Why is $\mathbb{Z}_n$ a field only if $n$ is prime?

Question

And $\mathbb{Z}_n$ is the set of residue classes modulo $n$.

Answer 1

If $n$ is composite say $n=ab$ then we get $[a][b]=[n]=[0]$ ; $[a],[b]\neq [0]$;hence not even an Integral Domain

Answer 2

Let $[a] \in \Bbb Z_n$ denote the residue class of $a \in \Bbb Z$. In order for $\Bbb Z_n$ to be a field, all non-zero elements must have multiplicative inverses. That is, we must be able to say that for every $a$ with $[a] \neq [0]$, there exists an $x \in \Bbb Z$ such that $[a][x] = [1]$.

In other words: for every $a$ with $n \nmid a$, the equation $ax \equiv 1 \pmod n$ must have a solution $x \in \Bbb Z$. That is, the equation $ax = 1 + ny$ must have a solution with $x,y \in \Bbb Z$. That is, there must exist $x,y \in \Bbb Z$ such that $$ ax - ny = 1 $$ whenever $n \nmid a$. By Bézout's identity, this is equivalent to saying that $\gcd(a,n) = 1$ whenever $n \nmid a$. That is, $\Bbb Z_n$ is a field if and only if $n$ is prime.