MHB Field extensions and Galois goup

[mathmari]

Hey!

Let $p$ b a prime, $n\in \mathbb{N}$ and let $f=x^{p^n}-1\in \mathbb{F}_p[x]$ be irreducible. Let $a\in \overline{\mathbb{F}}_p$ be a root pf $f$.

We have that $[\mathbb{F}_p(a):\mathbb{F}_p]=p^n$, where $\{1,a,\ldots , a^{p^n-1}\}$ is a basis of $\mathbb{F}_p(a)/\mathbb{F}_p$, so $\mathbb{F}_p(a)=\{c_0+c_1a+\ldots +c_{p^n-1}a^{p^n-1} : c_i\in \mathbb{F}_p\}$. There are $p$ choices for each $c_i$. So, there are $p\cdot p\cdot \ldots \cdot p=p^{p^n}$ choices for $(c_0, c_1, \ldots c_{p^n-1})$. Therefore, we have that $|\mathbb{F}_p(a)|=p^{p^n}$, right? (Wondering)

I want to show that $\mathbb{F}_p(a)$ contains all the roots of $f$.

I holds that every finite extension of a finite field is normal, right? But how can we prove this? (Wondering)

Then I have shown that for each $b\in \mathbb{F}_{p^n}$, $a+b$ is a root of $f$. I have also shown that $\mathbb{F}_{p^n}\leq \mathbb{F}_p(a)$ and $n=p^i$ for some $i\in \{0, 1, \ldots , n\}$.

Then I want to show that $\text{Gal}(\mathbb{F}_p(α)/\mathbb{F}_{p^n} )$ is cyclic and let $\tau$ be a generator. I want to calculate also the order of $\tau$ as a function of $i$.

In the book there is the following corollary:
View attachment 6352

We have that $|\mathbb{F}_{p^n}|=p^n$ and $|\mathbb{F}_p(a)|=p^{p^n}$. From the above we have that $\mathbb{F}_p(α)/\mathbb{F}_{p^n} $ is Galois, $n\mid p^n$, which is true since $n=p^i$. From the theorem we also have that $\text{Gal}(\mathbb{F}_p(α)/\mathbb{F}_{p^n} )$ is cyclic and is generated by the automorphism $\tau (x)=x^{p^n}$.

How can we calculate the order of $\tau$ ? (Wondering)

Then independent from that I want to calculate a simple expression of $\tau^k(a)$.

I have done the following:
Since $\tau$ is a generator of $\text{Gal}(\mathbb{F}_p(α)/\mathbb{F}_{p^n} )$ that means that $\tau$ is a $\mathbb{F}_{p^n}$-automorphism of $\mathbb{F}_p(a)$ that maps to a root of $f$ to an other root, right? (Wondering)

So, we have that $\tau (a)=a+b$. Therefore, we get the following:
$$\tau^2(a)=\tau(a+b)=\tau (a)+\tau (b)=a+b+b=a+2b \\ \tau^3(a)=\tau(\tau^2(a))=\tau (a+2b)=\tau (a)+\tau (2b)=a+b+2b=a+3b \\ \ldots \\ \tau^k(a)=a+kb$$ Is this correct? (Wondering)
The order of $\tau$ is the smallest integer $m$ such that $\tau^m(a)=a\Rightarrow a+mb=a$. Does this imply that $m=p$ ? (Wondering)

 

[mathmari]

There is a typo at $f$. It should be $f=x^{p^n}-x-1$.

 

[mathmari]

Let $p$ b a prime, $n\in \mathbb{N}$ and let $f=x^{p^n}-x-1\in \mathbb{F}_p[x]$ be irreducible. Let $a\in \overline{\mathbb{F}}_p$ be a root pf $f$.

We have that $[\mathbb{F}_p(a):\mathbb{F}_p]=p^n$, where $\{1,a,\ldots , a^{p^n-1}\}$ is a basis of $\mathbb{F}_p(a)/\mathbb{F}_p$, so $\mathbb{F}_p(a)=\{c_0+c_1a+\ldots +c_{p^n-1}a^{p^n-1} : c_i\in \mathbb{F}_p\}$. There are $p$ choices for each $c_i$. So, there are $p\cdot p\cdot \ldots \cdot p=p^{p^n}$ choices for $(c_0, c_1, \ldots c_{p^n-1})$. Therefore, we have that $|\mathbb{F}_p(a)|=p^{p^n}$, right? (Wondering)

I want to show that $\mathbb{F}_p(a)$ contains all the roots of $f$.

To show that we have to show that the extension $\mathbb{F}_p(a)/\mathbb{F}_p$ is normal, right? (Wondering)

Since $|\mathbb{F}_p(a)|=p^{p^n}$ we get that for each $e\in \mathbb{F}_p(a)$ it holds that $e^{p^{p^n}}=e \Rightarrow e^{p^{p^n}}-e=0$.
So, each element of $ \mathbb{F}_p(a)$ is a root of $x^{p^{p^n}}-x=0$.
From that we don't get yet that $\mathbb{F}_p(a)$ is the spitting field of $x^{p^{p^n}}-x\in \mathbb{F}_p$, do we? (Wondering)