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I need to find all irreducible polynoms of degree at most 4 in $\mathbb{Z}_2$

This is my result:

$$ x^4 + x^1 + 1 \\ x^4 + x^2 + 1\\ x^4 + x^3 + 1 \\ x^4 + x^3 + x^2 + x^1 + 1 \\ $$

Is this correct?

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    $\begingroup$ Degree at most $4$ you mean? $\endgroup$
    – lhf
    Dec 1, 2017 at 11:12
  • $\begingroup$ Yes, I've edited this. $\endgroup$ Dec 19, 2017 at 13:54

1 Answer 1

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$x^4 + x^2 + 1= (x^2 + x + 1)^2$ is not irreducible.

An irreducible polynomial of degree $4$ defines a field extension of $\mathbb Z_2$ of degree $4$. Its $2^4$ elements are the roots of $x^{2^4}-x$ because of Lagrange's theorem in group theory.

Therefore, the irreducible polynomials of degree $4$ are those that divide $x^{2^4}-x$ over $\mathbb Z_2$: $$ x^{16}-x=x (x + 1) (x^2 + x + 1) (x^4 + x + 1) (x^4 + x^3 + 1) (x^4 + x^3 + x^2 + x + 1) $$ and so there are exactly three of them.

To factor $x^{16}-x$, factor $x^{15}-1$, starting with the factorization into cyclotomic polynomials: $$ x^{16}-x = x(x^{15}-1)=x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x) $$ which gives $$ x^{16}-x = x(x - 1)(x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) $$ We then need to further factor these mod 2. The last factor is the only one that is reducible.

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  • $\begingroup$ Please explain why $x^{2^4}$! $\endgroup$ Dec 1, 2017 at 15:10
  • $\begingroup$ You are more than welcome to explain in this post how you factorized $x^{16}-x$. $\endgroup$ Dec 2, 2017 at 20:13

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