Recent zbMATH articles in MSC 12J25 https://www.zbmath.org/atom/cc/12J25 2021-04-16T16:22:00+00:00 Werkzeug Eliminating tame ramification generalizations of Abhyankar's lemma. https://www.zbmath.org/1456.12005 2021-04-16T16:22:00+00:00 "Dutta, Arpan" https://www.zbmath.org/authors/?q=ai:dutta.arpan "Kuhlmann, Franz-Viktor" https://www.zbmath.org/authors/?q=ai:kuhlmann.franz-viktor Before stating the theorems which are proved in this paper, we have to recall some properties of valued fields. For every valued field $$(K,v)$$, $$v$$ extends to the algebraic closure $$K^{ac}$$ of $$K$$. Now $$(K,v)$$ is henselian if $$v$$ extends in a unique way to every algebraic extension. Every valued field $$(K,v)$$ embeds in a minimal henselian valued field $$(K^h,v)$$, which is uniquely determined up to isomorphism and is called its Hensel closure. The extension $$(K^h|K,v)$$ is immediate, i.e. $$vK^h=vK$$ and $$K^hv=Kv$$, where $$vK$$ denotes the valuation group of $$(K,v)$$ and $$Kv$$ its residue field. An algebraic extension $$(L|K,v)$$ of henselian valued fields is tame if, for every finite subextension $$(E|K,v)$$, the characteristic of $$Kv$$ doesn't divide $$(vE:vK)$$, $$Ev|Kv$$ is separable and $$[E:K]=(vE:vK)[Ev:Kv]$$. Next, we have the following sequence of valued fields: $$K\subseteq K^h\subseteq K^i\subseteq K^r\subseteq K^s\subseteq K^{ac}$$. Here $$K^s$$ is the separable closure of $$K$$, $$K^r$$ is called its ramification field and $$K^i$$ its inertia field. These fields satisfy the following properties: $$vK^i=vK$$, $$K^iv=K^rv$$, the extensions $$K^iv|Kv$$, $$K^r|K$$ and $$K^i|K$$ are Galois, $$\mathrm{Gal}(K^iv|Kv)\simeq \mathrm{Gal}(K^i|K^h)$$, and if the characteristic of $$Kv$$ is $$p>0$$, then $$\mathrm{Gal}(K^s|K^r)$$ is a pro-$$p$$-group, $$\mathrm{Gal}(K^r|K^i)$$ doesn't contain any element of order $$p$$ and $$K^sv|K^iv$$ is purely inseparable. In the following the extension $$L|K$$ is algebraic, and $$F|K$$ is an arbitrary extension. First, assume that $$L\subseteq K^r$$. The authors prove that $$LF\subseteq F^r$$, $$v(LF)=vL+vF$$. Further, $$LF\subseteq F^i\Leftrightarrow vL\subseteq vF$$. Next, assume that $$L\subseteq K^i$$. Then $$LF\subseteq F^i$$, $$(LF)v=Lv\cdot Fv$$. Furthermore, $$LF\subseteq F^h\Leftrightarrow Lv\subseteq Fv$$. The authors deduce that if the rational rank of $$vK$$ is $$1$$, $$(L\cdot K^h|K^h,v)$$ is tame, $$(vL:vK)$$ and $$(vF:vK)$$ are finite, then $$(v(LF):vL)$$ is the lcm of $$(vL:vK)$$ and $$(vF:vK)$$. In particular, $$v(LF)=vF$$ if, and only if, $$(vL:vK)$$ divides $$(vF:vK)$$ ($$\Leftarrow$$ is an Abhyankar's lemma). At the end of the paper they construct an example which shows that this result fails for rational rank $$2$$. In the next theorem, they assume that $$L|K$$ is a normal extension and that the characteristic of $$Kv$$ is $$p>0$$. The greatest subgroup of $$(vL)$$ which contains $$vK$$ and such that $$p$$ doesn't divide the order of any non zero element modulo $$vK$$ is denoted by $$(vL)_{p'}$$. First they prove that the quotient group $$v(LF)/((vL)_{p'}+vF)$$ is a $$p$$-group, and they deduce that $$v(LF)/vF$$ is a $$p$$-group if, and only if, $$(vL)_{p'}\subseteq vF$$. Next, they assume that $$(vL)_{p'}=vK$$ and they show that the maximal separable subextension of $$(LF)v|(Lv)^sFv$$ is a $$p$$-extension. They give an example where the previous extension is nontrivial, although $$Lv=(Lv)^s=Kv$$. They also gives examples of several situations. We can quote a case where $$K=L^d=L^i\subsetneq L^r=L$$, but $$F=(FL)^h\subsetneq (FL)^i=(FL)^r=FL$$; so $$F=FL^i\subsetneq (FL)^i$$ and $$F\subsetneq L^rF=(FL)^i$$. In another example we have $$K=L^h\subsetneq L^i=L^r=L$$, $$F\subsetneq LF^h=(LF)^i=(LF)^r=LF$$ (so $$F\subsetneq L^iF=(LF)^d=LF$$). They also investigate examples where $$F$$ is a rational function field. Reviewer: Gerard Leloup (Le Mans)