# Thread: Definition of a Z-Module

1. ## 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 $\displaystyle \mathbb{Z}$ module (see attached)

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

Let R = $\displaystyle \mathbb{Z}$ module

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

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

For any $\displaystyle n \in Z$ and $\displaystyle a \in A$ 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 $\displaystyle \mathbb{Z}$ module, and the module axioms show that this is the only possible action of Z on A making it a (unital) $\displaystyle \mathbb{Z}$ module. Thus every abelian group is a $\displaystyle \mathbb{Z}$ 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) $\displaystyle \mathbb{Z}$ 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

2. ## 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.

3. ## Re: Definition of a Z-Module

Originally Posted by Drexel28
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 $\displaystyle 1\cdot a=a$ for all $\displaystyle a\in A$, and then the fact that $\displaystyle m\cdot a=a+\cdots+a$ (with m summands) follows from the fact that $\displaystyle m=1+\cdots+1$ (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 $\displaystyle a \in A$.

Thus the argument goes as follows:

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

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

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 $\displaystyle a \in A, n \in \mathbb{Z}$

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)

4. ## 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".