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I'm studying the construction of unramified extensions, and many references say that it's enough to attach the $p^n-1$ primitive root of unity to $\mathbb{Q}_p$ in order to obtain the unique degree $n$ unramified extension of the $p$-adics.

What I don't understand, is why the minimal polynomial of it, so a cyclotomic polynomial, has degree $n$. I'd say that the cyclotomic polynomial has degree $\phi(p^n-1)$, but $\phi(p^n-1)\neq n$ in general.

So where I'm overlooking something?

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  • $\begingroup$ Well, take $n=1$... In that case, the residue field $\mathbb F_p$ contains all the roots of $x^{p-1}-1 = 0$; hence the polynomial factors completely in $\mathbb Q_p$. $\endgroup$ Commented Feb 14, 2016 at 15:49
  • $\begingroup$ Sorry but I really don't get the meaning of both comments. What is a primitive root of $n$? If you mean order $n$ I'm aware I'm attaching a primitive root of order $p^n-1$ and not $n$. $\endgroup$ Commented Feb 14, 2016 at 15:52
  • $\begingroup$ Sorry I meant it doesn't say to add a primitive $n$th root of one, it say specifically to add "a primitive root of order $p^n-1$". So I am guessing a factor of $x^{p^n-1}-1$ that has degree $n$. Does that make more sense now? $\endgroup$ Commented Feb 14, 2016 at 15:54
  • $\begingroup$ and I guess this is different from what I intend. So a primitive root of order $p^n-1$ what is supposed to be? a Generator of the multiplicative group of $\mathbb{F}_{p^n}$? $\endgroup$ Commented Feb 14, 2016 at 15:57
  • $\begingroup$ The finite field $k=\mathbb F_{p^n}$ is the splitting field of $x^{p^n-1} -1$ - it is the extension of degree $n$ over $\mathbb F_p$ (and the finite field $k$ is the residue field of the corresponding unramified extension of $\mathbb Q_p$). $\endgroup$ Commented Feb 14, 2016 at 15:57

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The document says to add a "primitive root of order $p^n-1$" not a "primitive $n$-th root of one". So it must be an irreducible factor of $x^{p^n-1}-1$ that has degree $n$, but not $x^n-1$.

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  • $\begingroup$ Right, so a root of $x^{p^n-1}-1$. If it said of order $n$ then it would be a root of $x^n-1$. $\endgroup$ Commented Feb 14, 2016 at 16:03
  • $\begingroup$ Oops, where did your comment go? I guess you figured it out. $\endgroup$ Commented Feb 14, 2016 at 16:04
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    $\begingroup$ That's right, $\xi$ is the root of an irreducible factor of $x^{p^n-1}-1$ and that irreducible factor has to have degree $n$. $\endgroup$ Commented Feb 14, 2016 at 16:18
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    $\begingroup$ You seem to confusing irreducibility over $\Bbb Q$ (as applies, for instance to the cyclotomic polynomials) with irreducibility over $\Bbb Q_p$. For instance, $\Phi_4(X)=X^2+1$ is $\Bbb Q$-irreducible but factors into linears as a $\Bbb Q_5$-polynomial. $\endgroup$ Commented Feb 15, 2016 at 0:16
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    $\begingroup$ Right the factor of degree $n$ has to be irreducible over $\Bbb Q_p$. $\endgroup$ Commented Feb 15, 2016 at 0:19

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