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Let $L/K$ be finite extension of number fields, $O_K \to O_L$ of as schemes $\operatorname{Spec}(O_L) \to \operatorname{Spec}(O_L)$ the induced map of rings of integers. For $\frak{p} \ $$ \subset O_K$ prime ideal of $K$ one can talk about ramification behavior of this prime. Assume $\frak{p}$ is ramified, ie there exist a $\frak{P}_i$ over $\frak{p}$ such that in $O_{L_{\frak{P}_i}}$ the $\frak{p}O_{L_{\frak{P}_i}}$ decomposes as $(\frak{P}_i)^{e_i}$ with $e_i >1$.

Question: To which amount it is possible to "resolve" this ramification at $\frak{P}_i/\frak{p}$ by suitable "base change along another extension $M/K$" in following sense:

Namely, resolve in sense of to find appropiate extension $M/K$ such that for all (or at least one) $\frak{q} \ $ $\subset O_M$ lying over $\frak{p}$ and $\frak{Q}_i \ $ $\subset O_{M \otimes_K L}$ lying over $\frak{P}_i$ the extension $\frak{Q}_i/\frak{q}$ is unramified? (Note, in general $M \otimes_K L$ is not always a field, but by base change finite over $M$, so talking about ramification still makes sense.

Edited later: If that's not too restrictive, I want additionally require that such $M/K$ has additional property that $M \otimes_K L$ is a field. At all, that's not too restrictive: this can be eg trivially archived if $[M:K]$ coprime to $[L:K]$.

Is there something known for general ramification? The hope was that for "suitable" base change a prime over the ramified prime turns to be unramified, or say "less" ramified. Eg, from wild ramification to tame ramification.

Also, the fact that this is possible for tame ramified extensions, suggests that there is possibly some interesting "geometry" behind? Is there a "geometric picture" one should keep in mind thinking of tame ramifications contrasting them from say "wild ramified primes"? Clearly, these are rather precisley defined, but I'm struggling with developing a geometric intuition on them.

user267839
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  • The notion of "base change" might need to be refined in this setting, since I could take $M = L = ML$, in which case the conclusion is trivial. Maybe you want a criterion on a field extension $M/K$ to guarantee your conclusion, which is really the content of Abhyankar's lemma? – CJ Dowd Dec 16 '24 at 20:12
  • @CJDowd: Yes right, sorry for confusion. – user267839 Dec 16 '24 at 20:19
  • @CJDowd: I forgot that for finite extensions $M \otimes_K L$ is usually non isomorphic to $ML$. So everywhere where I wrote $ML$ I actaully meant $M \otimes_K L$, so "usual" base change, sorry for it. Note this is not always a field, but still finite over $M$, so one can still talk about ramification... – user267839 Dec 16 '24 at 20:31
  • @user1386629: Thank you! Two questions: Firstly, when you write that one can find infinitely many Gal extensns $M/K$ totally disjoint from $L$ so that "the localizations $M_u/K_v$ include corresponding extensions for $L$" by "include" you mean just that in sense of that $L_w/K_v$ is naturally nested as intermediate ext, ie forms tower $M_u/L_w/K_v$ of local fld, is that what you mean there? – user267839 Dec 17 '24 at 15:33
  • @user1386629: Secondly, you mentioned the etale coho grps $H^*(X/\overline{K},\mathbf{Q}_p)$ with Galois action. Could briefly sketch how the Galois action is declared (it seems that there are several ways to declare one). Firstly, I assume you mean $\bar{X}:= X \times_K \bar{K}$. Is the Galois action here induced from Galois action on 2nd factor, ie if $a \in Gal(K)$, then it acts on $\bar{X}$ via $id_X \times a$ and then induces by functoriality of coho action on the coho groups? Is this the Galois action you are refering to there? – user267839 Dec 17 '24 at 15:33
  • @user1386629: Or do mean the action induced by "relative Frobenius"? Recall when one considers Frobenii for schemes one can factor absolute Frobenius into relative Frob (...which fixes base field) and "coefficient Frob" which acts so to say "on scalars". Which one do you use here to induce action of the coho grps? – user267839 Dec 17 '24 at 16:04
  • @user1386629: One nitpick about the idea that one can find by means of Chinese Reminder thm finite infinitely many $M/K$ totally disjoint from $L$ (=linealy disjoint? ie such that $M \otimes L\cong ML$) so that the localizations $M_u/K_v$ include such an intermediate extensn isomorphic to $L_w/K_v$. The first part is easy, just take any $M/K$ with gegree coprime to degree of $L/K$. The other seems a bit more subtle. Could maybe sketch the idea how the Chinese remainder thm is going to be applied here to get this? My naive idea – user267839 Dec 17 '24 at 17:04
  • that we can twist $K_v \to M_u$ by $L_w$, then the tensor prod $M_u \otimes L_w$ factors (CRT determines the factors explicitly of we known the minimal pol of est $M/K$) and naturally contains $L_w, M_u$. How CRT helps to to find the desired $M_u$ including $L_w$? – user267839 Dec 17 '24 at 17:04

1 Answers1

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(Note: this may not be what you want, but it answers your question.)

Let $M$ be the normal closure of $L/K$. By the primitive element theorem, we have $L\cong K[x]/(f(x))$ for some irreducible and separable polynomial $f \in K[x]$. Thus

$$M \otimes_K L \cong M[x]/(f(x)) \cong M^n$$

Where we used the fact that $f$ splits over $M$ into distinct linear factors together with the Chinese remainder theorem.

The integral closure of $\mathcal O_M$ inside $M^n$ is just $(\mathcal O_M)^n$, this is an étale $\mathcal O_M$-algebra, it is unramified everywhere.

Note that we have not used any assumption about $L/K$ (other than being a finite extension of number fields.)

If you're interested in extending Abhyankar's lemma to wildly ramified extensions in a more non-trivial way, you should look at the paper Eliminating Wild Ramification by Epp.

Lukas Heger
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  • That's a nice argument if we not insist on that $M \otimes_K L$ should be a field, as in the construction in your answer $M \otimes_K L$ is never a field. Do you maybe know how to proceed if we want to construct an extension $M/K$ such that at every prime $\frak{Q}_i$ of $O_M$ over a fixed prime $\frak{p}$ of $O_K$ the extension $M \otimes_K L / M$ is unramified but if(!) we additionally insist on $M \otimes_K L$ to be a field? – user267839 Jan 06 '25 at 16:36
  • We can of course still use primitive element theorem, so the question seemingly depends on finding irreducible $f \in K[X]$ with desired splitting behavour at $\frak{p}$. It seems that there also Chinese remainder shall help, but not see how precisley. Do you have an idea? – user267839 Jan 06 '25 at 16:43