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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.

Hanul Jeon
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  • 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
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    @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
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    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
  • @Mathmo123 Oh, I don't think $(\sqrt{-5})$ (I only consider the principal ideal generated by 2.) I realize that the answer sometimes hidden in obvious thing, again. Thanks – Hanul Jeon Jan 29 '16 at 14:18
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    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

1 Answers1

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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.)

Community
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user26857
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  • 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
  • And in case anyone reads later, $(p,X)=(p)$ – rschwieb Jan 29 '16 at 21:53
  • I have of course upvoted this excellent answer, but why is it community wiki ? Also, since the example does not allow for the usual tools (like Krull's Hauptidealsatz) because of non-noetherianness, I would be very grateful if you were so kind as to succinctly sketch the proofs of some properties of this ring, like non-noetherianness, description of its spectrum, non factorialness, etc. I have the feeling that it would be a worthwhile investment of your time, since your post could then be used for future reference as a counter-example to many assertions. Anyway, thanks for your answer. – Georges Elencwajg Jan 30 '16 at 09:30
  • @GeorgesElencwaj It's a CW answer since the OP didn't actually asked this. – user26857 Jan 30 '16 at 09:50
  • You are right that it's good to have at hand most of the properties of this remarkable ring, but here is not the right place to make an exhaustive investigation for the same reason I've made this answer CW. – user26857 Jan 30 '16 at 09:58
  • Thanks for the explanations, user26857. – Georges Elencwajg Jan 30 '16 at 10:14