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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. –  mixedmath Dec 14 '11 at 16:31
    
@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? –  Daniel Freedman Dec 14 '11 at 16:33
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@Daniel: There was a passionate user on meta who thought that mbox was outdated, and should use \mathrm or another math-text form instead. –  mixedmath Dec 14 '11 at 16:39

2 Answers 2

If $p$ divides $m$, say $m = p^k m_0$, then $x^m - 1 = (x^{m_0} - 1)^{p^k}$, so the splitting field of $x^m - 1$ is the same as that of $x^{m_0} - 1$. The extension is still separable. The derivative check doesn't work because the polynomial is not irreducible.

As for your follow-up question: if $\omega$ is a primitive root of unity, $\sigma(\omega)$ and $\omega$ have the same minimal polynomial $\Phi$, so $K(\sigma(\omega)) = K(\omega)$. Since $\Phi$ divides $x^m - 1$, $\sigma(\omega)$ is also a root of unity, and since it generates the whole splitting field, it is a primitive root.

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To your first question, yes. $mx^{m-1}$ behaves poorly if $p \mid m$.

To your follow-up question, one should note that $\sigma$ permutes roots of minimal polynomials. Why? Suppose $a_0 + a_0\beta + ... + a_j \beta^j = 0$, where $a_i$ are in the base field and $\beta$ is a root of the irreducible polynomial. Then $\sigma (\beta^j) = \sigma(\beta)^j$ and $\sigma(a_i) = a_i$, and so $a_0 + a_0 \sigma (\beta) + ... + a_j \sigma(\beta)^j = 0$ as well. So $\sigma(\beta)$ is another root of the minimal polynomial.

In addition, when the extension is finite, each root can be sent to any other root. I won't prove that here. The cyclotomic polynomials are more special, because they are cyclic. Since the Galois Group is cyclic, it has a generator. One can extend $\omega \to \omega^j$ (but only among the primitive roots) to an automorphism of the field quickly, and this generates all the automorphisms.

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