# A nuance in the definition of UFD

We know that an integral domain $R$ is a UFD (unique factorization domain) if each nonzero, nonunit element of $R$ can be written as a product of irreducibles, and the factorization is unique if the following sense: If $r = q_1\cdots q_n = q'_1\cdots q'_m$ (given two factorization of $r$ into irreducibles), then $n = m$ and for each $i$ there is $j$ such that $a_i$ and $a_j$ are associates.

But does the converse also hold? Say, $r = q_1\cdots q_n$. If $q'_1\cdots q'_n$ is an element of $R$ such that $q'_1$ and $q_1$ are associates, then $q'_1\cdots q'_n = r$. It it true, or no?

If $(q'_1) = (q_1)$, then we can write $q'_1\cdots q'_n$ as $u_1\cdots u_nq_1\cdots q_n$(where $u_i$ is a unit in $R$ such that $q_i = u_iq'_i$). And they are equal if an only if $u_1\cdots u_n = 1$. I see no reason for that to always hold. Is there one?

• You just obtained a necessary and sufficient condition for your claim to be true.
– Pedro
Commented Sep 19, 2016 at 0:11
• There's none. If you replace each $q_i$ with an associate irreducible element, you'll only get an element that's associate to $r$. Commented Sep 19, 2016 at 0:22
• It doesn't really make sense to talk about the converse of a definition. But the claim of your second paragraph is false: take $R = \Bbb{Z}$, $n = 1$, $r = q_1=1$ and $q_1' = -1$. Commented Sep 19, 2016 at 0:22

As @Rob Arthan pointed out (and as you noticed yourself), there is no reason to believe that $u_1...u_n=1$.
For example, $6=2 \cdot 3$. And even though $2 \sim (-2)$ and $3 \sim 3$, we know that $$2 \cdot 3 \neq (-2) \cdot 3$$