$\frac{1+\sqrt{m}}{2}; (19)^{1/3}$  are algebraic integral 
This question is from a very old exercise sheet, I do not know what the notation $\mathbb{Z}_{\mathbb{C}}$ means: 
  
  
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*$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  

Your approach seems too complicated in both cases. Try these simpler approaches:


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*Let $\alpha =\frac{1+\sqrt{m}}{2}%$. Then $2 \alpha -1 = \sqrt{m}$. Can you now find a monic equation for $\alpha$? Make sure you use the hypothesis on $m \bmod 4$.

*Let $\beta=\frac{1+a+a^{2}}{3}$. Note that $\beta = \frac{a^3-1}{3(a-1)}=\frac{6}{a-1}$. Then $a=\frac{6}{\beta}+1$. Now use that $a^3=19$ to find a monic equation for $\beta$.
