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Let $L = \Bbb{Q}(\zeta_m)$ where we write $m = p^k n$ with $(p,n) = 1$. Let $p$ be a prime of $\Bbb{Z}$ and $P$ any prime of $\mathcal{O}_L$ lying over $p$.

Notation: We write $I = I(P|p)$ to denote the inertia group and $D = D(P|p)$ the decomposition group.

Now we have a tower of fields

$$\begin{array}{c} L \\ |\\ L^E \\| \\ L^D \\| \\ \Bbb{Q}\end{array}$$

and it is clear that $L^E = \Bbb{Q}(\zeta_n)$ so that $E \cong \Bbb{Z}/p^k\Bbb{Z}^\times$.

My question is: I want to identify the decomposition group $D$. So this got me thinking: Do we have a decomposition into direct products $$D \cong E \times D/E?$$ This would be very convenient because $D/E$ is already known to be finite cyclic of order $f$ while $E$ I have already stated above. Note it is not necessarily given that $(e,f) = 1$ so we can't invoke Schur - Zassenhaus or anything like that.

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1 Answer 1

up vote 4 down vote accepted

You can write your $L$ as the compositum of $\mathbb{Q}(\zeta_{p^k})$ and $\mathbb{Q}(\zeta_{n})$. Since $(p,n)=1$, the two are disjoint over $\mathbb{Q}$, and so the Galois group of $L$ is isomorphic to the direct product of the two Galois groups, one of which is $E$. Let's call the other subgroup $H$. Now, every element $g$ of $G$ is uniquely a product of an element $\epsilon$ of $E$ and an element $h$ of $H$. $E$ is contained in $D$, so $\epsilon h$ fixes $P$ if and only if $h$ does. In other words, the decomposition group $D$ is generated by $E$ and the decomposition group of $P$ in ${\rm Gal}(\mathbb{Q}(\zeta_{p^kn})/\mathbb{Q}(\zeta_{p^k}))=H$, so is indeed a direct product.

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Dear Alex, thanks for your answer. I get everything in your answer except the last line where it seems to me that you are claiming the decomposition group of the product is the product of the decomposition groups. Can you elaborate a little more on that? Thanks. –  user38268 Jan 15 '13 at 13:56
    
Dear Benjamin, I have edited in a brief explanation. Let me know if it's still unclear. –  Alex B. Jan 15 '13 at 14:12
    
Dear Alex, upon reading your answer again I think the crux of the matter is this: Why can we write every element $g \in G$ uniquely as a product of an element $\epsilon \in E$ and $h \in H$? Candidates for $\epsilon$ and $h$ to me seem like $g$ restricted to each of the subfields, but it doesn't seem to make sense to me... –  user38268 Jan 15 '13 at 16:33
    
Dear Benjamin, it is a general theorem of Galois theory that if you have two Galois extensions $F/K$, $M/K$ whose intersection is $K$, then the Galois group of the compositum $FM$ is isomorphic to the direct sum of the two Galois groups. Does that address your question? –  Alex B. Jan 15 '13 at 18:01
    
I'm afraid I don't quite understand your confusion. Do you agree that I have shown that $D$ is the internal direct product of E and the decomposition subgroup of $P$ in $H$? If you do, then how does that not answer your question? –  Alex B. Jan 16 '13 at 10:24

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