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So this is part of an exercise sheet where I thought I had figured it out but turns out I didn't.

Theory from lecture:

We know for a prime $p$ the finite field $\mathbb{F}_{p^n}$ is isomorphic to $\mathbb{F}_{p}[x]/(m)$ where $m$ is some monic irreducible (over $\mathbb{F}_{p}$) polynomial of degree n. We also know the multiplicative group $\mathbb{F}_p^*$ has order $p-1$ for a finite field $\mathbb{F}_p$ and $p$ prime.

Part 1 (solved): For $p^n=9$ find such a minimal polynomial.

We have $\mathbb{F}_{9}=\mathbb{F}_{3²}\simeq \mathbb{F}_{3}[x]/(m_2)$ where $m_2$ has degree 2 and is monic and irreducible over $\mathbb{F}_3$.
I guessed $m_2(x) = x²+1$ which is irreducible over $\mathbb{F}_3$ since $m_2(0) = 0²+1=1\neq0$, $m_2(1) = 1²+1=2\neq0$ and $m_2(2)=2²+1=2\neq0$. Remains to check a root $\alpha$ of $m_2$ is really a generator of $\mathbb{F}_3^*$. We therefore check the order of $\alpha$:
Firstly, note $\alpha$ a root $\implies$ $m_2(\alpha)=0=\alpha²+1 \iff \alpha² = - 1 = 2 \in\mathbb{F}_3$
Then,
$\alpha³=\alpha\alpha²=2\alpha$
$\alpha⁴=(\alpha²)²=4=1 \in \mathbb{F}_3$. Hence, the order of $\alpha$ is $4\neq8\implies\alpha$ is not a generator.
Taking an educated guess and trying $(\alpha+1)$ yields
$(\alpha+1)²=\alpha²+2\alpha+1=2\alpha$ (since $\alpha²=-1$)
$(\alpha+1)³=(\alpha+1)(\alpha+1)²=2\alpha(\alpha+1)=2\alpha²+2\alpha=2\alpha-2$
$(\alpha+1)⁴=((\alpha+1)²)²=4\alpha²=-4=-1\in\mathbb{F}_3$
Hence, $1=(-1)²=((\alpha+1)⁴)²=(\alpha+1)⁸\implies \alpha+1$ has order 8 and is hence a generator of $\mathbb{F}_3^*$. Now we need to find a minimal polynomial of $\alpha+1$. Above we find $(\alpha+1)²=2\alpha \iff (\alpha+1)²-2\alpha=0 \iff (\alpha+1)²-2(\alpha+1)+2=0$ hence $\alpha+1$ is a root of $m_2'=x²-2x+2=x²+x+2 \in\mathbb{F}_3$. This is our desired polynomial.

Part 2 (need help): For $p^n=16$ find such a minimal polynomial.

Like above we find $\mathbb{F}_{16}=\mathbb{F}_{2⁴}\simeq \mathbb{F}_{2}[x]/(m_4)$ where $m_4$ has degree 4 and is monic and irreducible over $\mathbb{F}_2$.
I guessed $m_4(x) = x⁴+x+1$ which is irreducible over $\mathbb{F}_2$ since $m_4(0) =1\neq0$ and $m_4(1) = 1\neq0$. Remains to check a root $\alpha$ of $m_4$ is really a generator of $\mathbb{F}_{16}^*$, hence has order 15. We therefore check the order of $\alpha$:
Firstly, note $\alpha$ a root $\implies m_4(\alpha) = \alpha⁴+\alpha+1=0 \iff \alpha=\alpha⁴+1 \in \mathbb{F}_2$ Then,
$\alpha²=(\alpha⁴+1)²=\alpha⁸+2\alpha⁴+1=\alpha⁸+1$
$\alpha³=\alpha\alpha²=\alpha⁹+\alpha$
$\alpha⁴=(\alpha²)²=\alpha^{16}+1$
$\alpha⁵=\alpha^{17}+\alpha$ but $\alpha^{15}=(\alpha⁵)³=(\alpha^{17}+\alpha)³=\alpha^{51}+3\alpha^{35}+3\alpha^{19}+\alpha³\neq1\implies \alpha$ not a generator? Please tell me there is a more intelligent way than guessing and computing all the orders of $\alpha+1$ and $\alpha-1$?

