3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $\mathbb Q_p$. $\endgroup$ – peter a g Feb 14 '16 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 $n$. $\endgroup$ – Luigi M Feb 14 '16 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? $\endgroup$ – Gregory Grant Feb 14 '16 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 $\mathbb{F}_{p^n}$? $\endgroup$ – Luigi M Feb 14 '16 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 $\mathbb Q_p$). $\endgroup$ – peter a g Feb 14 '16 at 15:57
3
$\begingroup$

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 $x^n-1$.

$\endgroup$
  • $\begingroup$ Right, so a root of $x^{p^n-1}-1$. If it said of order $n$ then it would be a root of $x^n-1$. $\endgroup$ – Gregory Grant Feb 14 '16 at 16:03
  • $\begingroup$ Oops, where did your comment go? I guess you figured it out. $\endgroup$ – Gregory Grant Feb 14 '16 at 16:04
  • 1
    $\begingroup$ 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 $n$. $\endgroup$ – Gregory Grant Feb 14 '16 at 16:18
  • 1
    $\begingroup$ 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. $\endgroup$ – Lubin Feb 15 '16 at 0:16
  • 1
    $\begingroup$ Right the factor of degree $n$ has to be irreducible over $\Bbb Q_p$. $\endgroup$ – Gregory Grant Feb 15 '16 at 0:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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