0
$\begingroup$

I'm reading this document: http://www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/ff/ffchap4.pdf

Given a finite field $F_q$, a cyclotomic polynomial $Q_n(x)$ with $gcd(q, n) = 1$, and $d = ord_n(q)$.

I want to proof that after factorizing the $Q_n(x)$ to irreducible polynomials, all of these factors have degree d.

I have found this document http://www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/ff/ffchap4.pdf.

and i don't quite understand the proof of Theorem 8.12. It says:

Now let K be the finite field $F_q$, assume gcd(q, n) = 1, such that primitive nth roots of unity over $F_q$ exist. Let η be one of them. Then

η ∈ $F_{q^k}$ ⇔ $η^{q^k}$ = η ⇔ $q^k$ ≡ 1 mod n.

The smallest positive integer for which this holds is k = d, so η is in $F_{q^d}$ but not in any proper subfield. Thus the minimal polynomial of η over $F_q$ has degree d. Since η was an arbitrary root of $Q_n(x)$, the result follows, because we can successively divide by the minimal polynomials of the roots of $Q_n(x)$.

I dont get the Step

η is in $F_{q^d}$ but not in any proper subfield.

to

Thus the minimal polynomial of η over $F_q$ has degree d

I feel like this is something elementary that i'm missing.

$\endgroup$
  • 1
    $\begingroup$ The multiplicative group of $\Bbb{F}_{q^d}$ is cyclic of order $q^d-1$, so it contains $n$th primitive roots of unity if and only if $n\mid q^d-1$. Here $d$ is the smallest positive integer with this property. This means that the extension $[\Bbb{F}_{q}(\eta):\Bbb{F}_q]$ has degree $d$. $\endgroup$ – Jyrki Lahtonen May 22 '18 at 16:18
  • 1
    $\begingroup$ But I have answered variants of this same (essentially) question too many times: 1,2, 3 $\endgroup$ – Jyrki Lahtonen May 22 '18 at 16:22
  • $\begingroup$ But why can't there be a polynomial with root η in $F_q$ with degree smaller than d? And why does it exist with degree d? $\endgroup$ – WaltJD May 22 '18 at 17:12
  • $\begingroup$ Sorry, I missed those parts. The gcd-condition implies that the roots of $Q_n(x)$ in any extension field $K$ still have order $n$. See here. This means that we must have $n\mid K^*$ (by Lagrange's theorem from elementary group theory). But $n\nmid q^k-1$ when $0<k<d$. Degree $d$ works, because when the order of a cyclic group is a multiple of $n$, then that group has elements of order exactly $n$. $\endgroup$ – Jyrki Lahtonen May 22 '18 at 17:21
  • $\begingroup$ What exactly do you mean by $n\mid K^*$. $\endgroup$ – WaltJD May 22 '18 at 18:11

Your Answer

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

Browse other questions tagged or ask your own question.