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Question:Prove that $x^{p^n} -x +1$ is irreducible over $\mathbb{F_p}$ only when n=1 or n=p=2

I know it is a duplicate question. However, someone gave some nice hints on this problem and I want to follow the hints he gave.

Hint (1): Note that if $\alpha$ is a root, then so is $\alpha+a$ for any $a \in \mathbb{F_{p^n}}$

Hint (2): Show that this implies that $F_p(\alpha)$ contains $\mathbb{F_{p^n}}$

Hint (3): Show that $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p^n}}]=p$

Hint(1) and Hint(2) are done:

Hint(1) is easy to prove, just use two properties: $(\alpha+\beta)^p=\alpha^p+\beta^p$ and $\forall a \in \mathbb{F_{p^n}}$, $a$ is a root of $x^{p^n}-x+1$.

Hint(2): Since $F_p(\alpha)$ is a Galois Extension, we have $F_p(\alpha)$ contain all the roots of $x^{p^n} -x +1$, thus contain $\mathbb{F_{p^n}}$.

However, I stuck on Hint(3), and actually Hint(3) is very important, once we get $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p^n}}]=p$, we will get $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p}}]=n \cdot p$ ,and we know $[\mathbb{F_{p}(\alpha)}:\mathbb{F_{p}}]=p^n$, thus we can finish the proof.

Can anyone give me any idea about how to prove hint(3)?

scitamehtam
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  • Closely related. – Jyrki Lahtonen Feb 18 '15 at 06:26
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    This is actually explained in Thomas Andrews' answer to the linked question. Because $\alpha$ is a zero of your polynomial it follows that $$\alpha^{p^n}=\alpha-1.$$ In terms of $\phi, \phi(z)=z^{p^n},$ this means that $\phi(\alpha)=\alpha-1$. But $\phi$ is the generator of the Galois group $Gal(\Bbb{F}p(\alpha)/\Bbb{F}{p^n})$. Therefore the conjugates of $\alpha$ are $\phi(\alpha)=\alpha-1$, $\phi(\phi(\alpha))=\alpha-2$ et cetera. So the set of conjugates consists of the elements $\alpha-j$, $j=0,1,2,\ldots,p-1$. There are exactly $p$ of them, so the extension has to be of degree $p$. – Jyrki Lahtonen Feb 18 '15 at 13:49
  • Your conclusion about hint 2 depends on the assumption that the polynomial must be irreducible. you'd better mention that – Display name Nov 22 '21 at 00:59

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