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Suppose {$K_i/\mathbb{Q}$} is a finite set of finite galois extensions of $\mathbb{Q}$ with Galois groups $G_i$.

Suppose we know the ramifications of $K_i$ quite well (e.g., their decomposition groups, inertia groups at some primes),

  1. What can we say about the ramifications of the compositum field of $K_i$ (e.g., the ramification index, inertia degree of some primes)? Any References?

  2. Particularly, when $K_1\cap K_2=\mathbb{Q}$, we know that $K_1K_2$ has Galois group $G_1\times G_2$. Is the corresponding decomposition group (resp. inertia group) of the form $D_1\times D_2$ (resp. $I_1\times I_2$)? (This is wrong in general, see Álvaro Lozano-Robledo's answer for a counterexample)

  3. How about the case if we remove the requirement that $K_i/\mathbb{Q}$ are Galois?

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    $\begingroup$ @Alex: however the ramification degree is NOT multiplicative while taking the compositum. You can observe what is going on locally over $\mathbb{Q}_p$, any two distinct totally ramified Galois extensions of degree $p$ generate the maximal abelian elementary $p$-extension, which contains an unramified part. $\endgroup$
    – Maurizio Monge
    Commented Jul 27, 2011 at 7:40
  • $\begingroup$ Maurizio's point is illustrated in the following answer: mathoverflow.net/questions/15666/… $\endgroup$
    – Chandan Singh Dalawat
    Commented Jul 27, 2011 at 10:37
  • $\begingroup$ See Ribenboim, Classical Theory of Algebraic Numbers, p. 263. $\endgroup$
    – Watson
    Commented Jan 5, 2017 at 9:15

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For (2), the answer is no, not in general. Here is a simple example: take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{-5})$. Then both $K_1/\mathbb{Q}$ and $K_2/\mathbb{Q}$ are (totally) ramified at $p=2$ and $K_1\cap K_2=\mathbb{Q}$, but $F=K_1K_2$ is not totally ramified at $2$. In other words, the extension $\mathbb{Q}(\sqrt{-5},i)/\mathbb{Q}(\sqrt{-5})$ is unramified at $2$ (in fact, $F$ is the Hilbert class field of $K_2$, so it is unramified everywhere).

On the other hand, if you take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{2})$, then $F=K_1K_2$ is totally ramified at $2$ over $\mathbb{Q}$. (Here $F=\mathbb{Q}(\zeta_8)$.)

In both cases, $I_1=D_1=G_1$ and $I_2=D_2=G_2$ (in your notation) but in the first case the inertia in the compositum has order $2$ and in the second case it has order $4$. This shows that one needs to know more than the decomposition and inertia subgroups at a prime in each $K_i$ to understand the ramification index in the compositum.

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  • $\begingroup$ This is a purely local phenomenon at the prime $2$. In other words, although the quadratic extensions ${\mathbf Q}_2(\sqrt{-1})$ and ${\mathbf Q}_2(\sqrt{3})$ are both (totally) ramified over ${\mathbf Q}_2$, their compositum ${\mathbf Q}_2(\sqrt{-1},\sqrt3)$ is not totally ramified, for it contains the unramified quadratic extension ${\mathbf Q}_2(\sqrt{5})$. For more examples, see {\it Example} 51 in arxiv.org/abs/0711.3878v2. $\endgroup$
    – Chandan Singh Dalawat
    Commented Jul 28, 2011 at 4:45
  • $\begingroup$ Ribenboim (in Classical Theory of Algebraic Numbers, p. 263) is precisely saying that the decomposition and inertia groups of the compositum are products $D_1 \times D_2$ and $I_1 \times I_2$ resp. How does it match your answer? Related on MSE: math.stackexchange.com/questions/2084878. $\endgroup$
    – Watson
    Commented Jan 6, 2017 at 9:41
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    $\begingroup$ I think I identified a weak link in the argument from Ribenboim. My answer is in the Math.SE thread linked to by Watson. $\endgroup$
    – Jyrki Lahtonen
    Commented Jan 8, 2017 at 9:10

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