Exercise 1.36

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Define $\Z[\sqrt{-2}]$ as the set of all complex numbers of the form $a+b\sqrt{-2}$, where $a,b\in \Z$. Show that $\Z[\sqrt{-2}]$ is a ring. Define $\lambda(\alpha) = a^2+2b^2$ for $\alpha = a+b\sqrt{-2}$. Use $\lambda$ to show that $\Z[\sqrt{-2}]$ is a Euclidean domain.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Note $\sqrt{-2} = i \sqrt{2}$, and $A = \Z[\sqrt{-2}]$.

Let $\alpha = a + b \sqrt{-2}, \beta =  c + d\sqrt{-2} \in A$ :

$\bullet$ $1 = 1 + 0 \sqrt{-2} \in A$.

$\bullet$ $\alpha - \beta = (a + b \sqrt{-2})- (c + d\sqrt{-2}) = (a-c) + (b-d) \sqrt{-2} \in A$.

$\bullet$  $\alpha \beta= (a + b \sqrt{-2}) (c + d\sqrt{-2}) = (ac -2 bd) + (ad+bc) \sqrt{-2} \in A.$

So $A$ is a subring of $(\mathbb{C},+,	imes)$ : $\Z[\sqrt{-2}]$ is a ring.

Let $z = a + b\sqrt{-2}$ be any complex number, and define integers $a_0, b_0 \in \Z$ such that $\vert a-a_0 \vert \leq 1/2, \vert b-b_0 \vert \leq 1/2$ (it suffice to take for $a_0$ the nearest integer of $a$, that is $a_0 =\left \lfloor a + \frac{1}{2}ight floor$). Let $z_0 = a_0 + b_0 \sqrt{-2}$.

As $\lambda(z) = z \overline{z} = a^2 + 2 b^2$, then 
$$\lambda(z-z_0) = (a-a_0)^2 + 2(b-b_0)^2 \leq \frac{1}{4} + 2 	imes \frac{1}{4} = \frac{3}{4}<1.$$

Conclusion : for any $z \in \mathbb{C}$, there exists $z_0 \in A$ such that $\lambda(z-z_0) < 1$.

Let $(z_1,z_2) \in A 	imes A, z_2 
e 0$. We apply the preceding result to the complex $z_1/z_2$ : there exists $q \in A$ such that $\left \vert \frac{z_1}{z_2} - q ight \vert \leq 1$. Let $r = z_1 - qz_2$. Then $z_1 = qz_2 +r, \lambda(r) < \lambda(z_2)$.

So $\Z[\sqrt{-2}]$ is a Euclidean domain.
\end{proof}

Exercise 1.36

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Exercise 1.36

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