A proof that $\mathbb{Z}_{p}$ is an integral domain if and only if $p$ is prime.

Question

I have already seen a proof of this statement here on stackexchange but I would like someone to verify whether "my proof" is correct and also if you could point out issues with the writing style. thanks in advance!

Statement. $\mathbb{Z}_{p}$ is an integral domain $\Leftrightarrow$ $p$ is prime.

Proof.$(\Rightarrow)$ Suppose $\mathbb{Z}_{p}$ is an integral domain. Assume (by way of contradiction) that $n$ is not prime. Then $n=ab$ for some $a,b\in \mathbb{Z}$ where $1<a$ and $b<n$, meaning $ab\equiv 0\pmod n$. This is a contradiction since $\mathbb{Z}_{p}$ is assumed to be an integral domain and therefore does not contain any zero divisors. Therefore $p$ must be prime.

$(\Leftarrow)$ Assume $p$ is prime. Then $a$ does not divide $n$ for any $1<a<n$ where $a\in \mathbb{Z}_{p}$. Thus $\mathbb{Z}_{p}$ does not contain any zero divisors and is therefore an integral domain.

Answer 1

I have a possibly more rigorous proof for the $\Leftarrow$ proof:

Assume $p$ is prime. Now, consider $ab \equiv 0 \pmod p$. This means, $ab=np$ for some $n \in \Bbb{Z}$. Thus, $p \mid ab$, so by the definition of a prime element, either $p \mid a$ or $p \mid b$. Therefore, either $a \equiv 0 \pmod p$ or $b \equiv 0 \pmod p$. This proves that $\Bbb{Z}_p$ has no zero divisors. It is trivial to show that it is a commutative ring with identity, so this means that $\Bbb{Z}_p$ is an integral domain.

Answer 2

$(\Rightarrow)$ Your proof is correct, but you start with $p$, then you start calling it $n$ and in the end it's $p$ again. Don't do that.

$(\Leftarrow)$ There is the same problem here. Besides, when you wrote $1<a<n$, you should have written $1\leqslant a<b$. Finally, you should explain how you passed from $a\nmid p$ to the assertion that $\mathbb Z_p$ has no zero divisors.