ring of group endomorphisms of $M$. But there is only one possible ring homomorphism from $mathbb{Z}$ to any ring $S$, given by sending $1 in mathbb{Z}$ to the multiplicative unit in $S$, so $eta$ above is uniquely specified hence adds no additional data to the abelian group structure of $M$.
Hence, free abelian groups are the same thing as free $mathbb{Z}$-modules. The translation between the definition you gave for free $R$-modules and a formulation given in terms of direct sums can be obtained by observing that having a free $R$-module $F$ on a set of generators $A subset F$ is equivalent to being able to write down an isomorphism of $R$-modules:
$$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$
answered Nov 26 at 2:28
Saad Slaoui
1315
1315
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
add a comment |
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
Thank you for your answer, and sorry for the delay. So, an R-module F is called a free $R$-module if $F$ is isomorphic to a direct sum of copies of $R$: that is $$Fcong R_1 oplus ... oplus R_n oplus ...$$ where $R_i=langle x_i rangle cong R, forall iin {1,2,...,n,...}$. And $X:={x_1,...,x_n}$ is called a basis of $F$. Correct?
– Chris
Nov 27 at 1:34
add a comment |
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