Let $F$ be a finite field such that $|F|=q$, and $f(x)\in{F[x]}$ is irreducible with $deg(f)=d$, $g(x)=x^{{q}^{n}}-x$. Then prove that $f$ divides $g$ if and only if $d$ divides $n$.
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Search for this site. This must be one of the most common questions about finite fields. 1) Elements of $L=\Bbb{F}{q^n}$ are the zeros of $x^{q^n}-x$. 2) The zeros of a degree $d$ irreducible polynomial belong to $K=\Bbb{F}{q^d}$. 3) $K\subseteq L$ if and only if $d\mid n$. – Jyrki Lahtonen Mar 22 '14 at 06:09
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@ Jyrki Lahtonen: But we don't know whether $q$ is prime or not. If $|F|=q^{n}$ for some prime $q$, then as you said the statement is very common in finite fields but the situations here is different. – Mar 22 '14 at 06:13
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None of the facts that I listed depends on $q$ being prime. Is one of them something you haven't met before? But to my surprise I could not quickly find an exact duplicate, so you are right there. I do recall seeing many answers use this piece of logic :-) – Jyrki Lahtonen Mar 22 '14 at 06:15
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Yes, this problem is uncommon for me and I haven't seen it before. You are absolutely right in most of the text book the logic is as given by you but the situation of the problems are as I commented earlier. – Mar 22 '14 at 06:21
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The $d$ roots of $f(x)$ are conjugates, that is, if $f(\alpha) = 0$, then so are $f\left(\alpha^{q^1}\right), f\left(\alpha^{q^2}\right), \ldots, f\left(\alpha^{q^{d-1}}\right)$ all equal to $0$. Furthermore, $$\left(\alpha^{q^{d-1}}\right)^q = \alpha^{q^{d}} = \alpha,$$ that is, $\alpha^{q^d} - \alpha = 0$ and so $\alpha$ is also a root of $x^{q^d} - x$. We conclude that $f(x)$ is a divisor of $x^{q^d} - x$, and just a little further thought shows that in fact $f(x)$ is a divisor of $x^{q^d-1} - 1$.
So we have that
$f(x)$ is a divisor of $x^{q^d}-x$.
$x^{q^d}-x$ is a divisor of $x^{q^n}-x = g(x)$ if and only if $x^{q^d-1} - 1$ is a divisor of $x^{q^n-1} - 1$
$x^{q^d-1} - 1$ is a divisor of $x^{q^n-1} - 1$ if and only if $q^d-1$ is a divisor of $q^n-1$.
See the answers to $x^a - 1$ divides $x^b - 1$ if and only if $a$ divides $b$ if you don't know this result.$q^d-1$ is a divisor of $q^n-1$ if and only if $d$ is a divisor of $n$.
Same reason as the previous claim.
We conclude that $f(x)$ is a divisor of $g(x) = x^{q^n}-x$ if and only if $d =\deg(f)$ is a divisor of $n$.
Dilip Sarwate
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