0

How can I factorize $x^8-x$ over the fields $F_2$ and $F_4$?

rubik
  • 9,522
Ovomaltine
  • 151

2 Answers2

4

This is an answer to "Is there a strategy to find this out?" referring to the factorization shown in Jack D'Aurizio's answer.

In this case, the factorization is easy to discover by simple calculations. First, we note that there are no repeated roots since $x^8-x$ and its formal derivative $8x^7-1 = -1$ are relatively prime polynomials. Next, we have the obvious factorization $$\begin{align}x^8-x &= x(x-1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)\\ &= x(x+1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) \end{align}$$ where the factors of $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ (if any) must be irreducible polynomials of degree $2$ or more; we have already accounted for all the degree-$1$ factors of $x^8-x$. Now, the only irreducible quadratic is $x^2+x+1$ (all other quadratic polynomials have $x$ or $x+1$ as factors), and it is easy to verify (by inspection, almost!) that $x^2+x+1$ does not divide $x^6 + x^5 +\cdots+ x + 1$. Turning to cubic polynomials, we can discard $x^3+1$ and $x^3+x^2+x+1$ as possible candidates since they both have $x+1$ as a factor. So we are left with irreducible cubic $x^3+x^2+1$ and $x^3+x+1$ as possibilities, and it is with much joy and satisfaction that we verify that their product does indeed equal $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ giving us that $$x^8-x = x(x-1)(x^3+x^2+1)(x^3+x+1).\tag{1}$$

Dilip Sarwate
  • 26,411
0

Hint: We have that $x^8-x$ splits as $$ x(x+1)(x^3+x^2+1)(x^3+x+1)$$ over $\mathbb{F}_2$ and $\gcd(x^8-x,x^4-x)=x^2+x=x(x+1)$.

Jack D'Aurizio
  • 361,689
  • Ok, in the first line we have the factorization over F2. Is there a strategy to find this out? What do mean exactly with the gcd of x^8-x and x^4-x in this context? – Ovomaltine Feb 06 '15 at 15:40
  • @Ovomaltine: over $\mathbb{F}_p$, $\gcd(p(x),x^{p^h}-x)$ gives the product of all the irreducible polynomials over $\mathbb{F_p}$, having degree $d\mid h$, that divide $p(x)$. – Jack D'Aurizio Feb 06 '15 at 15:46
  • And what about the factoriziation over F4? – Ovomaltine Feb 06 '15 at 15:47
  • 1
    Ovomaltine, the factorization over $F_4$ is the same as it is over $F_2$. Those cubics have no zeros in $F_4$ (those zeros are in $F_8\setminus F_2$), so they remain irreducible over $F_4$. A more general result is that an irreducible polynomial of degree $m$ over $F_2$ remains irreducible over $F_{2^n}$, iff $\gcd(m,n)=1$. Proving that more general version takes a bit more theory (e.g. basic Galois theory). – Jyrki Lahtonen Feb 06 '15 at 16:16