# Is there a domain which is not UFD but has a maximal principal ideal?

Maximal ideals in $\Bbb Z[\sqrt{-5}]$, which are not UFD, are not principal. I wonder, however, a maximal ideal could be principal. Is there known example? Also, I wonder the existence of UFD that has an maximal principal ideal. I would appreciate your help.

• I can't tell if this is necessary but $\Bbb Z[x]$ is a UFD, in case you thought otherwise. – rschwieb Jan 29 '16 at 13:58
• Do you mean to require the ring to be a domain? If not, then $\Bbb Z/n\Bbb Z$ works for any composite n – rschwieb Jan 29 '16 at 13:59
• @rschwieb Yes, I confuse a ring and a domain because I have not learn about non-domain ring deeply. I am going to modify them. – Hanul Jeon Jan 29 '16 at 14:02
• In $\mathbb Z[\sqrt {-5}]$, the ideal $(\sqrt{-5})$ is maximal and principal. In fact, $\mathbb Z[\sqrt {-5}]$ has infinitely many principal prime ideals, and all these primes are also maximal (since $\mathbb Z[\sqrt {-5}]$ is a Dedekind domain) – Mathmo123 Jan 29 '16 at 14:11
• If $p\equiv 11,13,17,19\pmod{20}$ is prime in $\mathbb Z$ then $p\mathbb Z[\sqrt{-5}]$ is a maximal principle ideal. – Thomas Andrews Jan 29 '16 at 14:48

There are integral domains whose all maximal ideals are principal and which are not UFDs.

An example is $R=\mathbb Z+X\mathbb Q[X]$. It is a Bézout domain which is not a PID (see here), so it can't be a UFD. (Note that $R$ is not noetherian since the ideal $X\mathbb Q[X]$ is not finitely generated.) Its maximal ideals are of the form $pR$, with $p\in\mathbb Z$ prime, and $fR$, with $f\in\mathbb Q[X]$ irreducible and $f(0)=1$.

Another example is a valuation ring with value group $\mathbb Z\times\mathbb Z$. (See here.)

• What form are the maximal ideals? Are they all just $(p, X)$ for $p$ prime? Just wondering what the trick is to seeing if this is a complete set, if so. – rschwieb Jan 29 '16 at 21:33
• @rschwieb There are also those generated by the irreducible polynomials $f\in\mathbb Q[X]$ with $f(0)=1$. – user26857 Jan 29 '16 at 21:38
• Hm yes, there is more variety. But anyhow, what's the trick to seeing all maximal ideals are principal? – rschwieb Jan 29 '16 at 21:43
• @rschwieb There is no trick: some contain $X$ and look like you noticed, and some don't and look like I said. – user26857 Jan 29 '16 at 21:48
• OK, so just proof by exhaustive description. Thanks – rschwieb Jan 29 '16 at 21:50