PlanetMath: primitive element of biquadratic field

In other words, $\sqrt{m}+\sqrt{n}$ is a primitive element of $\Q(\sqrt{m},\sqrt{n})$ .

Proof. We clearly have $\Q(\sqrt{m}+\sqrt{n}) \subseteq \Q(\sqrt{m},\sqrt{n})$ . For the reverse inclusion, it is equivalent to show that $\sqrt{m}+\sqrt{n}$ does not belong to any of the quadratic subfields of $\Q(\sqrt{m},\sqrt{n})$ , which are $\Q(\sqrt{m})$ , $\Q(\sqrt{n})$ , and $\Q(\sqrt{mn})$ .

Suppose that $\sqrt{m}+\sqrt{n}\in\Q(\sqrt{m})$ . Then $\sqrt{n}\in\Q(\sqrt{m})$ . Thus, $\Q(\sqrt{n})=\Q(\sqrt{m})$ , which is proven to be false here. By a similar argument, $\sqrt{m}+\sqrt{n}\notin\Q(\sqrt{n})$ .

Suppose that $\sqrt{m}+\sqrt{n}\in\Q(\sqrt{mn})$ . Let $a,b,c,d\in\Z$ with $\gcd(a,b)=\gcd(c,d)=1$ , $b\neq 0$ , and $d\neq 0$ such that

$\displaystyle \sqrt{m}+\sqrt{n}=\frac{a}{b}+\frac{c}{d}\sqrt{mn}.$

(1)

Now, we perform some basic algebraic manipulations.

$\displaystyle bd\sqrt{m}+bd\sqrt{n}$	$\displaystyle =ad+bc\sqrt{mn}$
$\displaystyle bd\sqrt{m}+bd\sqrt{n}-ad$	$\displaystyle =bc\sqrt{mn}$
$\displaystyle (bd\sqrt{m}+bd\sqrt{n}-ad)^2$	$\displaystyle =(bc\sqrt{mn})^2$
$\displaystyle b^2d^2m+2b^2d^2\sqrt{mn}-2abd^2\sqrt{m}+b^2d^2n-2abd^2\sqrt{n}+a^2d^2$	$\displaystyle =b^2c^2mn$
$\displaystyle b^2d^2m+b^2d^2n+a^2d^2-b^2c^2mn$	$\displaystyle =2abd^2(\sqrt{m}+\sqrt{n})-2b^2d^2\sqrt{mn}$

Now, we use equation (

) to eliminate the $\sqrt{m}+\sqrt{n}$ and obtain

$\displaystyle b^2d^2m+b^2d^2n+a^2d^2-b^2c^2mn$

$\displaystyle =2abd^2\left( \frac{a}{b}+\frac{c}{d}\sqrt{mn} \right) -2b^2d^2\sqrt{mn}.$

Now, we perform some more basic algebraic manipulations.

$\displaystyle b^2d^2m+b^2d^2n+a^2d^2-b^2c^2mn$	$\displaystyle =2a^2d^2+2abcd\sqrt{mn}-2b^2d^2\sqrt{mn}$
$\displaystyle b^2d^2m+b^2d^2n-a^2d^2-b^2c^2mn$	$\displaystyle =2bd(ac-bd)\sqrt{mn}$

Since $\sqrt{mn}\notin\Q$ , $b\neq 0$ , and $d\neq 0$ , we must have $ac-bd=0$ . Thus, $\frac{c}{d}=\frac{b}{a}$ . (Note that we have $a\neq 0$ since $ac=bd\neq 0$ .) Using this in equation (

), we obtain$$ \sqrt{m}+\sqrt{n}=\frac{a}{b}+\frac{b}{a}\sqrt{mn}.$$

Now we perform similar calculations as before.

$\displaystyle ab\sqrt{m}+ab\sqrt{n}$	$\displaystyle =a^2+b^2\sqrt{mn}$
$\displaystyle ab\sqrt{m}+ab\sqrt{n}-a^2$	$\displaystyle =b^2\sqrt{mn}$
$\displaystyle (ab\sqrt{m}+ab\sqrt{n}-a^2)^2$	$\displaystyle =(b^2\sqrt{mn})^2$
$\displaystyle a^2b^2m+2a^2b^2\sqrt{mn}-2a^3b\sqrt{m}+a^2b^2n-2a^3b\sqrt{n}+a^4$	$\displaystyle =b^4mn$
$\displaystyle a^2b^2m+a^2b^2n+a^4-b^4mn$	$\displaystyle =2a^3b(\sqrt{m}+\sqrt{n})-2a^2b^2\sqrt{mn}$
$\displaystyle a^2b^2m+a^2b^2n+a^4-b^4mn$	$\displaystyle =2a^3b\left( \frac{a}{b}+\frac{b}{a}\sqrt{mn} \right)-2a^2b^2\sqrt{mn}$
$\displaystyle a^2b^2m+a^2b^2n+a^4-b^4mn$	$\displaystyle =2a^4+2a^2b^2\sqrt{mn}-2a^2b^2\sqrt{mn}$
$\displaystyle a^2b^2m+a^2b^2n-a^4-b^4mn$	$\displaystyle =0$