In UFD every irreducible element is prime

Question

I have seen the proof given earlier in a similar question here. But I have a confusion in that proof. If $p$ be the irreducible element of the UFD $R$ with $p|ab$ , then, $ab=pc$ , for some $c \in R$. Then at one point we can write $a,b,c$ as the unique product of irreducible elements since $R$ is UFD. Here my question is that why c is a non-unit(as a result of which we were able to factorize it since it is there in the defn only that for factorization the element must be non-unit). Also in some other sites I saw that in this proof they have assumed $c$ to be non-unit, but in a book which I follow in my university, it's written that it $c$ must be a non-unit. There is no such assumption there.But in that book the reason is not written, they have asked to prove it to the readers. Any explanation will be of great help. Thankyou.