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I am looking for a way to characterize all infinite PID s having exactly one unit i.e. invertible element ( finite PID s are not interesting , they are all fields ) . The only such example I know of is $\mathbb Z_2[x]$ . Towards characterizing such domains, first of all if $D$ is such a PID then it cannot be a field and its characteristic is $2$ as $1=-1$ in it, so it contains a copy of $\mathbb Z_2$. Moreover, every element of $D\setminus \mathbb Z_2$ is transcendental over $\mathbb Z_2$, since other-wise if $a \in D\setminus \mathbb Z_2$ is algebraic i.e. integral over the field $\mathbb Z_2$ ( https://en.wikipedia.org/wiki/Integral_element ) then $\mathbb Z_2[a]$ would be an inetgral domain which is an integral extension over a field ( namely $\mathbb Z_2$ ) implying that $\mathbb Z_2[a]$ is a field, but in that case $\mathbb Z_2[a]$ contains at least two units namely $1,a$, then so does $D$, contradiction ! But other than these I have not been able to derive any further consequences in the way of characterizing , and it is to be noted that in all the previous arguments we used nothing more than that $D$ is an integral domain, we haven't seriously used the PID nature of $D$ which if used is bound to give some further restrictions, but I can't figure out any. Any reference or link relating to this question will be highly appreciated. Please help. Thanks in advance. (Note that if we instead wanted such a UFD there would be plenty of examples to go about, namely polynomials in any $n$-variables over $\mathbb Z_2$ ... but seeking PID strikes out such examples.)

Martin Sleziak
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    Ah sorry. I overlooked that. I was just answering since you seemed to wonder what to use PID for. Have you tried to look at rings of the form $\mathbb {Z}_2 [x_1,\ldots,x_n]/p $ for a prime ideal $p $? This would be the next thing I do. – Maik Pickl Aug 23 '16 at 15:14
  • @MaikPickl : Its okay ... umm know I haven't tried ... have you found any such example ? and do you have any other properties that such domain should satisfy in mind ? –  Aug 23 '16 at 15:16
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    Such a ring $R$ must contain a copy of $\Bbb{Z}_2[x]$. Moreover, its Krull dimension must be equal to $1$. Now, by Noether normalization lemma, it follows that $R$ is an integral extension of $\Bbb{Z}_2[x]$, and that the fraction field of $R$ is a finite extension of $\Bbb{Z}_2(x)$. Hence, we have to look for such a ring among integral closures of $\Bbb{Z}_2[x]$ inside finite extensions of $\Bbb{Z}_2(x)$. These rings are necessarily all Dedekind domains, so we have to look for one of them having trivial class group (and this is pretty hard, though). – Crostul Aug 23 '16 at 15:31
  • @Crostul : Why should it contain a copy of $\mathbb Z_2[x]$ ? And why is the Krull dimension $1$ ? Could you please elaborate ? Thanks for your remarks –  Aug 24 '16 at 04:01
  • I am tempted to ask : Let $D$ be an infinite PID with exactly one invertible element , then is $D$ an Euclidean domain ? –  Aug 24 '16 at 04:02
  • @Crostul : okay I get the $\mathbb Z_2[x]$ thing , since $a\in D \setminus \mathbb Z_2$ is transcendental so $\mathbb Z_2[a] \cong \mathbb Z_2[x]$ ... but could you please elaborate the Krull dimension thing ? –  Aug 24 '16 at 04:22
  • Every PID has Krull dimension equal to 1. This is a general fact one should know from basic commutative algebra. – Crostul Aug 24 '16 at 13:56
  • @Crostul : oh yes sorry , I was being stupid again :p –  Aug 24 '16 at 14:27
  • @Crostul : You are applying Noether normalization lemma ... but what is the field $k$ and the finitely generated algebra $A$ on which you are applying it ? Or are you applying some different version of it ? –  Aug 24 '16 at 14:36
  • The field is $\Bbb{Z}_2$, and the algebra is our unknown PID. – Crostul Aug 24 '16 at 19:19
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    @Crostul : But then , to apply Noether normalization , how do you know that the PID is a "finitely generated" $\mathbb Z_2$ algebra ? –  Aug 25 '16 at 02:31
  • @Crostul : hello ... you there ? –  Aug 25 '16 at 13:35
  • @SaunDev I am sorry, but I have no idea on why such a ring should be a finitely generated $\Bbb{Z}_2$-algebra. – Crostul Aug 25 '16 at 13:40
  • @Crostul : Then how did you apply Noether normalization ? –  Aug 25 '16 at 13:41
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    W.J. Heinzer, M. Roitman, "Principal ideal domains and euclidean domains having 1 as the only unit", Communications in Algebra 29, 5197--5208 (2007) considers precisely this. They don't give a classification, so don't answer your question, but they give examples of PIDs with one unit other than $\mathbb{F}$ and $\mathbb{F}[x]$, and prove that there are no such examples that are euclidean domains. – Jeremy Rickard Aug 26 '16 at 08:50
  • @JeremyRickard : Could you please give an open access of the article or atlease the DOI please –  Aug 26 '16 at 09:49
  • http://www.tandfonline.com/doi/abs/10.1081/AGB-100106813 is not open access, but there's a copy downloadable at https://www.researchgate.net/publication/232895723_Principal_ideal_domains_and_euclidean_domains_having_1_as_the_only_unit – Jeremy Rickard Aug 26 '16 at 10:07
  • @JeremyRickard : In the paper they characterize such rings which are "finitely generated " ... what do they mean by finitely generated in this context ? –  Aug 27 '16 at 10:45
  • They mean finitely generated as a ring. Or in other words (since their rings are all $\mathbb{F}_2$-algebras), quotient rings of polynomial rings $\mathbb{F}_2[x_1,\dots,x_n]$ in finitely many variables. – Jeremy Rickard Aug 29 '16 at 09:22

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