It is known that the ring of integer is a Dedekind domain which means that it is a UFD iff it is a PID. Since $-7\equiv1$ mod $4$, we have that $\mathscr{O}_{\mathbb{Q}(\sqrt{-7})}=\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$. Now I read something in the sense of: if $\alpha:=\frac{1+\sqrt{-7}}{2}$ has an irreducible minimal polynomial mod $2$ and mod $3$, then we have a PID; I don't know anything about that. I think I have stated that wrong since the minimal polynomial is $f_{\alpha}=x^2-x+2$ which is reducible mod $2$.

Dr. Math: We pick an arbitrary complex number $x + iy\in\mathbb{Z}[\alpha]$, and we must find a suitable lattice point:

$$z = r + s\alpha = (r+s/2) + i(s\sqrt{7})/2.$$

It is natural to try to have the real and imaginary parts of $(x + yi - z)$ as small as possible.

Let's start with the imaginary part $ y - s\sqrt{7}/2$. We take $s$ as the closest integer to $2y/\sqrt{7}$. This will give us the following: \begin{align*} | 2y/\sqrt{7} - s | &\leqslant 1/2\\ | y - s\sqrt{7}/2 | &\leqslant \sqrt{7}/4. \end{align*} Now, we turn to the real part $x - r - s/2$. If we select $r$ as the integer closest to $(x - s/2)$, we will have: \begin{align*} | x - r - s/2 | \leqslant 1/2. \end{align*} Putting both relations together, we get: $$ N(x + yi - z) = (x - r - s/2)^2 + (y - s\sqrt{7}/2)^2 \leqslant 1/4 + 7/16 < 1$$

as desired. Hence, Euclidean domain, so PID, so UFD.

Is this proof correct and can it be applied in all cases of showing that $\mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right]$, $\square\neq d\in\mathbb{N}$ is a Euclidean domain?

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    $\begingroup$ It's a Euclidean domain. $\endgroup$ Dec 16, 2018 at 20:22
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    $\begingroup$ @LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such? $\endgroup$
    – Algebear
    Dec 16, 2018 at 20:26
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    $\begingroup$ I didn't find the exact place yet but the conclusion that all $\mathbb Z[ 1+\sqrt d / 2]$ for $d \equiv 1 \pmod 4$ are euclidean is false. Something like this should work for $7$ i think, but there should be some part that breaks down once $d$ is large. $\endgroup$ Dec 16, 2018 at 21:10
  • $\begingroup$ @AlexJBest: You're right and an example for what you say is $ d = -19 $. $\endgroup$
    – hellHound
    Dec 17, 2018 at 17:47
  • $\begingroup$ The Minkowski bound for $\Bbb Q(\sqrt{-7})$ is ~1.68..., if you have this machinery available to you. $\endgroup$ Dec 22, 2018 at 11:30


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