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Let $\mathbb{F}$ be a field with $\operatorname{char}(\mathbb{F})=p>0$.

The derivative of a polynomial $P(x)={\displaystyle \sum\limits _{i=0}^{n}a_{i}x^{i}}\in\mathbb{F}[x]$ is $P'(x)=\sum\limits _{i=1}^{n}a_{i}ix^{i-1}\in\mathbb{F}[x]$.

I wish to prove that for $f\in\mathbb{F}[x]$:$f'=0\implies f(x)=g(x^p)$ where $g(x)\in\mathbb{F}[x]$.

I tried going by this definition of the derivative here but all I got was $p\mid ia_i$ for all $i=1,2,..,n$.

Any ideas ?

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    The following tangential comment might interest some readers. The error correction process for the Reed-Solomon codes used in CD and DVD players requires evaluation simultaneous evaluation of $\Lambda(x)$ and $x\Lambda^\prime(x)$ at the $255$-th roots of unity in $\mathbb F_{2^8}$ of characteristic $2$, and takes advantage of the fact that $x\Lambda^\prime(x)=\Lambda_1x+\Lambda_3x^3+\Lambda_5x^5+\cdots$ to save a lot of calculation by arranging the computation of $\Lambda(\alpha)$ to produce $\alpha\Lambda^\prime(\alpha)$ as a side result.2012-04-20

2 Answers 2

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Hint: While technically correct, the notion of divides, i.e. $\mid\;$, is not the best way to phrase things here; instead, since $\operatorname{char}(F)=p$, what you have is $ia_i=0$ for all $i=1,2,\ldots,n$. What does that tell you?

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    For elements of$a$finite field of size $p^n$, raising to the $p^n$th power doesn't do anything, i.e. $a^{p^n}=a$ for $a\in\mathbb{F}_{p^n}$. In fact, this is the defining property for elements of $\mathbb{F}_{p^n}$. So for example, look at the field like the rational functions in $T$ over $\mathbb{F}_p$, which is written $\mathbb{F}_p(T)$. This is a field of characteristic $p$. In it, we have $a^p=a$ for $a\in \mathbb{F}_p\subset \mathbb{F}_p(T)$, but $T^p\neq T$. So raising to a power divisible by $p$ doesn't always act differently, only sometimes.2012-04-20
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The main point you have to understand here is that the statement $p |ia_i $ doesn't make sense because $p$ is not in $F$ and also there is no reasonable definition of "divisibility" in a field.

The correct thing to say is that $ia_i=\bar i\cdot a_i\in F$, with $\bar i\in \mathbb F_p\subset F$.
Hence if $p$ does not divide $i$ in $\mathbb N$ (where divisibility makes sense!), we have $\bar i\neq 0\in F$ and since $\bar i\cdot a_i=0\in F$ we deduce $a_i=0\in F$.
From this you get $f(x)=g(x^p)$ where $g(y)=\sum_i a_{ip}y^i$.

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    Dear @Belgi, in$a$ formal sense it is true that there is a notion of divisibility in any ring: you can decree that $a$ divides $b$ if there exists $c$ with $b=ac$ . What I wrote in my answer is that this notion is not *reasonable* in a field, where any non-zero element divides every element. It is also useless and even treacherous in rings with zero-divisors. Divisibility on the other hand is admirably suited for going back and forth between a UFD $R$ and its field of fractions $K$, especially when considering the polynomial rings $R[X]$ and $K[X]$ : think Gauss, Eisenstein,...2012-04-20