Irreducible polynomials over fields

Question

Verify if the following polynomial over the indicated field is irreducible.

$p=x^4+2x^2-x+1 \in \mathbb{Z_7}[x]$

Because $p$ is degree $4$, it can only be factorized as the following cases:

i) $p=h \cdot g$, where either $h$ or $g$ is degree $1$ and the other is $3$

ii) $p=h \cdot g$, where $h$ and $g$ are degree $2$

For i) I can simply calculate the zeros of $p$ and check if there is any $h$ or $g$ with degree $1$.

We are in $\mathbb{Z_7}[x],$ so $x \in \{0,1,2,3,4,5,6\}$.

$p(0)=1$

$p(1)=3$

$p(2)=2$

$p(3)=6$

$p(4)=5$

$p(5)=6$

$p(6)=5$

Because $p$ has no zeros, it can't be factorized like I said in i). Now there is only the case ii) to verify.

For this I can suppose $p=x^4+2x^2-x+1=(ax^2+bx+c)(dx^2+ex+f)$

$\Leftrightarrow x^4+2x^2-x+1=(ad)x^4+(ae+bd)x^3+(af+be+cd)x^2+(bf+ce)x+cf $

$\begin{cases} ad=1 \\ ae+bd=0 \\ af+be+cd=2 \\ bf+ce=-1=6 \\ cf=1 \ \end{cases} \Leftrightarrow \begin{cases} ad=1 \\ ae=-bd \\ af+be+cd=2 \\ bf=6-ce \\ cf=1 \ \end{cases}$

Is there a simple way to solve this system of equations? I've start reading some stuff about irreducible polynomials and some times when I get to this part I get stuck in some systems.

Answer 1

Since we checked that $p$ has no root in $\mathbb{F}_7$, we just have two possible configurations:

1. $p$ is the product of two quadratic (monic) irreducible factors.
   In such a case $\mathbb{F}_7[x]/(p(x))$ is isomorphic to $\mathbb{F}_{49}\times\mathbb{F}_{49} $;
2. $p$ is irreducible. In such a case $\mathbb{F}_7[x]/(p(x))$ is isomorphic to $\mathbb{F}_{7^4}$.

In the former case, by raising some element of the ring to the $48$-th power we either get a polynomial which is not coprime with $p(x)$, or the element $1$. Let us reduce $x^{48}\pmod{p(x)}$ and $\pmod{7}$ in order to check that we are in the latter case. $$ x^4 = -2x^2+x-1\tag{A} $$ $$ x^8 = 3x^3-3x^2+2x-3\tag{B} $$ $$ x^{12} = 2x^3+x^2+3x+1\tag{C} $$ $$ x^{24} = -x^3+x^2+3\tag{D} $$ $$ x^{48} = -x^3-2x^2+x+3.\tag{E} $$ $-x^3-2x^2+x+3$ is coprime with $p(x)$ and differs from $1$.
It follows that $p(x)$ is irreducible over $\mathbb{F}_7$. This is more or less what softwares do to prove or disprove the irreducibility of a polynomial over some finite field.

For the general case, have a look at the Cantor-Zassenhaus and Berlekamp's algorithms.

A brute-force approach is to consider all the $\frac{7^2-7}{2}=21$ monic irreducible quadratic polynomials over $\mathbb{F}_7$ and check that they do not divide $p(x)$.

Answer 2

Since your original polynomial has leading coefficient 1, in any factorization you can always multiply each factor by a constant and assume that the leading coefficients of the factors are 1. So you can assume that $a=d=1$. That should make the system easier to solve.