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I'm curious how to prove $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ is a PID. Before I get started proving this, I want to know a correct direction. Is it a good way to prove this by showing that the ring has a Dedekind-Hasse norm $N$ such that $N(a+b\frac{1+\sqrt{-3}}{2})=a^2+ab+b^2$. (Is it true?)

Moreover, is this ring a Euclidean domain? And if you know an article or page investigating this particular ring, please do me a favor tell me the link. Thank you in advance .

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2 Answers 2

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It goes generally the other way : you cross your finger and pray to find an euclidian function, so that your ring is euclidian, which implies that the ring is a PID.

Why wouldn't you consider $N(a + b \sqrt{-3} ) := a^2 + 3 b^2$ for $(a,b)\in\mathbf{Z}[\frac{1}{2}]$ and try to show that it is a euclidian function, using the fact that $\mathbf{Z}[\frac{1+\sqrt{-3}}{2}] = \{\frac{a}{2}+\frac{b}{2}\sqrt{-3}\;|\;a,b\in\mathbf{Z}, 2\textrm{ divides }b-a\}$ ?

Fact. Note $R$ your ring. Let $\alpha,\beta\in R$ such that $\beta\not=0$. Then you can find $\gamma,\delta\in R$ such that $\alpha = \gamma \beta + \delta$ and $N(\delta)<N(\beta)$. So $N$ is a euclidian function for $R$, showing that $R$ is a euclidian ring.

Proof. Note $\omega = \frac{1+\sqrt{-3}}{2}$. Write $\alpha = a_1 + a_2 \omega$ with the $a_i$'s in $\mathbf{Z}[\frac{1}{2}]$. Write $\beta = b_1 + b_2\omega$ in the same fashion. Now in $\mathbf{Q}(\sqrt{-3})$ you can write $\frac{\alpha}{\beta} = c_1 + c_2 \sqrt{-3}$ where $$c_1 = \frac{(2a_1+a_2)(2b_1 + b_2)+3a_2 b_2}{(2b_1 + b_2)^2 + 3 b_2^2}$$ and $$c_2 = \frac{a_2 (2b_1 + b_2)-b_2 (2a_1 + a_2)}{(2b_1 + b_2)^2 + 3b_s2^2}.$$ Choose $q_2 \in\mathbf{Z}$ such that $|2c_2 - q_2|\leq\frac{1}{2}$, which implies $|c_2 - \frac{q_2}{2}|\leq\frac{1}{4}$. Now choose $t\in\mathbf{Z}$ to be the closest integer to $c_1 - \frac{q_2}{2}$ and let $q_1 = 2t + q_2$. We have $|c_1 - \frac{q_1}{2}|\leq\frac{1}{2}$. Set $\gamma := \frac{q_1}{2} + \frac{q_2}{2} \sqrt{-3}$. Clearly $q_1 - q_2 = 2t$ is divisible by $2$ so that $\gamma \in R$. Now, let $\theta := (c_1 - \frac{q_1}{2}) + (c_2 - \frac{q_2}{2})\sqrt{-3}$ and let $\delta := \theta \beta$. Note that $\theta = \frac{\alpha}{\beta}-\gamma$ so that $\delta = \alpha - \gamma \beta$ is in $R$, and we have $\alpha = \gamma \beta + \delta$. We have one thing to show to conclude : that $N(\delta)<N(\beta)$. To do this, see that $N(\theta) = (c_1 - \frac{q_1}{2})^2 + 3(c_2 - \frac{q_2}{2})^2 \leq (1/2)^2 + 3(1/4)^2 = \frac{4+3}{16} < 1$ and remember that $N$ is a multiplicative function, as $N(z) = z\overline{z}$ for each $z$. $\square$

Remark. The fact that a euclidian domain is a PID is standard. Mimic the proof you've surely have already seen for $\mathbf{Z}$.

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  • $\begingroup$ I don't get it. Isn't $\mathbb{Z}[\sqrt{3}]$ strictly smaller than $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$? $\endgroup$
    – Rubertos
    Feb 24, 2015 at 20:03
  • $\begingroup$ It is, yes, but there is not problem. $\endgroup$
    – Olórin
    Feb 24, 2015 at 20:06
  • $\begingroup$ I will soon edit my answer, with a proof of the fact that $N$ is a euclidian function indeed. $\endgroup$
    – Olórin
    Feb 24, 2015 at 20:10
  • $\begingroup$ Done, edited my answer $\endgroup$
    – Olórin
    Feb 24, 2015 at 20:25
  • $\begingroup$ I think it would have been much great to say $N(\alpha)$ is indeed an absolute value of $\alpha$. $\endgroup$
    – Rubertos
    Feb 25, 2015 at 15:20
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Write $\omega=1/2+1/2\sqrt{-3}$. I will give an alternative way to show that the norm function $N(x+iy)=x^2+y^2$ is a Euclidean function for $\mathbb Z[\omega]$.

Let $\alpha,\beta\in\mathbb Z[\omega]$ with $\beta\neq 0$. We need find $q,r\in\mathbb Z[\omega]$ such that $$ \alpha=q\beta+r $$ and $N(r)<N(\beta)$. Note that since $N$ is multiplicative and defined on all of $\mathbb C$, the above is equivalent to showing that $N(\alpha/\beta-q)<1$.

Write $\alpha/\beta=x+iy$ with $x,y\in\mathbb Q$. Let $a=\lfloor x\rfloor$ and $b=\lfloor y/\sqrt{3}\rfloor$. Let $r=a+b\sqrt{3}i$. Note that $$ N(\alpha/\beta-r)=N(x+iy-a-b\sqrt{3}i)=(x-a)^2+(y-b\sqrt{3})^2. $$ Since $(y-b\sqrt 3)^2=3(y\sqrt 3-b)^2\leq 3\cdot 1/4=3/4$ and $(x-a)^2\leq 1/4$. we see that $N(\alpha/\beta-r)\leq 1$. So if $N(\alpha/\beta-r)<1$ we are done. If $N(\alpha/\beta-r)=1$ then note that we must have $x=m+1/2$ and $y/\sqrt{3}=n+1/2$ with $m,n\in\mathbb Z$. Note also that since $\sqrt{-3}=2\omega-1$, we have $$ \alpha/\beta=m+1/2+(n+1/2)\sqrt{-3}=m+1/2+(n+1/2)(2\omega-1)=m-n+(2n+1)\omega. $$ Thus $\alpha/\beta\in\mathbb Z[\omega]$, so in that case we can just take $r=\alpha/\beta$.

We conclude that $N$ is indeed a Euclidean function, and therefore $\mathbb Z[\omega]$ is a Euclidean domain, and hence in particular a PID.

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