Zero divisors in $[a]_n[b]_n=[0]_n$

Question

Precisely,what must be true about n for there to be zero divisors in $ℤ_$ (i.e. elements $[a]_n$ and $[b]_n$ such that $[a]_n[b]_n=[0]_n$ but $[a]_n,[b]_n≠[0]_n$? State your theorem as an 'if and only if' statement. [hint: Consider gcd(a,n) and/or gcd(b,n)]-------This is the problem I posted before.

I figured that $\mathbb{Z}_2$,$\mathbb{Z}_3$,$\mathbb{Z}_4$,$\mathbb{Z}_5$,$\mathbb{Z}_7$ don't have zeor divisors. I think for there to be zero divisiors in$\mathbb{Z}_n$, $n$ should not be prime numbers. But I am not sure what to do with $\mathbb{Z}_4$ because 4 is not prime. It doesn't go with my assumption.

Also, in $\mathbb{Z}_6$, $gcd(2,6)=2$, $gcd(3,6)=3$. In $\mathbb{Z}_8$,$gcd(2,8)=2$, $gcd(4,8)=4$.. so on. I found that $gcd(a,n)=a$ and $gcd(b,n)=b$. So is this fact should be on my statement?

Help! Due tomorrow!

Answer 1

$\textbf{Hint:}$ We have $[a]_n [b]_n = [0]_n$ if and only if $n|ab$ in $\mathbb{Z}$. What does this imply if $n$ is prime? And if $n$ is not prime?

Answer 2

If $n$ is compositive, say $ab=n$, then the classes $[a]=x$ and $[b]=y$ will be nonzero, but $xy=0$.

If $n=p$ a prime, we know that $p\mid ab\implies p\mid a$ or $p\mid b$. This means that in $\Bbb Z/(p)$, $$xy=0 \implies x=0 \text{ or }\; y=0$$ Thus there are no zero divisors. Moreover, since $p\not\mid a$ is equivalent to $(p,a)=1$, by Bezout we have some $m,n$ for which $an+mp=1$, thus $an=1\mod p$ and every non-zero element has an inverse.

Thus, for $n$ composite $\Bbb Z/(n)$ is not even an integral domain, and for $n=p$ a prime $\Bbb Z/(p)$ is a field.