Why $[\mathbb{F}_{p}(\alpha): \mathbb{F}_{p^n}] = p$?

Question

I was reading the second answer of the following question here Why $x^{p^n}-x+1$ is irreducible in ${\mathbb{F}_p}$ only when $n=1$ or $n=p=2$ :

Prove that $f(X) = X^{p^n} - X + 1$ is irreducible over $\mathbb F_{p}$ if and only if either $n = 1$ or $n = p = 2.$

And the book gave the following hint: Note that if $\alpha$ is a root, then so is $\alpha + a$ for any $a \in \mathbb F_{p^n}.$ Show that this implies $\mathbb F_{p}(\alpha)$ contains $\mathbb F_{p^n}$ and that $[\mathbb F_p(\alpha) : \mathbb F_{p^n}] = p$

Here is the answer I am referring to:

I have another solution that might be easier to follow.

Let $\alpha$ be a root of $q(x)=x^{p^n}-x+1$. Note that $\alpha + a$ is also a root of $q(x)$ for all $a \in \mathbb{F}_{p^n}$. Consider cyclic muplicative group $\mathbb{F}_{p^n}^{\times} = \mathbb{F}_{p}(\theta)$ for some generator $\theta$, then $\alpha + \theta$ and $\alpha$ are roots of $q(x)$, so they belong to $\mathbb{F}_{p}(\alpha)$ which shows that $\theta \in \mathbb{F}_{p}(\alpha)$, hence $\mathbb{F}_{p^n} \subset \mathbb{F}_{p}(\alpha)$. We have $\mathbb{F}_{p} \subset \mathbb{F}_{p^n} \subset \mathbb{F}_{p}(\alpha)$.

If $p(x)$ is irreducible over $\mathbb{F}_p$, then $[\mathbb{F}_{p}(\alpha):\mathbb{F}_{p}] = p^n$, hence $|\mathbb{F}_{p}(\alpha)|=p^{pn}$. Consider the endomorphism $\sigma$: $\mathbb{F}_{p}(\alpha) \to \mathbb{F}_{p}(\alpha)$ which sends $\alpha \to \alpha^{p^n}$ (why it is a endomorphism?). Consider subgroup of automorphism $H = \langle \sigma \rangle$. $H$ fixes $\mathbb{F}_{p^n}$ (Why?), so we have $[\mathbb{F}_{p}(\alpha): \mathbb{F}_{p^n}]=|H|=p$ ($\sigma^p$ is identity map). Then $[\mathbb{F}_{p}(\alpha):\mathbb{F}_{p}] = [\mathbb{F}_{p}(\alpha): \mathbb{F}_{p^n}][\mathbb{F}_{p^n}:\mathbb{F}_{p}]$ which means $p^{n}=pn$ and this only happens when $n=1$ or $n=p=2$.

My questions are:

1- Why is the field containing all the roots of the given polynomial is $\mathbb{F}_{p}(\alpha)$ and not $\mathbb{F}_{p^n}(\alpha)$?

2- Why $[\mathbb{F}_{p}(\alpha): \mathbb{F}_{p^n}] = p$?

Could someone clarify these points to me please?

Answer 1

Your questions are ok in the sense that the answers to both your questions are negative: It is not always true that $\Bbb{F}_p(\alpha)$ would be a splitting field, nor is it always true that we would have $[\Bbb{F}_p(\alpha):\Bbb{F}_{p^n}]=p$. Simply because it is often possible to choose the zero $\alpha$ in such a way that $\Bbb{F}_p(\alpha)$ does not contain the field $\Bbb{F}_{p^n}$ as a subfield. Meaning that the degree $[\Bbb{F}_p(\alpha):\Bbb{F}_{p^n}]$ is non-sensical!

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The hint is correct: if $\alpha$ is any root of $f(x)$, all the roots are of the form $\alpha+z$, where $z\in\Bbb{F}_{p^n}$. This implies that the splitting field $E$ must contain $\Bbb{F}_{p^n}$ and one of the zeros $\alpha$. Obviously $\alpha\notin\Bbb{F}_{p^n}$, so $E=\Bbb{F}_{p^n}(\alpha)$.

It is also correct that $[E:\Bbb{F}_{p^n}]=p$. This is because the Galois group $Gal(\Bbb{F}_{p^n}(\alpha)/\Bbb{F}_{p^n})$ is generated by the Frobenius automorphism $F:z\mapsto z^{p^n}$. As the old answers explain, the orbit of $\alpha$ under iterates of $F$ consists of the elements $\alpha+j, j\in\Bbb{F}_p$. There are $p$ of those, so $[\Bbb{F}_{p^n}(\alpha):\Bbb{F}_{p^n}]=p$. These together do imply that $E=\Bbb{F}_{p^{np}}$ and the claimed reducibility result follows.

