I was trying to solve some exercise from my abstract algebra course, and at some point, I assumed $\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}_2$ is an integral domain, because I thought to remember that $\mathbb{Z}_p$ is integral domain iif $p$ is a prime number. Am I right? If I am, how can I prove it to add it to my execise? If not, could you give me a counter example? Any help will be appreciated, thanks in advance.
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 $1&lt;a$ and $b&lt;n$, 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 $1&lt;a&lt;n$ where $a\in \mathbb{Z}_{p}$. Thus $\mathbb{Z}_{p}$ does not contain any zero divisors and is therefore an integral domain.
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