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I am reading a paper, and have a next expresion Let $\sigma(X) \in F_{2^m}[X]$ with deg $(\sigma(X)) = \delta$, then writing $\sigma_1$ for the even part of $\sigma$ and $X\sigma'$ for the odd part of $\sigma$, obtain: $\sigma(X) = \sigma_1(X)+X\sigma'(X)$,

My question: How to prove this?

pdta: I know that $X$ is a odd function but I don't know $\sigma'(X)$, then I am not sure whether $X\sigma'(X)$ is odd or even function, ...

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Hint: Write down any polynomial, field does not matter. Do the calculation of $\sigma_1$ and $\sigma'$. The general case will jump out at you. –  André Nicolas Jan 12 '12 at 17:21
    
@AndréNicolas There is the issue that the derivative of a polynomial in $\mathbb F_{p^n}[x]$ is not defined in the usual calculus sense with limits etc but rather as a formal derivative obtained by replacing $x^i$ by $ix^{i-1}$. If the OP has not seen (or does not remember) this, the notion of $\sigma^{\prime}$ may well be puzzling him. –  Dilip Sarwate Jan 12 '12 at 17:59
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I think that characteristic 2 is important because \sigma'(X) only have indeterminate with power 2, then \sigma'(X) is always even, or not? I have other question in the same paper say: $\sigma_1(X)$ and $\sigma'(X)$ are relative prime hwo i can proof this please –  juaninf Jan 12 '12 at 18:05
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>"$\sigma_1(X)$ and $\sigma^{\prime}(X)$ are relative prime." Are you sure you have stated the result correctly? If $\sigma(X) = X^4 + X^3 + X + 1$, then $\sigma_1(X) = X^4 + 1 = (X+1)^4$ and $\sigma^{\prime}(X) = X^2 + 1 = (X+1)^2$ are not relatively prime. –  Dilip Sarwate Jan 13 '12 at 0:19

1 Answer 1

The formal derivative of the polynomial $f(x) = \sum_i f_ix^i \in \mathbb F_q[x]$ obtained by replacing each $x^i$ by $ix^{i-1}$. Here $q=2^m$, and so the field has characteristic $2$. Thus we have $$ f^{\prime}(x) = \sum_{i} if_i x^{i-1} = f_1 + f_3x^2 + f_5 x^4 + $$ and so $xf^{\prime}(x)$ is the odd part of $f(x)$, by which is meant the sum of terms of odd degree in $f(x)$. If $f^{(0)}(x) = \sum_i f_{2i}x^{2i}$ is the even part of $f(x)$, then we have $f(x) = f^{(0)}(x) + xf^{\prime}(x)$.

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