TITLE: Decomposition and Inertia groups and Frobenius automorphism computations
QUESTION [5 upvotes]: Let $K = \mathbb{Q}[\sqrt{5}, \sqrt{-1}]$. Calculate the Frobenius automorphisms $\left(\frac{K/\mathbb{Q}}{p}\right)$ for $p$ prime distinct from $2$ and $5$ (which are the only primes that ramify in $K$). Calculate the decomposition and inertia groups for $2$ and $5$.
This is related to an exercise in Milne; it would help me a lot to understand better the concepts if someone is willing to do this particular case. Thanks in advance!

REPLY [3 votes]: Instead of thinking in terms of decomposition group and inertia group, sometimes it's easier to think in terms of the corresponding fields (via the usual Galois correspondence).  For abelian Galois groups (as is the case for $K$ here), it's particularly simple:
The decomposition group corresponds to the largest subextension of $K/\mathbb{Q}$ over which $p$ splits completely.  (This subextension is usually called the decomposition field.)
The inertia group corresponds to the largest subextension of $K/\mathbb{Q}$ over which $p$ is unramified.  (This subextension is usually called the inertia field.)
(Note: In the non-abelian case, it's a little more complicated.  The decomposition and inertia groups are only determined up to conjugacy by the lower prime $p$; to get a specific subgroup, we need to choose a prime $P$ in $K$ lying above $p$.  In the above statements, we would need to change "largest subextension" to "largest subextension such that $P$ lies over the subextension".)
Now we can tackle $K = \mathbb{Q}(\sqrt{5}, \sqrt{-1})$.  Consider first the case $p \ne 2,5$.  Note that $K/\mathbb{Q}$ has 3 proper subextensions: $\mathbb{Q}(\sqrt{-1})$, $\mathbb{Q}(\sqrt{-5})$, $\mathbb{Q}(\sqrt{5})$.  Check that $p$ either splits in all 3 of these subextensions, in which case the decomposition field is all of $K$ (so the decomposition group is trivial), or else $p$ splits in exactly one of the subextensions, in which case the decomposition field is that subextension (the decomposition group is the corresponding subgroup of order 2).  The Frobenius automorphism is a generator of the decomposition group, which is uniquely determined in these cases since the decomposition groups all have order 1 or 2.
I'll leave the ramified cases $p=2,5$ to you for now.