(Proof explanation) In integral domain let there be a prime ideal $(a)$ then $a$ is irreducible. 
Theorem Let $D$ be an integral domain and let $(a)$ be a prime ideal of $D$, then $a$ is irreducible.

Proof: Assume that $1\neq a=bc$, we have to show that either $b=1$ or $c=1$. Because $bc=a\in(a)$ we have that $b\in(a)\vee c\in(a)$. Assume that $b\in(a)$, then $b=ad$ for some $d\in D$ thus $a=bc=adc$ which shows that $dc=1$ "Because $D$ is integral domain, it follows that $c=1$" similarly for $c\in(a)$ we have that $b=1$.
Now, my question is why should $dc=1$ imply $c=1$? What if it is the case that $c$ is inverse of $d$? The OP is Proposition 6 in http://homepages.math.uic.edu/~marker/math330/div.pdf
 A: It's not true, in a general integral domain, that $dc = 1$ implies $c = 1$; what one does have is that $c$ and $d$ are units, that is, divisors of $1$, and this is enough to show $a$ irreducible, since by definition the divisors of an irreducible such as $a$ are its associates, i.e., elements which are of the form $ua$, where $u$ is a unit.
In fact, it is easy to find integral domains in which $cd = 1$ without $c = 1$ or $d = 1$; indeed, 
$cd = 1 \Longrightarrow (-c)(-d) = 1, \tag 1$
so we could take
$c = d = -1; \tag 2$
going a little further, consider the Gaussian integers
$\mathcal Z = \{a + bi \mid a, b \in \Bbb Z; i^2 = -1 \}; \tag 3$
since
$\mathcal Z \subset \Bbb C, \tag 4$
$\mathcal Z$ is clearly an integral domain; but the units in $\mathcal Z$ are $\pm 1$, $\pm i$.
Note Added in Edit, Sunday 2 September 2018 6:21 PM PST:  I response to our OP Michal Dvorak's comment below, I provide the following remarks:
A unit in a commutative unital ring $R$ is a divisor of $1_R$; that is, $u$ is a unit if there exists $v \in R$ with $uv = vu = 1$; clearly, units are invertible.
Units in $R$ are to be carefully distinguished from the multiplicative identity; since we call a ring $R$ with $1_R \in R$ such that
$r1_R = 1_R r = r, \; \forall r \in R \tag 5$
a unital ring or ring with unit, it can get a little confusing.  But $1_R$ is unique, whereas there may be many units $u$ in the sense that $u \mid 1_R$, as the example $\mathcal Z$ shows.  
I hope these remarks clarify the situation.  End of Note.
