ring theory – Mathtuition88

It turns out that to prove $\mathbb{Z}[\sqrt{-2}]$ is a Principal Ideal Domain, it is easier to prove that it is a Euclidean domain, and hence a PID.

(Any readers who have a direct proof that $\mathbb{Z}[\sqrt{-2}]$ is a PID, please comment below, as it would be very interesting to know such a proof. 🙂 )

Proof:

As mentioned above, we will prove that it is a Euclidean domain.

Let $a, b\in\mathbb{Z}[\sqrt{-2}], b\neq 0$ .

We need to show: $\exists q, r\in \mathbb{Z}[\sqrt{-2}]$ such that $a=bq+r$ , with $N(r)<N(b)$ .

Consider $\frac{a}{b}=c_1+c_2 \sqrt{-2} \in \mathbb{Q}[\sqrt{-2}]$ . Define $q=q_1+q_2 \sqrt{-2}$ where $q_1, q_2$ are the integers closest to $c_1, c_2$ respectively.

Then, $\frac{a}{b}=q+\alpha$ , where $\alpha=\alpha_1+\alpha_2 \sqrt{-2}$ .

$a=bq+b\alpha$ .

Take $r=b\alpha$ .

$\begin{aligned} N(r)&=N(b\alpha)\\ &=N(b)\cdot N(\alpha)\\ &=N(b)\cdot (\alpha_1^2+2\alpha_2^2)\\ &\leq N(b)\cdot ({\frac{1}{2}}^2+2(\frac{1}{2})^2)\\ &=N(b)\cdot (\frac{3}{4})\\ &<N(b) \end{aligned}$

Check out recommended Abstract Algebra books: Recommended Books for Math Undergraduates

Tag: ring theory

Z[Sqrt(-2)] is a Principal Ideal Domain Proof