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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?

Emptymind
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  • I just wanted to point out that your answers are always so valuable to me and they are to the point @JyrkiLahtonen I was actually waiting for your replies to my questions through the whole day yesterday – Emptymind Mar 25 '24 at 13:35
  • Unlucky timing. Don't worry about it. – Jyrki Lahtonen Mar 25 '24 at 14:05

2 Answers2

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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!

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 :-)

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 $z\in M$.

Jyrki Lahtonen
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  • But I notice that $\alpha ^ p$ is also a root, how will this help me in proving that $[E: \mathbb F_{p^n}] = p$ with the $E$ you mentioned in the second paragraph? – Emptymind Mar 25 '24 at 19:31
  • @Emptymind $\alpha^p$ is, indeed, a root. So it must be one of the elements $\alpha+z, z\in\Bbb{F}{p^n}$. The polynomial $g(x)=x^{p^n}-x=f(x)-1$ is additive, so $$g(\alpha^p-\alpha)=g(\alpha^p)-g(\alpha)=[f(\alpha^p)-1]-[f(\alpha)-1]=-1+1=0.$$ The zeros of $g(x)$ are exactly the elements of $\Bbb{F}{p^n}$, so $\alpha^p-\alpha\in\Bbb{F}_{p^n}$. – Jyrki Lahtonen Mar 25 '24 at 20:19
  • The result $[E:\Bbb{F}{p^n}]=p$ is a consequence of the facts that A) $E=\Bbb{F}{p^n}(\alpha)$ and B) the $\Bbb{F}_{p^n}$-conjugates of $\alpha$ are $\alpha^{p^n}=\alpha+1$ and all the other elements $\alpha+j, j=0,1,2,\ldots,p-1$. – Jyrki Lahtonen Mar 25 '24 at 20:22
  • I did not get $B)$ in what you said ..... But from what I observed, can I conclude that $$[E: \mathbb F_{p^n}] \leq p$$ or no? – Emptymind Mar 25 '24 at 20:57
  • Why the $\mathbb F_{p^n}$ conjugates of $\alpha$ look like this? And how will this help me in the proof? – Emptymind Mar 25 '24 at 21:01
  • The degree of the minimal polynomial of $\alpha$ over $\Bbb{F}{p^n}$ is equal to the extension degree $[E:\Bbb{F}{p^n}]$. And you get the $\Bbb{F}_q$ conjugates by applying the generator of the Galois group $z\mapsto z^q$ iteratively. – Jyrki Lahtonen Mar 25 '24 at 21:09
  • I was thinking about completing my argument by saying that then that of $1, \alpha, \alpha^2,\dots, \alpha^p$ are linearly dependent but then I do not know how to complete – Emptymind Mar 25 '24 at 21:11
  • what is $q$ in relation to $p^n$? – Emptymind Mar 25 '24 at 21:14
  • $q=p^n$. Did you take a look at this? It's the same method. – Jyrki Lahtonen Mar 25 '24 at 21:17
  • Can not I say that since $[\mathbb F_{p^n}(\alpha): \mathbb F_{p^n} ] \leq p$(is this something correct I can conclude) and from the linear dependence argument the required follows? – Emptymind Mar 25 '24 at 21:18
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    Forget linear dependence. You want to use Galois theory. Or, alternatively, properties of finite fields. If $K=\Bbb{F}_q$ and $\beta$ is algebraic over $K$, the degree of the extension $K(\beta)/K$ is equal to the smallest positive integer $\ell$ such that $\beta^{q^\ell}=\beta$. That's what Lubin (and all the other answerers actually) is using. – Jyrki Lahtonen Mar 25 '24 at 21:21
  • (cont'd) With $q=p^n$ and $\alpha$ any zero of $f(x)$ we get at $\ell=p$. Done. – Jyrki Lahtonen Mar 25 '24 at 21:22
  • Yeah this is exactly what I do not understand, why specifically this condition? Is there a proof of this fact? – Emptymind Mar 25 '24 at 21:25
  • I looked at Lubin's answer and this line was the thing that I actually do not understand "the number of $k$-conjugates of $\alpha$ is precisely $p$, and the irreducible factor $g$ of $F$ is of degree $p$", is there a proof for this? – Emptymind Mar 25 '24 at 21:27
