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Let $M = \mathbb{Z}[\phi] \setminus \{0\}$ be the multiplicative monoid of the ring $\mathbb{Z}[\phi]$ with $\phi = \frac{1+\sqrt{5}}{2}$ the golden ratio. We define the equivalence relationship $x\sim y$ iff $x =uy$ for some unit.

Is there a good description of the quotient $M/\sim$? My initial thought was to do $a+b \phi \sim a+b$ since $\phi \sim 1$ but this is just plain wrong.

Is this also known for other integral rings $O_K$?

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    $\begingroup$ Can you add some info on $\phi$. Is it a polynomial unknown, or does it satisfy some special properties? Sorry if this is common knowledge that I don't have. $\endgroup$
    – Dirk
    Sep 10, 2019 at 9:53
  • $\begingroup$ Yes, sure. I meant it to be the other lower case $\phi$, standing for the golden ratio. I'll edit. $\endgroup$
    – Adi Ostrov
    Sep 10, 2019 at 10:21
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    $\begingroup$ Az $\mathbb{Z}[\phi]$ is a PID, $M/\sim$ is just the monoid of nonzero ideals of $\mathbb{Z}[\phi]$. It is the free monoid generated by the prime ideals of $\mathbb{Z}[\phi]$. Am I missing something? $\endgroup$
    – GH from MO
    Sep 10, 2019 at 11:36
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    $\begingroup$ To say that M/~ is a free abelian monoid is to say that the domain is a UFD. $\endgroup$ Sep 12, 2019 at 23:07

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For general (commutative) rings $R$, we can construct $M = R \setminus \{0\}$ and $\sim$ as you did. It's not too hard to see (as GH mentioned) that in this general situation, $a \sim b$ if and only if they generate the same ideal, so $M/\sim$ will be isomorphic to the monoid of nonzero principal ideals of $R$.

Again, as GH mentioned, the fact that $\mathbb{Z}[\phi]$ is a PID means that this is free of infinite rank, generated by the primes. For general number rings $\mathcal{O}_K$, it should have finite index in some free monoid (that of the nonzero integral ideals) by finiteness of the class number. Being inside a free monoid should tell you it's torsion free and that no non-identity elements have inverses, but I don't know if we can get anything interesting from this, or if we can get anything extra from the finite index.

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    $\begingroup$ It will be true for domains, but not quite for general rings: then generating the same ideal does not imply differing by a unit, so this unit equivalence is finer. There is a paper of Anderson and Valdes-Leon discussing this in great detaio and with bunch of examples: projecteuclid.org/euclid.rmjm/1181072068 $\endgroup$ Sep 13, 2019 at 1:53
  • $\begingroup$ Ah, I see. If a = rb and b = sa, then a = rsa. Cancellation would then give rs=1, but we don't have cancellation for non-domains. I'm trying to think of an example, but haven't come up with one yet. $a\sim b$ does mean $(a) = (b)$ though, so you at least get a surjective map to that monoid. $\endgroup$
    – Jon Aycock
    Sep 13, 2019 at 2:11
  • $\begingroup$ There are some in the linked paper at least. $\endgroup$ Sep 13, 2019 at 2:19
  • $\begingroup$ I looked before, even going so far as to search the paper for the word "example," and found nothing. You're definitely right again though. There's an explicit one in Example 2.3 at the bottom of page 443 (the 5th page of the article), using the language of definition 2.1 from the bottom of page 441, if anyone else has some trouble finding it too. $\endgroup$
    – Jon Aycock
    Sep 13, 2019 at 2:58
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    $\begingroup$ I apologize, that's the one I encountered too - I should have said it explicitly. $\endgroup$ Sep 13, 2019 at 3:05

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