8
$\begingroup$

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?

$\endgroup$
3
  • 4
    $\begingroup$ @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. $\endgroup$ Jul 27, 2011 at 7:40
  • $\begingroup$ Maurizio's point is illustrated in the following answer: mathoverflow.net/questions/15666/… $\endgroup$ Jul 27, 2011 at 10:37
  • $\begingroup$ See Ribenboim, Classical Theory of Algebraic Numbers, p. 263. $\endgroup$
    – Watson
    Jan 5, 2017 at 9:15

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ 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
    Jan 6, 2017 at 9:41
  • 1
    $\begingroup$ I think I identified a weak link in the argument from Ribenboim. My answer is in the Math.SE thread linked to by Watson. $\endgroup$ Jan 8, 2017 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.