Showing that Z[Sqrt[-3]] is a Euclidean Domain

• Nov 22nd 2010, 09:56 PM
roninpro
Showing that Z[Sqrt[-3]] is a Euclidean Domain
Hello everyone,

I'm having a bit of trouble showing that $\displaystyle \mathbb{Z}[\sqrt{-3}]$ is a Euclidean domain. My attempt is:

For this ring, define the norm $\displaystyle N(a+b\sqrt{-3})=a^2+3b^2$. We need to show that there is a division algorithm, so let $\displaystyle \alpha=a+b\sqrt{-3}$ and $\displaystyle \beta=c+d\sqrt{-3}$. We will compute $\displaystyle \alpha/\beta$ as complex numbers and fish for an appropriate quotient and remainder. We have
$\displaystyle \displaystyle \frac{\alpha}{\beta}=\frac{ac+3bd}{c^2+3d^2}+\frac {bc-ad}{c^2+3d^2}\sqrt{-3}$
Now choose a quotient $\displaystyle q\in \mathbb{Z}[\sqrt{-3}]$ such that $\displaystyle |Re(q)-\frac{ac+3bd}{c^2+3d^2}|\leq 1/2$ and $\displaystyle |Im(q)-\frac{bc-ad}{c^2+3d^2}|\leq \sqrt{3}/2$. Then we have $\displaystyle \alpha/\beta=q+r/\beta$, where $\displaystyle r/\beta$ is the leftover "fractional" term. From the choice of $\displaystyle q$, we have $\displaystyle N(r/\beta)\leq 1$, and $\displaystyle N(r)\leq N(\beta)$.

At this point, I can't shake the possible equality. I would appreciate any suggestions on this issue.
• Nov 23rd 2010, 01:50 AM
tonio
Quote:

Originally Posted by roninpro
Hello everyone,

I'm having a bit of trouble showing that $\displaystyle \mathbb{Z}[\sqrt{-3}]$ is a Euclidean domain.

No wonder: it is not. For example, check $\displaystyle 4=2\cdot 2=(1+\sqrt{-3})(1-\sqrt{-3})$ ...(Wink)

Tonio

My attempt is:

For this ring, define the norm $\displaystyle N(a+b\sqrt{-3})=a^2+3b^2$. We need to show that there is a division algorithm, so let $\displaystyle \alpha=a+b\sqrt{-3}$ and $\displaystyle \beta=c+d\sqrt{-3}$. We will compute $\displaystyle \alpha/\beta$ as complex numbers and fish for an appropriate quotient and remainder. We have
$\displaystyle \displaystyle \frac{\alpha}{\beta}=\frac{ac+3bd}{c^2+3d^2}+\frac {bc-ad}{c^2+3d^2}\sqrt{-3}$
Now choose a quotient $\displaystyle q\in \mathbb{Z}[\sqrt{-3}]$ such that $\displaystyle |Re(q)-\frac{ac+3bd}{c^2+3d^2}|\leq 1/2$ and $\displaystyle |Im(q)-\frac{bc-ad}{c^2+3d^2}|\leq \sqrt{3}/2$. Then we have $\displaystyle \alpha/\beta=q+r/\beta$, where $\displaystyle r/\beta$ is the leftover "fractional" term. From the choice of $\displaystyle q$, we have $\displaystyle N(r/\beta)\leq 1$, and $\displaystyle N(r)\leq N(\beta)$.

At this point, I can't shake the possible equality. I would appreciate any suggestions on this issue.

.
• Nov 23rd 2010, 06:09 AM
roninpro
Ah, yes - the statement of the problem in the book was a little bit confusing. I think that they meant to put $\displaystyle \alpha=\frac{1+\sqrt{-3}}{2}$ and to show $\displaystyle \mathbb{Z}[\alpha]$ is a Euclidean domain. I'll go back and look at it again.