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Let $K$ be a field of characteristic $p$. Let $a \in K$. Then the derivative homomorphism of $x^{p^n}-a$ is zero for $n$ positive integer. Does that mean that $x^{p^n}-a$ has a unique multiple root in $K$? How can i think of this rigorously?

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There is also a neat characterization of finite purely inseparable extensions as those obtained by adjoining $p$-power roots of base field scalars; cf. section 2 here. –  anon Sep 23 '12 at 0:36
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3 Answers

up vote 5 down vote accepted

The answer is yes, but I don't agree with Qiaochu that your argument is correct (or maybe I didn't understand your argument).

To see that the fact that $d (x^{p^n}-a)/dx=0$ is not enough to imply that $x^{p^n}-a$ has at most one root, note that $d(x^{2p}-x^p)/dx=0$ and $x^{2p}-x^p$ has two roots; $0$ and $1$.

Here is a correct proof: Suppose there were two roots of $t^{p^n}-a$; call them $x$ and $y$. Then $x^{p^n} = y^{p^n}$ so $x^{p^n} - y^{p^n}=0$. Since we are in characteristic $p$, this means that $(x-y)^{p^n}=0$. Since we're in a field, $x-y=0$, or $x=y$. We have shown that $t^{p^n}-a$ only hs one root, as desired.

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how about specifically for the polynomial $x^{p^n}-a$? –  Manos Dec 5 '11 at 20:04
    
Does the new edit clear things up? –  David Speyer Dec 5 '11 at 20:09
    
Yes, thank you :-) –  Manos Dec 5 '11 at 20:13
    
Yes, sorry, I was a little confused. –  Qiaochu Yuan Dec 7 '11 at 16:51
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Yes. If $b$ is any element of an extension of $K$ such that $b^{p^n} = a$, then $x^{p^n} - b^{p^n} = (x - b)^{p^n}$ by standard properties of the Frobenius homomorphism $x \mapsto x^p$. Your argument is already rigorous.

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HINT $\rm\ f = g(x^p)\ \Rightarrow\ f{\:'} = 0\:.\:$ But $\rm\: g\in \mathbb F_p[x]\ \Rightarrow\ g(x^p) = g(x)^p\: $ has the same roots as $\rm\:g\:.\:$ So a counterexample arises by choosing $\rm\:g\:$ with at least two roots, e.g. $\rm\:g = x\:(x-1)\:$ as in DS's post.

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