  • Is this what you are trying to say to me and what Lubin was trying to say in his answer "$x \to x^p$ is an automorphism sending $r$ to $r-a$ for any root $r$ of the polynomial. This operation is cyclic of order $p$, so that one can get from any root to any other by applying the automorphism several times. The Galois group thus acts transitively on the roots, which is equivalent to irreducibility."? – Emptymind Mar 25 '24 at 21:34
  • Is the result you stated in your last 2 comments is stated in any book? – Emptymind Mar 25 '24 at 21:36
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    A proof is based on this (picked it because it is an excerpt from D&F). The size of the field $\Bbb{F}{p^n}(\alpha)$ is $p^m$, where $m$ is the smallest positive integer such that A) $z^{p^m}=z$ for all $z\in\Bbb{F}{p^n}$ and B) $\alpha^{p^m}=\alpha$. Condition A) holds if and only if $n\mid m$ (by that result from D&F again), and condition B) then dictates $m=pn$ as the smallest multiple of $n$ as per Lubin and others. – Jyrki Lahtonen Mar 25 '24 at 21:37
  • But, yeah, you can think about it in terms of the action of the Galois group as well. I'm sure D&F explain the Galois theory of finite fields. All the books at that level do. – Jyrki Lahtonen Mar 25 '24 at 21:40
  • Let me try to rationalize it :) – Emptymind Mar 25 '24 at 21:46
  • I am confused, what exactly is the Frobenius map? Does it take $\alpha $ to $\alpha^p$ or $\alpha^q$, if it takes it to power $q,$ then how I will get the conjugates you mentioned? – Emptymind Mar 26 '24 at 03:18
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    The way I use the term Frobenius automorphism depends on what I think of as the base field. When thinking about the Galois group of the extension $L/K$, both finite field, I think of the Frobenius map as the natural generator of $Gal(L/K)$. That is, the mapping $z\mapsto z^{|K|}$. The mapping $z\mapsto z^p$ is then the generator of $Gal(L/\Bbb{Z}_p)$. Sorry about not making that clear. Anyway, an element is a fixed point of a cyclic group $G$ of automorphisms if and only if it is a fixed point of a generator of $G$. – Jyrki Lahtonen Mar 26 '24 at 04:30
  • But then how you will get the conjugates? You are writing $\alpha^{p^n}$ above to get the conjugates. – Emptymind Mar 26 '24 at 07:59
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    The conjugates = the automorphic images = the set of zeros of the minimal polynomial over $K$. That's basic Galois theory. – Jyrki Lahtonen Mar 26 '24 at 09:11
  • Whenever $L/K$ is a Galois extension, and $\alpha\in L$ you get the degree of the extension $K(\alpha)/K$ by figuring out how many distinct $K$-conjugates $\alpha$ has. The point is that in the case of finite fields the mapping $z\mapsto z^{|K|}$ is the generator of the Galois group for every choice of $K$. Here it is easy to figure out the number of conjugates ($=p$) when $K=\Bbb{F}_{p^n}$, but it is more taxing when we choose $K=\Bbb{F}_p$, which is the game I am playing. Don't get stuck with some slightly inaccurate posting somebody else made. – Jyrki Lahtonen Mar 27 '24 at 04:52
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    (cont'd) I may be willing to claim that I understand this better than D&F, don't tempt me :-) – Jyrki Lahtonen Mar 27 '24 at 04:53
  • You should :))) you understand the subject very well! @Jyrki Lahtonen – Emptymind Mar 27 '24 at 08:32
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  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, $[\mathbb{F}_p(\alpha) : \mathbb{F}_{p^n}] = |H| = p$.

Valentin Massicot
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