Divisibility of polynomials in finite fields

Question

Is the following known (or even correct)?

Let $q = p^n$ be a prime power.

Let $f(x), g(x) \in F_q[x]$ with $f(x) = x^m + ax^r + b$ with $a,b \ne 0$ and $g(x) = x^{mt} + cx^{rt} + d$ with $c,d \ne 0$ such that $f|g$

Then $t = p^i$ for some $i \ge 0$ and $c = a^{t}, d = b^{t}$.

ie. if the ${\bf exponents}$ match up nicely, then divisibility implies that $g$ is a Frobenius style power or $f$.

Note that $f$ and $g$ being trinomials is not the important part, but only that they have the same number of terms and the corresponding exponents divide as indicated.

Note that if $t = p^i$ then $f^{t} = x^{mt} + a^{t}x^{rt} + b^{t}$ holds. I'm looking for the converse or something close to it.

OK. I've fixed up the question a bit to get away from this solution. How about just trinomials dividing trinomials to start? — Dan, Nov 02 '18 at 02:59
$x^2-2x+1$ divides $x^4-2x^2+1$. $x^2-4x+4$ divides $x^4-8x^2+16$. — Gerry Myerson, Nov 02 '18 at 08:18

Jyrki Lahtonen · Answer 1 · 2018-11-02T05:16:32.167

1

Nope. The following mechanism producing counterexamples springs to mind.

Consider for example the irreducible polynomial $f(x)=x^2+x+1$ from $\Bbb{F}_2[x]$. Its zeros, say $\alpha$ and conjugates, are third roots of unity, and therefore satisfy the equation $\alpha^7=\alpha$. Implying that they are also zeros of $f(x^7)$. Implying that $$f(x)\mid f(x^7)=x^{14}+x^7+1.$$

Actually this divisibility relation holds in all characteristics (even zero). For exactly the same reason.

More generally, this happens to ALL irreducible trinomials over ANY finite field. The zeros of any irreducible polynomial $f(x)$ are roots of unity of some order, say $\ell$, and consequently $f(x)\mid f(x^{1+k\ell})$ for all $k$. The polynomial $f(x^{1+k\ell})$ is obviously a trinomial with the exponents matched up as prescribed. All you need to do is to select $k$ such that $1+k\ell$ is not a power of the characteristic.

edited Nov 02 '18 at 05:16

answered Nov 02 '18 at 04:50

Jyrki Lahtonen

133,153

And the trinomial need not even be irreducible. As long as its constant term is non-zero, all its zeros are roots of unity of some order. We can use the least common multiple of those orders in the role of $\ell$ above. – Jyrki Lahtonen Nov 02 '18 at 05:36
Thanks. What if we force $t=p^i$? Do the coefficients then have to lign up as $c = a^{p^i}$ and $d = b^{p^i}$? – Dan Nov 02 '18 at 14:28
Did you not see my examples in the comments, Dan? – Gerry Myerson Nov 02 '18 at 21:37
Are you still here, Dan? – Gerry Myerson Nov 04 '18 at 09:12

Divisibility of polynomials in finite fields

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