Any unit has an irreducible decomposition The proposition is the following:

Let $R$ be a principal ideal domain. Then every $a \in R$ with $a \neq 0$ has an irreducible decomposition, that is, there is a unit $u$ and irreducible elements $p_1, \dots, p_n$ such that $a = up_1\dots p_n$.

We learned that an irreducible element is non unit, non zero, can not be written as a product of two non unit, and that the unit is the element that has a multiplicative inverse in ring theory.
But how come in the proof of this proposition, it says: "Any unit clearly has an irreducible decomposition" ? Why? How can a unit have an irreducible decomposition? Isn't irreducible a non unit? 
For example, $1$ in $(\mathbb{Z},+, \cdot)$, it is a unit. What is its irreducible decomposition?

In the definition of an "irreducible decomposition" $a = up_1\dots p_n$, it is possible to have $n=0$.  Then you have no irreducible factors $p_i$ at all and just have a unit $u$, so you are saying $a=u$.  So, for instance, the irreducible decomposition of $1$ is just $1=1$ (with $n=0$ and $u=1$).
