7
$\begingroup$

Let $F=\mathbb Z/(2)$. The splitting field of $x^3+x^2+1\in F[x]$ is a finite field with eight elements.

my attempt of solution:

If $\alpha$ is a root in this polynomial in its splitting field, then I would like to prove that $F(\alpha)$ is the splitting field.

what I get is $x^3+x^2+1=(x-\alpha)(x^2+(1+\alpha)x+(\alpha +\alpha^2))$.

I'm trying to find the root of $x^2+(1+\alpha)x+(\alpha +\alpha^2)$, maybe it's a multiple of $\alpha$.

I need help!

thanks

  • 0
    I'm trying to prove that $\alpha^4 \neq \alpha$ If the equality holds, we'd have $\alpha^4-\alpha=0$ implies $\alpha(\alpha^3-1)=0$ implies $\alpha=0$ or $\alpha^3=1$. What is the characteristic of $F(\alpha)$? the characteristic of $F(\alpha)$ is more that 3 in order to get a contraction with the fact that $\alpha^3=1$?2012-12-02

3 Answers 3

3

Hint: In a field of characteristic $2$, the map $x\mapsto x^2$ is an automorphism. (If $\alpha$ is one root, then $\alpha^2$ and $\alpha^4$ are also roots; why are these three different? Note that $\alpha^8=\alpha$ again)

Hint for alternative solution: What can you say about $F[x]/(x^3+x^2+1)$ as $F$ vector space and as ring? (It is a field where $[x]$ is an obvious root of our polynomial and a threedimensional vector space, hence with $8$ elements; why is there no smaller splitting field?)

  • 0
    @RafaelChavez $a^2=\alpha$ implies $\alpha=1$ or $\alpha=0$ (the fixed points of the Frobenius are precisely the elmenets of the prime field)2012-12-03
1

Hint: Prove there are no roots, deduce the polynomial is irreducible of degree $3$ and conclude that the quotient is of degree $3$. Now recall that if $V$ is a vector space of dimension $n$ over $F$ then $V\cong F^{n}$ and in particular $|V|=|F^{n}|=|F|^{n}$

0

Hint: If $f(x)$ is an irreducible polynomial, what do you know about the relationship between its roots?

  • 0
    @Hurkyl it's a content of the next chapter of my book2012-12-02