I'm reading Ch.II, $\S$ 7 of Neukirch's Algebraic Number Theory and I'd be really grateful if someone could help me understand the following:

Let $K$ be a complete valued field wrt a non-archimedean valuation $v$. Neukirch proves that the composite of two finite unramified extensions of $K$ is again unramified.

How does it follow from this that the composite of all unramified extensions of $K$ inside a fixed algebraic closure $\bar{K}$ is again unramified (as he is assuming on p. 154)?

The problem for me is that there could be an infinite number of such subextensions to consider.

Many thanks for your answers.


By definition (see 7.1 in p. 153 of Neukirch's book), an arbitrary algebraic extension $L/K$ is unramified if it is the union of finite unramified subextensions. So the compositum of all finite unramified extensions of $K$ is, by definition, unramified.

  • $\begingroup$ I think I see it now-- is the maximal unramified extension of K inside the algebraic closure taken to be the union of all finite unramified subextensions of K which also equals the composite of all such extensions? So is this is what Neukirch is referring to when talking about the composite of all unramified extensions of K? $\endgroup$ – Josh F. Mar 27 '13 at 14:18
  • $\begingroup$ I think Neukirch's definition would be clearer if instead of "...union of finite unramified subextensions" he wrote "... compositum of finite unramified subextensions." If you have an infinite extension $L/K$ which is the compositum of finite extensions $F/K$, then you can write it in particular as a union $L=\bigcup F_i$, with $F_i\subseteq F_{i+1}$. $\endgroup$ – Álvaro Lozano-Robledo Mar 27 '13 at 17:44
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    $\begingroup$ My complain is that a union $F_1\cup F_2$ of two extensions of $K$ need not be a field. If they are in a chain $K\subset F_1\subset F_2$, then of course their union is a field, namely $F_2$. $\endgroup$ – Álvaro Lozano-Robledo Mar 27 '13 at 17:46
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    $\begingroup$ For this definition to be consistent, one must show that the compositum of two finite unramified extensions is (finite) unramified. This is of course true. Then an infinite unramified extension is a filtering inductive limit of finite unramified extensions. $\endgroup$ – user18119 Mar 28 '13 at 16:20
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    $\begingroup$ Many thanks for your answers, much appreciated. $\endgroup$ – Josh F. Apr 2 '13 at 14:42

I think that he is trying to employ the lemma of Zorn. And to show that the composite of two unramified extensions is still unramified amounts to verifying the condition required for that lemma. The result follows directly from the lemma.
Maybe this should be put into the comment-form? Thanks in any case.

  • $\begingroup$ Many thanks for your answer- but I don't see how Zorn's Lemma can be applied in this case. First, you would need to show that any chain of finite unramified extensions of K lying in the algebraic closure of K has an upper bound (which I assume would be their union) that is also a finite unramified extension of K? $\endgroup$ – Josh F. Mar 27 '13 at 11:59
  • $\begingroup$ Maybe it is to show that any chain of unramified (not necessarily finite) extensions is still unramified? I am not so sure now. Sorry for that. $\endgroup$ – awllower Mar 27 '13 at 12:25

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