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Let the $ \mathbb F_{p}$ denote the finite field of $\mathbb Z/ p \mathbb Z$ and $\overline{ \mathbb F_{p}}$ its algebraic closure.

Now let $f(x)=X^p- b \in \overline{ \mathbb F_{p}}[x]$. I want to $f(x)$ has exactly one root. I know all is derivative are zeroes. How does that prove it has exactly one root?

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All derivatives? The first derivative must be constant, no? –  Salech Alhasov Apr 1 '12 at 11:14
    
Well, the first derivative is $pX^{p-1}$, which is the zero polynomial. –  Mohan Apr 1 '12 at 11:47

2 Answers 2

up vote 4 down vote accepted

Hint: There is an $\alpha\in \overline{\mathbb{F}_p}$, such that $f(\alpha)=0$. Show $f(X)=(X-\alpha)^p$.

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Working with derivatives is not a good idea here because the characteristic of the fields is positive.

As a hint: Do you know about freshman's dream?

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No, I don't know about it. –  Mohan Apr 1 '12 at 12:20
    
It says that $(x + y)^n = x^n + y^n$. This is not true in general, but if everything takes place in a field of characteristic $p > 0$ and $n$ is a power of $p$, then it is true. Can you now prove that freshman's dream is true in this case? Can you use it to solve your problem? –  Lennart Apr 1 '12 at 12:29
    
Yes, I can show that.Then I can prove $X^p-b=(X-a)^p$ where $a$ is one solution of the equation $X^p-b$. –  Mohan Apr 1 '12 at 12:33

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