Multiplying in Z/mZ: Solving m=3 & m=7 Questions

[Fernando Revilla]

I quote a question from Yahoo! Answers

Multiply in Z\mZ question: m= 3, 7 also which m is Z/mZ a field? m=7? m=3? show some steps.?
multiply in Z\mZ
question: m= 3, 7 also which m is Z/mZ a field?
m=7?
m=3?

I have given a link to the topic there so the OP can see my response.

 

[Fernando Revilla]

I suppose you want to prove that if $m$ prime, then $\mathbb{Z}/m\mathbb{Z}$ is a field. For all $m\geq 2$ integer, we know that $\mathbb{Z}/m\mathbb{Z}=\{\bar{0},\bar{1},\ldots,\overline{m-1}\}$ is a finite, conmutative and unitary ring. But we also know that a finite integral domain is a field, so we only need to prove that if $m$ prime, there are no divisors of zero.

Suppose $\bar{k}\bar{s}=\bar{0}$, then $ks$ is multiple of $m$ or equivalently $m|ks$. If $m$ prime, $m|k$ or $m|s$ which implies $\bar{k}=\bar{0}$ or $\bar{s}=\bar{0}$.

For example, in the particular case $m=7$ the inverses are
$$(\bar{1})^{-1}=\bar{1},\;(\bar{2})^{-1}=\bar{4},\;(\bar{3})^{-1}=\bar{5},\;(\bar{4})^{-1}=\bar{2},\;(\bar{5})^{-1}=\bar{3},\;(\bar{6})^{-1}=\bar{6}$$