Using the fundamental theorem of arithmetic, it's easy to prove this proposition:

Proposition. Every divisor of $mn$ can be written as the product of a divisor of $m$ to a divisor of $n$.

My question: How heavily does the proposition rely on the fundamental theorem of arithmetic? Is there anyway to prove it with this theorem or one of its equivalents? What happens in rings which prime factorization is not unique?


Your property is $(2)$ in the list below of properties of domains that are equivalent to uniqueness of factorizations into atoms (= irreducibles).

Nonunits satisfying $(2)$ are called primal. One easily checks that atoms are primal $\iff$ prime. Products of primes are also primal. So "primal" may be viewed as a generalization of the notion "prime" from atoms to composites (recall a nonunity $\,\rm p\,$ is prime if $\rm\,p\mid ab\,\Rightarrow\, p\mid a\,$ or $\rm\,p\mid b)$.

$\rm(1)\ \ \ gcd(a,b)\:$ exists for all $\rm\:a,b\ne 0\ \ \ $ [GCD domain]

$\rm(2)\ \ \ a\mid BC\:\Rightarrow a=bc,\ b\mid B,\ c\mid C\ \ $ [Schreier refinement, Euler's four number theorem]

$\rm(3)\ \ \ a\,\Bbb Z + b\, \Bbb Z\, =\, c\,\Bbb Z,\:$ for some $\rm\,c\quad\ \ $ [Bezout domain]

$\rm(4)\ \ \ (a,b)=1,\ a\mid bc\:\Rightarrow\: a\mid c\qquad\ \ $ [Euclid's Lemma]

$\rm(5)\ \ \ (a,b)=1,\ \dfrac{a}{b} = \dfrac{c}{d}\:\Rightarrow\: b\mid d\quad\ \ $ [Unique Fractionization]

$\rm(6)\ \ \ (a,b)=1,\ a,b\mid c\:\Rightarrow\: ab\mid c$

$\rm(7)\ \ \ (a,b)=1\:\Rightarrow\: a\,\Bbb Z\cap b\,\Bbb Z\, =\, ab\,\Bbb Z $

$\rm(8)\ \ \ gcd(a,b)\ \ exists\:\Rightarrow\: lcm(a,b)\ \ exists$

$\rm(9)\ \ \ (a,b)=1=(a,c)\:\Rightarrow\: (a,bc)= 1$

$\rm(10)\ $ atoms $\rm\, p\,$ are prime: $\rm\ p\mid ab\:\Rightarrow\: p\mid a\ \ or\ \ p\mid b$

Which of these properties sheds the most intuitive light on why uniqueness of factorization entails? If I had to choose one, I would choose $(2),$ Schreier refinement, a.k.a. Euler's four number theorem (Vierzahlensatz),$\,$ or $\,$ Riesz interpolation. If you extend this by induction it implies that any two factorizations of an integer have a common refinement. For example if we have two factorizations $\rm\: a_1 a_2 = n = b_1 b_2 b_3\:$ then Schreier refinement implies that we can build the following refinement matrix, where the column labels are the product of the elements in the column, and the row labels are the products of the elements in the row

$$\begin{array}{c|ccc} &\rm b_1 &\rm b_2 &\rm b_3 \\ \hline \rm a_1 &\rm c_{1 1} &\rm c_{1 2} &\rm c_{1 3}\\ \rm a_2 &\rm c_{2 1} &\rm c_{2 2} &\rm c_{2 3}\\ \end{array}$$

This implies the following common refinement of the two factorizations

$$\rm a_1 a_2 = (c_{1 1} c_{1 2} c_{1 3}) (c_{2 1} c_{2 2} c_{2 3}) = (c_{1,1} c_{2 1}) (c_{1,2} c_{2 2}) (c_{1,3} c_{2 3}) = b_1 b_2 b_3$$

This immediately yields the uniqueness of factorizations into primes (atoms). It also works more generally for factorizations into coprime elements, and for factorizations of certain types of algebraic structures (abelian groups, etc).

See also this answer for further closely related properties.


A counterexample is $R = Z[\sqrt{-5}]$. The ring $R$ is a Dedekind domain (which is certainly among the luckier things that can happen to you if you are not a unique factorization domain. In fact, there is even something like a unique factorization for ideals in them).

In R, we have $$6 = 2 \cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5}).$$ Thus, $(1-\sqrt{-5})$ divides 6. But $(1-\sqrt{-5})$ neither divides 2 nor 3.

You can show this by using norms $$N(a+b\sqrt{-5}) = a^2+5b^2.$$ These are multiplicative in R, meaning that $ N(xy)=N(x)N(y)$ for $x,y\in R$. Thus, $x \mid z$ in $R$ implies $N(x) \mid N(y)$ in $\Bbb Z$. But $N(1-\sqrt{-5}) = 1+5 = 6$ doesn't divide $N(2)=4$, nor does it divide $N(3)=9$.


(*) Every divisor of $mn$ can be written as the product of a divisor of $m$ to a divisor of $n$.

In a domain where property (*) holds, every irreducible element is prime.

Indeed, let $i$ be an irreducible element and suppose $i$ divides $mn$. Then $i=uv$, where $u$ divides $m$ and $v$ divides $n$. Since $i$ is irreducible, we must have $i \sim u$ or $i \sim v$. In any case, $i$ divides $m$ or $n$, and so $i$ is prime.

Hence, a domain that has factorization into irreducibles and property (*) must be a UFD.

Thus, property (*) is very close to providing uniqueness of factorization, but not existence.


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