# An integral domain with the factorization property and gcd for every two elements is a UFD

Theorem 0.6.1 of Roman's book Field Theory says:

Let $R$ be an integral domain for which the factorization property holds (factorization property means that every non zero non unit can be written as a product of irreducibles). The following conditions are equivalent:

1) $R$ is a UFD

2) Every irreducible element of $R$ is prime

3) Any two non-zero elements of $R$ have a greatest common divisor.

I showed that 1) implies 2) and that 2) implies 3), but I don't see why 3) implies 1)

Question: What is the proof that 3) implies 1) ?

Thank you

• It may be easier to show $1 \iff 2$ and $2\iff 3$ - although I'm sure a direct proof exists – Mathmo123 Jul 16 '14 at 15:09
• @Mathmo123 I also don't see while 3) implies 2) – Amr Jul 16 '14 at 15:17

Hint Irreducible $\,p\nmid a,\ p\mid ab\,\Rightarrow\, p\mid ab,pb\,\Rightarrow\,p\mid (ab,pb) = (a,p)b = b,\,$ so irreducibles are prime, which, then easily yields uniqueness of factorizations into irreducibles, so reversing all implications.

• @user26857 That's what is proved in the first line, i.e. $(3)\Rightarrow(2)\,$ (and, recall that, primes are always irreducible) – Bill Dubuque Jul 16 '14 at 16:44
• Hi. Why is $(ab,pb)=(a,p)b$ ? – Amr Jul 16 '14 at 22:07
• @Amr That's the fundamental GCD Distributive Law. See this answer for a few proofs, using unique factorization, or Bezout's Identity, or the universal property of the gcd. – Bill Dubuque Jul 16 '14 at 22:19
• @Amr Please let me know what was confusing so I can avoid that in the future. – Bill Dubuque Jul 16 '14 at 22:29
• Your answer is great. It was just me who was confused of something not related to your answer. Thank you :) – Amr Nov 5 '14 at 12:55

Hint (I'll let you prove $2) \implies 1)$): To see that 3) implies 2), suppose that $a$ is irreducible, you want to prove that $a$ is prime, equivalently that $A/(a)$ is an integral domain. Suppose $xy$ is divisible by $a$. By irreducibility of $a$, either $\gcd(x,a) = 1$ or $\gcd(x,a) = a$. In the latter case you're done (because then $a = \gcd(x,a)$ divides $x$). In the former case, apply a well known result (apparently it's not named after Gauss in English-speaking countries?) that $\gcd(x,a) = 1$ and $a \mid xy \implies a \mid y$ (just use the definition of gcd if you don't know this result).

• You should probably say explicitly what you mean by a prime ring since that terminology is not commonly used in a first algebra course. The lemma in the former case is not named after Gauss but, rather, Euclid. – Bill Dubuque Dec 24 '14 at 15:11
• That was not clear because prime ring is in fact used by some authors! – Bill Dubuque Dec 24 '14 at 15:33
• But it does - please follow the above link to learn that denotation of "prime ring". Then I think you'll understand the point of my original remark. – Bill Dubuque Dec 24 '14 at 15:39