Inertia field of a compositum.

My question is regarding composite field extensions. The problem is motivated but not explicitly stated in the chapter 4 exercises of Marcus' Number Fields, in particular ex. 31 c)

We first provide the set-up:

Let $K$ be an extension of $\mathbb{Q}$ st $[K:\mathbb{Q}]=p^m$, where $p$ is prime $\in\mathbb{Z}$. Let $q\neq p\in\mathbb{Z}$ such that $q$ is ramified in $K$. Fix $Q$ to be a prime of $\mathcal{O}_K$ lying over $q$, and let $e$ be the ramification index of $q$ in $Q$. Earlier exercises tell us that $e$ divides $q-1$, and hence, that $\mathbb{Q}(\zeta_q)$ has a unique subfield of degree $e$ over $\mathbb{Q}$. (To see this note that as $q$ is prime, |Gal$(\mathbb{Q}(\zeta_q)/\mathbb{Q})$| = $q-1$, and as this is a cyclic group and $e$ divides $q-1$ we have a unique subfield, which can be determined by considering the multiplicative group of integers modulo $q$). We denote this subfield by $L$, and note that it is clearly a finite extension of $\mathbb{Q}$.

We can now state the problem:

Consider the compositum field $KL$, and let $U$ be a prime of $\mathcal{O}_{KL}$ lying over $Q$. My question is this, when we consider the inertia field of $KL$, where does it lie in the field extensions diagram? i.e. does it necessarily contain $K$? Does it necessarily contain $L$? Or is it a subfield of one or both of these fields?

Let me be clear in saying that the inertia field I am referring to is specifically the subfield of $KL$ fixed by the inertia subgroup of $U$, that is $$\{\sigma \in G\:|\:\sigma(\alpha)\equiv\alpha\pmod{U}\: \forall \alpha\in\mathcal{O}_{KL}\},$$ where $G=\text{Gal}(KL/\mathbb{Q})$.

• I see no reason why it should contain K or L, maybe think about how Frobenius elements permute. – Slime Online Feb 13 '15 at 17:43

Judging by the way the question is phrased (and this is certainly the case in the question in Marcus' textbook to which the OP refers, where $K/\mathbb{Q}$ is abelian) we may assume that $K/\mathbb{Q}$ is Galois. Also assume that $q\geq3$ to avoid trivialities.

Let $(KL)^{I_U}$ denote the inertia field of $U$, where in turn $I_U$ is the inertia group as defined in the question.

Now $q$ is totally ramified in $\mathbb{Q}(\zeta_q)/\mathbb{Q}$, hence in $L/\mathbb{Q}$, and so in particular $I_U$ is non-trivial and $(KL)^{I_U}$ cannot contain $L$ (see for example Ramification in a tower of extensions). Also $q$ is ramified in $K$ by hypothesis, and so once again the action of $I_U$ upon the sub-extension $K$ must be non-trivial.

So $(KL)^{I_U}$ contains neither $K$ nor $L$. However it can definitely be a subfield of one or both of them.

Here is an illustrative (though far from universally representative!) example:

Let $p=5$, $q=11$ and let $C_{p^2q}$ be the cyclotomic field obtained from the 275-th roots of unity. Notice $p|(q-1)$ which is essential here.

Consider the fixed field $K$ of the Sylow-2-subgroup of the Galois group Gal$(C_{p^2q}\mid\mathbb{Q})$. This has degree $p^2=25$: it has Galois group equal to the product of two cyclic groups of order $5$ and is ramified of degree $e=5$ over $q=11$. For completeness we mention it is ramified of degree $5$ over $p=5$ as well, and that the (unique because it is an abelian extension) inertia groups over $p=5$ and $q=11$ are distinct.

$L$ is the fixed field of the cyclotomic field $C_q$ of $q$-th roots of unity under the action of its Sylow-2-subgroup, an extension of $\mathbb{Q}$ of degree $e=5$. By construction in this case $L\subseteq K$ and so $KL=K$. The inertia group $I_U$ therefore is just the inertia group of $K$ at $Q=U$, which from above is a cyclic group of order $5$ isomorphic to Gal$(L/\mathbb{Q})$.

So finally we see that $(KL)^{I_U}$ is the maximal subextension of $K$ which is unramified above $q$, which MAGMA gives as the (totally real) splitting field of $x^5-10x^3-5x^2+10x-1$ over $\mathbb{Q}$, ramified only over $5$. It is clear this contains neither $K$ nor $L$, though it is a subfield of $K$.