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

Let $a\in \overline{\mathbb{F}}_p$ (algebraic closure of $\mathbb{F}_p$) be a root of $f$.

We have that $\mathbb{F}_p(a)$ contains all the roots of $f$, for each $b\in \mathbb{F}_{p^n}$, $a+b$ is a root of $f$ and $\mathbb{F}_{p^n}\subseteq \mathbb{F}_p(a)$.

How can we show that $n=p^i$ for any $i\in \{0, 1, \ldots , n\}$?

I want to show that $Gal(\mathbb{F}_p(a)/\mathbb{F}_{p^n})$ is cyclic and let $\tau$ be a generator.

Since $a+b$ is a root of $f$ for each $b\in \mathbb{F}_{p^n}$, we have that $f$ is separable.

So, $\mathbb{F}_p(a)$ is the splitting field of the separable polynomial $f\in \mathbb{F}_p[x]$.

Therefore the extension $\mathbb{F}_p(a)/\mathbb{F}_p$ is Galois, with $|Gal(\mathbb{F}_p(a)/\mathbb{F}_p)|=[\mathbb{F}_p(a):\mathbb{F}_p]=\deg f=p^n$.

How could we continue?

user26857
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Mary Star
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1 Answers1

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We know from your other topic that this is a Galois extension. Now consider $|K|=|\Bbb F_p(a)| = p^k(=p^{p^n})$, then $[K:\Bbb F_p]=k \; (=p^n)$. We also know that the defining characteristic of elements of $K$ is that $x^{p^k}-x=0$, and $k$ is the smallest positive integer for which this holds for all $\alpha\in K$. But then the field automorphism $x\mapsto x^p$ has order $k$ in the Galois group. But as $|\text{Gal}(K/\Bbb F_p)|=k$, we see that this automorphism generates the group.

Note: Just as in your previous topic, there is absolutely nothing special about the polynomial you chose: this is true for all finite extensions of finite fields.

Adam Hughes
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  • We have that $\mathbb{F}_p(a)$ contains all the roots of $f$. Does this mean that it is a spitting field of $f$ or that a $K\leq \mathbb{F}_p(a)$ is a splitting field of $f$? – Mary Star Dec 05 '16 at 17:22
  • Why do we not have that $[\mathbb{F}_p(a):\mathbb{F}_p]=\deg f=p^n$ ? – Mary Star Dec 05 '16 at 17:24
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    @MaryStar the polynomial is irreducible, so $K$ is a smallest degree field containing a root of $f$, so it will not be a proper subfield. I just renamed the field $K$ because I didn't want to typeset $\Bbb F_p(a)$ a million times. :) – Adam Hughes Dec 05 '16 at 17:24
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    @MaryStar to your second question: you do, that bit was a typo, sorry! :) – Adam Hughes Dec 05 '16 at 17:25
  • Ah ok :-) $$$$ What do you mean by "the field automorphism $x\mapsto x^p$ has order $k$ in the Galois group " ? – Mary Star Dec 05 '16 at 17:33
  • Ah we have to show that $\sigma^k(x)=x$ when $\sigma (x)=x^p$, right? – Mary Star Dec 05 '16 at 17:33
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    @MaryStar yes, that's baked in though: we know that all elements of the field satisfy $x^{p^k}-x=0$, and all the elements of the field are roots, so we cannot have $p^k$ solutions to a degree $<p^k$ polynomial, which would be the case if $\sigma^j(x)=x$ for $j<k$. – Adam Hughes Dec 05 '16 at 17:35
  • We have that the automorphism has order $k$, does this mean that $\langle \sigma \rangle$ has order $k$ ? And since this is a subgroup of the galosi group G and it has the same order they must be the same groups? – Mary Star Dec 05 '16 at 17:42
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    @MaryStar yes exactly – Adam Hughes Dec 05 '16 at 17:45
  • At the previous comment you say that the elements satisfy $x^{p^k}-x=0$ souldn't it be $x^k-x=0$ ? – Mary Star Dec 05 '16 at 17:53
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    @MaryStar Remember $k=p^n$ is the degree of the extension. – Adam Hughes Dec 05 '16 at 17:55
  • I got stuck right now... Why do we have that $|K|=p^{p^n}$ ? – Mary Star Dec 05 '16 at 18:12
  • Doesn't it have $p^i$ elements for some $i$ ? – Mary Star Dec 05 '16 at 18:12
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    @MaryStar it's a dimension $k$ vector space over a field with $p$ elements. So you multiply $p$ by itself $k$ times. That's just set theory. And yes, $i=p^n$ is a fine integer – Adam Hughes Dec 05 '16 at 18:22
  • How do we know that it is a dimenson $k$ vector space over a field with $p$ elements? – Mary Star Dec 05 '16 at 19:58
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    @MaryStar that is the definition of the degree of a field extension. – Adam Hughes Dec 05 '16 at 20:01