Confusion with definition of irreducible. An integer $p$ is said to be irreducible if whenever $p=ab$ then $a $ or $b$ is $1$ or $-1$.  
Then we define an irreducible element $p$ in a commutative ring $R$ with unity as:  

$1)$ $p \neq 0$ and $p$ isn't unit, and
  $2)$ whenever $p=ab$ then one of $a$ and $b$ is a unit.  

I can't understand $2)$. Shouldn't it be that one of $a$ and $b$ is unity instead of unit, if we see this from definition of an integer...  
I can't understand. Please help.
 A: If the definition of unity you have helpfully put in your comments is applied in the integers, consider the number $7=1\times 7=7\times 1=-7\times -1=-1\times -7$
It clearly has factorisations involving $-1$, and $-1$ is a unit but not a unity: note that $-1\times 7=-7\neq 7$.
For the integers, therefore, we already need a definition of irreducible which goes beyond "unity" to the beginnings of the idea of "unit" - we need to include $-1$ as well as $1$ otherwise the definition does not work as we would like and every line needs a comment on $-1$ as an exception.
The idea of unit is generalised in other contexts to all elements with a multiplicative inverse. The point is that the definition is not useful unless we pick out the units in this way.
A: Saying "one of a and b is a unit" is the same as saying "one of a and b is +1 or -1." Note: the multiplicative identity (typically 1) and the negative multiplicative identity (-1) are ALWAYS units. A unit is simply an invertible element of a ring.
