# primitive element of biquadratic field

###### Theorem.

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

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 $\gcd(a,b)=\gcd(c,d)=1$, $b\neq 0$, 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.

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

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

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

Now, we perform some more basic algebraic manipulations.

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

Since $\sqrt{mn}\notin\mathbb{Q}$, $b\neq 0$, 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 (1), we obtain

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

Now we perform calculations as before.

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

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.

$\begin{array}[]{rl}a^{2}m-a^{4}-mn+a^{2}n&=0\\ a^{2}(m-a^{2})-n(m-a^{2})&=0\\ (m-a^{2})(a^{2}-n)&=0\end{array}$

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 PrimitiveElementOfBiquadraticField 2013-03-22 17:54:17 2013-03-22 17:54:17 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Theorem msc 11R16 PrimitiveElementTheorem