Galois Theory and automorphisms

Question

Let $F = \mathbb{F}_{2}$ and $K=F(\alpha)$, where $\alpha$ is root of $1 + x + x^{2}$. Show that the function $\sigma: K \longrightarrow K$ given by $\sigma(a + b\alpha) = a + b + b\alpha$ for $a,b \in F$ is an $F$-automorphism of $K$.

I know that the automorphisms are determinated by their action on a generating set, but I don't know how to use this for show that $\sigma$ is an automorphism of $Gal(K/F)$. Any hint?

Answer 1

First you need to show that it fixes $F.$ This is clear since $\sigma(a) = a.$

Then you need to show it is an automorphism. Since $K$ is the field with four elements $a+b\alpha$ for $a,b\in \mathbb F_2,$ it's clear that this maps $K\to K.$ So you need to show it is bijective and a homomorphism.

For the homomorphism part, we have $$\sigma(a+b\alpha + c + d\alpha) = \sigma ((a+c)+ (b+d)\alpha) = a+c+b+d +(b+d)\alpha.$$ You need to show this is equal to $\sigma(a +b\alpha)+\sigma(c+d\alpha).$ Similarly for multiplication (which is a bit harder and where you need to use that $\alpha^2+\alpha +1=0$ and that the field is characteristic 2) and identity (but that follows simply from $\sigma$ fixing $F$).

Then you need to show it's bijective. It's the matrix $\pmatrix {1&1\\0&1}...$ can you invert that in $\mathbb F_2$? Worst comes to worst (for all of these) there are only four elements you need to worry about.

Answer 2

$u,v\in K, c\in F, \sigma(u+v)=\sigma(u)+\sigma(v), \sigma(cu)=c\sigma(u)$.

$\sigma(\alpha^2)=\sigma(1+\alpha)=1+1+\alpha=\alpha$, $\sigma(\alpha)\sigma(\alpha)=(1+\alpha)(1+\alpha)=1+\alpha+\alpha+\alpha^2=1+\alpha^2=1+1+\alpha=\alpha$.

Let $a,b,c,d\in F$, $\sigma(a+b\alpha)\sigma(c+d\alpha)=(a+b\sigma(\alpha))(c+d\sigma(\alpha))$ $=ac+(ad+(ad+bc)\sigma(\alpha)+bd\sigma(\alpha)^2=ac+(ad+bc)(1+\alpha)+bd\alpha$ $=ac+ad+bc+(ad+bc+bd)\alpha$.

$(a+b\alpha)(c+d\alpha)=ac+(ad+bc)\alpha+bd\alpha^2=ac+(ad+bc)\alpha+bd(1+\alpha)=ac+bd+(ad+bc+bd)\alpha$.

This implies that $\sigma((a+b\alpha)(c+d\alpha))=\sigma(ac+bd+(ad+bc+bd)\alpha)=ac+bd+ad+bc+bd+(ad+bc+bd)\alpha$ $=ac+ad+bc+(ad+bc+bd)\alpha=\sigma(a+b\alpha)(\sigma(c+d\alpha))$.