Show that if $m$ is prime then the product of two non-zero elements of $\mathbb Z/m\mathbb Z$ is again non-zero I am having some trouble concluding this. 
I have the following:
$[a][b]=[ab]$ by the definition of multiplication in $\mathbb Z/m\mathbb Z$
By the remainder theorem, we know that $a=qm+r$ and $b=km+s$ where $q,k,r,s\  \epsilon\  \mathbb Z$
Additionally,  $0\lt r\lt m$ and  $0\lt s\lt m$ 
*It is greater than (not greater than or equal to) because at the onset of this problem, we have defined $[a]$ and $[b]$ to be non-zero
So, with those stated:
$[a][b]=[qm+r][km+s]=[(qm+r)(km+s)]$ 
which is equal to $[m^2qk+qms+mkr+rs]$ 
which is simply equal to $[rs]$
This is where I run into trouble. I am aware that I need to show that $[rs]$ is not equal to zero (i.e. $r*s$ should not be a divisor of $m$ or some multiple of $m$...say $\alpha m$.)
I thought that I could maybe setup a contradiction by stating something like:
"Assume $r*s=\alpha* m$" and then go on to show this can only be the case if $r$ or $s$ is not coprime with m (we know that $r$ and $s$ must be coprime with $m$ if $m$ is prime because $r$ and $s$ are both less than $m$ and greater than $0$...which would be a contradiction!)
Unfortunately, I cannot figure out how to explicitly demonstrate this. Any help would be appreciated! 
edit: Using the recommended Euclid's Lemma, while retaining the structure of my argument, we have the following:
Assume $r*s=\alpha* m$
This means that $m|r*s$. Recall, $m$ is prime and therefore, by Euclid's Lemma, we have either $m|r$ or $m|s$. This produces a contradiction for the following reason:
$r$ and $s$ have both been defined to be between $0$ and $m$. A number cannot evenly divide a number that is smaller than itself but greater than $0$.
Therefore, $rs\neq \alpha*m$ which means that $[rs] \neq[0]$ (i.e. it is non-zero). 
 A: If $[rs]$ is zero then $m \mid rs$ but $m$ is prime so $m \mid r$ or $m \mid s$ (no loss of generality) let's assume that $m\mid r$ then $m \mid a = qm +r$, i. e., $[a]$ is zero, wich is a contradition.
A: Slightly different approach: 
When $m$ is prime, $m\Bbb Z$ will be a maximal ideal in $\Bbb Z$. (This is easy using Bezout's identity.) 
Thus $\dfrac {\Bbb Z}{m\Bbb Z}$ is a field,  and in particular an integral domain. (That an integral domain modulo a maximal ideal is a field is also a well known result.)
A: The answer is much easier. 
Take $~x~$ a non multiple of $~m~$. Then the additive cyclic group generated by $~[x]~$ is cyclic of a order dividing $~m~,~$ thus $~m~$ since $~m~$ is prime. 
Therefore $~n[x]=[nx]=0=[m]~$ if and only if $~n~$ multiple of $~m~$. 
A: We know that if $\mathbb{Z}/m\mathbb{Z}$ contains nonzero $a,b$ such that $ab = 0$, then $\mathbb{Z}/m\mathbb{Z}$ is not a field. We also know that $\mathbb{Z}/m\mathbb{Z}$ is a commutative ring with identity element, so it suffices to show that for nonzero $a$, the function defined by $x \mapsto ax$ is injective. Since $\mathbb{Z}/m\mathbb{Z}$ is finite, this also implies the bijectivity of the function, which means there is some $x$ such that $ax=1$, proving the statement.
A: You can also think of it as the following:
Since $\mathbb{Z}/m\mathbb{Z}$ is isomorphic to $\mathbb{Z_m}$ and $m $ is prime, hence $\mathbb{Z_m}$ is field and we know that a field has no zero divisors, whence the result follows.
