# The splitting field of $x^3+x^2+1$ over ${\Bbb Z}/(2)$

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

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The other zeros are $\alpha^2$ and $\alpha^2 + \alpha + 1$. – Cocopuffs Dec 1 '12 at 19:20
Why? I can't realize why they are roots. Thank you for your answer. – user42912 Dec 1 '12 at 19:23
Just plug them in: for example, $(\alpha^2)^3 + (\alpha^2)^2 + 1 = (\alpha^3 + \alpha^2 + 1)^2 = 0$ – Cocopuffs Dec 1 '12 at 19:32
You get $\alpha^2 + \alpha + 1$ by squaring $\alpha^2$ again and then reducing – Cocopuffs Dec 1 '12 at 19:34
I didn't understand why, because: $(\alpha^3 +\alpha^2 +1)^2=(\alpha^2)^3+(\alpha^2)^2+(1)^2+2\alpha^5+2\alpha^2+2\alpha^3=\alpha^6+ \alpha^4 +1+2\alpha^5+2\alpha^2+2\alpha^3$ – user42912 Dec 1 '12 at 20:02

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?)

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since $x^3+x^2+1$ is irreducible, then $F[x]/(x^3+x^2+1)$ is a field? What can I do with this information? thank you for your answer – user42912 Dec 1 '12 at 18:46
Using the automorphism, you should find two other roots for $x^3+x^2+1$. – Joel Cohen Dec 1 '12 at 18:58
@JoelCohen then can I say that $\alpha^2$ is a root of $x^3+x^2+1$, because of this automorphism? – user42912 Dec 1 '12 at 19:05
Yes. This automorphism (Frobenius) leaves the polynomial invariant, hence maps roots to roots. And indeed we have $(x^3+x^2+1)^2=x^6+x^4+1$. Thus the othre roots are $\alpha^2$ and $\alpha^4=\alpha^2+\alpha+a$. – Hagen von Eitzen Dec 1 '12 at 22:48
@HagenvonEitzen It miss only one problem, why these roots are distinct? – user42912 Dec 2 '12 at 3:01

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

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$\{1,\alpha,\alpha^2\}$ is a basis of $F(\alpha)$ – user42912 Dec 1 '12 at 18:26
But what can I do with this information? thank you for your answer. – user42912 Dec 1 '12 at 18:29
@RafaelChavez - can you write down all the elements of the field in terms of the basis you have identified - how many are there? – Mark Bennet Dec 1 '12 at 18:31
@MarkBennet yes, of course, there are 8 elements, so if I prove $F(\alpha)$ is the splitting we're done. – user42912 Dec 1 '12 at 18:33
There is a more direct relationship between the roots than simply knowing the others are a linear combination of powers of $\alpha$.... – Hurkyl Dec 1 '12 at 18:45

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 $|V|=|F^{n}|=|F|^{n}$

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