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.

Re: Definition of a Z-Module

Quote:

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)

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 = 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".