1

I am in the process of solving the following problem: $x^{p^n}-x+1$ is irreducible only when $n=1$ or $n=p=2$ over $F_p$.

The hint is to note that if $\alpha$ is a root, then $\alpha +a$, $a \in F_{p^n}$ is a root. I am supposed to use this to show that $F_p(\alpha)$ contains $F_{p^n}$ and that $[F_p(\alpha):F_{p^n}]=p$.

I was thinking about this problem and it got me very confused, mainly the first part. I am torn because assuming the polynomial is irreducible, $F_p(\alpha)$ is a finite extension so it must be Galois, so normal and so $F_p(\alpha)$ is a splitting field. This means that $F_p(\alpha)$ has all the zeroes of $f$.

My question is, are they of the "form" $\alpha +a$? This would imply $F_{p^n}$ is contained in the splitting field. My idea is just to say take two roots, subtract them, and you get an element of $F_{p^n}$ but i am not sure how to formalize this. Hopefully my confusion is clear: roots are unique, but their representations are not.

Community
  • 1
Sorfosh
  • 3,266
  • I think this question is a duplicate of this oldie. If there is something that is not clear to you after studying those answers, please let me know. Otherwise I'm inclined to vote to close this as a duplicate. It is a nice question and the reason for the exception with $p=n=2$ is a bit subtle. But the answers there are rather perfect IMO. – Jyrki Lahtonen Nov 26 '19 at 20:51
  • @JyrkiLahtonen Question is not about the problem per se, but rather about the confusions that were brought upon by the problem. I thought including it would be helpful to understand what I do not understand. – Sorfosh Nov 26 '19 at 20:54

1 Answers1

1

The map $x \to x^{p^n}$ is a Frobenius automorphism. It is an automorphism of any field of characteristic $p$. It's fixed field is the field $\mathbb{F}_{p^n}$.

if $\alpha$ is a root, then $\alpha+a$, is a root for every $a\in > \mathbb{F}_{p^n}$.

All equalities are equalities $\mod p$.
Let $\alpha$ be a root, and $a \in \mathbb{F}_{p^n}$. Write $q := p^n$. Then we have

$$(\alpha + a)^{q} = \sum_{i=0}^q \binom{q}{i} \alpha^i \cdot a^{q-i} = \alpha^q + a^q$$ since for every i with $0<i<q$, $p$ divides $\binom{q}{i}$ hence $\binom{q}{i} = 0 \mod p$.

Now, since $a\in \mathbb{F}^q$, we have $a^q=a$. Hence, $$(\alpha + a)^q -(\alpha + a) +1 = \alpha^q + a^q - \alpha - a +1 = \alpha^q - \alpha + 1 =0$$ QED

Olivier Roche
  • 5,319
  • 9
  • 16
  • I think you misunderstood my question, I know that all such elements are zeroes in $F_{p^n}$ but are they of the same form in $F_p(\alpha)$ ? – Sorfosh Nov 26 '19 at 20:14
  • @Sorfosh Why does their form matter? Maybe you want to know that all the roots are of the form $\alpha + a$ for $a \in \mathbb{F}_{q}$, but that's just a mathematical way of saying "the roots we've already found are the only roots". – Olivier Roche Nov 26 '19 at 20:19
  • besides, I'm working in the algebraic closure of $\mathbb{F}_p$ all along. – Olivier Roche Nov 26 '19 at 20:22
  • I understand that these are the roots, in $F_{p^n}(\alpha)$ however, it is not clear these are the same roots in $F_p(\alpha)$. – Sorfosh Nov 26 '19 at 20:23
  • As you pointed out, $\mathbb{F}_p(\alpha)$ must contain all roots of this polynomial since we assumed it is irreducible. – Olivier Roche Nov 26 '19 at 20:25
  • Hence, $\mathbb{F}_p(\alpha)$ contains $\mathbb{F}_q$, by the way. – Olivier Roche Nov 26 '19 at 20:27
  • I think i understand what you are saying. If we are in the algebraic closure of $F_p$ we know that the roots are $\alpha +a$ in there. But the algebraic closure contains $F_p(\alpha)$ so the roots there must also be of the form $\alpha+a$. – Sorfosh Nov 26 '19 at 20:28
  • I think what confused me is that you can have "different roots." $x^2+1$ has 1 as a root in $F_2$ but not in $C$. Whats important is that the fields have the same algebraic closure, correct? – Sorfosh Nov 26 '19 at 20:30
  • Yes, that's what makes every thing smooth. – Olivier Roche Nov 26 '19 at 20:30