# Find all irreducible polynomials of $F_5[X]$ of degree 4 and classify which are primitive

Find all irreducible polynomials of $F_5[X]$ of degree 4 (or lower)

and classify which are primitive

Frobenius map is being used. If $f(x)$ is irreducible is $f(bx)$

\begin{aligned} \mathbb{F_q} \to \mathbb{F_q} \\ x \to bx \text{ where }b\in \mathbb{F}_q \\ \alpha \to \alpha \end{aligned}

this is an isomoprhism with inverse $x \to b^{-1} x$ . If f(x) is irreducible then $f(bx)$ simply search for irreducibles. I know there are $125$ total polynomials

$F_5 \subset F_{25}$ irreducible quadratics and div by 2 $25-5 \in F_{25} \ F_{5}$ paired those to give 10 irreducible

Instructure when on to look at degree 2 polynomials

\begin{aligned} {F_5}[x] \to F_{5} [x] \\ x \to bx \\ x^2 + \alpha x + \beta \to (bx)^2 + \alpha \beta x + \beta /b^2 \end{aligned}

Now set $b = \alpha$ $$x^2+x+ \beta / \alpha^2$$

If find all polys of the form of

$$x^2 + x + \rho , \text{ for } \rho \in F_5$$ that are irred We can get all the others.

From what I can see is that we considered $x^2+x+1$

\begin{aligned} 1,2,3,4 \text{ not all roots} \\ 2,6 ,12,0 \\ 2,2,0 \end{aligned}

$x^2 +x+1,$ and $x^2+x+2$ each give 4 polynomials of $F_{25} / F_5$

also consider

\begin{aligned} \cancel{x^2+1} \\x^2+2 \\x^2+3 \\\cancel{x^2+4} \end{aligned}

start with $$x^2 +x + \rho$$ sub $$x \to \alpha^{-1} x$$

$$\alpha^{-2} (x^2+ \alpha x + \dots ) = \alpha^{-2} (x^2 + \alpha x + \alpha^2 \rho^2)$$ this is a class example which I had some trouble following. It seems like Magic. I want to break this down to a systematic way

Adding one or two more pages of notes will problably break this down to smaller questions

• Related. I believe I have used mobious function past to say how many there are but can't list them math.stackexchange.com/questions/1609700/…. – Tiger Blood Mar 7 '18 at 15:57
• There are $150$ irreducible (monic) polynomials of degree four. And $48$ of those are primitive. I don't think I want to list them all :-( – Jyrki Lahtonen Mar 7 '18 at 16:01
• I can list them in sage. guessing I am not sure how to go from from one to the other with mappings of some sort. Thanks @JyrkiLahtonen – Tiger Blood Mar 7 '18 at 16:05
• WA gives the answer, which is long... – lhf Mar 30 '18 at 2:08
• I find it very hard to follow any part of this question. It is written very incoherently and with no regard for grammar, spelling or style. Do you want an explanation and exposition of the shown methods used in your class? Or do you just want a systematic way to construct all primitive polynomials? The latter I can do, the former I cannot because I have no clue what is going on in your question. – Servaes Apr 3 '18 at 14:55

There are $5^4-5^2=600$ elements of the field $K=\Bbb{F}_{5^4}$ that don't belong to any proper subfield. Their minimal polynomials over the prime field are all irreducible of degree four. Four elements always share the same minimal polynomial, so there are a total of $600/4=150$ irreducible (monic) quartic polynomials in $\Bbb{F}_5[x]$. Observe that by uniqueness of the field $K$ all such polynomials are minimal polynomials of some element $\in K$.

You asked for transformations of getting from one to another. A most general such group of transformations is the Möbius group of fractional linear transformations. If $a,b,c,d\in\Bbb{F}_5$ are constants such that the determinant $ad-bc\neq0$, and $p(x)\in \Bbb{F}_5[x]$ is irreducible then by invertibility of the Möbius transformation $M:x\mapsto (ax+b)/(cx+d)$, the numerator $p_M$ of the rational function $$p(\frac{ax+b}{cx+d})=\frac{p_M(x)}{(cx+d)^{\deg p}}$$ is automatically also irreducible, and clearly $\deg p_M(x)=\deg p(x)$.

Two caveats:

• The transformations coming from $(a,b,c,d)$ and $\lambda(a,b,c,d)$, $\lambda\in\Bbb{F}_5^*$ obviously give rise to the same polynomial $p_M(x)$, so we want to go to a quotient group of invertible $2\times2$ matrices by scalars. This quotient group $\Gamma$ has $|GL_2(\Bbb{F}_5)|/|\Bbb{F}_5^*|=480/4=120$ elements.
• The polynomial $p_M(x)$ is not automatically monic, so we need divide it by its leading coefficient to get one.

With all this in place I had a bit of fun looking for a suitable census of those irreducible quartics. Also sprach Mathematica:

• The orbit of $p_1(x)=x^4+2$ has $30$ irreducible quartics. Here $p_1$ has a stabilizer of size four $\le\Gamma$. Namely the substitutions $x\mapsto ax, a\in\Bbb{F}_5^*$ because by Little Fermat $a^4=1$ for all those multipliers.
• The orbit of $p_2(x)=x^4+x+4$ has $60$ irreducible quartics.
• The orbit of $p_3(x)=x^4+x^2+2$ also has $60$ irreducible quartics.

Each polynomial $p_1,p_2,p_3$ was lexicographically the first in its orbit, so those orbits are distinct, and hence won't overlap (being orbits). Therefore we have accounted for all the $150$ irreducible quartic polynomials.

This is somewhat unsatisfactory. I would like to give an explanation as to why we have these three orbits rather than state it as a result of a brute force calculation. Ideally I want an argument involving the action of both the Galois group as well as $\Gamma$ on those elements of $K$. Hopefully I can return to this later.

The multiplicative group $K^*$ is cyclic of order $5^4-1=2^4\cdot3\cdot13$. Therefore the total number of generators is equal to $$\phi(2^4\cdot3\cdot13)=2^3\cdot(3-1)\cdot(13-1)=192.$$ Between them those $192$ generators have $192/4=48$ minimal polynomials, and these are exactly the quartic primitive polynomials. I don't know how to give a census of those. Möbius transformations won't necessarily take a primitive element to another, so those transformations cannot be used.

• Very unsatisfactory, but we got to start from somewhere. Might as well be this :-( – Jyrki Lahtonen Mar 29 '18 at 22:06
• Non-trivial elements of the Möbius group don't have fixed points in the set of interest, $\Bbb{F}_{5^4}\setminus\Bbb{F}_{5^2}$, so for a minimal polynomial to have a non-trivial stabilizer it must act on the zeros of an irreducible quartic as a power of Frobenius (Frobenius commutes with fractional linear transformations). This implies that only stabilizers of sizes $1,2$ or $4$ are possible. That still leaves a number of different ways of partitioning the set of 150 polynomials into orbits. I need to think about it a bit more... – Jyrki Lahtonen Apr 1 '18 at 19:31