12千葉大学 2018年 数学 第12問解答・解説

Independent derivation

Let $\theta=2\pi/9$. Then $\alpha=z+z^8=2\cos\theta$, so the triple-angle identity gives $$\alpha^3-3\alpha=2\cos(3\theta)=2\cos(2\pi/3)=-1.$$ Thus $f(x)=x^3-3x+1$ is a monic integer cubic satisfying $f(\alpha)=0$.

For the first other root, using $z^9=1$, $$\alpha^2-2=(z+z^8)^2-2=z^2+z^7=2\cos(4\pi/9).$$ The triple-angle identity gives $(\alpha^2-2)^3-3(\alpha^2-2)=2\cos(4\pi/3)=-1$, so $\alpha^2-2$ is a root of $f$.

This root differs from $\alpha$: otherwise $\alpha^2-\alpha-2=0$, so $\alpha=2$ or $\alpha=-1$, but $f(2)=f(-1)=3$. The three roots of $f$ sum to zero because its $x^2$ coefficient is zero. The remaining root is therefore $$-\alpha-(\alpha^2-2)=2-\alpha-\alpha^2.$$ Both requested roots are integer-coefficient polynomials in $\alpha$ of degree at most two.