[seqfan] Re: x^k-1 divisible by an irreducible polynomial in a finite field. A371164 and A106309

[Fred Lunnon]

  Arithmetic in your field  K  is by construction identical to arithmetic
of
polynomials in  |Q(x) mod q(x) mod p , which I presume was intended here.
So  ord(r) =  min(k)  by definition; proof is awkward to pin down
convincingly,
essentially because it is vacuous!    WFL



On Tue, Mar 26, 2024 at 11:25 PM <israel at math.ubc.ca> wrote:

> This is relevant to A371164 and A106309. I think this is correct, but I'd
> like confirmation from someone better acquainted with finite fields than I
> am.
>
> Suppose the monic polynomial q(z) of degree d >1 is irreducible over a
> finite field F (the integers modulo a prime p, if that makes a
> difference).
> I want to find the least positive integer k such that z^k - 1 is divisible
> by q(z) over F. If I take the extension field K = F[r] where r is a root
> of
> q(z), is the answer the order of r in the multiplicative group K^x$ of K?
>
> For example, take F to be the integers mod 13, and
> q(z) = x^4 + 5 x^2 + x + 10.
> Maple (using the GF package) tells me the order of r in this case is
> 14280.  Indeed z^{14280}-1 is divisible by q(z) over F in this case,
> and no divisor of 14280 will work.
>
> Cheers,
> Robert
>
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