Proof that in an integral domain every prime element is irreducible

Question

So I'm trying to prove that in an integral domain every prime element is irreducible.

Let's suppose that $p$ is an element in my integral domain.

So now I can study $ p=ab $ (if is not in this form, $p=a*1 $ and this is an irreducible element).

1. $ p $ divides itself so p|p so $ p=ab.$
2. $p|a $ or p|b (because p is prime) and now, WLOG let's suppose that p|a.
3. If $p|a $ need to exist an element $ q $ such that $a=pq.$
4. I can multiply both sides with $ b $ so $ p=pqb.$
5. $p(1-qb)=0 $ so $ qb=1$ (this is true because we are working in an integrity domain and if $ a*b=0 $ a needs to be $0 $ or $ b$ needs to be $0$) and now I can say that b is a unit and $ p=ab $ is an irreducible element.

My question is, why b needs to be a unit element of my domain? Cannot be an invertible element? Like $4*(1/4)$=1 in $\mathbb{Q}$. And if $ b$ can be an invertible element, why does that mean that $ p $ is irreducible?

Sorry for my english, any help would be appreciated! By the way, I'm using Artin's book and $ a $ in $ R $ where $ R$ is a ring is an irreducible element if is not a unit and if it has no proper divisors.