arridadiyaat
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    If $g$ is a generator, what is wrong with $x-g?$ Or do you want the polynomial that has roots all generators? Then you want the cyclotomic polynomial, $\Phi_{p-1}.$ https://en.wikipedia.org/wiki/Cyclotomic_polynomial – Thomas Andrews May 02 '24 at 15:41
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    It seems like you want the minimal polynomial for a generator of $\mathbb F_{p^n}^\times$ with coefficients in $\mathbb F_p?$ That is not what your title says. – Thomas Andrews May 02 '24 at 15:44
  • Even so, the idea that there are minimal polynomials is very confusing. All polynomials such that $\mathbb F_{p^n}=\mathbb F_p[x]/(m(x))$ are minimal. – Thomas Andrews May 02 '24 at 15:53
  • You are using the minimal polynomial $x^4+x+1\in\Bbb{F}_2[x]$ in the wrong way. You are supposed to use it to turn higher powers of $\alpha$ into lower degree polynomials. See the list of consecutive powers of $\alpha$ near the end of this old answer of mine. What you denote by $\alpha$ is called $\gamma$ there, because I had used up $\alpha$ and $\beta$ earlier in that post :-) – Jyrki Lahtonen May 03 '24 at 04:04
  • I already posted an answer. I deleted it since you didn't accept it. I undeleted it now since it seems like your question is unresolved. –  May 06 '24 at 12:38

3 Answers3

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If $p$ is a primitive root mod $p^n-1,$ then the minimal polynomial is the cyclotomic polynomial $\Phi_{p^n-1}.$ Because said polynomial will be irreducible (some more theory). It looks like this almost never happens.

If not, then $\Phi_{p^n-1}$ is reducible over $\Bbb F_p.$

An irreducible factor of deg $n$ will have $\alpha $ as a root. And that's the minimal polynomial.

So in your second example, we have $\Phi_{15}=x^8-x^7+x^5-x^4+x^3-x+1.$ Since $2$ is not primitive mod $15$ ($2^4\equiv1$), it's reducible.

We find the factorization $(x^4+x+1)(x^4-x^3+1).$

Now one of those is going to be the minimal polynomial (whichever has $\alpha $ as a root).

There are $\varphi (15)=8$ different generators of $\Bbb F_{16}^*,$ and depending on which one is $\alpha $ we get either the first or second factor as the minimal polynomial.

Upshot: we need more information to find the actual minimal polynomial. We need to know more about $\alpha.$

suckling pig
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  • No generator is given. I strongly believe that the question is to find a $m$ such that $m$ is the minimal polynomial of a generator. And there are, in the given examples, $2$ such choices of $m$. –  May 06 '24 at 14:30
  • And your first $2$ sentences are unclearly written (should talk over $\mathbb{Q}$ or something?) or false. Also $p$ has two meanings. –  May 06 '24 at 14:32
  • Why would I talk over $\Bbb Q?$ We're in a finite field. And, the minimal polynomial is unique. @underflow – suckling pig May 06 '24 at 15:11
  • Read your first paragraph again. And yes, the minimal polynomial of any specific element is indeed unique. I never claimed something else. –  May 06 '24 at 15:13
  • $p$ doesn't have two meanings. There's only one $p.$ @underflow – suckling pig May 06 '24 at 15:15
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    I think I understand what you are saying now. I still don't understand why you distinguish between the irreducible and reducible case. Also if I've done no mistakes the only pairs of numbers $(n,p)$ where $n\geq1$ and $p$ is prime where $\Psi_{p^n-1}$ is irreducible over $F_p$ are $(n,p)=(1,2),(2,2),(1,3)$. –  May 06 '24 at 15:47
  • @underflow since $\alpha $ is a root of the cyclotomic polynomial, if it's irreducible it's the minimal polynomial. – suckling pig May 06 '24 at 16:02
  • Yes but if it's irreducible it's still true that a monic irreducible factor of it is the minimal polynomial since the only such factor will be itself. I don't see why you seperate the cases. –  May 06 '24 at 16:04
  • Last comment, because I think we're square now. If it's reducible we need to find which factor is the minimal polynomial. I thought you got that right. That's why I upvoted. @underflow – suckling pig May 06 '24 at 16:09
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Let $q=p^n-1$. Any non trivial factor of $\Psi_q(x)\in F_p[x]$ will do the trick where $\Psi_q(x)$ is the $q$-th cyclotomic polynomial.

