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It is the first part of an exercise in the book “Number Fields” written by Marcus. Let $K$ be a sub field of $\mathbb{Q}(w)$ with $w=exp(2i\pi/m).$ Identify $\mathbb{Z}_m^\star$ with the Galois group of $\mathbb{Q}(w)$ over $\mathbb{Q}$ in the usual way, and let $H$ be its subgroup fixing $K$ point wise. For a prime $p\in\mathbb{Z}$ not dividing $m$, let $f$ denote the least positive integer such that $p^f\equiv\pm1\mod{m}.$ Show that $f$ is the inertial degree $f(P|p)$ for any prime $P$ of $K$ lying over $p.$

I know that $f(P|p)$ is the order of the frobenius automorphism $\phi(P|p)=\phi$. Since the Galois group is abelian it depends only on $p.$ I also know that $p$ does not ramify in the bigger field, since it does not divide the discriminant, thus it does not ramify in $K$ too. I think I have to prove that $f$ is the order of $\phi$, but I’m not visualising how to do it, in particular I don’t see how to use the hypothesis on $f.$

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The Galois morphism determined by $w\to w^p$ (so also $w^k\to (w^k)^p$) is a lift of $\text{Frob}_p$, acting on a corresponding residual field, so the Frobenius $\phi=\phi_p$ corresponds to $p\in(\Bbb Z/m)^\times$.

The order $f'$ of $p$ (modulo $m$) in this ring is the minimal $f'$ such that $p^{f'}$ is plus one modulo $m$.

This is also the order of $\phi_p$ in the Galois group of the given cyclotomic extension.

$\square$


Example using sage: Let $m=13$, and $p=5$. Then $p^2=25=26-1=-1$ modulo $m$, and $p^4=1$ modulo $m$.

With above notations, $f=2$, $f'=4$.

The cyclotomic field extension is done with respect to the cyclotomic polynomial $\Phi_{13}$ of order $13$, which splits as follows, when considered over $\Bbb F_p$:

sage: K.<a> = CyclotomicField(13)
sage: a.minpoly()
x^12 + x^11 + x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1
sage: F = GF(5)
sage: R.<x> = PolynomialRing(F)
sage: factor( R( x^12 + x^11 + x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 ) )
(x^4 + x^3 + 4*x^2 + x + 1) * (x^4 + 2*x^3 + x^2 + 2*x + 1) * (x^4 + 3*x^3 + 3*x + 1)

In sage, one can also require the residual field:

sage: K.residue_field( K.ideal(5).factor()[0][0] )
Residue field in abar of Fractional ideal (5, a^4 - 2*a^3 - 2*a + 1)

sage: _.degree()
4

(The last factor in the decomposition, (x^4 + 3*x^3 + 3*x + 1), corresponds to the chosen second generator a^4 - 2*a^3 - 2*a + 1.)

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  • $\begingroup$ So the condition $f\equiv\pm1$ is wrong, isn’t it? To be clear, in the text it’s written “$\overline{p^f}\in H,$ where the bar denotes the congruence class mod $m.$” But isn’t possible $p^f\equiv-1\mod{m}$ writing in this way? $\endgroup$ – MatP May 28 '18 at 19:54

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