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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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    >"$\sigma_1(X)$ 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.2012-01-13

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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 $f(x) = f^{(0)}(x) + xf^{\prime}(x)$.