Generalized Euclid's lemma for factorial rings.

Question

Let $A$ be a factorial ring. Is the following result true in this generality: if $a|bc$ and $a$, $b$ are coprime then $a|c$? It would be great if this were true, but I am not at all true. I think the classical proof should go through

Answer 1

As we are working in a factorial ring (what I learned to call a unique factorization domain), we have that every element can be written as a product of irreducibles and if two such products are equal then there is a bijection between terms with each an associate of the other.

Furthermore, for definition of coprime, we mean that if $a$ is coprime to $b$ that no divisor of $a$ is also a divisor of $b$ and vice versa (except for a unit). I.e. in the factorizations, none of $a$'s factors are associate to any of $b$'s factors.

Let $a = a_1a_2\dots a_n$, $b = b_1b_2\dots b_o$, $c=c_1c_2\dots c_p$

$a|bc \Rightarrow ka = bc\Rightarrow k_1k_2\dots k_ma_1a_2\dots a_n=b_1b_2\dots b_oc_1c_2\dots c_p$

Since $a$ is coprime to $b$, that means that none of $a_1,a_2,\dots,a_n$ are associates to $b_1,b_2,\dots,b_o$, however each of $a_1,a_2,\dots,a_n$ must be associate to something on the righthandside of the equation, implying that each of $a_1,a_2,\dots,a_n$ are associates to some collection of $c_1,c_2,\dots,c_p$ (not necessarily all).

As such, by a permutation of indices and by grouping of units, we may assume without loss of generality that $a_1 = uc_1, a_2 = c_2, a_3 = c_3,\dots, a_n = c_n$ with $n\leq p$ and $u$ a unit.

Thus, $a(u\prod_{i=n+1}^p c_i) = c$ and thus $a|c$.

Answer 2

Yes, it remains true in any UFD (or any gcd domain), but you cannot use the classical proof via the gcd Bezout-identity since UFDs needn't be Bezout. Here are a couple alternative proofs.

$\qquad a\mid bc\,\Rightarrow\,a\mid ac,bc\,\Rightarrow\, a\mid (ac,bc) = \overbrace{(a,b)}^{\large =\,1}c = c\, $ by the gcd Distributive Law.

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Or, by induction on $\,n\,$ = #prime factors of $\,a.\,$ Clear if $\,n=0\,$ since then $\,a\,$ is a unit so $\,a\mid c.\,$ Else $\,a = pd\,$ for a prime $\,p\,$ and $\,(p,b)=1,\,\ p\mid bc\,\Rightarrow\, p\mid c.\,$ Thus $\,d\mid b(c/p)\,$ so by induction $\,d\mid c/p,\,$ thus $\,a=pd\mid c.$

Answer 3

If by "factorial ring" you mean a unique factorisation domain, then the proof is easy. Just note that if $p \mid a$ and $p$ is prime, then $p \nmid b$ by hypothesis, so $p \mid c$.

This is because by definition a prime element is a $p$ such that if $p \mid bc$ then either $p \mid b$ or $p \mid c$. Also note that in a UFD every irreducible element, i.e. every element which is not the product of two non-invertible elements, is prime.