Exercise 1.1.9 (Groups $\mathbb{Q}(\sqrt{2}), (\mathbb{Q}(\sqrt{2}))^\times$)

Proof. Let $G=\{x\in ℝ\mid \exists <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{Q},\exists <!--\; FUNCTION\; APPLICATION\; -->b\in \mathbb{Q},x=a+b\sqrt{2}\}$.

(a)

We show that $G$ is a subgroup of $(ℝ,+)$.

• By definition, $G\subseteq ℝ$, and $0=0+0\sqrt{2}\in G$, so $G\ne \varnothing$.

• If $x\in G$ and $y\in G$, then there are $a,b,c,d\in \mathbb{Q}$ such that $x=a+b\sqrt{2},y=c+d\sqrt{2}$. Then

  $$x-y=(a+b\sqrt{2})-(c+d\sqrt{2})=(a-c)+(b-d)\sqrt{2}=u+v\sqrt{2},$$

  where $u=a-c\in \mathbb{Q}$ and $v=b-d\in \mathbb{Q}$. So $x-y\in G$.

So $G$ is a subgroup of $ℝ$ under addition.

(b)

Let $$G^{\ast }=\{x\in ℝ-\{0\}\mid \exists <!--\; FUNCTION\; APPLICATION\; -->a\in \mathbb{Q},\exists <!--\; FUNCTION\; APPLICATION\; -->b\in \mathbb{Q},x=a+b\sqrt{2}\}.$$

We show that $G^{\ast }$ is a subgroup of $(ℝ-\{0\},\times )$.

• $1=1+0\sqrt{2}\in G^{\ast }$, so $G^{\ast }\ne \varnothing$.

• If $x\in G^{\ast }$ and $y\in G^{\ast }$, then $x\ne 0,y\ne 0$, and

  $$x=a+b\sqrt{2},y=c+d\sqrt{2},$$

  where $a,b,c,d$ are rational numbers. Then $\mathit{𝑥𝑦}\ne 0$, and

  $$\mathit{𝑥𝑦}=(a+b\sqrt{2})(c+d\sqrt{2})=(\mathit{𝑎𝑐}+2\mathit{𝑏𝑑})+(\mathit{𝑏𝑐}+\mathit{𝑎𝑑})\sqrt{2}=r+s\sqrt{2},$$

  where $r=\mathit{𝑎𝑐}+2\mathit{𝑏𝑑}\in \mathbb{Q}$, and $s=\mathit{𝑏𝑐}+\mathit{𝑎𝑑}\in \mathbb{Q}$. Therefore $\mathit{𝑥𝑦}\in G^{\ast }$.

• If $x\in G^{\ast }$, then $x=a+b\sqrt{2}\ne 0$, where $a,b\in \mathbb{Q}$. Then $(a,b)\ne (0,0)$ (otherwise $a+b\sqrt{2}=0$). Therefore $a-b\sqrt{2}\ne 0$: indeed, if $a-b\sqrt{2}=0$, then $b=0$, otherwise $\sqrt{2}=a∕b\in \mathbb{Q}$, which is false, and $a=-b\sqrt{2}=0$. Thus

  $$a^2-2b^2=(a+b\sqrt{2})(a-b\sqrt{2})\ne 0.$$
  $$\begin{array}{llll}\hfill x^{-1}& =\frac{1}{a+b\sqrt{2}}& \hfill & \\ \hfill & =\frac{a-b\sqrt{2}}{(a-b\sqrt{2})(a+b\sqrt{2})}& \hfill & \\ \hfill & =\frac{a-b\sqrt{2}}{a^2-2b^2}& \hfill & \\ \hfill & =A+B\sqrt{2},& \hfill & \end{array}$$

  where $A=\frac{a}{a^2-2b^2}\in \mathbb{Q}$ and $B=-\frac{b}{a^2-2b^2}\in \mathbb{Q}$. So $x^{-1}\in G$, and $x^{-1}\ne 0$, therefore $x^{-1}\in G^{\ast }$

So $G^{\ast }$ is a subgroup of $ℝ-\{0\}$ under multiplication.