# Isomorphism between Galois Groups

Let $q$ prime and $n$ natural number coprime to $q$. Futhermore let $\zeta_n$ be the primitive $n$-th root.

I want to know how to prove using Hensel's Lemma that for Galois groups $$\text{Gal}(\mathbb{Q}_q[\zeta_n]/\mathbb{Q}_q) \cong \text{Gal}(\mathbb{F}_q[\zeta_n]/\mathbb{F}_q)$$ holds?

My first idea was to consider the cyclotomic polynomial $\Phi_n$ in $\mathbb{F}_q[X]$. Does it split there?

• So what is $p$? – awllower Feb 1 '18 at 4:06
• The splitting of $\Phi_n$ over $\Bbb Q_q$ should be the same as the splitting of $\Phi_n$ over $\Bbb F_q$. – Lubin Feb 1 '18 at 4:34
• @Lubin: Well, obviously the minimal polynomial $f$ of $\zeta_n$ divides $\Phi_n$. But I don't see why Hensel's Lemma provides the equivalence of these two statements: $f$ is irreducible in $\mathbb{Q}_p[X]$ if and only if $\bar{f}$ is irreducible in $\mathbb{F}_p[X]$ (here $\bar{f}$ is the image of $f$ under canonical reduction map). – Tim Grosskreutz Feb 1 '18 at 19:10

Let $(\mathfrak o,\mathfrak m)$ be a complete (noetherian) local ring with residue field $\mathfrak o/\mathfrak m=k$. Let $f(X)\in\mathfrak o[X]$, with $\bar f=\gamma\eta\in k[X]$, and $\gcd(\gamma,\eta)=1$. Then there are $g,h\in\mathfrak o[X]$ with $\deg(g)=\deg(\gamma)$, $\bar g=\gamma$, $\bar h=\eta$, and $gh=f$.
It’s particularly easy to prove when $\mathfrak o$ is a discrete valuation ring, your situation, but in any case, you prove it by using the standard strategy of improving a factorization modulo $\mathfrak m^r$ to a factorization modulo $\mathfrak m^{r+1}$, using the hypothesis that $\gamma$ and $\eta$ are relatively prime at a crucial point. And you have to pay attention to the degree of the approximations to $g$ at every stage, for otherwise you’ll find yourself defining a power-series factorization. And one other thing: you can get around the degree aspect of the statement and proof by demanding at the outset that $\,f$, $\gamma$, and $\eta$ be monic, and finding monic $g$ and $h$; but one of the most useful applications requires allowing $\,f$ to have highest coefficient in $\mathfrak m$.