Answer to Question #104439 in Abstract Algebra for Sourav Mondal

Question

Show that Z[√-2] is not a UFD.

Answer 1

Actually, $\mathbb{Z}[\sqrt{-2}]$ is UFD.

Proof:

We will prove that $\mathbb{Z}[\sqrt{-2}]$ is Euclidean domain. It will be enough because every Euclidean domain is a unique factorization domain.

If there exists a Euclidean norm on $\mathbb{Z}[\sqrt{-2}]$ , then the integral domain $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain. (definition)

So, we will prove, that in the integral domain $\mathbb{Z}[\sqrt{-2}]$ , the function $N(\alpha)=\alpha \overline{\alpha}$ is a multiplicative Euclidean norm.

(i) $\forall r\in \mathbb{Z}[\sqrt{-2}] /\{0\} : N(r)\geq0$

$N(a+b\sqrt{-2})=(a+b\sqrt{-2})(a-b\sqrt{-2})=a^2+2b^2\geq 0$

(ii) $\forall a,b\in \mathbb{Z}[\sqrt{-2}] , b \cancel{=}0, \ \exists r,q\in \mathbb{Z}[\sqrt{-2}] : a=bq+r \ (r=0$ or $N(r)<N(b))$

Let $a+b\sqrt{-2}, \ c+d\sqrt{-2} \in \mathbb{Z}[\sqrt{-2}] , \quad a,b,c,d\in \mathbb{Z}.$

$\exists e,f\in \mathbb{Q}: \quad \frac{a+b\sqrt{-2}}{c+d\sqrt{-2}}=\frac{(ac+2bd)+(bc-ad)\sqrt{-2}}{c^2+2d^2}=e+f\sqrt{-2}$

Consider $r,s\in \mathbb{Z}: \ |e-r|\leq1/2, |f-s|\leq1/2$

Then $a+b\sqrt{-2}=(c+d\sqrt{-2})(e+f\sqrt{-2})=$

$=(c+d\sqrt{-2})(r+s\sqrt{-2}+(e-r)+(f-s)\sqrt{-2})=$

$=(c+d\sqrt{-2})(r+s\sqrt{-2})+(c+d\sqrt{-2})((e-r)+(f-s)\sqrt{-2})$

We need to check that $N(c+d\sqrt{-2})>N((c+d\sqrt{-2})((e-r)+(f-s)\sqrt{-2})$ .

$N((c+d\sqrt{-2})((e-r)+(f-s)\sqrt{-2}))= N(c+d\sqrt{-2})N((e-r)+(f-s)\sqrt{-2})\leq N(c+d\sqrt{-2})(1/4+2\times 1/4)=3/4 N(c+d\sqrt{-2})< N(c+d\sqrt{-2})$

Therefore, $N$ is Euclidean norm $\Rightarrow$ $\mathbb{Z}[\sqrt{-2}]$ is Euclidean domain $\Rightarrow$ $\mathbb{Z}[\sqrt{-2}]$ is UFD.

This material can be found here: https://math.stanford.edu/~ksound/Math152A10/Solution6.pdf