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The following is given in the text that I have: In $GF(2^8)$ [Galois Field] let: $$h(x)=x^8+x^4+x^3+x+1$$ $$x^8 \bmod h(x)= [h(x)-x^8]$$ I basically don't understand the second step I think $h(x) mod x^8=[h(x)-x^8]$, is the text mistaken or am I ? If I am please explain this to me.

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By definition, $x^8\equiv x^8-h(x) \bmod h(x)$, namely $h(x)\mid (x^8+h(x)-x^8)$. On the other hand, since $2=0$ we have $h(x)-x^8=x^8-h(x)$.

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  • $\begingroup$ since 2=0? I didn't quite get it. $\endgroup$ – mathmaniage Apr 15 at 12:39
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    $\begingroup$ The characteristic of the field $GF(2^8)$ is equal to $2$. By definition then, $1+1=0$. $\endgroup$ – Dietrich Burde Apr 15 at 12:40
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The leading terms of $x^8$ and $h(x)$ are the same so that the quotient is $1$, and both

$$x^8=h(x)+(x^8\bmod h(x))$$ and $$h(x)=x^8+(h(x)\bmod x^8)$$ hold.

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  • $\begingroup$ but is h(x)modx^8=x^8mod h(x) $\endgroup$ – mathmaniage Apr 15 at 12:50
  • $\begingroup$ @mathmaniage: this is precisely shown by my answer ! $\endgroup$ – Yves Daoust Apr 15 at 12:51
  • $\begingroup$ I don't know, but, is the answer no? I came to the conclusion: (h(x)modx^8)=-(x^8modh(x)) $\endgroup$ – mathmaniage Apr 15 at 12:55
  • $\begingroup$ @mathmaniage: you are in $FG(2^8)$. $\endgroup$ – Yves Daoust Apr 15 at 12:57

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