0

Let $L/K$ be finite extension of number fields with $O_K \to O_L$ the induced map of rings of integers. Let pick a prime $\frak{p}$ $ \subset O_K$ and a prime $\frak{P}$ $ \subset O_L$ lying over $\frak{p}$, and let complete $L/K$ with respect to valuations associated to these primes, so consider now the associated finite extension $L_w/K_v$ of local fields with valuations $v,w$ where latter extends the former.

We consider the local ramification behavior of $L_w/K_v$. Assume $L_w/K_v$ is ramified, ie $e_{L_w/K_v} >1$.

In this question I was wondering if it is possible to perform an appropriate "base change" $K \to M$ in order to "resolve" the ramification at $K_v$ in sense of that there exist an appropriate extension $M$ linearly independent from $L$ (ie $M \otimes_K L \cong ML$) such that the composition $ML /L/K=ML/M/K$ induces tower of local field extensions $(ML)_u/M_t/K_v= (ML)_u/L_w/K_v$ such that $(ML)_u/M_t$ becomes unramified.

Having Abhyankar's lemma in mind such "resolution" not always exists; it seems that we need to impose additional assumtions on the given ramification; for instance assume that ramification of $L_w/K_v$ is tame.
That was the content of the linked question.

Now this question adresses following point:

In comments below the linked question user1386629 sketched few weeks ago certain argument basing heavily on Chinese reminder theorem (...so far I remember correctly)) to show that one is always able to find infintely many extensions $M/K$ totally disjoint from $L$ (...here I guess he/she meant by "totally disjoint" actually "linealy disjoint", ie satisfying $M \otimes_K L \cong MK$) "resolving" indeed the ramification in $L_w/K_v $ presumably in sense as elaborated above, namely that it give rise to induced composition/"tower" of local extensions $(ML)_u/L_w/K_v= (ML)_u/M_t/K_v$ such that the intermediate extension $M_t/K_v$ "absorbs" fully the ramified part of $L_w/K_v$ in sense of that $M_t/K_v$ contains intermediate extension isomorphic to $L_w/K_v$ and such that $(ML)_u/M_t$ is unramified;
so intuitively that $M_t/K_v$ "absorbs" whole ramifying part from $L_w/K_v$ such "base change" $(ML)_u/M_t$ is unramified.

Unfortunately he/she decided to delete the comments adressing this argument shortly before I managed punctually to profoundly ponder about this argument. Does anybody has an idea how to reconstruct this argument user1386629 was refering to? If I remember correctly it was based substantially on Chinese reminder theorem, but can't reconstruct it any more details. Any ideas what kind of argument was presented there?

Edit (added later): I found this and this discussion also very helpful towards the posed question.

user267839
  • 9,217
  • @Lullaby: Why Galois closure $M$ of $K$ can be expressed as $K[x]/P(x)$? By primitive element theorem this works (over char $0$) is extension $M/K$ is finite, but thats in general not the case for algebr closure. – user267839 Mar 15 '25 at 11:14
  • @Lullaby: right, sorry, I see. But then I still not got two points: Firstly, why in Galois closure $M$ of $L/K$ all ramification of the latter extension get killed? And secondly, what you mean by "auxilary primes" in construction of $N$ ( ...at which $Q(x)$ should be choosen to be Eisenstein)? – user267839 Mar 16 '25 at 15:10
  • (cont) Also, is there an algorithmical/methodical way to pick such $Q(x)$? – user267839 Mar 16 '25 at 15:12
  • @Lullaby: ok thank you, but one little nitpick about the meaning of the statement "All ramification in $L/K$ is killed off in the Galois closure $M$ of $K$" (...I guess you meant there Gal closure of $L/K$): Do you mean by this statement just that $L \otimes_K M / M$ becomes unramified, is it correct? . – user267839 Mar 17 '25 at 11:40
  • ...so in sense of "after base change" as in usual framework of algebraic geometry? – user267839 Mar 17 '25 at 12:38
  • @Lullaby: A little update about one important point: So far I see you nowhere in your argumentation invoke explicitly Chinese remainder theorem to which user1386629 refered in quoted thread. Just to clarify: Do you maybe tacitly make use of it in the construction of $Q(x)$ in order to make it "close" to $P(x)$ at primes dividing the discriminant ideal? Or, where shall CRT else appear in the consideration above? – user267839 Mar 19 '25 at 10:49
  • (sorry for annoying mentioning the CRT, but I would really like to understand where the latter is crucially involved in above construction) – user267839 Mar 19 '25 at 10:55

0 Answers0