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I'm working out in detail the well known example of a PID which is not euclidean, $A=\mathbb{Z}\left[\frac{1+i\sqrt{19}}{2}\right]$. I've already shown it is not euclidean, and now I'm trying to show it is a PID.

The path I'm trying to take is the following :

  1. show it is almost euclidean in the sense that for all $a,b \in A$ with $b$ non-zero, then there exists $q,r$ s.t. $r=0$ or $\mid r \mid < \mid a\mid $ and either $a=bq +r$ or $2a = bq+r$ ; where $\mid \cdot \mid$ denotes the usual norm on $\mathbb{C}$

  2. Take $I$ an ideal of $A$, and $a\in I$ non-zero minimising the norm. Then by the "division" defined in point 1. we prove $I=(a)$.

There's just two things I can't quite figure out.

The first point I sort of assumed would be easy because a point in $A$ would be close enough to a lattice point in $\mathbb{Z}[i]$ so that you could do the division there and then figure out something to make it work in $A$. Turns out when I tried to write it down I don't actually know how to go about it.

The second thing I'm struggling with is the following : Assuming I've proven the first part, take $a \in I$ non-zero minimising $\mid\cdot\mid$, and let $x\in I$ be given. Then there exists $q,r$ with either $r=0$ or $\mid r\mid < \mid a \mid$ such that either $x=aq+r$ or $2x=aq+r$. Now one can easily check that $r\in I$ so that $r$ must be 0, for other wise $\mid a\mid $ would not be minimal. Hence either $x=aq$ or $2x=aq$. The first case then gives us the result, i.e. $I\subset (a)$ whence $I=(a)$. But if the second case, i.e. if $2x=aq$, I can't reach the same conclusion, nor can I find a contradiction. I've tried proving that $q$ is divisible by 2, using 1, but again I've hit a wall, I've been pondering about this on and off for a few days when I have time and I can't seem to crack this problem. I've seen a bunch of solutions I could understand but they didn't specifically go this exact route and that's fine, I'm just very curious on how to resolve the issues here, because I know this should work I just can't see why. Any hint is appreciated.

t_kln
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  • How do you know “this should work”? – KCd Dec 07 '22 at 13:48
  • @KCd I know step 1 and 2 should be enough to prove that $A$ is a PID because I'm following the guideline of an exercise. The exercise essentially divides into two parts, the first one showing it's not euclidean and the second part about A being a PID ( 1/ proving a general lemma for euclidean rings, 2/ showing A/2A and A/3A are fields, 3/ applying the lemma from question 1 to deduce using 2/ it can't be euclidean and then the next question is my 1. point, and the one after that is showing that it's a PID) – t_kln Dec 07 '22 at 13:55
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    @KCd It follows from Hurwitz's division algorithm, which follows from a simple Dirichlet approximation, e.g. for an elementary exposition see Lemma $11$ in Fendel, Prime producing polynomials and principal ideal domains, Math. Mag 1985, here or here. The method is $(2)$ is a special case of the Dedekind-Hasse PID criterion - a generalization of the Euclidean (algorithm) descent to PIDs. – Bill Dubuque Dec 07 '22 at 14:52
  • For generalizations see wortk of Lenstra, Lemmermeyer, Clark and others (off the top of my head). – Bill Dubuque Dec 07 '22 at 14:55
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    Compare to Lenstra's sketched proof in this answer. – Bill Dubuque Dec 07 '22 at 15:01

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