Question

Let d be an integer that is not a perfect square. Show that$\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of$\mathrm{ℂ}$ .

Short answer

It is proved that $\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of$\mathrm{ℂ}$ .

Step-by-step solution

Subfield

It is given that $\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$, and we have to show $\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of$\mathrm{ℂ}$ .

If $\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ satisfies the following conditions for be a fieldthen,$\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ is subset of $\mathrm{ℂ}$ then, we can say$\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of $\mathrm{ℂ}$.

Conditions of a field

(1). Let $a=a_1+b_1\sqrt{d},b=a_2+b_2\sqrt{d}\in \mathrm{\mathbb{Q}}\sqrt{d}$.

$\begin{array}{rcl}& \Rightarrow & a+b=\left(a_1+b_1\sqrt{d}\right)+\left(a_2+b_2\sqrt{d}\right)\\ & =& \left(a_1+a_2\right)+\left(b_1+b_2\right)\sqrt{d}\\ & =& a+b\sqrt{d}\in \mathrm{ℂ}\end{array}$

(2). Let $a=a_1+b_1\sqrt{d},b=a_2+b_2\sqrt{d}\in \mathrm{\mathbb{Q}}\sqrt{d}$.

localid="1653927106135" $\begin{array}{rcl}& \Rightarrow & ab=\left(a_1+b_1\sqrt{d}\right)\left(a_2+b_2\sqrt{d}\right)\\ & =& \left(a_1{a}_2+d{b}_1{b}_2\right)+\left(b_1{a}_2+a_1{b}_2\right)\sqrt{d}\\ & & \end{array}$

It implies that$ab\in \mathrm{\mathbb{Q}}\sqrt{d}$.

(3). For zero vector $0\in \mathrm{\mathbb{Q}}$.

$\begin{array}{rcl}0& =& 0+0\sqrt{d}\\ & =& 0\end{array}$

It implies that$0\in \mathrm{\mathbb{Q}}\sqrt{d}$.

(4). From first condition, we can write $a+b\in \mathrm{\mathbb{Q}}\sqrt{d}$ where, $b=a_2+b_2\sqrt{d};\forall a_2,b_2\in \mathrm{\mathbb{Q}}$ . Let $b=-a_1+\left(-b_1\right)\sqrt{d};\forall -a_1,-b_1\in \mathrm{\mathbb{Q}}$ then it gives,

localid="1653927185388" $\begin{array}{rcl}& \Rightarrow & a+b=\left(a_1+b_1\sqrt{d}\right)+\left(-a_1-b_1\sqrt{d}\right)\\ & =& \left(a_1-a_1\right)+\left(b_1-b_1\right)\sqrt{d}\\ & =& 0\end{array}$

Hence, the set$\mathrm{\mathbb{Q}}\sqrt{d}$ is a subring of $\mathrm{ℂ}$.

(5). From the second condition, we can write $ab\in \mathrm{\mathbb{Q}}\sqrt{d}$ where, $b=a_2+b_2\sqrt{2};\forall a_2,b_2\in \mathrm{\mathbb{Q}}$ .

Let localid="1653927742592" $b=\frac{1}{a_1+b_1\sqrt{d}};\forall \frac{1}{a_1},\frac{1}{b_1}\in \mathrm{\mathbb{Q}}$ then it gives,

$\begin{array}{rcl}& \Rightarrow & ab=\left(a_1+b_1\sqrt{d}\right)\left(a_2+b_2\sqrt{d}\right)\\ & =& \left(a_1+b_1\sqrt{d}\right)\frac{1}{\left(a_1+b_1\sqrt{d}\right)}\\ & =& 1\end{array}$

Thus, it implies that$a=\frac{1}{b}$.

Now, it is also written as:

$\begin{array}{rcl}a& =& \frac{1}{b}\\ & =& \frac{1}{a_1+b_1\sqrt{d}}\end{array}$

After rationalizing, it gives,

localid="1653927367452" $\begin{array}{rcl}a& =& \frac{\left(a_1-b_1\sqrt{d}\right)}{\left(a_1+b_1\sqrt{d}\right)\left(a_1-b_1\sqrt{d}\right)}\\ & =& \frac{a_1-b_1\sqrt{d}}{a_1^2-b_1^2{\left(\sqrt{d}\right)}^2}\\ & =& \frac{a_1}{a_1^2-b_1^{2}d}+\frac{-b_1\sqrt{d}}{a_1^2-b_1^{2}d}\end{array}$

It implies that,

$\frac{a_1}{a_1^2-b_1^{2}d}+\frac{-b_1\sqrt{d}}{a_1^2-b_1^{2}d}=a\in \mathrm{\mathbb{Q}}\left(\sqrt{d}\right)$

where localid="1653927398641" $\frac{a_1}{a_1^2-b_1^{2}d},\frac{-b_1}{a_1^2-b_1^{2}d}\in \mathrm{\mathbb{Q}}$ .

Hence, $\mathrm{\mathbb{Q}}\left(\sqrt{d}\right)=\left\{a+b\sqrt{d}\overline{)a,b\in \mathrm{\mathbb{Q}}}\right\}$ is a subfield of $\mathrm{ℂ}$.