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Math Help - Definition of a Z-Module

  1. #1
    Super Member Bernhard's Avatar
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    Definition of a Z-Module

    On page 337 Dummit and Foote define a left R-Module (see attached) and then on page 339 they define and describe a  \mathbb{Z} module (see attached)

    The definition of a  \mathbb{Z} module is as foillows:

    Let R =  \mathbb{Z} module

    Let A be any abelian group (finite or infinite) and write the operation of Aas +.

    Make A into a  \mathbb{Z} module as follows:

    For any  n \in Z and  a \in A define na as follows:

    If n > 0 then na = a + a + ... ... + a (n times)

    If n = 0 then na = 0

    If n < 0 then na = -a - a - ,,, ,,, -a (-n times)

    where 0 is the identity of the additive group.


    D&F then write the following:

    This definition of the action of the integers on A makes A into a  \mathbb{Z} module, and the module axioms show that this is the only possible action of Z on A making it a (unital)  \mathbb{Z} module. Thus every abelian group is a  \mathbb{Z} module.


    My problem/question is with the statement: "the module axioms show that this is the only possible action of Z on A making it a (unital)  \mathbb{Z} module."

    While this seems intuitive how do we know that this is true? How would one go about showing this is true?

    Can anyone help with this issue?

    Peter
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Definition of a Z-Module

    Basically because we are requiring $1\cdot a=a$ for all $a\in A$, and then the fact that $m\cdot a=a+\cdots+a$ (with $m$ summands) follows from the fact that $m=1+\cdots+1$ ($m$ summands) and applying the distributivity law.
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Definition of a Z-Module

    Quote Originally Posted by Drexel28 View Post
    Basically because we are requiring $1\cdot a=a$ for all $a\in A$, and then the fact that $m\cdot a=a+\cdots+a$ (with $m$ summands) follows from the fact that $m=1+\cdots+1$ ($m$ summands) and applying the distributivity law.

    Thanks Drexel28!

    You write: "Basically because we are requiring 1\cdot a=a for all a\in A, and then the fact that m\cdot a=a+\cdots+a (with m summands) follows from the fact that m=1+\cdots+1 (m summands) and applying the distributivity law."

    So basically you are saying the definition of the action of Z on A is determined in a Z-module by the requirement that 1a = a for all  a \in A .

    Thus the argument goes as follows:

     1 \in \mathbb{Z} so a  \mathbb{Z} -module is a unital module  \ \Longrightarrow    1n = n for all  n \in \mathbb{Z}

    Thus we must have (1 + 1)a = 1a + 1a = 2a because of distributivity and addition of integers for all  a \in A

    Then it follows similarly that (1 + 1 + 1)a = 1a + (1a + 1a) = 1a + 2a = 3a

    and so on so that

    (1+ 1 + ... ... +1) (n summands) = na for any  a \in A, n \in \mathbb{Z}

    Similar for n = 0 or n negative.

    Thus the group action of Z on A must be defined as Dummit and Foote do on page 339 (see attached)
    Last edited by Bernhard; August 3rd 2013 at 12:47 AM.
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  4. #4
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    Re: Definition of a Z-Module

    Mhm. In other words, "Z-module" and "Abelian group" refer to the exact same objects. If one writes the product of an abelian group multiplicatively, then:

    na = an.

    In other words the action of Z on abelian groups was introduced to you as soon as you started to examine cyclic groups, disguised as "exponent rules".
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