TITLE: How can I compute the discriminant of the field $\mathbb{Q}(\sqrt[3]{28})$?
QUESTION [8 upvotes]: $$\newcommand{\Q}{\mathbb{Q}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\al}{\alpha}
\newcommand{\bcal}{\mathcal{B}}
\newcommand{\qroot}{\sqrt[3]}
\newcommand{\froot}{\sqrt[4]}
$$
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
        $$\gamma + c\beta =  \frac{1154}{3} - \frac{65}{3} = 363$$.
        $$ \therefore \beta^3 - \beta^2 - 65\beta - 363 = 0$$
$$ \therefore \beta \in \mathcal{O}[\qroot{28}] \hspace{5pt} \square$$

REPLY [7 votes]: (1.1) It is ok, the confirmation using computer algebra support, sage is as follows:
sage: R.<X> = PolynomialRing(QQ)
sage: K.<t> = NumberField( X^3-28 )
sage: K
Number Field in t with defining polynomial X^3 - 28
sage: t.minpoly()
x^3 - 28
sage: K(1).trace(), K(t).trace(), K(t^2).trace()
(3, 0, 0)

Here and in the sequel:
$$ t=\sqrt[3]{28}\ .
$$
(1.2) The norm of an element of the shape
$$ \xi = a+bt+ct^2 $$
is the product 
$$ N\xi=(a+bt+ct^2)(a+b\epsilon t+c\epsilon^2 t^2)(a+b\epsilon^2 t+c\epsilon t^2)\ ,$$
where $\epsilon$ is a primitive root of unit of order three. Explicitly:
$$
\begin{aligned}
N\xi 
&= a^3+b^3t^3 +c^3t^6\color{red}{-} 3abct^3 \\
&= a^3 + 28 b^3+ 28^2 c^3 \color{red}{-} 3\cdot 28abc\ .
\end{aligned}
$$
(Edited sign...)
For instance:
sage: (1+t+2018*t^2).norm()
6442872498805
sage: 1^3 + 28 + 2018^3*28^2 - 3*28*2018*1*1 
6442872498805

So i use the basis $\{1,t,t^2\}$ above. Which is not the integral basis,
sage: K.integral_basis()
[1/3*t^2 + 1/3*t + 1/3, t, 1/2*t^2]

(2) 
It is enough to compute the norm of the element:
$$
\begin{aligned}
N\beta 
&=
N\left(\frac 13(1+7t+t^2)\right)
\\
&=
\frac 1{3^3}(1+28\cdot 7^3+28^2\cdot 1-3\cdot 28\dot 7)
\\
&=\frac 1{3^3}(1+9604+784-588)=\frac 13\cdot 9801\\
&=3\cdot 11^2\ .
\end{aligned}
$$
So $\beta$ is integral. (The norm is in $\Bbb Z$.)
We can also ask the computer for this (and its minimal polynomial):
sage: beta = ( 1 + 7*t + t^2 ) / 3
sage: beta.norm().factor()
3 * 11^2
sage: beta.minpoly()
x^3 - x^2 - 65*x - 363

Yes, the computation in (2), as posted is ok.
(3) Let us ask first for the result:
sage: K.discriminant().factor()
-1 * 2^2 * 3 * 7^2

Using the given basis, 
$$
\frac 13(1+t+t^2)\ ,\ 
t\ ,\
\frac 12 t^2\ ,
$$
The field has three complex embeedings in $\Bbb C$, 
defined on generator by 
$t\to t=\sqrt[3]{28}$,
respectively by 
$t\to \epsilon t$,
respectively by 
$t\to \epsilon^2 t$.
So we have by definition to compute the square of the determinant:
$$
\begin{aligned}
D
&=
\begin{vmatrix}
\frac 13(1+t+t^2) & t & \frac 12t^2\\
\frac 13(1+\epsilon t+\epsilon^2 t^2) & \epsilon t & \frac 12\epsilon^2 t^2\\
\frac 13(1+\epsilon^2 t+\epsilon t^2) & \epsilon^2 t & \frac 12\epsilon t^2
\end{vmatrix}
\\[2mm]
&=
\frac 12\cdot \frac 13
\begin{vmatrix}
1+t+t^2 & t & t^2\\
1+\epsilon t+\epsilon^2 t^2 & \epsilon t & \epsilon^2 t^2\\
1+\epsilon^2 t+\epsilon t^2 & \epsilon^2 t & \epsilon t^2
\end{vmatrix}
\\[2mm]
&=
\frac 12\cdot \frac 13
\begin{vmatrix}
1 & t & t^2\\
1 & \epsilon t & \epsilon^2 t^2\\
1 & \epsilon^2 t & \epsilon t^2
\end{vmatrix}
\\[2mm]
&=
\frac 12\cdot \frac 13\cdot t^3
\begin{vmatrix}
1 & 1 & 1\\
1 & \epsilon  & \epsilon^2\\
1 & \epsilon^2  & \epsilon
\end{vmatrix}
\\[2mm]
&=
\frac 12\cdot \frac 13\cdot 28
\cdot\underbrace{(1-\epsilon)^3}_{-3i\sqrt 3}
\cdot\underbrace{(\epsilon+\epsilon^2)}_{=-1}
\\[2mm]
&=
2\cdot 7\cdot i\sqrt 3
\ .
\end{aligned}
$$
The square is the discriminant.
$\square$