$F_p(a)$ contains all the roots of $f$

Question

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 $\mathbb{F}_p$) is a root of $f$.

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

$$$$

If $a\in \mathbb{F}_p$ then $a^p=a$ and then

$a^{p^n}=(a^p)^{p^n-1}=a^{p^n-1}=(a^p)^{p^n-2}=a^{p^n-2}=\ldots =a$

So, $f(a)=a^{p^n}-a-1=-1\neq 0$.

Thus it must be $a\notin \mathbb{F}_p$.

Could you give me a hint how we could show that $\mathbb{F}_p(a)$ contains all the roots of $f$ ?

See this and this question for more about this family of polynomials. It is rarely irreducible. And all the zeros will be in $\Bbb{F}_{p^m}$ with $m=pn$. Irrespective of whether it is irreducible or not. — Jyrki Lahtonen, Dec 05 '16 at 16:51

score 2 · Accepted Answer · answered Dec 05 '16 at 14:21

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Let $|F_p(a)|=g$. Then as $F_p(a)^\times$ is a group of order $g-1$, all elements of $F_p(a)$ satisfy $x^g-x=0$. Because $F_p$ is a field, this polynomial, which has all distinct roots as its derivative is identically $-1$ has distinct roots in the algebraic closure, in particular, all elements of $F_p(a)$ are roots of it. However, since--in particular--$a\in F_p(a)$ the minimal polynomial for $a$ divides $x^g-x$. So since all the roots of the minimal polynomial of $a$ are roots of $x^g-x$, a fortiori all the roots of the minimal polynomial of $a$ are in $F_p(a)$.

answered Dec 05 '16 at 14:21

Adam Hughes

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Not very clear. You showed $F_p(a)$ is the splitting field of $P(x)= x^g-x$ over $\mathbb{F}_p$. At the end you are assuming $x^{p^n}-x-1$ is irreducible over $\mathbb{F}_p$ ? – reuns Dec 05 '16 at 14:34
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$x^{p^n}-x-1$ is indeed irreducible. I just forgot to mention it. @user1952009 – Mary Star Dec 05 '16 at 14:36
Since $|\mathbb{F}_p(a)|=g$ we have that $|\mathbb{F}_p(a)^{\star}|=g-1$. So, $\forall x\in \mathbb{F}_p(a)^{\star}$ it holds that $x^{g-1}=1$, So, $x^g=x \Rightarrow x^g-x=0$. This holds also for $0$. Therefore each element of $\mathbb{F}_p(a)$ is a root of $x^g-x$. So, $\mathbb{F}_p(a)$ is a splitting field of $x^g-x$. This is a splitting field of $x^g-x$ over $\mathbb{F}_p(x)$, or not? – Mary Star Dec 05 '16 at 15:14
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@MaryStar You mean over $\Bbb F_p$ not over $\Bbb F_p(x)$, but yes. – Adam Hughes Dec 05 '16 at 15:16
I understand!! Thank you!! :-) – Mary Star Dec 05 '16 at 16:02
Does the downvoter want to comment on what they think is wrong with my answer? – Adam Hughes Dec 05 '16 at 17:30

score 1 · Answer 2 · answered Dec 05 '16 at 14:27

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Hint:

For any $\;t\in\Bbb F_p\;$ , we have that

$$(a+t)^{p^n}-(a+t)-1=a^{p^n}-a-1+t^{p^n}-t=0$$

answered Dec 05 '16 at 14:27

DonAntonio

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And how do we know that $a+t$ for $t\in \mathbb{F}_p$ are all the roots? – Mary Star Dec 05 '16 at 14:37
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The degree of $f$ is $p^n$. Since $t\in \mathbb{F}_p$ there are $p$ such $t$. Therefore the number of $a+t$ is $p$ or not? – Mary Star Dec 05 '16 at 14:42
@MaryStar Yes...that was a hint... – DonAntonio Dec 05 '16 at 18:01

$F_p(a)$ contains all the roots of $f$

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