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This should hopefully be a very simple question:

"Suppose $\mbox{char}K = 0$ or $p$, where $p\not| \ m $. The $m$th cyclotmic extension of $K$ is just the splitting field $L$ over $K$ of $X^m - 1$"

Is the condition that $p \not | \ m$ there to guarantee $X^m - 1$ is separable over $K$? (the derivative is $mX^{m-1}$, which certainly shares a factor of degree $\geq 1$ with $X^m - 1$ if $m$ is a multiple of $p$)

EDIT: A follow up question:

$L/K$ is Galois. Why does an element $\sigma$ in $\mbox{Gal}(L/K)$ send primitive roots to primitive roots?

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    By the way, it's interesting that you use mbox. There was an interesting conversation on the meta about that.2011-12-14
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    @mixedmath: I use mbox too, what's the alternative? As for the question, can you see that $ L = K(\omega)$, where $\omega$ is a primitive mth root of unity?2011-12-14
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    @Daniel: There was a passionate user on [meta](http://meta.math.stackexchange.com/questions/3299/tex-usage-in-stackexchange) who thought that mbox was outdated, and should use \mathrm or another math-text form instead.2011-12-14

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