Is $x^4 + 2x^2 - x + 1$ irreducible in $\mathbb Z_7[x]$?

Question

How would one do this? I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2. How would I prove that it is or it isn't? There are a lot of polynomials with this conditions.

Answer 1

The product of all the monic, irreducible polynomials over $\mathbb{F}_7$ wih degree $\leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $\mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1\pmod{f(x)}$, for instance through the Brauer chain $x^4\to x^8\to x^{16}\to x^{32}\to x^{48}$. $$ x^4 \equiv -2x^2+x-1\pmod{f(x)} $$ $$ x^8 \equiv 3x^3-3x^2+2x-3\pmod{f(x)} $$ $$ x^{16} \equiv x^3+3x^2+2x-1\pmod{f(x)} $$ $$ x^{32} \equiv -x^3+2x^2+x-3\pmod{f(x)} $$ $$ x^{48}-1 \equiv -x^3-2x^2+x+2 \pmod{f(x)}$$ give $$\gcd(x^{48}-1,x^4+2x^2-x+1)=\gcd(x^3+2x^2-x-2,x^4+2x^2-x+1) $$ $$ = \gcd(x^3+2x^2-x-2,-x-3) $$ but $x=4$ is not a root of $x^3+2x^2-x-2$ since $7\nmid 90$ and we have finished.

Answer 2

You can also try to factor it as a product of two polynomials of degree $2$. Such a factorization must have the form

\begin{align}x^4 + 2x^2 - x + 1 & = (x^2+ax+b)(x^2-ax+b^{-1})\\ & =x^4+(b+b^{-1}-a^2)x^2+(ab^{-1}-ab)x+1\end{align} Hence the relations $$b+b^{-1}-a^2=2, ab^{-1}-ab=-1,$$ which can be easily seen to have no solution in $\mathbb{Z}_7$.