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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 $p | i a_i$ ? –  Lierre Apr 20 '12 at 18:04
3  
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2 Answers 2

up vote 4 down vote accepted

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 $a_i=0$ ? –  Belgi Apr 20 '12 at 18:10
    
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... –  Zev Chonoles Apr 20 '12 at 18:11
    
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 ? :) –  Belgi Apr 20 '12 at 18:16
    
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 $a_{np}=0$. –  Zev Chonoles Apr 20 '12 at 18:23
    
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 ?) –  Belgi Apr 20 '12 at 18:23

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. –  Zev Chonoles Apr 20 '12 at 18:44
    
Thank you for pointing this out, indeed it sould be written in a different way as you stated. +1 –  Belgi Apr 20 '12 at 20:00
    
Zev note is a good point too as there is a reasonable definition for divisibility in any ring really. –  Belgi Apr 20 '12 at 20:01
    
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) –  Georges Elencwajg Apr 20 '12 at 20:32
    
(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. –  Georges Elencwajg Apr 20 '12 at 20:33

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