cubic unit with positive norm must be positive
问题内容
Let $a$ be a positive integer, not a cube, so that $\alpha=\sqrt[3]a$ is irrational, and write $$R={\mathbb Z}[\alpha]=\{\,x+y\alpha+z\alpha^2\ |\ x,y,z\in\mathbb{Z}\,\}\ .$$ Let $\beta=x+y\alpha+z\alpha^2$ be a unit in $R$ with norm (product of conjugates) equal to $1$ (and not $-1$). Then $\beta>0$.
Proof. The reciprocal of $\beta$ is $$x^2+\alpha^2y^2+\alpha^4z^2-\alpha xy-\alpha^2 xz-\alpha^3 yz\ ,\tag1$$ and completing the square, we find $$\frac4\beta=(2x-\alpha y-\alpha^2z)^2+3(\alpha y-\alpha^2z)^2>0\ ,$$ so $\beta>0$.
My question. One doesn't usually expect completing the square to work out as neatly as that. So is it just an amazing coincidence? Or am I missing some simple fundamental reason for this result?
回答 (1)
The two conjugates of $a$ are $a\zeta$ and $a\overline{\zeta}$ where $\zeta\in\mathbb C$ is a primitive third root of $1$.
Hence, $$\frac{1}{\beta} = (x+y(\alpha\zeta) + z(\alpha\zeta)^2)\overline{(x+y(\alpha\zeta) + z(\alpha\zeta)^2)}>0$$