# Show that $\mathscr{O}_{\mathbb{Q}(\sqrt{-7})}$ is a UFD

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?

• It's a Euclidean domain. Dec 16, 2018 at 20:22
• @LordSharktheUnknown Yes, but how can I show that? Is there a method for showing such? Dec 16, 2018 at 20:26
• 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. Dec 16, 2018 at 21:10
• @AlexJBest: You're right and an example for what you say is $d = -19$. Dec 17, 2018 at 17:47
• The Minkowski bound for $\Bbb Q(\sqrt{-7})$ is ~1.68..., if you have this machinery available to you. Dec 22, 2018 at 11:30