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Let $q=p^n-1$. Any monic irreducible factor (or equivalently any monic factor of degree $n$) of $\Psi_q(x)\in F_p[x]$ will do the trick (and those are the only ones) where $\Psi_q(x)$ is the $q$-th cyclotomic polynomial.

So in your example $p=3$ and $n=2$ so $q=8$ and $$\Psi_q(x)=x^4+1=(x^2+x+2)(x^2+2x+2)$$ hence $x^2+x+2$ and $x^2+2x+2$ are the only ones which solve the question.

In the other example $p=2$ and $n=4$ so $q=15$ and $$\Psi_q(x)=x^8-x^7+x^5-x^4+x^3-x+1=(x^4+x+1)(x^4+x^3+1).$$

  • Why is the OP talking about the cyclic group of $\mathbb F_p^\times$ and "minimal polynomials," when all such polynomials are minimal. Better to wait until OP clarifies their question rather than jumping so quickly. I'm thinking the question is really for $m(x)$ a minimal polynomial over $\mathbb F_p$ of some generator of $\mathbb F_{p^n}^\times,$ but that is just a guess. – Thomas Andrews May 02 '24 at 15:57
  • "I'm thinking the question is really for m(x) a minimal polynomial over Fp of some generator of F×pn, but that is just a guess." That's how I've understood the question and that's the question I answered. –  May 02 '24 at 15:58
  • You haven't shown the image of $x$ in each $F_3[x]/(m(x))$ is a generator. It is true, but you haven't shown it. This is not true for all $m(x)$ a factor of $\Phi_q(x)$ in general over $\mathbb F_p.$ – Thomas Andrews May 02 '24 at 16:03
  • @ThomasAndrews Wdym? I talked about irreducible factors. Also the proof is a simple counting argument... Basically they have to be roots of the cyclotomic polynomial which is of degree $\phi(q)$ and there exist $\phi(q)$ generators since the multiplicative group is cyclic. –  May 02 '24 at 16:05
  • And the degree of the factors have to all be $n$ 'cause a generator generates the whole field. –  May 02 '24 at 16:13
  • Irreducibility is not enough. The number of irreducible monics of degree $n$ over $\mathbb F_p$ is $$\frac1n\sum_{d\mid n}\mu(n/d)p^d.$$ The number of such which generators as roots is $\frac1n\phi(p^n-1).$ These are not, in general, the same. – Thomas Andrews May 02 '24 at 16:20
  • @ThomasAndrews I wasn't talking about a general irreducible, just an irreducible factor of the $q$-th cyclotomic polynomial. –  May 02 '24 at 16:23
  • hi! thanks for making the effort, I am really lost and would love some more details if you can make it. :) so i can follow because i found $x²+2x+2$ and $x²+x+2$ myself, but you found them via $x⁴+1$ which i don't understand where it comes from. same thing for the second case: did you just guess $x^8−x^7+x^5−x^4+x^3−x+1$ and how do you find the divisors? – arridadiyaat May 06 '24 at 12:48
  • @arridadiyaat See here for cyclotomic polynomials https://en.wikipedia.org/wiki/Cyclotomic_polynomial and to factor them I used Wolfram Alpha https://www.wolframalpha.com/input?i=factor+x%5E4%2B1+mod+3 . All of this can be done by hand. To get the cyclotomic polynomials by hand see for example the "easy cases" subsection in the wiki link. To calculate the product by hand you'd need to solve for example $x^4+1=p(x)q(x)$ where $p$ and $q$ are monic polynomials of degree $2$. This can for example be done by comparing coefficients. –  May 06 '24 at 12:59