Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
@Daniel: There was a passionate user on meta who thought that mbox was outdated, and should use \mathrm or another math-text form instead.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

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.