Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When is the cyclotomic polynomial $f(x)$ over a finite field $\mathrm{F}_q$ also the minimal polynomial of some element $\alpha \in \mathrm{F}_q$?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

In general, a polynomial $f\in F[X]$ is the minimal polynomial of its roots over $F$ if and only if $f$ is irreducible over $F$. Cyclotomic polynomials provide no exception.

However, while cyclotomic polymomials are irreducible over $\mathbb Q$, not all of them are irreducible when considered as a polynomial over a finite field. For example, over $\mathbb F_2$ the $7$th cyclotomic polynomial $\Phi_7 = X^6 + X^5 + X^4 + X^3 + X^2 + X + 1$ factors into two factors of degree $3$: $$ \Phi_7 = (X^3 + X + 1)(X^3 + X^2 + 1) $$ Hence $\Phi_7\in\mathbb F_2[X]$ is not irreducible and therefore not the minimal polynomial of its roots.

Note: This was written as a response to an earlier version of the question.

share|improve this answer
    
Please explain it for the following example. Given finite field $\mathrm{F}_q$, I have a decomposition $x^n-1=\prod_{i=0}^{m-1}f_i(x)$ over $\mathrm{F}_q$, where all $f_i(x)$ are cyclotomic. I am not sure that all $f_i(x)$ are minimal polynomials for some elements from extension $[\mathrm{F}_q^r : \mathrm{F}_q]$, where $r$ is a smallest number that $n$ divides $q^r - 1$. What is wrong in that? –  Piotr Semenov Oct 21 '13 at 11:41
1  
@PiotrSemenov: I've added information on cyclotomic polynomials over finite fields to my answer. –  azimut Oct 21 '13 at 11:54
    
Thanks! So I did a mistake while suggesting that cyclotomic polynomials are also irreducible in case of finite fields. –  Piotr Semenov Oct 21 '13 at 12:18
1  
+1: In some sense an even smaller counterexample is $$\Phi_3(X)=X^2+X+1=(X-2)(X-4)\in\Bbb{F}_7[X].$$ Or even $\Phi_3(X)=(X-1)^2\in \Bbb{F}_3[X].$ –  Jyrki Lahtonen Oct 22 '13 at 7:00
1  
@JyrkiLahtonen: That's true. For my example, I decided to go with an example over $\mathbb F_2$. Here, the smallest reducible one is $\Phi_4 = (X + 1)^2$, which might be a bit misleading. So I ended up with $\Phi_7$. –  azimut Oct 22 '13 at 7:10

One can show that the cyclotomic polynomial $\Phi_n(X)$ is irreducible over $\mathbf F_p$ precisely when $p$ has multiplicative order $\varphi(n)$ modulo $n$. This follows from the theory of cyclotomic extensions of $\mathbf Q$.

In particular, if $(\mathbf Z/n\mathbf Z)^\times$ is not cyclic (i.e. unless $n$ is an odd prime power, twice an odd prime power, or $n=2$ or $4$), then $\Phi_n(X)$ is reducible modulo every prime $p$, despite being irreducible over $\mathbf Q$!

share|improve this answer
    
@JyrkiLahtonen Of course! How did I even write that. Thanks for the heads up! :) –  Bruno Joyal Oct 22 '13 at 10:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.