# Calculate a galois group

I am trying to calculate the galois group $\operatorname{Gal}( \mathbb{Z}_q (\vartheta_p) : \mathbb{Z}_q)$, where $p$ and $q$ are different primes, $\mathbb{Z}_q$ $q$-adic ring, $\vartheta_p$ a primitive $p$-th root of unity, but I just get, that it is embedded in $\mathbb{Z}/(p)$ and no equality. How could one solve this?

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Do you mean "embedded in $(\mathbb{Z}/(p))^\times$?" It may not be an equality because $\mathbb{Z}_q$ contain some roots of unity; in particular, the $(q-1)$st roots of unity. This follows from Hensel's lemma and $\mathbb{Z}_q/(q) \cong\mathbb{Z}/(q)$. One strategy for computing the Galois group of cyclotomic extensions is to 1. observe the embedding you of the Galois group you name, and 2. prove the $p$th cyclotomic polynomial is irreducible; comparing the two observations shows the embedding is surjective. – tghyde May 2 '13 at 22:07
However, I doubt $\Phi_p(x)$ will be irreducible over $\mathbb{Z}_q$, since the factorization of $\Phi_p(x)$ modulo $q$ depends on the decomposition of the prime $q$ in the cyclotomic extension $\mathbb{Q}(\zeta_p)$. I believe Hensel's lemma may be generalized to lift monic polynomial factorizations mod $q$ to $\mathbb{Z}_q$... – tghyde May 2 '13 at 22:32

Sorry for posting my thoughts in the comments instead of just answering the question, but I found the references I was looking for in my notes.

Here is the version of Hensel's lemma I want to use:

Let $f(x)\in \mathbb{Z}_q[x]$ be not divisible by $q$. Suppose $f(x) \equiv \overline{g}(x)\overline{h}(x) \bmod q$ where $(\overline{g}(x), \overline{h}(x)) = 1$. Then we can find lifts $g(x)$ and $h(x)\in \mathbb{Z}_q[x]$ such that $f = gh$ and $\deg(g) = \deg(\overline{g})$.

Let $\Phi_p(x)$ be the $p$th cyclotomic polynomial. By the theory of cyclotomic fields, $\Phi_p$ factors as a product of $(p-1)/f$ monic irreducible polynomials of degree $f$ modulo $q$ where $f$ is the order of $q$ in $(\mathbb{Z}/(p))^\times$. These irreducible polynomials will be pairwise relatively prime in $\mathbb{Z}[x]/q\mathbb{Z}[x]$. Then Hensel implies this factorization lifts to a factorization of $\Phi_p(x)$ in $\mathbb{Z}_q[x]$.

Adjoining $\zeta_p$ to $\mathbb{Q}_q$ will then be a degree $f$ extension with Galois group isomorphic to the Galois group of the local extension $\mathbb{F}_q(\zeta_p)/\mathbb{F}_q$. This Galois group is cyclic, generated by the Frobenius element $x \mapsto x^q$.

Thus, unless I have myself confused, $$\operatorname{Gal}(\mathbb{Q}_q(\zeta_p)/\mathbb{Q}_q) \cong \mathbb{Z}/f\mathbb{Z},$$ generated by $\sigma(x) = x^q$. Where $f$ is the order of $q$ in $(\mathbb{Z}/(p))^\times$.

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