$\mathbb{Z}_p$ with $p$ prime is integral domain

Question

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.

Answer 1

Being an integral domain means: if $a,b\in R$ are such that $ab=0$ then $a=0$ or $b=0$.

In case of $\mathbb{Z}/n\mathbb{Z}$ this translates to: if $a,b\in\mathbb{Z}$ are such that $n|ab$ then $n|a$ or $n|b$. Then the classical number theory shows that this works only for prime $n$.

Answer 2

You're right. $\Bbb Z_p$ is an integral domain if $p$ is prime. Otherwise, it's not.

First, the units of $\Bbb Z_p$ are the elements $a$ with $gcd(a,p)=1$ and so are the nonzero elements of $\Bbb Z_p$. For the proof use Euclid's algorithm and Bezout.

Moreover, units are not zero divisors, since if $a$ is a unit with $ba=0$, the

$b= b1 = b(aa^{-1}) = (ba)a^{-1} = 0 a^{-1}=0$.