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This is a follow up question to this. Say we have Galois extensions $L ,M$ of $\Bbb{Q}$ with $L \cap M = \Bbb{Q}$ and consider the composite $K = LM$. I want to prove that $$\begin{array}{cccccc} f : &D(P|p)& \stackrel{\simeq}{\longrightarrow} &D(P_M|p) \times D(P_L|p)\\ &\sigma& \mapsto &(\sigma|_M,\sigma|_L)& \end{array}$$ where $p$ is a prime in $\Bbb{Z}$, $P$ a prime lying over $p$ in $LM$ and $P_M$, $P_L$ primes that lie over $p$ in $\mathcal{O}_L$ and $\mathcal{O}_M$. We can put everything into a picture:enter image description here

The map $f$ is a group homomorphism and is injective because if $\sigma$ is the identity on $M$ and $L$ then it is the identity on $ML = K$.

My Problem is: Showing surjectivity of $f$.

For suppose we are given a pair $(\sigma,\tau) \in D(P_M|p) \times D(P_L|p)$. Then by considering $(\sigma,\tau)$ as an element of $\text{Gal}(M/\Bbb{Q}) \times \text{Gal}(L/\Bbb{Q})$ we know that there is $\sigma' \in \text{Gal}(K/L)$ and $\tau'\in \text{Gal}(K/M)$ such that $\sigma'|_M = \sigma \hspace{2mm} \text{and} \hspace{2mm} \tau'|_L = \tau.$ Now I want to say that in fact $(\sigma,\tau) = f(\sigma'\tau')$ but the problem is I am not guaranteed that $\sigma'\tau'$ is actually an element on $D(P|p)$. I know $\sigma'$ will have to fix $P$ because it is in $E(P|p) \subseteq D(P|p)$ but what about $\tau'$?

Question: Can we say $\tau$ fixes $P_L$ $\iff$ $\tau'$ fixes $P$? Or phrased differently, does an isomorphism $\text{Gal}(K/M) \cong \text{Gal}(L/\Bbb{Q})$ induce an isomorphism $$D(P|P_M) \cong D(P_L|p)?$$

I have tried to find this result in many places (Neukirch, Milne, Janusz, Marcus) but nowhere do I see it mentioned.

Edit: In the case that $L = \Bbb{Q}(\zeta_n)$ and $M = \Bbb{Q}(\zeta_{p^k})$ and $K = \Bbb{Q}(\zeta_{p^k n})$ then the map $f$ will be an isomorphism by counting orders of the groups involved. The group $D(P|p)$ has order $e(P|p)f(P|p)$ while the direct product has order $e(P|p)f(P|p)$ as well.

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1 Answer 1

This map is not always surjective for example consider $L = \mathbb{Q}(\sqrt{-1})$ and $K = \mathbb{Q}(\sqrt{-7}).$ Then $D_3(LK/\mathbb{Q}) \cong \mathbb{Z}/2$ but $D_3(K/\mathbb{Q}) \times D_3(L/\mathbb{Q}) \cong \mathbb{Z}/2 \times \mathbb{Z}/2.$

To see why this fails note that $\mathbb{Q}_3(\sqrt{-1}) = \mathbb{Q}_3(\sqrt{-7}),$ so the completions of $L$ and $K$ inside of the $P$-adic completion of $LK$ do not have trivial intersection and hence the diamond isomorphism theorem is not applicable.

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Thanks for your answer! Though I haven't studied completions yet so I don't really get your second answer. Do you know if the query in my question (the one in bold) is true in general? –  user38268 Jan 16 '13 at 8:08

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