# Multiplicative monoid of ring modulo units

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 $$O_K$$?

• 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.
– Dirk
Sep 10, 2019 at 9:53
• Yes, sure. I meant it to be the other lower case $\phi$, standing for the golden ratio. I'll edit. Sep 10, 2019 at 10:21
• 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? Sep 10, 2019 at 11:36
• To say that M/~ is a free abelian monoid is to say that the domain is a UFD. Sep 12, 2019 at 23:07

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.
• 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. Sep 13, 2019 at 2:11