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 $G=\text{Gal}(KL/\mathbb{Q})$.

  • 2
    $\begingroup$ I see no reason why it should contain K or L, maybe think about how Frobenius elements permute. $\endgroup$ – 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 $K$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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