# A characterisation of tame ramification

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

Blockquote 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 $L=T(\sqrt[m_1]{a_1}\dots \sqrt[m_r]{a_r})$ , such that $(m_i,p)=1$ .

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 suspicious. First of all, what's the right statement? There are at least two ways:

(1) $K$ is a Heselian field, for $a_i \in K$ Let $L=K(\sqrt[m_1]{a_1}\dots \sqrt[m_r]{a_r}) \qquad (m_i,p)=1 \qquad p=char(\kappa)$ . 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 $L=\sqrt[m]{a}$ form?

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A little linguistic help, for non-experts in English. "tame ramification" and "tamely ramified". Adjective and adverb. –  GEdgar Jun 7 '11 at 15:26

The statement is just as Neukirch writes: an extension of the form $L = T(a_1^{1/m_1}, \ldots, (a_r)^{1/m_r})$, with $T/K$ unramified and the $a_i \in T$, is tamely ramified.

I don't see why you find it suspicious.

The proof is straightforward. First of all, we can enlarge $T$ by adjoining all the $m_i$th roots of $1$ (this gives an unramified extension of $T$, since the $m_i$ are prime to $p$, and since an unramified extension of an unramified extension is unramified, it also yields an unramified extension of $K$).

Now the extension $L$ is Galois (it is a compositum of Kummer extensions), so it suffices to show that if $T$ contains all the $m$th roots of $1$ (with $m$ prime to $p$), then $T(a^{1/m})$ is tamely ramified for any $a \in T$. But this is clear:

1. Such an extension is cyclic of degree dividing $m$ (by Kummer theory), and so the inertia subgroup of the Galois group is also cyclic of order dividing $m$.

2. Since $p \nmid m$, the inertia subgroup has order prime to $p$, and so the extension is tame, as claimed.

I'm not sure what you mean in your last sentence.

By the way, this proposition is more-or-less how one thinks of tamely ramified extensions: they are what you get by making unramified extensions together with extracting prime-to-$p$ roots of arbitrary elements.

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Thank you! Your answer: "an extension of the form $L=T(\sqrt[m_1]{a_1},\dots \sqrt[m_r]{a_r} )$, with $T/K$ unramified and the $a_i \in T$, is tamely ramified." answered my question (1), but do you know in this case, whether $T$ is the maximal unramified subextension of $L/K$. –  Li Zhan Jun 8 '11 at 2:54
@Li Zhan: Dear Li Zhan, It depends on the $a_i$. For example, if $a_i$ is a unit in the ring of integers of $T$, then $T(a_i^{1/m_i})$ is unramified over $T$. I suggest that you consider some concrete examples, e.g. $K = T = \mathbb Q_p$, and $L = \mathbb Q_p(a^{1/m})$ for various choices of $m$ and $a$. In this case, $L$ is totally ramified over $\mathbb Q_p$ precisely when the $p$-adic valuation of $a$ is coprime to $m$. Regards, –  Matt E Jun 8 '11 at 4:21

To answer your second question, if $K$ is a local field (or more generally a complete DVR) and $L/K$ is finite and tamely ramified, then $L = T(\sqrt[e]{a})$ where $e$ is the ramification index of $L/K.$

To see this, let $\pi$ be a uniformizer in the ring of integers of $L.$ Then $L = T(\pi)$ and $irr(\pi,T)(X) = \sum_{i=0}^{i=e} a_i X^i$ is an Eisenstein polynomial over $\mathcal{O}_T.$ Let $$f(X) = X^e - \frac{a_0}{\pi^e}$$ and $$g(X) = \frac{irr(\pi,T)(\pi^eX)}{\pi^e}.$$ Then $f,g\in\mathcal{O}_L[X]$ and $f \equiv g \mod \mathcal{M}_L[X].$

It follows that $1$ is a root of the coordanatewise reduction of $f$ to $\mathcal{O}_L/\mathcal{M}_L.$ Furthermore, as $(e,p) = 1,$ we have

$$f'(1) \equiv e1^{e-1} \mod \mathcal{M_T} \not\equiv 0 \mod \mathcal{M_L}.$$

Therefore by Hansel's lemma, there exists a root $\gamma$ of $f$ in $\mathcal{O}_L.$

Considering the element $\gamma\pi,$ we observe it is a root of the Eisenstien polynomial $X^e-a_0$ over $\mathcal{O}_T$ and thus a uniformizer in $\mathcal{O}_L.$

It follows

$$L = T(\gamma\pi) = T(\sqrt[e]{a_0}).$$

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First, I apologize that I did not make the question clear. Here is the setup of the inverse direction: I have a field $K$, an unramified extension of $K$, which is $T$, and then one adjoints some radicals of elements from $T$, this field is denoted by $L$. The difference is that I do not know $T$ is the maximal unramified extension in priori. Second, I cannot see how you get the long expression in the sixth line: $e=[L:T]=\dots\dots$. –  Li Zhan Jun 8 '11 at 3:02