The integers of $\mathbb{Q}(\sqrt{2})$

Let $\alpha=\sqrt{2}$ . The quadratic number field $\mathbb{Q}(\sqrt{2})$ is the field

$${\mathbb{Q}}(\alpha)=\{x+\alpha y \vert x,y\in\mathbb{Q}\},$$

which we denote by $K$ for short. The integers of $K$ is the ring

$${\mathbb{Z}}[\alpha]=\{a+\alpha b \vert a,b\in {\mathbb{Z}}\} ={\mathbb{Z}}+\alpha {\mathbb{Z}}.$$

Definition 2.1.11   An element $p=a+\alpha b\not= 0$ is a prime (element of $\mathbb{Z}[\alpha ]$ ) if it has the following property: for all $r,s\in \mathbb{Z}[\alpha ]$ , $p\vert rs$ then $p\vert r$ or $p\vert s$ .

Is there an analog of Theorem 2.1.9 for primes in this field?

Yes. Before describing them, we need to introduce the notion of an associate.

Definition 2.1.12   If $R$ is an commutative ring with unit $1$ and $u\in R$ has the property that $uv=1$ , for some $v\in R$ then we call $u$ a unit of $R$ and call $v$ the inverse of $u$ . . The set of units of $R$ is denoted $R^\times$ .

Let $x,y\in R$ . We say $x$ is an associate of $y$ if $x=uy$ , for some $u\in R^\times$ . The set of associates of $x$ is denoted $xR^\times$ .

There is an analog of the fundmental theorem of arithmetic for $\mathbb{Z}[\alpha ]$ . In other words, each element in $\mathbb{Z}[\alpha ]$ can be uniquely (up to order and ``unit'' factors of the form $\pm (1+\sqrt{2})^k$ , $k\in \mathbb{Z}$ ) factored into a product of prime elements. In other words, $\mathbb{Z}[\sqrt{2}]$ is a unique factorization domain.

Example 2.1.13 (a)   If $R=\mathbb{Z}$ then $R^\times =\{1,-1\}$ .

(b) If $R=\mathbb{Z}[i]$ then $R^\times =\{1,-1,i,-i\}$ .

(c) if $R=\mathbb{Z}[\alpha ]$ then

$$R^\times =\{ \pm (1+\sqrt{2})^k \vert k\in \mathbb{Z}\}.$$

The proof of this goes beyond the scope of this book. See for example, Hardy and Wright, [HW], §14.5. (See the exercises below for further examples.)

The following characterization of primes is known.

Theorem 2.1.14   An element $a+\alpha b\not= 0$ is a prime element of $\mathbb{Z}[\alpha ]$ if and only if either

(a) $a=0$ and $b=\pm 1$ ,

(b) $b=0$ and $\vert a\vert$ is a prime integer which is $\equiv \pm 3 ({\rm mod} 8)$ , or

(c) $ab\not= 0$ and $a^2-2b^2$ is a prime integer which is $\equiv 1 ({\rm mod} 8)$ ,

(d) $a+\alpha b$ is an associate of an element as in (a), (b) or (c).

Remark 2.1.15   If we replace $\alpha=\sqrt{2}$ by the root of some higher degree polynomial in $\mathbb{Z}[x]$ then there is, in general, no analog of Theorem 2.1.14 for $\mathbb{Q}(\alpha)$ .

In fact, the search for an analog is one of the motivations for the very deep Langlands philosophy [Art], §1.

If we replace $\alpha=\sqrt{2}$ by the root of some higher degree polynomial in $\mathbb{Z}[x]$ then there is, in general, no analog of the fundamental theorem of arithmetic. The fundmental theorem of arithmetic fails for the integers in $\mathbb{Q}(\sqrt{-5})$ and for the integers in $\mathbb{Q}(\sqrt{10})$ , for example. (See Hardy and Wright, §§14.6 for more details.)