7
$\begingroup$

I am a beginner student of Algebraic Number Theory and I am starting to learn ramification theory (of global fields). My question asks for motivation for a definition I was given.

Let $K$ be an algebraic number field, $\mathcal{O}_{K}$ its ring of integers, $L/K$ a Galois extension and $\mathcal{O}_{L}$ the integral closure of $\mathcal{O}_{K}$ in $L$.

I know that the group $G=Gal(L/K)$ acts transitively on the set of prime ideals $\mathfrak{P}_{i}$ of $\mathcal{O}_{L}$ above a prime $\mathfrak{p}$ of $\mathcal{O}_{K}$ and it's just a natural thing to consider the decomposition group (of one of these ideals) $G^{Z}(\mathfrak{P})=\{\sigma\in G\:|\:\sigma(\mathfrak{P})=\mathfrak{P}\}$, which is the stabilizer of $\mathfrak{P}$ under this action.

Now, in the paper I am following, together with the decomposition group, it was defined the group \begin{equation} G^{T}(\mathfrak{P})=\{\sigma\in G\:|\:\sigma(\alpha)\equiv\alpha\mod \mathfrak{P}\:\:\forall\alpha\in\mathcal{O}_{L}\}, \end{equation}

and this one I want to understand better.

I was told that each element of $G^{Z}(\mathfrak{P})$ induces an automorphism in the quotient $\mathcal{O}_{L}/\mathfrak{P}$, which is pretty reasonable. This $G^{T}(\mathfrak{P})$ looks like the subgroup of elements of $G^{Z}(\mathfrak{P})$ that induce the identity in the quotient $\mathcal{O}_{L}/\mathfrak{P}$. From my spying on other books and papers, i recognize this group as the so called $\textbf{Inertia group}$.

My question is basically:

What does the Inertia group tells us? When we look at the index $(G:G^{Z}(\mathfrak{P}))$, it gives us a notion of "how many primes did $\mathfrak{p}$ split into in $\mathcal{O}_{L}$". What about the inertia group? What is its meaning? And it is something as natural as considering the stabilizer of a group action?

$\endgroup$
  • $\begingroup$ Are you familiar with the inertial degree $f$? If $\mathfrak{q}$ in $\mathcal{O}_L$ lies over $\mathfrak{p}$ in $\mathcal{O}_K$, then we define $f(\mathfrak{q}| \mathfrak{p}) = (\mathcal{O}_L/\mathfrak{q} : \mathcal{O}_K/ \mathfrak{p})$. $\endgroup$ – Marcus M Sep 10 '15 at 16:52
  • $\begingroup$ Yes. And i know about the fundamental identity. But I don't know exactly how is the inertial degree related to the inertia group, even though I can imagine it being the galois group of the extension of residue fields or something like this. $\endgroup$ – Shoutre Sep 10 '15 at 17:08
  • $\begingroup$ Okay, I'll write up something quick that'll explain it a bit. $\endgroup$ – Marcus M Sep 10 '15 at 17:31
10
$\begingroup$

Let $\mathfrak{q}$ be a prime of $\mathcal{O}_L$ lying over $\mathfrak{p}$, a prime in $\mathcal{O}_K$.

Let $E$ be the inertial group, i.e. $$ E = \{\sigma \in G : \sigma (\alpha) \equiv \alpha \mod \mathfrak{q} \}.$$

Let $D$ be the decomposition group, i.e. $$ D = \{\sigma \in G : \sigma(\mathfrak{q}) = \mathfrak{q} \}. $$

Moreover, let $L_E$ and $L_D$ denote the fixed fields of $E$ and $D$ respectively, and let $\mathfrak{q}_E = \mathfrak{q} \cap L_E$ and define $\mathfrak{q}_D$ in a similar manner. It'll turn out that you have the following: \begin{align*} (L:L_E) &= e\\ (L_E:L_D) &= f\\ (L_D:K) &= r \end{align*} where $r$ is the number of primes that lie over $\mathfrak{p}$.

Moreover, you'll have that $\mathfrak{q}$ is totally ramified over $\mathfrak{q}_E$, i.e. $e( \mathfrak{q} | \mathfrak{q}_E ) = e$ and $f(\mathfrak{q} | \mathfrak{q}_E ) = 1$.

We also have $e( \mathfrak{q}_E | \mathfrak{q}_F ) = 1$ and $f(\mathfrak{q}_E | \mathfrak{q}_F ) = f$.

Finally, we have $e( \mathfrak{q}_F | \mathfrak{p} ) = 1$ and $f(\mathfrak{q}_F | \mathfrak{p} ) = 1$.

Now, let $K'$ be an arbitrary intermediate field, i.e. $K \leq K' \leq L$ and let $\mathfrak{p}' = \mathfrak{q} \cap K'$. The following is taken from Daniel Marcus' Number Fields: we then have

  • $L_D$ is the largest intermediate field $K'$ s.t. $e(\mathfrak{p}' | \mathfrak{p}) = f(\mathfrak{p}' | \mathfrak{p}) = 1$.

  • $L_D$ is the smallest $K'$ s.t. $\mathfrak{q}$ is the only prime lying over $\mathfrak{p}'$.

  • $L_E$ is the largest $K'$ s.t. $e(\mathfrak{p}' | \mathfrak{p}) = 1$.

  • $L_E$ is the smallest $K'$ s.t. $\mathfrak{q}$ is totally ramified over $\mathfrak{p}'$.

In summary, $E$ and $D$ allow us to decompose the splitting into subfields where each step ($r,e,$ and $f$) occurs in a separate place

$\endgroup$

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.