Let $\gamma$ be the natural map of $\mathbb Z$ into $\Bbb Z_n$, given by $\gamma(m)=r$ where $r$ is the remainder given by the Division Algorithm when $m$ is divided by $n$, show that $ \gamma$ is a homomorphism.
Solution:
As I solve this problem, I show that
$$\gamma(s+t)=\gamma(s)+\gamma(t),$$ for $s,t \in\mathbb{Z}$.
Using the Division algorithm,
$$s=q_1n+r_1$$ and $$t=q_2n+r_2,$$ where $0\le r_i\lt n$, for $ i=1,2$.
If $ r_1+r_2=q_3n+r_3$, for $0\leq r_3\lt n$.
Adding $s$ and $t$, I got
$$s+t=(q_1+q_2+q_3)n+r_3.$$
Here, I just don't know what is next. I'm having trouble in making conclusion in the last line of my solution. Please help me.
Best Answer
To prove that it is a $\mathbb{Z}$-module homomorphism, we have to first prove that it is a homomoprhism of groups, and second that it respects the $\mathbb{Z}$-module operation, i.e. $\gamma(\alpha t)=\alpha\gamma(t)$ for every $\alpha\in\mathbb{Z}$. I would do it like the following.
First of all, recall that $$\mathbb{Z}_n=\dfrac{\mathbb{Z}}{n\mathbb{Z}}=\{x+n\mathbb{Z}\mid x\in\mathbb{Z}\},$$ i.e. the group of lateral classes of $n\mathbb{Z}$ in $\mathbb{Z}$, where the sum is defined in the natural way as $$(a+n\mathbb{Z})+(b+n\mathbb{Z})=a+b+n\mathbb{Z}.$$
In your case, then, $$\gamma(s)=s+n\mathbb{Z}, \quad \gamma(t)=t+n\mathbb{Z}$$ and $$\gamma(s+t)=s+t+n\mathbb{Z}.$$
By plugging these definitions in the definition of the operation above, we obtain the first part of the statement, i.e. it is a groups homomorphism.
For the second part, we have to recall the definition of the (natural) action of $\mathbb{Z}$ over $\mathbb{Z}_n$, i.e. $$\begin{aligned} \mathbb{Z}\times\mathbb{Z_n}&\longrightarrow\mathbb{Z}_n\\ (\alpha, s+n\mathbb{Z})&\longmapsto \alpha(s+n\mathbb{Z})=(\alpha s)+n\mathbb{Z} \end{aligned} $$
For this and from the definition of $\gamma$, then, $$\alpha \gamma(s)=\alpha(s+n\mathbb{Z})=(\alpha s)+n\mathbb{Z}=\gamma(\alpha s).$$
And so also the second condition is verified.