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

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. 🙂 )


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}.


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


Author: mathtuition88


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: