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

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.

• Just try to solve for the coefficients. if it factored, we'd have $p(x)=(x^2+ax+b)\times (x^2+cx+d)$
– lulu
Jan 2, 2019 at 22:12

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.
• That's it. I spent awhile looking for a trick answer, but didn't see any. An alternative way of organizing the calculation would be to first build a list of remainders of $x^0,x^7\equiv-1+3x+3x^2+3x^3,x^{14}$ and $x^{21}$ modulo $f(x)$. Raising to seventh power modulo $f(x)$ is linear, so $$x^{49}\equiv-1+3x^7+3x^{14}+3x^{21}\pmod{f(x)}$$ et cetera. This may help with a higher degree polynomial. I'm afraid I don't have a good feel for the difference in complexity for I usually do this in characteristic two only :-) Jan 2, 2019 at 23:19
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 $$\mathbb{Z}_7$$.