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.
On page 337 Dummit and Foote define a left R-Module (see attached) and then on page 339 they define and describe a module (see attached)
The definition of a module is as foillows:
Let R = module
Let A be any abelian group (finite or infinite) and write the operation of Aas +.
Make A into a module as follows:
For any and 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 module, and the module axioms show that this is the only possible action of Z on A making it a (unital) module. Thus every abelian group is a 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) 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
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 for all , and then the fact that (with m summands) follows from the fact that (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 .
Thus the argument goes as follows:
so a -module is a unital module 1n = n for all
Thus we must have (1 + 1)a = 1a + 1a = 2a because of distributivity and addition of integers for all
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
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)
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 = a^{n}.
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".