Exercise 10.1.4 (Frobenius automorphism on F3)

Question

Work out the values of the Frobenius automorphism on the field $\mathbb{F}_3(\sqrt 2)$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
By 	extbf{Lemma 10.1.2} we have that $n = [\mathbb{F}_3(\sqrt 2): \mathbb{F}_3] = 2$, hence we have that
$|\mathbb{F}_3(\sqrt 2)| = 3^2 = 9$. Hence we have that $\mathbb{F}_3(\sqrt 2) = \{0, 1, 2, \sqrt 2, 2 \sqrt 2, 1+ \sqrt 2, 1+ 2\sqrt2, 2+ \sqrt2, 2+2\sqrt2\}$.
The Frobenius map for our field is then (with $p = 3$) $	heta: r \mapsto r^3$.
Hence we get:
\begin{equation*}
    \begin{aligned}
        0 & \mapsto 0\
        1 & \mapsto 1\
        2 & \mapsto 2\
        \sqrt2 & \mapsto 2\sqrt2\
        1+\sqrt2 & \mapsto 1 + 2\sqrt2\
        2+\sqrt2 & \mapsto 2 + 2\sqrt2\
        2\sqrt2 & \mapsto \sqrt2\
        1+2\sqrt2 & \mapsto 1 + \sqrt2\
        2+2\sqrt2 & \mapsto 2 + \sqrt2\
    \end{aligned}
\end{equation*}

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} 
The polynomial $m(t) = t^2 - 2 =t^2+1 $ has no root in $\F_3$. Since $\deg(m) = 2$, it is irreducible over $\F_3$. Consider the field $M = \mathbb{F}_3[t]/\langle t^2 - 2 angle$, and $\alpha =  t + \langle t^2 - 2 angle$. Then $\alpha$ is a root of $t^2 - 2 = t^2 +1$, and 
$$M = \F_3(\alpha) ( = \F_3(\sqrt{2})  = \F_3(\sqrt{-1}) = \{0,1,-1,\alpha, - \alpha, 1 - \alpha, 1 + \alpha, -1 - \alpha, -1 + \alpha\}.$$

The Frobenius automorphism is given by 
$$
\mathrm{Frob}
\left\{
\begin{array}{ccl}
\F_3(\alpha)& 	o & \F_3(\alpha)\
x & \mapsto &x^3
\end{array}
ight.
$$
If $a,b \in \F_3$, then $a^3 =a, b^3 = b$. Moreover $\alpha^3 = - \alpha$. Since $\mathrm{Char}(M) = 3$,
 $$(a + b\alpha)^3 = a^3 + b^3 \alpha^3 = a - b\alpha.$$
 So $\mathrm{Frob}$  is similar to complex conjugation, where $\alpha$ plays the role of $i$, because $\alpha^2 = -1$.
 $$
 \begin{array}{c|ccccccccc}
 x & 0&1&-1&\alpha& - \alpha &1 - \alpha& 1 + \alpha& -1 - \alpha& -1 + \alpha\
 \hline\
 \mathrm{Frob}(x) &0&1&-1&-\alpha& + \alpha &1 + \alpha& 1 - \alpha& -1 + \alpha& -1 - \alpha
 \end{array}
 $$
 (same results as Hassaan Naeem's computations.)
\end{proof}

Exercise 10.1.4 (Frobenius automorphism on F3)

Answers

Comments

Comments

Exercise 10.1.4 (Frobenius automorphism on F3)

Answers

Comments

Comments

Add answer