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We know that $X^{15}-1=(X+1)(X^4+X+1)(X^4+X^3+X^2+X+1)(X^2+X+1)(X^4+X^3+1)$ over $F_2$ where the polynomials come from the conjugacy classes in $F_{16}$:

{1}, {$a,a^2,a^4,a^8$}, {$a^3,a^6,a^9,a^{12}$}, {$a^5,a^{10}$} and {$a^7,a^{14},a^{13},a^{11}$} respectively.

So $a$ is a 15th root of unity.

But we don't understand how these minimal polynomials are computed in the first place? And e.g. why $a$ is not a root of $X^4+X+1$?

Thanks!

Cecilie
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  • It may well be that $a$ is a root of $X^4+X+1$. How did you conclude that is not the case? The alternative is that $a$ is a root of $X^4+X^3+1$, for the roots of other factors are roots of unity of order $<15$. If you decide that $a$ is a root of, say, $X^4+X+1$, then you can just expand e.g. $$ (X-a^7)(X-a^{11})(X-a^{13})(X-a^{14})=X^4+X^3+1 $$ using the field table here. I denoted your $a$ by $\gamma$ there. – Jyrki Lahtonen Sep 10 '15 at 11:58
  • Related questions 1, 2. – Jyrki Lahtonen Sep 10 '15 at 12:01
  • Okay, thank you, this was very helpful! – Cecilie Sep 10 '15 at 12:25
  • Okay, so maybe we need some more help.

    You say that for $q=8=2^3$ we have the relation $a^3=a+1$, and likewise when $q=16=2^4$ we have $a^4=a+1$. Why is that? Is it a general thing that when working in $F_{2^n}$ we have $a^n=a+1$?

     – Cecilie Sep 10 '15 at 15:45

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