Ignoring units in unique factorization and divisibility theory Wikipedia defines UFDs as 

A unique factorization domain is an integral domain R in which every non-zero element can be written as a product of a unit and prime elements of R.

Why is there an insistence on the unit? 
Isn't a prime times a unit another prime? So one could absorb the unit into a prime, and just define the UFD by saying that every non-zero element factors as a product of prime elements. 
Would this be an equivalent definition?
 A: Consider for motivation divisibility theory in the ring of integers $\Bbb Z.\,$ We can reduce such to positive integers by ignoring signs, i.e. by factoring out the group of units $\,U = \langle -1\rangle = \{\pm1\}\,$ from its multiplicative group $\,\Bbb Z^*,\,$ i.e. we consider elements congruent iff they are associate (equal up to unit factors, so here up to sign).
The same method works in any domain when we wish to ignore units. The quotient monoid is known as the reduced monoid and it is the standard place to begin study of factorization in general domains (at least for those properties that purely multiplicative, i.e. monoid-theoretic).
Many properties of domains are purely multiplicative so they can be described purely in terms of monoid structure. Let $R$ be a domain with fraction field $K.$ Let $R^*$ and $K^*$ be the multiplicative groups of units of $R$ and $K$ respectively. Then $G(R)$, the divisibility group of $R$, is the quotient group $K^*/R^*.\,$ This group nicely encodes many familiar factorization properties. For example:

*

*$R$ is a UFD$\iff G(R)$ is a sum of copies of $\Bbb Z$.


*$R$ is a gcd-domain $\iff G(R)$ is lattice-ordered (${\rm lub}(x,y)$ exists)


*$R$ is a valuation domain $\iff G(R)$ is linearly ordered


*$R$ is a Riesz domain $\iff G(R)$ is a Riesz group, i.e. an ordered group satisfying the Riesz interpolation property: if $a,b\le c,d$ then $a,b\le x\le c,d$ for some $x.$ A domain $R$ is called Riesz if every element is primal, i.e. $\,A\mid B\,C \Rightarrow\, A=bc,\ b\mid B,\ c\mid C,\,$ for some $\,b,c\in R.$
For more on divisibility groups see the following surveys:
J.L. Mott, Groups of divisibility: A unifying concept for integral domains and partially ordered groups, Mathematics and its Applications, no. 48, 1989, pp. 80-104.
J.L. Mott, The group of divisibility and its applications, Conference on Commutative Algebra (Univ. Kansas, Lawrence, Kan., 1972), Springer, Berlin, 1973, pp. 194-208. Lecture Notes in Math., Vol. 311. MR 49 #2712
A: I think this is what it's aiming at. Let's suppose we have the collection of equivalence classes of associates in front of us. Select a representative from each one, and call the set of representatives $\mathscr P$.
Every nonzero element of $R$ can be written as a finite product of powers of members of $\mathscr P$, times a unit.
Now imagine you have a different prime factorization in terms of primes that aren't these representatives, so that they differ from the representatives by a unit. You can compress all those extra units into a single one up front, and presto, convert it into a factorization purely in terms of representatives.
Thinking of it this way allows you to move from factorizations using random prime elements to a "nice" factorization in terms of a minimal set of representatives, modulo a unit.
