Let $L/K$ be a (infinite) Galois extension, $A \subset K$ a Dedekind domain with field of fractions $K$ and $B$ its integral closure in $L$. Then for every nonzero prime ideal $\mathfrak{P}$ of $B$ and $\mathfrak{p}$ of $A$, there is a group homomorphism $D(\mathfrak{P}|\mathfrak{p}) \to {\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p}))$. I read somewhere that this homomorphism is surjective, and I am trying to deduce this from the case where $L/K$ is finite (which is a standard fact from algebraic number theory). I figured the easiest way to do this deduction would be to write (using the usual arguments)

$$D(\mathfrak{P} | \mathfrak{p}) \cong \varprojlim_{F \in \mathcal{F}} D_{F/K}(\mathfrak{P} \cap F | \mathfrak{p})$$ and $${\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p})) \cong \varprojlim_{F \in \mathcal{F}} {\rm Aut}(\kappa(\mathfrak{P} \cap F)/\kappa(\mathfrak{p}))$$ where $\mathcal{F}$ is the collection of intermediate fields $K \subset F \subset L$ such that $F/K$ is finite and Galois. So the homomorphism we want to prove is surjective is given in the $F$-coordinate by the one in the finite case, $$D_{F/K}(\mathfrak{P} \cap F | \mathfrak{p}) \to {\rm Aut}(\kappa(\mathfrak{P} \cap F)/\kappa(\mathfrak{p}))$$ which we know is surjective. But this doesn't seem like enough to deduce the result I want. In particular, pulling back each coordinate under this map will not necessarily yield a valid element of the inverse limit. Is there some way to fix this issue? In particular, I think it should be possible to take $\mathcal{F}$ to be a smaller collection of finite Galois subextensions with the property that for each $F_1, F_2 \in \mathcal{F}$, either $F_1 \subset F_2$ or the other way around. Then, because $I_{F_2/K}(\mathfrak{P} \cap F_2|\mathfrak{p})$ surjects onto $I_{F_1/K}(\mathfrak{P} \cap F_1|\mathfrak{p})$ under the restriction map, we know that if we are given $\sigma_2 \in D_{F_2/K}(\mathfrak{P} \cap F_2|\mathfrak{p})$ that reduces mod $\mathfrak{P} \cap F_2$ to a particular $\tau_2$ which restricts to $\tau_1$ on $\kappa(\mathfrak{P} \cap F_1)$, and $\sigma_1 \in D_{F_1/K}(\mathfrak{P} \cap F_1 | \mathfrak{p})$ reduces mod $\mathfrak{P} \cap F_1$ to $\tau_1$, then we can modify $\sigma_2$ by composing with the appropriate element of $I_{F_2/K}(\mathfrak{P} \cap F_2 | \mathfrak{p})$ so that $\sigma_2$ restricts to $\sigma_1$. The problem is, we can only do this one preimage at a time, and the collection $\mathcal{F}$ is not necessarily countable. Is there some kind of induction/Zorn's lemma argument to fix this? Also, is there a simpler or more direct way to prove this fact?


Pass from $K$ to the fixed field of the decomposition group $L^{D(\mathfrak{P}|\mathfrak{p})}$ so that we may assume $\rm{Gal}(L/K)=D(\mathfrak{P}|\mathfrak{p})$.

Let $I(\mathfrak{P}|\mathfrak p)=\ker(D(\mathfrak{P}|\mathfrak p) \to {\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p}))$ be the inertia group and let $L^{I(\mathfrak P|\mathfrak{p})}$ be the inertia field. Then $L':=L^{I(\mathfrak{P}|\mathfrak{p})}$ is Galois over $K$ (as we assumed that $\rm{Gal}(L/K)=D(\mathfrak{P}|\mathfrak{p})$) and if we let $\mathfrak{P}'=\mathfrak{P} \cap L'$, then we have $\kappa(\mathfrak{P}')=\kappa(\mathfrak{P})$ and the homomorphism $\rm{Gal}(L/K) \to {\rm Aut}(\kappa(\mathfrak{P})/\kappa(\mathfrak{p})) $ factors over $\rm{Gal}(L/K) \to \rm{Gal}(L'/K)$ which is surjective. But now $\mathfrak{p}$ is unramified in $L'/K$ which means that we can reduce to the unramified case. But in the unramified case, the map $\rm{Gal}(L'/K) \to {\rm Aut}(\kappa(\mathfrak{P}')/\kappa(\mathfrak{p}))$ is actually an isomorphism, which can be easily concluded from the finite degree case, without the problems that arise if you just try to show surjectivity directly with a limit argument.

  • $\begingroup$ Why are we allowed to pass to the fixed field of the decomposition group? In particular, since the extension of residue fields could be infinite and not separable, we can't do the usual degree calculation to show that $\kappa(\mathfrak{P} \cap L^D) = \kappa(\mathfrak{p})$. $\endgroup$ – babu_babu Jan 17 at 16:02
  • $\begingroup$ Similarly, in the infinite case when we also make no separability assumption on the residue fields, how do we know having full inertia group implies you have trivial residue field extension? $\endgroup$ – babu_babu Jan 17 at 16:08

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.