6
$\begingroup$

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

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

$\endgroup$
  • 4
    $\begingroup$ 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.) $\endgroup$ – David C. Ullrich Apr 9 '18 at 14:59
  • 1
    $\begingroup$ 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). $\endgroup$ – Watson Apr 10 '18 at 7:14
0
$\begingroup$

Here are some thoughts:

  • Exsitence is easy to prove using induction 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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What about UFD s that aren't even PID s? $\endgroup$ – John Cataldo Apr 9 '18 at 15:17
  • $\begingroup$ That's a god question, and I'm not sure there's a general method. I mean, there are things like that if $R$ is a UFD, then so is $R[x]$, but that's not useful in many circumstances. $\endgroup$ – Jordan Hardy Apr 9 '18 at 17:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.