In galois field- how does modular arithmetic

Question

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.

Dietrich Burde · Accepted Answer · 2019-04-15T12:42:02.377

1

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 $h(x)-x^8=x^8-h(x)$.

edited Apr 15 '19 at 12:42

answered Apr 15 '19 at 12:37

Dietrich Burde

130,978

since 2=0? I didn't quite get it. – mathmaniage Apr 15 '19 at 12:39
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The characteristic of the field $GF(2^8)$ is equal to $2$. By definition then, $1+1=0$. – Dietrich Burde Apr 15 '19 at 12:40

score 0 · Answer 2 · answered Apr 15 '19 at 12:48

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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.

answered Apr 15 '19 at 12:48

but is h(x)modx^8=x^8mod h(x) – mathmaniage Apr 15 '19 at 12:50
@mathmaniage: this is precisely shown by my answer ! – Apr 15 '19 at 12:51
I don't know, but, is the answer no? I came to the conclusion: (h(x)modx^8)=-(x^8modh(x)) – mathmaniage Apr 15 '19 at 12:55
@mathmaniage: you are in $FG(2^8)$. – Apr 15 '19 at 12:57

In galois field- how does modular arithmetic

2 Answers2