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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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    What does mean $p | i a_i$ ?2012-04-20
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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

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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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    $i=0$ or $a_i=0$ ?2012-04-20
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    Exactly! :) $\text{}$ But remember, $i$ is not an element of the field $\mathbb F$, but is just a positive integer, so $i=0$ really only tells you that...2012-04-20
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    With this hint I got that:$a_1=0,...a_{p-1}=0,a_{p+1}=0...$ hence $P(x)=a_px^p+a_{2p}x^{2p}+...$ right ? :)2012-04-20
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    Well, I should mention that for any $i\geq0$, either $a_i=0$ or $p\mid i$, *or both* - so it still might be the case that some of the $a_{np}=0$.2012-04-20
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    Thanks for the hint! by the way, can we say something about $x^p$ ? it's probably not true but does $x^p=x^0=1$ ? (why ?)2012-04-20
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    You're right, that's not true, since $x$ is a variable - it doesn't satisfy any relations. In fact, it isn't even true that for every element $a$ in a field of characteristic $p$, that $a^p=a$. A good way to remember that is that $a^p=a$ is the **defining property** of being an element of $\mathbb{F}_p$.2012-04-20
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    hmm..so $x^p$ is just a notation for $x*x*...x$ p times (and we consider $p\in\mathbb{N}$) ?2012-04-20
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    That's right. Well, that's what exponentiation means in general, not just for $x$ :) It just so happens that raising to a power divisible by $p$ can have unusual effects on elements of a field of characteristic $p$. But since $x$ is not an element of the field, just an "indeterminate", it acts normally.2012-04-20
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    "raising to a power divisible by p can have unusual effects on elements of a field " can you please say what do you mean ?2012-04-20
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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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    Well, I said that at first too, but $p=\underbrace{1+\cdots+1}_{p\text{ times}}$ still makes sense in $F$, and since $p=0$ in $F$ and $0\mid m$ implies $m=0$ in any ring, $p\mid ia_i$ is still saying that $ia_i=0$ in $F$. But of course, I agree that $\mid$ is not the right thing to use in this situation.2012-04-20
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    Thank you for pointing this out, indeed it sould be written in a different way as you stated. +12012-04-20
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    Zev note is a good point too as there is a reasonable definition for divisibility in any ring really.2012-04-20
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    Dear @Zev, if $1\in \mathbb N$ is the usual Peano integer and if $p=1+...+1\in \mathbb N$ then, no, $p=1+1+...+1$ is not in $F$ because $\mathbb N$ is in no sense included in $F$: for example $F$ might be finite and $\mathbb N$ is infinite. On the other hand it is true that $\bar p=p*1_F=1_F+...+1_F=0_F\in F$ where I have denoted by $*$ the scalar multiplication of an element of $\mathbb Z$ with an element of the $\mathbb Z$-algebra $F$. The sentence you use "p=0 in F" is an abuse of language short-circuiting these precise notations. (To be continued)2012-04-20
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    (Continued) This is perfectly admissible for people, like you, who know exactly what they mean but beginners should be absolutely precise when confronted with these subtle concepts for the first time.2012-04-20
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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