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I am aware that irreducibles and primes are equivalent in PIDS. However, are they equivalent in UFDs?

Would the following reasoning work?

1) Suppose $f$ is irreducible. Then the prime factorization of $f$ must be $f$ itself. Hence $f$ is a prime itself.

2) Suppose $f$ is prime. That is, if $f \mid ab$ then $f \mid a$ or $f \mid b.$ Suppose $f = gh$ such that $g, h$ are not units. Then $f \mid gh.$ If $f \mid g$ then $g = fk$ It follows that $h$ must be a unit. Hence $f$ must be irreducible.

Does this work? So would it follow that in any $k[X_1, \ldots, X_n]$ for $k$ a field, the principal ideal generated by an irreducible is a prime ideal as $k[X_1, \ldots, X_n]$ is a UFD?

green frog
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1 Answers1

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Yes, they are equivalent in UFDs. Most texts have this fact, and it is also in the wiki article. This is also mentioned in the section of the UFD article.

Prime implies irreducible always. If $p$ is prime and $p=ab$, then $p$ divides one of the things on the right hand side. After you pick which one it divides, you cancel $p$ from both sides and the thing you didn't divide with $p$ is a unit.

The 'hard' part is the other way, that irreducible implies prime in a UFD. (In fact, it is true in just a GCD domain.)

As far as I know, UFD's are defined using factorizations into irreducible elements, not usually primes. Assuming that elements have a unique prime factorization trivializes the task, as you noted: if a product of primes equals an irreducible element $r$, then there can only be one prime.

If you use the definition that every element has a factorization into irreducibles then you should use one of the numerous existing solutions for this question on the site, like this one.

rschwieb
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