Example on Z-modules .... Dummit & Foote, Page 339 ....

[Math Amateur]

I am reading Dummit and Foote's book: "Abstract Algebra" (Third Edition) ...

I am currently studying Chapter 10: Introduction to Module Theory ... ...

I need some help with an aspect of Dummit and Foote's example on Z-modules in Section 10.1 Basic Definitions and Examples ... ...Dummit and Foote's example on Z-modules reads as follows:

In the above example we read the following:

" ... ... This definition of an action on the integers on $A$ makes $A$ into a $\mathbb{Z}$-module, and the module axioms show that this is the only possible action of $\mathbb{Z}$ on $A$ making it a (unital) $\mathbb{Z}$-module ... ... "Can someone please explain how/why the module axioms demonstrate that this is the only possible action of $\mathbb{Z}$ on $A$ making it a (unital) $\mathbb{Z}$-module ... ... ? How do we know it is the only possible such action ... ... ?Hope someone can help ...

Peter

 

[member 587159]

Hey peter.

Take any action $\mathbb{Z} \times A \to A: (n,a) \mapsto n.a$ satisfying the module axioms.

It is easy to see that

$n.a = (1+1+ \dots +1).a = 1.a + \dots + 1.a = a + \dots + a$ if n > 0.

The other cases can be treated in the same way. Since I assumed nothing but the action axioms, it follows that if there is an action, it must look like the action that is written is your book.

Hope this helps.

 

[fresh_42]

We have $1.a=a$ and $0.a=0$. Then $n.a=((n-1)+1).a=(n-1).a+1.a=(a+\ldots+a)_{n-1\text{ times }} \cdot a + a=n\cdot a$ by induction. The same works for $-a$ if we use $0=0.a$ to show $(-1).a=-a\,.$

Remark: Here $n.a$ stands for an operation of $\mathbb{Z}$ on $A$ and $n\cdot a$ for $n$ additions, resp. subtractions of $a$, because at prior, $n\cdot a$ isn't defined, so it is merely a short hand notation of $a+\ldots +a$ ($n$ times), which is defined. The assertion is basically that the two don't have to be distinguished, and we can speak of the operation, which makes Abelian groups (additively written) naturally into a $\mathbb{Z}-$module.