0

The following is the statement from Algebraic Number Theory by Neukirch (Chapter 2 Proposition (7.7) p155)

Suppose $K$ is Henselian field, $p=char(\kappa)$, the character of the residue field of $K$. A finite extension $L\K$ is tamely ramified if and only if the extension $L/T$, ($T$ is the maximal unramified subextension of $L/K$) is generated by radicals $\quad L=T( \sqrt[m_1]{a_1},\dots, \sqrt[m_r]{a_r})$, such that $gcd(m_i, p)=1$.

For "$\Rightarrow$" direction, the proof given in the book is correct, but it should be pointed out that $"a_i"$s come from $T$.

The proof of "$\Leftarrow$" direction is highly suspect. First of all, what's the right statement? There are at least two ways:

(1) $K$ is a Henselian field, for $a_i \in K,$Let$ L=K( \sqrt[m_1]{a_1},\dots, \sqrt[m_r]{a_r}), \quad gcd(m_i,p)=1, \quad p=char(\kappa)$. Then $L/K$ is a tamely ramified extension.

(2) Same as (1) + $K$ is just the maximal unramified subextension of $L/K$ (i.e. $L/K$ is totally ramified ).

Does anyone know the proof of either statement? In addition, if $L/K$ happens to be a finite Galois extension (or maybe you only need simple extension), is it true $L=K(\sqrt[m]{a})$ form?

flag
2 
 I think that math.stackexchange.com would be a better fit for this question. Re correct statement: obviously, both are correct and equivalent: $L/K$ is tamely ramified if and only if $L$ is tamely ramified over the maximal unramified intermediate extension. But of course in the statement of Neukirch's theorem, you need to allow for the possibility of an unramified + a tamely ramified bit to get an equivalence, hence the presence of $T$. He could have said instead that an extension is totally tamely ramified if and only if it is of the given form, that would have been equivalent to his theorem. – Alex Bartel Jun 7 2011 at 4:38
Hi Alex Bartel, I cannot see the equivalence of two statement. As you said,$L$ is tamely ramified over the maximal unramified intermediate extension. But how do you know this maximal unramified subextension is just $K$? – Li Zhan Jun 7 2011 at 11:59
Since this question has been answered on math.SE (by M. Emerton, which is to say, wholly satisfactorily), I have voted to close it here as "no longer relevant". – Pete L. Clark Jun 7 2011 at 17:10

closed as no longer relevant by Pete L. Clark, Cam McLeman, Qiaochu Yuan, James Borger, Karl Schwede Jun 7 2011 at 23:52

Browse other questions tagged or ask your own question.