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I know that we can't define $dx/dy$ with this equation because $2y = 0$ with finite field of charateristic $2$. But with $GF(2^n)$ (has characteristic by $2$) $2=x$ not $0$. Do I misunderstand here?

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In $GF(2^n)$, $y+y = 0$ for all $y$. These elements are not integers; using something like “2” to denote a field element is merely a convention, and, as you demonstrate, a confusing one. In this case, $2y$ means $y+y$, not the product in the field of $y$ with the element conventionally represented by “2”.

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  • $\begingroup$ $y+y$ and $2 \times y$ are the same thing if you define $2$ as $1+1$ (where $1$ is the multiplicative identity), as you should. $\endgroup$ – fkraiem Oct 9 '17 at 0:48
  • $\begingroup$ so $3$ equals $1+1+1=1$ , right? it is so confused here for me between scalar multiplication and multiplication operation in field. $\endgroup$ – Do Minh Oct 9 '17 at 4:14
  • $\begingroup$ @DoMinh: yes, that's it. In $GF(2^n)$, addition really is XOR, so $y+y+y = y$ for all $y$. Also, subtraction is identical to addition. It takes a bit of time to get accustomed to it. First step, though, is to make the distinction between integers used to represent multiple additions, and integers used to represent field elements, which are completely different animals. $\endgroup$ – Thomas Pornin Oct 9 '17 at 13:31

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