## The uniqueness of the power series representation of certain fields with valuations.(English)Zbl 0019.04901

Let $$K=K^{(m)}$$ be a field on which a discrete valuation $$V$$ of rank $$m$$ is defined. The valuation $$V$$ determines a series of $$m$$ homomorphisms $$H^{(i)}$$ $$(i = 1,\ldots, m)$$ given as $H^{(m)}K^{(m)} = \{K^{(m-1)}, \infty\},\quad H^{(m-1)}K^{(m-1)}= \{K^{(m-1)}, \infty\}, \ldots , H^{(1)}K^{(1)}=\{\mathfrak K, \infty\},$
where $$\mathfrak K$$ denotes the field of residue classes of $$K$$ with respect to $$V$$. Each homomorphism $$H^{(i)}$$ determines a discrete valuation of rank 1 on $$K^{(i)}$$. A field $$K$$ is called “step perfect” with respect to $$V$$ if each field $$K^{(i)}$$ is perfect with respect to the valuation $$H^{(i)}$$. It is shown that each field $$K$$ of the type described possesses an immediate extension $$L$$ which is step perfect, i. e. the value group of $$L$$ coincides with the value group of $$K$$ and the respective fields of residue classes coincide, too. Let $$\chi^{(i)}$$ be the characteristic of $$K^{(i)}$$. Then there exists a unique step perfect extension $$\overline K$$ of $$K$$ (to within analytic isomorphisms) provided that $$\chi^{(m)} = \ldots = \chi^{(1)}= 0$$.
The proof of this theorem essentially depends an the following lemma: “If $$HK = \{\mathfrak K, \infty\}$$ is a discrete perfect homomorphism of rank 1, if $$N\subset K$$ is a subfield such that $$H(N^*)\subset \mathfrak K^*$$, and if $$\mathfrak K$$ is separable over $$H(N)$$ then there exists a subfield $$M$$ of $$K$$ with $$N\subset M$$ and $$H(M^*)= \mathfrak K^*$$. ($$N^*$$, $$\mathfrak K^*$$ denote the multiplicative groups of the fields $$N$$, $$\mathfrak K$$ respectively.)
Finally, the author shows that in general the uniqueness theorem breaks down if the condition concerning the characteristics does not hold, i. e. if $$\chi^{(i)}\neq 0$$.

### MSC:

 12J20 General valuation theory for fields

### Keywords:

discrete valuation; step perfect field
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