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Let $L$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ with uniformizer $\pi$. Let us consider the polynomial ring $L[x_1,x_2,\dotsc,x_l]$ in $l$-variables and $f_1, f_2, \dotsc, f_m \in L[x_1, \dotsc, x_l]$.

Let $S$ be the set of simultaneous zeros of the following system \begin{align} f_1(x_1,\dotsc,x_l) & {}=0 \\ \vdots & \\ f_m(x_1, \dotsc,x_l) & {}=0. \end{align} Then, clearly every element in $S$ is a point in an affine $l$-space over some field extension of $L$. So the coordinates of each points in $S$ generate a field extension. i.e., consider the field extension $L(S)$ obtained by adjoining the coordinates of each solution to $S$. So it looks like $L(S)$ is a subfield of $\bar L$.

Questions:

  1. Is (or when is) the extension $L(S)/L$ Galois?

  2. When is $L(S)/L$ totally ramified?


My Effort:

  1. $L(S)$ is an algebraic extension of $L$ because its elements are algebraic over $L$. In this multivariable case, we don't need separability of the roots/solutions because we are taking the coordinates only. Also two solutions $(x_1, \dotsc,x_l)$ and $(x_1',\dotsc, x_l')$ may have some common coordinates, say, $x_i=x_i'$ but this doesn't affect because both give the same extension, so we will take just one of the coordinates.

Let $L'$ be the Galois closure of $L(S)$ over $L$ i.e, the smallest field containing $L(S)$ that is Galois over $L$. Take any $\sigma \in \operatorname{Gal}(L'/L)$ and a solution $s \in S$, then its Galois conjugate $\sigma(s)$ is also a solution i.e., $\sigma(s) \in S$. So $\sigma(s)=t$ for some $t \in S$. So by restricting the domain of $\sigma$ to $L(S)$, we have $$\sigma(L(S)) \subset L(S).$$ Next, since $\sigma$ induces a permutation on $S$, then for each $\alpha \in L(S)$, $\sigma^{-1}(\alpha)=\beta$ for some $\beta \in L(S)$. So $\alpha=\sigma(\beta)$ and so we have $$L(S) \subset \sigma(L(S)).$$

Therefore, $\sigma(L(S))=L(S)$. Thus each $\sigma\rvert_{L(S)}$ is an automorphism. So it seems that $L(S)$ is Galois over $L$.

Am I correct? Am I missing something?

Any discussion please.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Stefan Kohl
    Commented Aug 20, 2021 at 19:13
  • $\begingroup$ The work you do seems mostly fine, aside from some strange terminology (you seem to say that $L(S)$ is a proper subfield of $\bar L$ to mean just that it is a subfield—in fact there's some argument in the chat that it's often all of $\bar L$; and you say that a permutation acts transitively on a set to mean just that it is, well, a permutation of that set), but this does not seem to be a research-level question. $\endgroup$
    – LSpice
    Commented Aug 21, 2021 at 16:49
  • $\begingroup$ @LSpice, thanks, I have excluded the word proper. But in Chat section, as you said, $L(S)=\bar L$ was in previous setting, slightly different. I am not sure whether in the current setting $L(S)$ is equal to $\bar L$ or not. Second, regarding the level of the question, I want to define a map $Gal(L(S)/L) \to (\mathcal{O}_K/\pi^n \mathcal{O}_K)^{\times}, \ n \geq 1$, once I know $Gal(L(S)/L)$ is Galois. Any help here ? $\endgroup$
    – MAS
    Commented Aug 21, 2021 at 16:58
  • $\begingroup$ Well, it depends on what you want; your current question doesn't mention any such map. Of course there is a trivial map $\operatorname{Gal}(L(S)/L) \to (\mathcal O_K/\pi^n\mathcal O_K)^\times$, so probably some conditions need to be imposed. \\ You have also kept the language "$\operatorname{Gal}(L'/L)$ acts transitively on $S$" where you seem just to mean "$\operatorname{Gal}(L'/L)$ acts on $S$". $\endgroup$
    – LSpice
    Commented Aug 21, 2021 at 17:34
  • $\begingroup$ @LSpice, If I want to define a group homomorphism $f:~Gal(L(S)/L) \to (\mathcal{O}_K/\pi^n\mathcal{O}_K)^{\times}$ which becomes injective. I am not sure how to do it. But I think we need to use the action of $Gal(L(S)/L)$ on $S$ to define the homomorphism. Assume that the zero set $S$ has $\mathcal{O}_L$-module structure. Now, set $f(\sigma)=\theta$ such that $\sigma(s)=\theta * s$ for $s \in S$, $\theta \in \mathcal{O}_L$. Here $*$ is the action of $\mathcal{O}_L$ on the zero set $S$. Does this make $f$ a group homomorphism ? $\endgroup$
    – MAS
    Commented Aug 21, 2021 at 17:51

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