Is $\mathbb{Q}(\sqrt{2})$ a Euclidean domain?

Question

How do I prove that $\mathbb{Q}(\sqrt{2})$ is or isn't a Euclidean domain?

So if $F$ is a field, then $F[X]$ is a Euclidean domain. I don't see why this means that $\mathbb{Q}(\sqrt{2})$ is a Euclidean domain, because we have $\sqrt{2}$ instead of $X$.

But every field is a Euclidean domain. The question usually is for $\mathbb{Z}[\sqrt{2}]$. — Dietrich Burde, Jun 04 '16 at 14:39
Any field is a Euclidean domain relative to any norm because $a=(ab^{-1})b+0$ and $N(0)=0\leq N(b)$. So one can theoretically raise the question but it is a dead end. — marwalix, Jun 04 '16 at 14:40
Okay it is a Euclidean domain because it is a field. What Euclidean function can I apply on it? — user343295, Jun 04 '16 at 14:43
Read my comment. It works with any Euclidean function (or norm) — marwalix, Jun 04 '16 at 14:44

score 2 · Accepted Answer · answered Jun 04 '16 at 14:40

2

Every field is a Euclidean domain with Euclidean function $E(x)=1$ (or any function really) because every division leaves a zero remainder.

answered Jun 04 '16 at 14:40

lhf

221,500

Thanks. For some reason I thought the function had to be surjective, so now I understand my problem. – user343295 Jun 04 '16 at 15:14

Is $\mathbb{Q}(\sqrt{2})$ a Euclidean domain?

1 Answers1