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Let $K$ be a field and $\mu_n$ a primitive $n$-th root of unity. Then I can embed $\operatorname{Gal}(K(\mu_n) / K) \subseteq (\mathbb{Z} / (n) )^*$. For $K=\mathbb{Q}$ there would be even an isomorphism. Is there a condition such that for $K=\mathbb{Z}/(p_1)$ and $n=p_2$, with $p_1,p_2$ different primes there would be an equality $$\operatorname{Gal}(K(\mu_{p_2}) / K) = \{\phi\in (\mathbb{Z} / (p_2))^* | \text{ condition }\} ?$$ Maybe some condition on $p_1$ and $p_2$?

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This question is the same as studying how the $p_2$th cyclotomic polynomial factors mod $p_1$. Once you find the degree of the minimal polynomial of $\mu_{p_2}$ from this you will be able to find the Galois group since $\mathbb{Z}/p_2\mathbb{Z}$ is cyclic. –  fretty Sep 1 '12 at 16:35
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1 Answer 1

Since $p_1 \neq p_2$, $P(x) = x^{p_2}-1 \in \Bbb F_{p_1}[x]$ is separable. Let $L = \Bbb F_{p_1}(\mu_{p_2}) = \Bbb F_{p_1^k}$ for some positive integer $k$. Then $L^\times$ is cyclic of order $p_1^k-1$, and since it contains the ($p_2$ distinct) $p_2$th roots of unity, we must have $p_2 \mid p_1^k-1$. On the other hand, $\mu_{p_2} \in \Bbb F_{p_1^d}$ for any positive $d$ such that $p_2 \mid p_1^d -1$ (Why?). Thus $k$ must be the smallest positive integer such that $p_2 \mid p_1^k - 1$, or equivalently, the smallest $k$ such that $p_1^k \equiv 1 \pmod {p_2}$. Hence $[L \, : \, \Bbb F_{p_1}]$ is exactly the order of $p_1$ mod $p_2$, and it follows that $\mathrm{Gal}(L/\Bbb F_{p_1}) \simeq \langle p_1 \rangle \subseteq (\Bbb Z/p_2 \Bbb Z)^\times$.

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Short version: the Galois group of a finite field extension $L/K$ is generated by the Frobenius automorphism $x \mapsto x^{|K|}$. –  Jack Schmidt Sep 1 '12 at 18:24
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