$$\newcommand{\Q}{\mathbb{Q}} \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\al}{\alpha} \newcommand{\bcal}{\mathcal{B}} \newcommand{\qroot}{\sqrt[3]} \newcommand{\froot}{\sqrt[4]} $$ I have a problem which consist in 3 problems. I solve part 1 and 2, but I want to make sure it is correct. Also I dont know how to solve part 3. Anyone ? Thanks
Consider the number field $E = \Q(\sqrt[3]{28})$.
Find $T_{E|\Q}(\alpha)$ and $N_{E|\Q}(\alpha)$ for every $\alpha \in E$.
Let $\mathcal{O}[\qroot{28}]$ be the set of all integral elements in $E$. Show that if $\beta = \frac{1}{3}(1 + 7\qroot{28} + 2\qroot{98})$, then $$\beta \in \mathcal{O}[\qroot{28}]$$
Consider the set $$ \mathcal{B} = \{\qroot{28},\qroot{98},\frac{1}{3}(1 + 7\qroot{28} + 2\qroot{98}) \}$$ Assuming that $\mathcal{B}$ is an integral basis for $\mathcal{O}(\qroot{28})$, Calculate the field discriminant of the number field $E$.
$\textbf{Solution for 1}$: One ordered basis for the field extension is $\bcal = \{1,\qroot{28},\qroot{98}\}$. Any element $\al \in E$ is of the form $$ \al = a + b\qroot{28} + c\qroot{98}$$ Multiplying $\al$ for each basis elements we obtain $$ \al\cdot1 = a + b\qroot{28} + c\qroot{98}$$ $$ \al\cdot\qroot{28} = a\qroot{28} + 2b\qroot{98} + 14c$$ $$ \al\cdot\qroot{98} = a\qroot{98} + 14b + 7c\qroot{28}$$ Hence, $$[\al]_\bcal = \begin{bmatrix} a & 14c & 14b \\ b & a & 7c \\ c & 2b & a \end{bmatrix}$$ and we obtain that $$T_{E|\Q}(\alpha) = 3a$$ $$N_{E|\Q}(\alpha) = Det([\al]_{\bcal}) =a^3 - 42 abc + 28 b^3 + 98 c^3\hspace{5pt} \square $$
$\textbf{Solution for 2:}$ Consider $\beta = \frac{1}{3}(1 + 7 \qroot{28} + 2 \qroot{98})$. Let $\gamma = \beta^3 - \beta^2$. Then we must find $c \in \mathbb{Z}$ such that $\gamma + c\beta \in \mathbb{Z}$. We obtain that
$$\gamma = \frac{1}{3}(1154 + 455\qroot{28} +130\qroot{98})$$
Hence, for $c = -65$ we obtain
$$\gamma + c\beta = \frac{1154}{3} - \frac{65}{3} = 363$$.
$$ \therefore \beta^3 - \beta^2 - 65\beta - 363 = 0$$
$$ \therefore \beta \in \mathcal{O}[\qroot{28}] \hspace{5pt} \square$$