Is the polynomial $x^8+x+1$ irreducible in $\mathbb{F}_2[x]$?

Question

Is the polynomial $f(x)=x^8+x+1$ irreducible inf $\mathbb{F}_2[x]$?

I know that if $x^8+x+1$ divides $x^{2^8}-x=x^{256}-x$, then it is irreducible over $\mathbb{F}_2$. I started using the division algorithm to see if this is true, but its too painful.

Is there a quick way to do this that I do not see? Can we show that $\mathbb{F}_2[x]/(f(x))$ is a field (or not)?

If $f$ is reducible, then at least one factor will have degree at most 4, and the constant term of every factor is $1$. That gives you $2^4-1=15$ possible factors to check, which should be pretty quick. — hmakholm left over Monica, Jun 02 '15 at 18:53
This is a special case of the more general result of this question. — Jyrki Lahtonen, Jun 03 '15 at 07:49

score 5 · Accepted Answer · answered Jun 02 '15 at 18:55

If $x^8 = -x-1$, then $x^{256}-x = x^{256}+x = (x+1)^{32}-x = x^{32} + x + 1$ $= (x+1)^4 + x + 1 = x^4 + x$.

So $x^{256}-x \equiv x^4 + x \pmod{x^8 + x + 1}$, and we can reduce no further. Therefore $x^8 + x + 1$ is not irreducible over $\mathbb{F}_2$.

Concretely, we have $x^8 + x + 1 = (x^2 + x + 1)(x^6 + x^5 + x^3 + x^2 + 1)$.

Is the polynomial $x^8+x+1$ irreducible in $\mathbb{F}_2[x]$?

1 Answers1