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What are direct methods for proving that a ring is a UFD in general without proving that it's a PID/Euclidean domain/field and using the fact that all those things are UFDs?

As an example, we can take $\mathbb{Z}[i]$ or $\mathbb{Z}[\sqrt{-2}]$ or other rings you come up with.

J. W. Tanner
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John Cataldo
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    Seems to me that if you're interested in this for some curious reason you might ask first how to show that $\Bbb Z$ is a UFD by another method... (no, you don't necessarily see the words "suclidean domain" in the standard proof, but that is how the standard proof for $\Bbb Z$ goes.) – David C. Ullrich Apr 09 '18 at 14:59
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    You might want to look at a PID (hence a UFD) which is not an euclidean domain, like here. Think also about $\Bbb Z[X]$ : it is not euclidean (not even a PID), but it is a UFD (however euclidean divisions by monic polynomials always work). – Watson Apr 10 '18 at 07:14

2 Answers2

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Here are some thoughts:

  • Exsitence is easy to prove using induction on the norm.

  • Uniqueness is the hard part, especially since it fails for most rings of the form $\mathbb Z[\sqrt d]$. For the rings you've mentioned, it can be proved by knowing the units and exactly how primes in $\mathbb Z$ decompose in $\mathbb Z[\sqrt d]$. There are three possibilities for a prime $p$: it remains prime, it is product of two non associate primes, it is a square. For the rings you've mentioned, this can be decided in ad hoc ways.

In the general case of the ring of integers in quadratic fields $\mathbb{Q}(\sqrt{n})$, the answer is not simple but is fascinating, See the book Primes of the Form $x^2+ny^2$, by David Cox.

lhf
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This would be extremely wasteful, and nobody would do it before showing they were Euclidean domains or PIDs, but you could show that they have class number 1 through some indirect means. $\mathbb Z[i]$ is a Dedekind domain so it is a UFD if and only if its class number is 1.

Jordan Hardy
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