What to call the following variant of tame ramification - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T13:51:02Zhttp://mathoverflow.net/feeds/question/48509http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/48509/what-to-call-the-following-variant-of-tame-ramificationWhat to call the following variant of tame ramificationKarl Schwede2010-12-06T22:46:33Z2010-12-06T23:19:59Z
<p>Suppose that $R \subseteq S$ is a generically separable extension of 1-dimensional normal domains (you can assume that $R$ is local if you'd like) of equal-characteristic $p > 0$ (for simplicity, as you'll see, it's not really needed).</p>
<p>Suppose that for each point $x \in \text{Spec} R$, there exists a point $y \in \text{Spec} S$ such that $R_x \subseteq S_y$ has tame ramification (ie, the residue field extension is separable and if we write, $r = us^n$ where $r$ and $s$ are uniformizers for $R_x$ and $S_y$ respectively, then $p \not | n$).</p>
<p><strong>Question:</strong> Does this condition have a name? (ie, weakly tame, domesticated :-), etc)</p>
<p>I think this condition is also the same as requiring that the trace map $\text{Tr}: S \to R$ is surjective. In the case that $R \subseteq S$ is generically Galois, Kerz and Schmidt have called this surjective trace condition <em>cohomologically tamely ramified</em>. Has anyone else seen a name for this surjective trace condition especially in the non-1-dimensional case?</p>