1
$\begingroup$

I read somewhere that :

A subfield of $F_{p^n}$ has order $p^d$ where $d\mid n$, and there is one such subfield for each $d$.

Let $q = p^n$

We have that any irreducible polynomial of degree $n$ over $F_p$ is a factor of $X^q - X \in F_p[X]$. For any positive integer $n$

Suppose, we have a subfield as in the quoted statement and an irreducible polynomial of degree $k$ over it, such that $ n = dk$.

Is this polynomial also a factor of $X^q - X$ ?

$\endgroup$
2
  • $\begingroup$ How's that "any polynomial of the form $X^q - X \in F_p[X]$"? There is only one such polynomial, and it is $X^q - X$. $\endgroup$ – Marc van Leeuwen Jun 28 '17 at 6:54
  • $\begingroup$ @MarcvanLeeuwen You're right, it's changed $\endgroup$ – Donno Jun 28 '17 at 7:01
3
$\begingroup$

One has $$ k \mid n \qquad\text{iff}\qquad p^{k} - 1 \mid p^{n} - 1 \qquad\text{iff}\qquad x^{p^{k}} - x \mid x^{p^{n}} - x. $$

Now $k$ divides $n = d k$, and your irreducible polynomial of degree $k$ divides $x^{p^{k}} - x$, and thus divides $x^{p^{n}} - x$.


Post Scriptum

I may have misunderstood the problem. On second thoughts it appears to ask the following.

Suppose $n = d k$, and let $F$ be a finite field of order $p^{d}$. Suppose the (monic) polynomial $f \in F[x]$ is irreducible in $F[x]$. Does $f$ divide $x^{p^{n}} - x$?

The answer is still yes in this formulation. If $\alpha$ is a root of $f$ in some extension of $F$, then $E = F[\alpha]$ is a finite field of order $(p^{d})^{k} = p^{n}$. If $g$ is the minimal polynomial of $\alpha$ over $F_{p}$, then $f \mid g$. Since $g$ has a root in $E$, $g$ divides $x^{p^{n}} - x$, and so does $f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.