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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

  • 1
    The other zeros are $\alpha^2$ and $\alpha^2 + \alpha + 1$.2012-12-01
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    Why? I can't realize why they are roots. Thank you for your answer.2012-12-01
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    Just plug them in: for example, $(\alpha^2)^3 + (\alpha^2)^2 + 1 = (\alpha^3 + \alpha^2 + 1)^2 = 0$2012-12-01
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    You get $\alpha^2 + \alpha + 1$ by squaring $\alpha^2$ again and then reducing2012-12-01
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    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$2012-12-01
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    Replace all your $2$ with $0$ since we are in $\mathbb{Z}/2\mathbb{Z}$2012-12-01
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    @Cocopuffs yes, of course, thank you very much :)2012-12-01
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    @Cocopuffs why $\alpha$ and $\alpha^2$ are distinct roots?2012-12-02
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    $X^2 - X = X(X - 1)$ has no zeros other than $0$ and $1$. $\alpha$ is neither of these2012-12-02
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    Why did you edit the title ? I think it was better the way it was since it had more information...2012-12-02
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    @Belgi It's a little subjective the reason I edited, but if you want to edit again, no problem :)2012-12-02
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    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

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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 answer2012-12-01
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    Using the automorphism, you should find two other roots for $x^3+x^2+1$.2012-12-01
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    @JoelCohen then can I say that $\alpha^2$ is a root of $x^3+x^2+1$, because of this automorphism?2012-12-01
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    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 $\alpha^4=\alpha^2+\alpha+a$.2012-12-01
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    @HagenvonEitzen It miss only one problem, why these roots are distinct?2012-12-02
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    how did you get $\alpha^8=\alpha$?2012-12-02
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    @RafaelChavez I didn't perform the polynomial division, but knowing that squaring three times must give one o fthe three roots, I figured that the first root is the only nice result. :)2012-12-03
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    @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
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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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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)$2012-12-01
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    But what can I do with this information? thank you for your answer.2012-12-01
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    @RafaelChavez - can you write down all the elements of the field in terms of the basis you have identified - how many are there?2012-12-01
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    @MarkBennet yes, of course, there are 8 elements, so if I prove $F(\alpha)$ is the splitting we're done.2012-12-01
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    There is a more direct relationship between the roots than simply knowing the others are a linear combination of powers of $\alpha$....2012-12-01
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    $\alpha²\in F(\alpha)$?2012-12-01
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    I'm trying to find the relation you said, I've already searched in two abstract algebra books, and I didn't find it.2012-12-01
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    All of the roots of any irreducible polynomial over any field are conjugates. (in fact, they form one entire conjugacy class)2012-12-01
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    @Hurkyl the problem is up to this point of the book I'm studying, The author doesn't mention what conjugation is, what I know about conjugation is the conjugation of complex numbers.2012-12-01
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    That's unfortunate. It would be by far the easiest way to carry out the step that you wanted to do next. :( Just to double check, have you heard the word "Frobenius" yet? Maybe you've learned the same thing in other words?2012-12-01
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    @Hurkyl no problem, thank you :)2012-12-01
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    @Hurkyl it's a content of the next chapter of my book2012-12-02