I think I can show that the following definitions of "tamely ramified" coincide. I thought it would be good to be sure. Sorry for the easy questions.

Let $O_K$ be a dvr with maximal ideal $\mathfrak p$, $K$ its fraction field and $k$ its residue field. Let $p\geq 0$ be the characteristic of $k$. Let $L/K$ be a finite Galois extension. Let $O_L$ be the integral closure of $O_K$ in $L$; this is a free $O_K$-module of rank $[L:K]$. We say that $L/K$ is tamely ramified if

  1. for any maximal ideal $\mathfrak q$ of $O_L$ lying over $\mathfrak p$, we have $$ (e_{\mathfrak q/\mathfrak p},p) = 1$$ and $l/k$ is separable, where $l$ is the residue field of $\mathfrak q$.

  2. for any maximal ideal $\mathfrak q$ of $O_L$ lying over $\mathfrak p$, we have that $p$ does not divide the order of the inertia group of $\mathfrak q$.

  3. for any maximal ideal $\mathfrak q$ of $O_L$ lying over $\mathfrak p$, we have that $l/k$ is separable and $p$ does not divide the degree of $L$ over the maximal unramified extension of $K$ in $L$.

Are 1, 2 and 3 really equivalent?

1 implies 2 is trivial once you know that the order of the inertia group at $\mathfrak q$ is $e_{\mathfrak q/\mathfrak p} [l:k]_{insep}$.

For 2 implies 1, you need that if $p$ does not divide $[l:k]_{insep}$, then $l/k$ is separable. Why is this true?

For 2 implies 3, you need again that $l/k$ is separable under the assumption that $p$ does not divide $[l:k]_{insep}$ and you need that $e_{\mathfrak q/\mathfrak p}$ equals the degree of $L$ over the maximal unramified extension of $K$ in $L$.

For 3 implies 1 you need the same.

Is the above notion really the correct notion of $L/K$ tamely ramified?

Is there also a notion of $L/K$ is tamely ramified at $\mathfrak q$? I'm guessing this would be defined by simply removing "for any" in the above definition.

Can it happen that some $\mathfrak q$ of $L$ is tamely ramified, but the others are not?

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    $\begingroup$ "Can it happen that some q of L is tamely ramified, but the others are not?" No, not when the extension is Galois. $\endgroup$ – awllower Feb 3 '13 at 18:18
  • $\begingroup$ Hey, I think I just realized this myself. I asked a new question to be sure. Maybe I should remove this question now.... $\endgroup$ – Tom Feb 3 '13 at 18:21
  • $\begingroup$ Wait, but you did not yet show the equivalences! At least you could explain for me? Thanks in any case. $\endgroup$ – awllower Feb 3 '13 at 18:23
  • $\begingroup$ Ow no, I meant "remove the other question". haha. Sorry. $\endgroup$ – Tom Feb 3 '13 at 18:26
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    $\begingroup$ Sine $k$ is of characteristic $p$, it is evident that every inseparable extension of $k$ must have degree divisible by $p$. I am sorry that I did not see this earlier... Hence all your questions are now answered! $\endgroup$ – awllower Feb 3 '13 at 18:34

I think you have yourself answered the most of the questions. But, for the sake of completeness, I would like to write an answer. When $p$ cannot divide $ [l:k]_{insep}$, there could be no inseparable extensions between $l$ and $k$, hence $l/k$ is separable, showing that $(i)$ and $(ii)$ are equivalent. Other implications have already been explicitly answered by you.
P.S. One could show that $l/k$ is separable by the fact that inseparable extensions occur only when the characteristic is $>0$, and when the degree is divisible by the prime characteristic. Hence the conclusion.
Notice that this degree needs not be a power of the prime, as indicated by QiL'8.

  • $\begingroup$ @QiL'8 I am not quite certain what "postscriptum" means. In any case, the degrees of inseparable extensions of a characteristic $p$ field must be a power of $p$, because of the considerations of minimal polynomials, right? Maybe I am not correctly acquiring this fact before. Sorry for that then. $\endgroup$ – awllower Feb 6 '13 at 2:13

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