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$L, E\supset K$ are number fields. $L/K$ is normal. And field $M=LE$. Assume $\Omega$ is a prime ideal of M and its intersections with $L, E, K$ are $\mathfrak B,\mathfrak q,\mathfrak p$.

$(1)$ Show that if $\mathfrak B/\mathfrak p$ is unramified then so is $\Omega /\mathfrak q$;

$(2)$ Is that right if $L/K$ is not normal?

$(3)$Show that proposition $(1)$ is still true and consider whether $(2)$ is true if we change all "unramified" into "totally ramified", "inertial" or "splitting completely".

I have studied theory of ramification. So I can understand methods about theory of ramification, Galois theory and commutative algebra. Thanks a lot!

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I'll give an outline of the arguments. Since $L/K$ is a Galois extension, so is $M/E$, and there is a natural map $$\mathrm{Gal}(M/E)\to \mathrm{Gal}(L/K)\\\sigma\mapsto \sigma\vert_L.$$

Crucially, this map is injective, since if $\sigma\vert_L$ is the identity, then $\sigma$ fixes $L$ and $E$ so is the identity on $LE$.

Moreover, this map takes the decomposition group $D_{\Omega/\mathfrak q}$ to $D_{\mathfrak B/\mathfrak p}$ and the inertia group $I_{\Omega/\mathfrak q}$ to $I_{\mathfrak B/\mathfrak p}$. For example, if $\sigma$ fixes $\Omega$, then $\sigma|_L$ fixes $\Omega\cap L=\mathfrak B$.

Since the the splitting properties of $\Omega/\mathfrak p$ are completely determined by these groups, we can deduce that if $\mathfrak B/\mathfrak p$ is unramified or splits completely, then so does $\Omega/\mathfrak q$.

Here are some hints for the other cases:

  1. $\mathbb Q(i,\sqrt{-5})/\mathbb Q(\sqrt{-5})$ is unramified at all primes.
  2. $2$ is unramified in $\mathbb Q(\zeta_3)/\mathbb Q$, and $\mathbb Q(\sqrt[3]2,\zeta_3)=\mathbb Q(\sqrt[3]2)\cdot\mathbb Q(\zeta_3\sqrt[3]2)$.
  3. $3$ is not inert in $\mathbb Q(\sqrt{5},i)/\mathbb Q$. In fact no primes are inert, since the extension is not cyclic.
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  • $\begingroup$ Thanks but why an element in $D_{\mathfrak B/\mathfrak p}$ can be lifted as an element in $D_{\Omega/\mathfrak q} $ $\endgroup$ – Li Li Mar 30 '16 at 13:42
  • $\begingroup$ It can't. The point is that $D_{\Omega/\mathfrak q}\hookrightarrow D_{\mathfrak{B/p}}$ via the map $\sigma\mapsto \sigma|_L$. If the latter group is trivial, then so is the former, but the converse need not be true. $\endgroup$ – Mathmo123 Mar 30 '16 at 15:14

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