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Recently I read about a theorem in a book, which goes as follows:

Theorem. Let $n$ be a positive integer and $(a_i)_{i=0}^{n-1}$ a sequence of non-negative integers. We define the polynomial $$P(x)=\sum_{i=0}^{n-1}x^{a_i}.$$ If the terms of the sequence $(a_i)$ form a complete residue system modulo $n$, then $$Q(x)=\sum_{i=0}^{n-1}x^{i}$$ is a divisor of $P$.

When we consider polynomials over $\mathbb{C}[x]$, this is relatively easy to prove by just looking at the $n$-th roots of unitys.

My question: Are there related results for polynomials over finite fields?

Bill Dubuque
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Joel Gerlach
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  • The same divisibility result holds over any field. A simple way to see that is to obseve that $$Q(x)\mid (x-1)Q(x)=x^n-1\mid x^{kn}-1$$ for all $k\in\Bbb{N}$. Consequently $Q(x)\mid x^A-x^B$ whenever $A\equiv B\pmod n$. Meaning that you can replace $a_i$ with its remainder modulo $n$. – Jyrki Lahtonen Nov 24 '24 at 04:33
  • The above idea is the same used here for $n=3$. Check out Bill Dubuque's answer. In the last part he explains the general case. Observe that the method does not require roots of unity. – Jyrki Lahtonen Nov 24 '24 at 04:39
  • One last comment. Roots of unity do reside in extensions of finite fields. That's because every non-zero element of every finite field is a root of unity! The notable complication is that in characteristic $p$ no $p$th roots of unity can exist, because $x^p-1=(x-1)^p$. Consequently no $m$th roots either whenever $p\mid m$. – Jyrki Lahtonen Nov 24 '24 at 04:44
  • All, I realize that even though the idea for the solution is in that post by Bill D, some may find it a stretch to call this a duplicate. If you feel that some aspects haven't been dissected there (or possibly elsewhere), I'm open to other opinions. This kind of closure decisions are reversible for a reason :-) – Jyrki Lahtonen Nov 24 '24 at 20:29
  • See the method of simpler multiples in the 2nd dupe for generalizations. Could you please cite the result you mention (what book, and what theorem / chapter / page)? – Bill Dubuque Dec 02 '24 at 23:18
  • There is no need to restrict to finite fields. Polynomial divisibilities in $,\Bbb Z[x],$ are universal - they map into every ring since the equation witnessing the divisibility persists in every ring (by the universal property of polynomial rings), e.g. by the first dupe, in $,\Bbb Z[x],$ we have $,x^2+x+1\mid x^{14}+x^2+1,$ so $,x^{14}+x^2+1 = (x^2+x+1)f(x),\ f(x)\in \Bbb Z[x].,$ This equation persists to be true when we evaluate it at $,x=r,$ for $,r,$ in any ring $,R.\ \ $ – Bill Dubuque Dec 02 '24 at 23:42
  • Beware that we need to restrict to commutative target rings if we allow more general polynomials than those in $,\Bbb Z[x],,$ since then the polynomial arithmetic may depend on commutativity, e.g. $,xy = yx,$ or $,\alpha x = x\alpha,,$ which need not remain true after evaluation into a noncommutative ring. $\ \ $ – Bill Dubuque Dec 02 '24 at 23:48

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