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Im once again struggling to see the equivalence of two definitions. In my abstract algebra book (abstract algebra by beachy and blair) it says that elements $d_1,d_2,....,d_n \in D$ where $D$ is a UFD is said to be relatively prime in $D$ if there is no irreducible element $p \in D$ such that $p \mid a_i$ for $i=1,2,...n$.

However in my other book (A classical introduction to modern number theory by ireland and rosen) it says that two elements $a$ and $b$ are relatively prime if the only common divisors are units.

I am sure both definitions can not be correct. For instance, suppose that $a_1$ and $a_2$ are relatively prime according to the first definition, then we may very well have the case where a nonunit $p_1$ that is not irreducible divides both $a_1$ and $a_2$, so according to definition 2 the elements are not relatively prime. In this case, the elements are both relatively prime and not at the same time which shows that the definitions are not equivalent.

Am I missing something here or are the authors simply being sloppy?

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  • $\begingroup$ If it was not a Unique Factorization Domain, you would be 100% correct. $\endgroup$ – Karl Kronenfeld Jun 27 '14 at 3:35
  • $\begingroup$ I noticed this from Pedros answer below, thanks :) $\endgroup$ – user117449 Jun 27 '14 at 3:36
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    $\begingroup$ BTW, it's great that you are comparing the definitions from two sources. $\endgroup$ – Karl Kronenfeld Jun 27 '14 at 3:37
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For instance, suppose that $a_1$ and $a_2$ are relatively prime according to the first definition, then we may very well have the case where a nonunit $p_1$ that is not irreducible divides both $a_1$ and $a_2$, so according to definition 2 the elements are not relatively prime. In this case, the elements are both relatively prime and not at the same time which shows that the definitions are not equivalent.

No. Any nonzero nonunit is a product of irreducibles in a UFD. In particular, if some nonunit nonzero element divides $a_1,a_2$, there is an irreducible element that divides them both, namely any irreducible factor of this common divisor.

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I think an irreducible element is not a unit by definition.

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