3

Express $x^8-x$ as a product of irreducibles in $\Bbb Z_2[x]$.

Work so far:

$$x^8 - x = x(x^7 - 1) = x(x - 1)(x^6+x^5+x^4+x^3+x^2+x+1)$$

From here, I think the Zeros of an Irreducible over a splitting field Theorem would be the method. I am unsure how to proceed

user26857
  • 52,094
jensen
  • 31
  • I believe that $x^8-x$ is identically zero in $\mathbb{Z}_2$. – Darío G Mar 16 '16 at 07:13
  • 2
    @user27454 There is a difference between the polynomial $x^8 - x$ and the function $x \mapsto x^8 - x$. It's just that over $\Bbb Z, \Bbb Q, \Bbb R$ and $\Bbb C$ you never notice the difference like you do in $\Bbb Z_n$. In other words, while functions are uniquely determined by what they assign every element to, polynomials are not. – Arthur Mar 16 '16 at 07:18
  • 1
    You could also go the other way, and find all cubics that are irreducible by using brute force, since if a cubic factors than it must have a linear factor. If the cubic has no linear factor, then its irreducible. – Nikolas Wojtalewicz Mar 16 '16 at 07:43

1 Answers1

6

Note that $$ (x^3+x^2+1)(x^3+x+1)=x^6+x^5+x^4+x^3+x^2+x+1 $$ over $\mathbb{Z}_2$, and those cubics are irreducible as they have no roots in $\mathbb{Z}_2$.

In general, $X^{q^n}-X$ is the product of all monic irreducibles over $\mathbb{F}_q$ with degree dividing $n$, so in this case, $X^8-X=X^{2^3}-X$ is the product of all monic irreducibles over $\mathbb{Z}_2$ with degree dividing $3$. You factored out the two monic irreducibles of degree $1$, so it made sense to look for the monic irreducibles of degree $3$, which would give the factorization.

Ben West
  • 12,366