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Let $K=F_q((X))$, from Artin-Schreier theory, for a cyclic extension $L/K$ of degree p, we have $L=K(x)$, $x^p-x=\alpha$. So when $L/K$ is totally ramified, could we find some $x$ s.t the corresponding $\alpha$ has valuation -1? or $\alpha$ has some bounded valuation independent of $L$? Thanks!

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    $\begingroup$ No. Take $\alpha = X^{-n}, (p,n)=1$. $\endgroup$ Dec 25, 2011 at 17:27
  • $\begingroup$ You need to use the quotient field $K[x^{-1}]$ of the ring $K$. Try for example $\alpha=1/x$. $\endgroup$ Dec 25, 2011 at 18:53
  • $\begingroup$ @Jyrki: Huh? $K$ is a field. Maybe you are confused by the notation. $\endgroup$ Dec 25, 2011 at 19:58
  • $\begingroup$ Sorry, I shouldn't have written so many $X$... $\endgroup$
    – user20167
    Dec 26, 2011 at 13:55

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Artin-Schreier theory gives a bijection between the set of all degree-$p$ cyclic extension of a field $K$ of prime characteristic $p$ and the set of all $\mathbf{F}_p$-lines in the $\mathbf{F}_p$-space $K^+/\wp(K^+)$, where $\wp$ is the endomorphism $x\mapsto x^p-x$ of the additive group $K^+$ (thought of as an $\mathbf{F}_p$-space).

When $K=k((T))$ for some finite extension $k|\mathbf{F}_p$ of degree $f=[k:\mathbf{F}_p]$, the space $K^+/\wp(K^+)$ can be explicitly computed, along with its filtration. This is done for example in arXiv:0909.2541, from which the following is extracted.

Denote by $\frak{o}$ the ring of integers of $K$, and by $\frak{p}\subset\frak{o}$ the unique maximal ideal of $\frak{o}$. Concretely, we have ${\frak{o}}=k[[T]]$ and ${\frak{p}}=T{\frak{o}}$.

The filtration on the additive group $K^+$ is given by the powers $({\frak{p}}^n)$ of $\frak{p}$, indexed by $n\in\mathbf{Z}$. It induces a filtration on the $\mathbf{F}_p$-space $\overline{K^+}=K^+/\wp(K^+)$. We denote the induced filtration by $\overline{{\frak{p}}^n}$; we have $\overline{\frak{p}}=\{0\}$, and the codimension at each step is given by $$ \{0\} \subset_1\overline{\frak{o}} \subset_f\overline{\frak{p}^{-1}} \cdots \subset_f\overline{{\frak{p}}^{-pi+1}} =\overline{{\frak{p}}^{-pi}} \subset_f\cdots \quad\subset\overline{K^+} $$ Here, $i$ is any integer $>0$, and an inclusion of $\mathbf{F}_p$-spaces $E\subset_rE'$ means that $E$ is a codimension-$r$ subspace of $E'$.

The unramified degree-$p$ extension of $K$ corresponds to the $\mathbf{F}_p$-line $\overline{\frak{o}}$ (which can be canonically identified with $\mathbf{F}_p$ using the trace map $k\to\mathbf{F}_p$).

So every other line $D$ in this infinite-dimensional $\mathbf{F}_p$-space $K^+/\wp(K^+)$ corresponds to a ramified cyclic degree-$p$ extension $L=K(\wp^{-1}(D))$ of $K$. If the level of $D$ in the displayed filtration is $m$ (say), in the sense that $m>0$ is the smallest integer such that $D\subset\overline{{\frak{p}}^{-m}}$, then the unique ramification break of $\mathrm{Gal}(L|K)$ occurs at $m$.

Addendum. On my screen, the \overline does't seem to have any effect, which is very confusing. It is best to read p.17 of the arXiv version referred to above, or 409 of the published version (Journal of the Ramanujan Mathematical Society 25 (2010) 4, 393--417).

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  • $\begingroup$ Dear Dalawat, thanks for your answer. So in this case, the different of a totally ramified p-extension can be arbitarily small, (I guess it doesn't happen in the mixed character case, say, $\mathbf{Z}_p$ ), and the smaller it becomes, (or the corresponding \textbf{level} grows?), this extension is more close to an inseparable extension? $\endgroup$
    – user20167
    Dec 26, 2011 at 13:54
  • $\begingroup$ On the contrary, as the level $m$ of $D$ grows, so does the valuation $v_L({\frak{D}}_{L|K})$ of the different ${\frak{D}}_{L|K}$ of $L=K(\wp^{-1}(D))$ over $K$; by Hilbert's formula $$v_L({\frak{D}}_{L|K})=(1+m)(p-1).$$ $\endgroup$ Dec 26, 2011 at 14:13

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