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On page 587 in Dummit and Foote, the authors say the polynomial $\frac{x^{16}-x}{x(x-1)(x^2+x+1)}$ can be factored into quartics over $\mathbb{F}_2$ as $(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$.

I am having trouble seeing this. When I divide the polynomial $x^{16}-x$ by $x(x-1)(x^2+x+1)$ using long division, I get $x^{12}+x^9+x^6+x^3+1$. However, I am not sure how to factor this polynomial into quartics. How do I do this?

ponchan
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    You could try every possible quartic. You could use a systematic method, such as Berlekamp's algorithm. You could multiply the given factors to see if they work. – Angina Seng Apr 06 '18 at 15:04
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    There are only three irreducible quartics over $\Bbb F_2$, namely $X^4+X+1$, $X^4+X^3+1$, and $X^4+X^3+X^2+X+1$. – Lubin Apr 06 '18 at 18:44
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    Over $\Bbb{F}_p$ we always have that $x^{p^n}-x$ is the product of all irreducible polynomials of degrees that are factors of $n$. You already listed all the irreducible linear and quadratics, so the remaining factors are all the irreducible quartics, listed by Lubin. – Jyrki Lahtonen Apr 06 '18 at 20:09
  • This is also close to being a duplicate of this old thread, but I honestly wonder why that was so highly voted. – Jyrki Lahtonen Apr 06 '18 at 20:10

2 Answers2

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The Berlekamp algorithm gives $$ x^{16}-x=(x^4 + x^3 + x^2 + x + 1)(x^4 + x^3 + 1)(x^4 + x + 1)(x^2 + x + 1)(x + 1)x. $$ So the statement in Dummit and Foote is correct.

Dietrich Burde
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Over $\mathbb Z$ we have $$ \begin{align} x^{16}-x &= x(x^{15}-1) \\&= x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x) \\&= x (x - 1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) \end{align} $$ and so your polynomial is $(x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$.

You only need to factor $(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$ mod $2$.

lhf
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