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Suppose that $\beta$ is a zero of $f(x)=x^4+x+1$ in some field extensions of $E$ of $Z_2$.Write $f(x)$ as a product of linear factors in $E[x]$

Attempt: In $\mathbb Z_2: \beta^4+\beta+1=0$

Going by the long division method, I was able to factorize $f(x)=x^4+x+1 = (x-\beta)(x^3+\beta x^2+ \beta^2x + \beta^3+1)$.

if $g(x)=x^3+\beta x^2+ \beta^2x + \beta^3+1$, then $g(1)=\beta^2 =g(-1)~;~$

Is there any other method than trial and error in this problem. Please note that this problem is listed in the field extension chapter and even before the chapter on algebraic extensions and finite fields

Thank you for your help.

MathMan
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  • What do you mean with $\mathbb{Z}_2$? The 2-adic integers or the field with two elements? – Dune Aug 10 '14 at 13:38
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    Something that could help is $f(x+y) = f(x) + f(y) + 1$. – Najib Idrissi Aug 10 '14 at 13:45
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    @Dune I think the problem means the field with two elements – MathMan Aug 10 '14 at 13:50
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    The polynomial is irreducible over the prime field, so its zeros are Galois conjugates. In other words they are gotten from each other by applying the Frobenius automorphism. Does that sound familiar? So $\beta^2$ is another zero. Get to work! Afterwards you can check your answer against the last part of this answer. No peeking to soon :-) The $\gamma$ there is what you call $\beta$. – Jyrki Lahtonen Aug 10 '14 at 13:53
  • Thanks. I am afraid though that I haven't yet reached the chapter on Galois theory in Gallian. I am on the field extension chapter (which also happens to be before Algebraic extensions ) in which I came across this problem. – MathMan Aug 10 '14 at 13:58
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    Ok, now I see that the theoretical tools are limited. Have you done the example that $x\mapsto x^2$ is an endomorphism of a commutative ring of characteristic two? – Jyrki Lahtonen Aug 10 '14 at 13:59
  • aah yes, I know this result – MathMan Aug 10 '14 at 14:00
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    Great! Then I can give you a hint. Let $g(x)=x^4+x$. Show that we always have $g(y^2)=(g(y))^2$ for all $y\in E$. You can find more zeros of $f$ starting from this. – Jyrki Lahtonen Aug 10 '14 at 14:02
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    $g(x) = x^4+x+1 \implies g(y^2) = (y^2)^4 + y^2 +1 = (y^4)^2+(y)^2 +1= (y^4+y+1)^2$

    (as char $\mathbb Z_2 =2$ and all terms with coefficients as multiples of $~2$ equal $~0~ ) \implies $ if $\beta$ is a root of $g(x)$, then $\beta^2$ is also a root.

    Thank you :-)

     – MathMan Aug 10 '14 at 14:25
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    Well, done VHP! – Jyrki Lahtonen Aug 10 '14 at 17:51

1 Answers1

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$\require{cancel}$For a problem with small coefficients like that, trial and error works. We have $$f(x+y) = (x+y)^4 + x + y + 1 = f(x) + f(y) + 1$$ This implies $$f(x+1) = f(x) + f(1) + 1 = 0 + 1 + 1 = f(x)$$ So if $x$ is a root of $f$, $x+1$ is another root of $f$. In particular $\beta+1$ is a root of $f$.

Besides $f(\beta^2) = (f(\beta))^2 = 0$, because the Frobenius is a field automorphism (as Jyrki Lahtonen told you in the comments). It follows that the roots are $\beta$, $\beta+1$, $\beta^2$, $\beta^2+1$.

Najib Idrissi
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  • Thank you for the answer :-) – MathMan Aug 10 '14 at 14:25
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    I like the Galois-theoretical treatment, as @JyrkiLahtonen has guided OP to, namely that the roots are $\beta^{2^m}$, $0\le m<4$. But your answer was really very efficient. – Lubin Aug 10 '14 at 14:26