PlanetMath: using the primitive element of biquadratic field

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using the primitive element of biquadratic field

(Application)

Let $m$ and $n$ be two distinct squarefree integers $\neq 1$ . We want to express their square roots as polynomials of

$\displaystyle \alpha\, := \,\sqrt{m}\!+\!\sqrt{n}$

(1)

with rational coefficients.

If $\alpha$ is cubed, the result contains no terms with $\sqrt{mn}$ :

$\displaystyle \alpha^3$	$\displaystyle = (\sqrt{m})^2+3(\sqrt{m})^2\sqrt{n}+3\sqrt{m}(\sqrt{n})^2+(\sqrt{n})^3$
	$\displaystyle = m\sqrt{m}+3m\sqrt{n}+3n\sqrt{m}+n\sqrt{n}$
	$\displaystyle = (m+3n)\sqrt{m}+(3m+n)\sqrt{n}$

Thus, if we subtract from this the product $(3m\!+\!n)\alpha$ , the $\sqrt{n}$ term vanishes: $$\alpha^3-(3m+n)\alpha = (-2m+2n)\sqrt{m}$$ Dividing this equation by $-2m\!+\!2n$ ($\neq 0$ ) yields

$\displaystyle \sqrt{m}\, = \,\frac{\alpha^3-(3m+n)\alpha}{2(-m+n)}.$

(2)

Similarly, we have

$\displaystyle \sqrt{n}\, = \,\frac{\alpha^3-(m+3n)\alpha}{2(m-n)}.$

(3)

The representations (2) and (3) may be interpreted as such polynomials as intended.

Multiplying the equations (2) and (3) we obtain a corresponding representation for the square root of $mn$ which also lies in the quartic field $\mathbb{Q}(\sqrt{m},\,\sqrt{n}) = \mathbb{Q}(\sqrt{m}\!+\!\sqrt{n})$ :

$\displaystyle \sqrt{mn}\, = \,\frac{\alpha^6-4(m+n)\alpha^4+(3m^2+10mn+3n^2)\alpha^2}{4(-m^2+2mn-n^2)}$

For example, in the special case $m := 2,\; n := 3$ we have $$\sqrt{2} = \frac{\alpha^3-9\alpha}{2}, \quad \sqrt{3} = -\frac{\alpha^3-11\alpha}{2}, \quad \sqrt{6} = \frac{-\alpha^6+20\alpha^4-99\alpha^2}{4}.\\$$

Remark. The sum (1) of two square roots of positive squarefree integers is always irrational, since in the contrary case, the equation (3) would say that $\sqrt{n}$ would be rational; this has been proven impossible here.

"using the primitive element of biquadratic field" is owned by pahio.

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