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In an article by Prof. Wilson, there's an example of a PID that isn't an Euclidean domain, $\mathbb{Z}[\frac12(1+\sqrt{-19})]$. (See also On a PID that is not an Euclidean domain to find another question about the same article). But I have some doubts...

He says that "we can assume that the imaginary part of $\frac{a}{b}$ lies between $\pm \frac{\sqrt{19}}{4}$". Now, how can I justify this assumption?

Moreover, why we can consider only two cases? How can I obtain the value $\frac{\sqrt{3}}{2}$? At the end I can't show that the distance between $\frac{a}{b}$ and an integer $k$ is less than $1$.

And why the imaginary part of $2\frac{a}{b}-\frac{1}{2}(1+\sqrt{-19})$ is sufficiently small that the complex number lies at a distance less than 1 from some ordinary integer?

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TheWanderer
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  • A link to the initial article would be most helpful. Even better would be if the relevant text from all sources were briefly posted as part of this question so that the possibility of the linked article(s) no longer being available would not harm the integrity of this question. – abiessu Nov 08 '14 at 16:47
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    @abiessu: I followed the links and added one going to the paper, giving us the specific ring under discussion. However I agree that more of the relevant problem statement should be given. No doubt some of the difficulty is that federico is unsure what is relevant to the points raised by the Question. – hardmath Nov 08 '14 at 16:51
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    @federico: It seems that you are asking about points raised from the second part of Wilson's class note, showing that $\mathbb{Z}[\frac12(1+\sqrt{-19})]$ is a PID (where the first part introduces some notation and shows that this is *not* a Euclidean domain). However you should include in your Question a pertinent definition for $\frac{a}{b}$, so that the issues can be explained to the benefit of future Readers. I'm voting to close for lack of context (off-topic) until this can be remedied. – hardmath Nov 08 '14 at 17:14

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Wilson mentions that if $\;I\le R\;$, then one choose $\;b\in I\;$ with $\;|b|\;$ as small as possible. Usual proofs then go assuming that $\;I\neq bR\implies \exists\,a\in I\setminus bR\;$ and etc.

Now, he also mentions that we consider $\;\frac ab\;$ , and since

$$\forall\,q\in R\;,\;\;a=a-bq\pmod I$$

instead of considering $\;\frac ab\;$ we can instead consider $\;\frac{a-bq}b\;$ , and taking a convenient $\;q\;$ we can then assume

$$-\frac{\sqrt{-19}}4\le\text{Im}\,\frac ab\le\frac{\sqrt{-19}}4$$

since if $\;a=\alpha+i\beta\;,\;\;b=x+iy\;,\;\;q=r+is$ , then

$$\text{Im}\frac ab=\frac{-\alpha y+\beta x}{x^2+y^2}$$

$$\text{Im}\left(\frac ab-q\right)=\frac{-\alpha x+\beta y}{x^2+y^2}-s$$

so

$$-\frac{\sqrt{-19}}4\le\text{Im}\,\left(\frac ab-q\right)=\frac{-\alpha x+\beta y}{x^2+y^2}-s\le\frac{\sqrt{-19}}4$$

and we can always choose $\;s\;$ , and thus $\;q\;$ so as to make the above last inequality true.

Timbuc
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