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.




  1. Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?


  2. Is it useful to make such comparisons?



Thank you in advance.





Reference: Abstract Algebra, Dummit and Foote, 3rd ed.










share|cite|improve this question


























    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.




    1. Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?


    2. Is it useful to make such comparisons?



    Thank you in advance.





    Reference: Abstract Algebra, Dummit and Foote, 3rd ed.










    share|cite|improve this question
























      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.




      1. Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?


      2. Is it useful to make such comparisons?



      Thank you in advance.





      Reference: Abstract Algebra, Dummit and Foote, 3rd ed.










      share|cite|improve this question













      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.




      1. Can we define similarly free modules (with the use of direct sum and with infinite cyclic modules)?


      2. 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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      asked Nov 26 at 1:38









      Chris

      836411




      836411






















          1 Answer
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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:



          $$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$






          share|cite|improve this answer





















          • 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













          Your Answer





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          up vote
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          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:



          $$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$






          share|cite|improve this answer





















          • 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

















          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:



          $$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$






          share|cite|improve this answer





















          • 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















          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:



          $$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$






          share|cite|improve this answer












          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:



          $$ bigoplus_{a in A} Ra to F, \ (r_a)_{a in A} mapsto sum_{a in A} r_a a. $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • 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




















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