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$$\newcommand{\Q}{\mathbb{Q}} \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\al}{\alpha} \newcommand{\bcal}{\mathcal{B}} \newcommand{\qroot}{\sqrt[3]} \newcommand{\froot}{\sqrt[4]} $$ 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 $E = \Q(\sqrt[3]{28})$.

  1. Find $T_{E|\Q}(\alpha)$ and $N_{E|\Q}(\alpha)$ for every $\alpha \in E$.

  2. 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 $$\beta \in \mathcal{O}[\qroot{28}]$$

  3. 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 $$\gamma + c\beta = \frac{1154}{3} - \frac{65}{3} = 363$$. $$ \therefore \beta^3 - \beta^2 - 65\beta - 363 = 0$$ $$ \therefore \beta \in \mathcal{O}[\qroot{28}] \hspace{5pt} \square$$

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(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.

$\square$

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  • $\begingroup$ Very impressive! Thank you! $\endgroup$ – Richard Clare Jun 14 '18 at 17:36
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    $\begingroup$ Also many thanks for posting the question, and the for the long computations done, i also learned a lot! (I also corrected a sign error... according to the decomposition of $x^3+y^3+z^3\color{red}{-}3xyz=(x+y+z)(\dots)$...) $\endgroup$ – dan_fulea Jun 14 '18 at 18:56
  • $\begingroup$ I posted a similar question here: math.stackexchange.com/questions/2819955/… in case you are interested. $\endgroup$ – Richard Clare Jun 14 '18 at 20:11
  • $\begingroup$ @RichardClare :: ok, i'll take a look at it... $\endgroup$ – dan_fulea Jun 15 '18 at 10:35

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