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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

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

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

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  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Dec 24 '14 at 15:11
  • $\begingroup$ That was not clear because prime ring is in fact used by some authors! $\endgroup$ – Bill Dubuque Dec 24 '14 at 15:33
  • $\begingroup$ 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. $\endgroup$ – Bill Dubuque Dec 24 '14 at 15:39

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