On the other hand, the degree of the extension $m:=[K:\Bbb{F}_p]$, $K=\Bbb{F}_p(\alpha)$, is in the dark, because it may depend on the choice of $\alpha$. It takes $p$ iterations of $F$ to get back to $\alpha$, so we always have $p\mid m$.

However, if $p\nmid n$, the polynomial $f(x)$ always has $p$ roots in $\Bbb{F}_{p^p}$. This can be seen by taking a peek at an old answer of mine, or more simply by the following argument. The polynomials $g_a(x):=x^p-x-a$, $a\in\Bbb{F}_p\setminus\{0\}$, are known to be irreducible over the prime field. If $\alpha$ is a root of $g_a(x)$, then by the familiar induction of $j$, we have $$ \alpha^{p^j}=\alpha+ja.\qquad(*) $$ When $p\nmid n$, we can select $a\in\Bbb{F}_p^*$ in such a way that $n\cdot a=-1$, so $a=-1/n$ if you don't mind the mild abuse of notation. The equation $(*)$ thus says that a root of $g_{-1/n}(x)$ is also a root of $f(x)$. As $g_{-1/n}(x)$ has all its $p$ roots in $\Bbb{F}_{p^p}\setminus\Bbb{F}_p$, the claim follows.

The appearance of $g_{-1/n}(x)$ can be explained by a trace condition. Pablo Rotondo already made the observation that it is a factor of $f(x)$ in a comment under the linked question. IIRC that tipped me off to some extent :-)

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If the hint in the book, indeed, suggests that $\Bbb{F}_p(\alpha)$ should be the splitting field, that is a bit careless. I suspect that whoever gave that hint had in mind a contrapositive argument like: Assume that $f(x)$ is irreducible. Let $\alpha$ be one of its roots. As all the extensions of finite fields are normal, it follows that $E=\Bbb{F}_p(\alpha)$ is the splitting field. $\ldots$. In other words, may be this part of the hint was intended to only apply, when $f(x)$ truly is irreducible.

As there are plenty of elements fulfilling the trace condition in my linked answer, I would be very surprised if it were not always possible to choose the root $\alpha$ in such a way $\Bbb{F}_p(\alpha)$ actually is the splitting field $E$. This always happens (desirable from the point of view of the hint) when $n=p$. The splitting field $E$ is a degree $p^2$ extension of the prime field. But no zeros of $f(x)$ can be in the (only!) subfield $M=\Bbb{F}_{p^p}$ as $f(z)=1$ for all the elments $z\in M$.

Answer 2

1. $\mathbb{F}_p(\alpha)$ is a splitting field of the polynomial $q$ (when $q$ in irreducible). I'm not quite sure but I think that from general finite field theory, we know that a polynomial on any finite field either has no root in that field or splits in that field. So in fact, $\mathbb{F}_p(\alpha)$ is a splitting field for $q$ (when $q$ is irreducible).

2. Here, $\sigma$ is $F^n$ where $F$ denotes the Frobenius map. It is an endomorphism since $F$ is an endomorphism as the characteristic of $\mathbb{F}_p(\alpha)$ is $p$. $H$ fixes $\mathbb{F}_{p^n}$ beacause $\mathbb{F}_{p^n}$ is the splitting field of $X^{p^n}-X$ over $\mathbb{F}_p$. To get that $[\mathbb{F}_p(\alpha):\mathbb{F}_{p^n}] = p$, we want to prove that $H$ is the Galois group of $\mathbb{F}_p(\alpha)$ over $\mathbb{F}_{p^n}$. It is clear from what we have said that $H$ is a subset of the Galois group. Let $\tau$ be an $\mathbb{F}_{p^n}$-automorphisme of $\mathbb{F}_p(\alpha)$. Again by general finite field theory, we know that $\tau$ is a power of the Frobenius map hence $\tau = F^m$ for an integer $m$. As $\mathbb{F}_{p^n}$ is the kernel of $F^{n}$, Euclidean division shows that $k$ must be a multiple of $n$ so we get that $\tau \in H$. So we have $H = Gal(\mathbb{F}_p(\alpha)/\mathbb{F}_{p^n})$. Thus, $[\mathbb{F}_p(\alpha) : \mathbb{F}_{p^n}] = |H| = p$.