Show that if $m$ is prime then the product of two non-zero elements of $\mathbb Z/m\mathbb Z$ is again non-zero

Question

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).

Answer 1

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.

Answer 2

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 $~m~$.

Answer 3

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.

Answer 4

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.

Answer 5

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.)