This question is from a very old exercise sheet, I do not know what the notation $\mathbb{Z}_{\mathbb{C}}$ means:
- $1\equiv 1 \mod 4 $ in $\mathbb{Z}$. Show that $\frac{1+\sqrt{m}}{2} \in \mathbb{Z}_{\mathbb{C}}$ 2. Let $a= (19)^{1/3}$. Show that $\frac{1+a+a^{2}}{3} \in \mathbb{Z}_{\mathbb{C}}$
1. It will now be shown that $\frac{1+\sqrt{m}}{2}$ is an algebraic integer, that means that it fulfills a polynomial with leading coefficient 1 (monic) and rest integral. Wikipedia article on algebraic integers says that it suffises the monic polynomial: $x^{2}-x+\frac{1-m}{4}$ This is true because: $(x-\frac{1}{2}(1+\sqrt{m})) (x-\frac{1}{2}(1-\sqrt{m})) = x^{2}-x+\frac{1-m}{4}$
2. Let $b = 19$, then $\frac{1+b+b^{2}}{3} = \frac{1+19+19^{2}}{3}$
Then one examines : $(x-(\frac{1+19+19^{2}}{3}))(x-\frac{-1-19-19^{2}}{3}) = (x-127)(x+127) = x^{2}- 16129$
now subtituting $x$ with $x^{3}$ gives: $x^{6}-16129$, and this is the monic polynomial for $a= (19)^{1/3}$
Is this done correctly? Please do tell