# In galois field- how does modular arithmetic

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.

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

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

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.

• but is h(x)modx^8=x^8mod h(x) – mathmaniage Apr 15 at 12:50
• @mathmaniage: this is precisely shown by my answer ! – Yves Daoust Apr 15 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 at 12:55
• @mathmaniage: you are in $FG(2^8)$. – Yves Daoust Apr 15 at 12:57