Let $q=p^n$ where $p$ is prime.

Q1) Why is $X^{q}-X\in \mathbb F_p[X]$ irreducible ? Don't we have $X^q-X=X(X^{q-1}-1)$ ?

Q2) Suppose it is irreducible, his degree is $p^n$, then $[\mathbb F_q:\mathbb F_p]=p^n$, not $n$, Am I right ? So is wikipedia wrong ?

Q3) How to show that $\mathbb F_q$ is really the splitting field of $\mathbb F_p$ ? We have that $\mathbb F_q=\mathbb Z/p^n\mathbb Z$. If $\alpha ^q-\alpha =0$, then $\alpha ^{p^n}=\alpha $. How to prove that $\alpha \in \mathbb F_q$ ?

I know that my question are a little bit stupid (and sorry for that), but those fields drives me crazy...


1 Answer 1


At some point, we've all had to learn this stuff and been confused by it, so don't worry! At least you're asking clear questions. I think these answers will be helpful as you continue to progress.

Q1: $X^q-X$ is definitely not irreducible over $\Bbb F_p$; you've pointed out that $X$ divides it, and indeed $X-a$ divides it for every $a\in\Bbb F_p$. However, the Wikipedia section you linked to doesn't claim it's irreducible; it just says that $\Bbb F_q$ is the splitting field of $X^q-X$ over $\Bbb F_p$. Splitting fields of polynomials are defined even for reducible polynomials. (For example, the splitting field of $(x+1)(2x-3)(4x+5)^2$ over $\Bbb Q$ is $\Bbb Q$ again.) I think this resolves Q2 as well, since it's not irreducible over $\Bbb F_p$.

Q3: Are you asking how to show that $\Bbb F_q$ really is the splitting field of $X^q-X$ over $\Bbb F_p$? (The "splitting field of $\Bbb F_p$" isn't a well-defined notion.) It depends on what you already know. For example, if you already know that the multiplicative group of every finite field is cyclic, then you know that every nonzero element of $\Bbb F_q$ has order dividing $q-1$, which means that they are all roots of $X^{q-1}-1$. This is basically enough to show that all the roots of $X^q-X$ are in $\Bbb F_q$. That shows that the splitting field of $X^q-X$ over $\Bbb F_p$ is contained in $\Bbb F_q$; but they must be equal, because $X^q-X$ has $q$ distinct roots, so its splitting field must have at least $q$ elements.

By the way, there's a commom misconception in this area that I want to point out explicitly, so you can avoid the pitfall in the future: $\Bbb F_q$ is not the same as $\Bbb Z/q\Bbb Z$! When $q$ is a prime, they are the same; but when $q=p^n$ is a prime power then they are very different rings. For example, $p$ times any element of $\Bbb F_q$ equals $0$, but the same is definitely not true in $\Bbb Z/q\Bbb Z$ (look at $1+q\Bbb Z$ for example). Also, $\Bbb Z/q\Bbb Z$ has zero-divisors (since $(p+q\Bbb Z)\cdot(p^{n-1}+q\Bbb Z)=0+q\Bbb Z$ for example) so is not a field, but $\Bbb F_q$ has no zero divisors and is a field.

  • $\begingroup$ First of all, thank you for your detailed answer. For Q2, I'm sorry but it still unclear. The polynomial $X^q-X$ has $q$ distinct roots, right ? But how can we conclude that the degree of the extension is really $n$ ? What is the minimal polynomial of an element of $\mathbb F_q$ over $\mathbb F_p$ ? $\endgroup$
    – MSE
    Dec 6, 2015 at 21:46
  • $\begingroup$ Moreover, if $\mathbb F_q$ is note $\mathbb Z/q\mathbb Z$, what is it ? $\endgroup$
    – MSE
    Dec 6, 2015 at 21:49
  • $\begingroup$ $\Bbb F_q$ is $\Bbb F_q$! It's a fundamental object worth studying, regardless of whether it's isomorphic to $\Bbb Z/q\Bbb Z$ or anything else we know. As for your first question - good question! There are definitely lots of irreducible polynomials of degree $n$ over $\Bbb F_p$, and $\Bbb F_q$ is the splitting field of any one of them over $\Bbb F_p$; but there's not one "right" polynomial to choose. See en.wikipedia.org/wiki/… for example $\endgroup$ Dec 7, 2015 at 5:07
  • $\begingroup$ I don't understand either why $[\mathbb F_{p^n}:\mathbb F_p]=n$. Your explanation is not clear. Could you give more information please ? $\endgroup$
    – user301068
    Jan 18, 2016 at 20:14
  • 1
    $\begingroup$ The best explanation depends on your definition of $\Bbb F_{p^n}$. But, for example, the degree of a field extension $[L:K]$ is equal to the dimension of $L$ as a $K$-vector space; and just by counting elements, a vector space of cardinality $p^n$ over $\Bbb F_p$ must have dimension $n$. $\endgroup$ Jan 19, 2016 at 1:30

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