primitive element of biquadratic field

Theorem.

Let $m$ and $n$ be distinct squarefree integers, neither of which is equal to $\mathrm{1}$. Then the biquadratic field $\mathrm{Q}\mathit{}\mathrm{(}\sqrt{m}\mathrm{,}\sqrt{n}\mathrm{)}$ is equal to $\mathrm{Q}\mathit{}\mathrm{(}\sqrt{m}\mathrm{+}\sqrt{n}\mathrm{)}$.

In other words, $\sqrt{m}+\sqrt{n}$ is a primitive element (http://planetmath.org/PrimitiveElement) of $\mathbb{Q}(\sqrt{m},\sqrt{n})$.

Proof.

We clearly have $\mathbb{Q}(\sqrt{m}+\sqrt{n})\subseteq \mathbb{Q}(\sqrt{m},\sqrt{n})$. For the reverse inclusion, it is equivalent (http://planetmath.org/Equivalent3) to show that $\sqrt{m}+\sqrt{n}$ does not belong to any of the quadratic subfields of $\mathbb{Q}(\sqrt{m},\sqrt{n})$, which are $\mathbb{Q}(\sqrt{m})$, $\mathbb{Q}(\sqrt{n})$, and $\mathbb{Q}(\sqrt{mn})$.

Suppose that $\sqrt{m}+\sqrt{n}\in \mathbb{Q}(\sqrt{m})$. Then $\sqrt{n}\in \mathbb{Q}(\sqrt{m})$. Thus, $\mathbb{Q}(\sqrt{n})=\mathbb{Q}(\sqrt{m})$, which is proven to be false here (http://planetmath.org/QuadraticFieldsThatAreNotIsomorphic). By a , $\sqrt{m}+\sqrt{n}\notin \mathbb{Q}(\sqrt{n})$.

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

$\sqrt{m}+\sqrt{n}={\displaystyle \frac{a}{b}}+{\displaystyle \frac{c}{d}}\sqrt{mn}.$ (1)

Now, we perform some basic algebraic manipulations.

$bd\sqrt{m}+bd\sqrt{n}$ $=ad+bc\sqrt{mn}$

$bd\sqrt{m}+bd\sqrt{n}-ad$ $=bc\sqrt{mn}$

${(bd\sqrt{m}+bd\sqrt{n}-ad)}^2$ $={(bc\sqrt{mn})}^2$

$b^2{d}^{2}m+2b^2{d}^2\sqrt{mn}-2ab{d}^2\sqrt{m}+b^2{d}^{2}n-2ab{d}^2\sqrt{n}+a^2{d}^2$ $=b^2{c}^{2}mn$

$b^2{d}^{2}m+b^2{d}^{2}n+a^2{d}^2-b^2{c}^{2}mn$ $=2ab{d}^2(\sqrt{m}+\sqrt{n})-2b^2{d}^2\sqrt{mn}$

Now, we use equation (1) to eliminate the $\sqrt{m}+\sqrt{n}$ and obtain

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

Now, we perform some more basic algebraic manipulations.

$b^2{d}^{2}m+b^2{d}^{2}n+a^2{d}^2-b^2{c}^{2}mn$ $=2a^2{d}^2+2abcd\sqrt{mn}-2b^2{d}^2\sqrt{mn}$

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

Since $\sqrt{mn}\notin \mathbb{Q}$, $b\ne 0$, and $d\ne 0$, we must have $ac-bd=0$. Thus, $\frac{c}{d}=\frac{b}{a}$. (Note that we have $a\ne 0$ since $ac=bd\ne 0$.) Using this in equation (1), we obtain

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

Now we perform calculations as before.

$ab\sqrt{m}+ab\sqrt{n}$ $=a^2+b^2\sqrt{mn}$

$ab\sqrt{m}+ab\sqrt{n}-a^2$ $=b^2\sqrt{mn}$

${(ab\sqrt{m}+ab\sqrt{n}-a^2)}^2$ $={(b^2\sqrt{mn})}^2$

$a^2{b}^{2}m+2a^2{b}^2\sqrt{mn}-2a^{3}b\sqrt{m}+a^2{b}^{2}n-2a^{3}b\sqrt{n}+a^4$ $=b^{4}mn$

$a^2{b}^{2}m+a^2{b}^{2}n+a^4-b^{4}mn$ $=2a^{3}b(\sqrt{m}+\sqrt{n})-2a^2{b}^2\sqrt{mn}$

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

$a^2{b}^{2}m+a^2{b}^{2}n+a^4-b^{4}mn$ $=2a^4+2a^2{b}^2\sqrt{mn}-2a^2{b}^2\sqrt{mn}$

$a^2{b}^{2}m+a^2{b}^{2}n-a^4-b^{4}mn$ $=0$

Since $b^2$ divides $a^4$ and $\gcd (a,b)=1$, we must have $b^2=1$. Plugging into the equation above yields

$$a^{2}m+a^{2}n-a^4-mn=0.$$

Now for yet some more algebraic manipulations.

Thus, $m=a^2$ or $n=a^2$, a contradiction. It follows that $\mathbb{Q}(\sqrt{m}+\sqrt{n})=\mathbb{Q}(\sqrt{m},\sqrt{n})$. ∎

Title	primitive element of biquadratic field
Canonical name	PrimitiveElementOfBiquadraticField
Date of creation	2013-03-22 17:54:17
Last modified on	2013-03-22 17:54:17
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	9
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 11R16
Related topic	PrimitiveElementTheorem