## Elimination of ramification. II: Henselian rationality.(English)Zbl 1443.12005

When one works on valued fields, one is sometimes led to extend isomorphisms (for example, when proving a back-and-forth property). That is, $$(F|K,v)$$ is an extension of valued fields, there is an isomorphism $$f$$ from $$(K,v)$$ onto a valued field $$(K',v')$$, and we need to extend $$f$$ from $$(F,v)$$ to some valued field which contains $$K'$$.
If $$(F|K,v)$$ is an immediate extension (i.e.$$vF=vK$$ and $$Fv=Kv$$, where $$vL$$ and $$vK$$ denote the valuation groups, $$Lv$$ and $$Kv$$ denote the residue fields), then there are cases where one can easily extend this isomorphism. The first one consists in extending to a transcendental element, this can be done by mean of what are called pseudo-Cauchy sequences. The second one concerns the case of algebraic extensions. It consists in extending to the Hensel closure, which is unique, up to isomorphism (we will give the definition later). So one can try to embed $$F$$ in a field of the form $$K(\mathcal{T})^h$$, where $$\mathcal{T}$$ contains a transcendence basis of $$F|K$$ and $$^h$$ denotes the Hensel closure.
The aim of this paper is to prove such a property (which is called Hensel rationality) in the case of a function field over a tame field.
This article has been “in preparation” for a long time, and its results have already been used several times.
If $$(K,v)$$ is a valued field and $$F$$ is a field extension of $$K$$, then there always exist extensions of $$v$$ to $$F$$. The valued field $$(K,v)$$ is said to be henselian if $$v$$ extends in a unique way to every algebraic extension of $$K$$. Every valued field can be embedded in a unique way (up to isomorphism) in a minimal henselian field, which is called its henselization, and is denoted $$(K^h,v)$$. The extension $$(K^h|K,v)$$ is algebraic and immediate. In order to define tame fields, we first define tame extensions. Let $$(F|K,v)$$ be an extension of valued fields. It is called tame if every finite subextension $$(L|K,v)$$ enjoys the following three properties. The integer $$(vL: vK)$$ is not divisible by the characteristic of $$Kv$$, the extension $$Lv|Kv$$ is separable and $$[L:K]=[Lv:Kv](vL:vK)$$. Now, we say that $$(K,v)$$ is a tame field if it is henselian, and every algebraic extension is tame.
In the following, $$(F|K,v)$$ is an immediate extension of valued fields where $$F|K$$ is a function field.
In the first result of this paper, $$(K,v)$$ is assumed to be tame. If the transcendence degree of $$F|K$$ is $$1$$, then there is $$x\in F$$ such that $$F^h=K(x)^h$$. In general, if $$(N|F,v)$$ is immediate and $$(N,v)$$ is a tame field, then there exist $$F_1\subseteq N$$ which is a finite extension of $$F$$, and a transcendence basis $$\mathcal{T}$$ of $$F_1|K$$, such that $$F_1^h=K(\mathcal{T})^h$$.
In the second result $$F|K$$ is separable and $$(K,v)$$ is separably tame, i.e.it is henselian and every separable algebraic extension of $$(K,v)$$ is tame. If the transcendence degree of $$F|K$$ is $$1$$, then there is $$x\in F$$ such that $$F^h=K(x)^h$$. In general, if $$(N|F,v)$$ is separable immediate and $$(N,v)$$ is a separably tame field, then there exist $$F_1\subseteq N$$ which is a finite separable extension of $$F$$, and a transcendence basis $$\mathcal{T}$$ of $$F_1|K$$, such that $$F_1^h=K(\mathcal{T})^h$$.
The author also proves that extensions $$(N,v)$$ of $$(F,v)$$, as in any of the above two theorems, always exist.
In the last theorem, $$(K,v)$$ does not admit any separable algebraic immediate extension, $$vK$$ is archimedean, $$(K^c,v)$$ denotes the topological completion of $$K$$ equipped with the topology induced by the valuation, $$(F|K,v)$$ is an immediate extension where $$F|K$$ is a function field of transcendence degree $$1$$. We assume that there is no embedding of $$(F|K,v)$$ in $$(K^c|K,v)$$. If there exist a tame extension $$(L|K,v)$$ and a transcendence basis $$\mathcal{T}$$ of $$F.L|L$$ such that $$(F.L)^h=L(\mathcal{T})^h$$, then there exists a transcendence basis $$\mathcal{T}'$$ of $$F|K$$ such that $$F^h=K(\mathcal{T}')^h$$.
[For Part I see Trans. Am. Math. Soc. 362, No. 11, 5697–5727 (2010; Zbl 1225.12008). ]

### MSC:

 12J10 Valued fields 13A18 Valuations and their generalizations for commutative rings 14E15 Global theory and resolution of singularities (algebro-geometric aspects)

Zbl 1225.12008
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### References:

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