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This is a basic question from Neukirch's Algebraic Number Theory, Prop. 7.2:

Fix a non-Archimedean local field $K$. Let $L/K$ and $K'/K$ be two extensions inside an algebraic closure $\bar{K}/K$ and let $L'=LK'$. Then one has $$L/K\text{ unramified} \implies L'/K'\text{ unramified}.$$

For an arbitrary extension, Neukirch defines an arbitrary extension $L/K$ to be unramified if $L$ is a union of finite unramified subextensions of $K$.

Now at the start of the proof it states that we may assume $L/K$ to be finite. Why is this?

This is the only reason I can think of: suppose $L=\bigcup_{i\in I}L_i$ for some collection $L_i$ of finite unramified subextensions of $K$, then the composite $LK'$ is the union $\bigcup_i (L_iK)$.

I've tried showing this but I can't do it without also assuming that $L=\bigcup_{i\in I}L_i$ is a directed union.

I'd be very grateful for some help.

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    $\begingroup$ If $x\in LK'$, then, by definition, $x =\sum_{j=1}^n l_jk_j$ for some elements $l_j\in L$, $k_j\in K$. The $l_j$ all lie in some $L_i$. Hence $ x\in L_iK\subset\cup_i(L_iK)$. $\endgroup$
    – Mathmo123
    May 20, 2016 at 14:14
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    $\begingroup$ Can you tell me why the $l_j$ all lie in some $L_i$? regards $\endgroup$
    – eddie
    May 20, 2016 at 17:19
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    $\begingroup$ All $l_j$ are assumed to be algebraic, so they all lie in a finite extension, namely their splitting field over $L$. $\endgroup$
    – Pax
    May 20, 2016 at 20:38

1 Answer 1

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I think the section around proposition 7.2 is indeed somewhat problematic. After making the definition of a not necessarily finite unramified extension one should first check that for finite extensions it agrees with the original definition.

To do this, one already needs the proposition 7.2 and its corollary 7.3 (the compositum of two unramified extensions is unramified) for finite extensions. Then one can show:

Lemma 1: Let $L|K$ be a finite extension of non-archimedean henselian valued fields. Then the following are equivalent:

  • $L|K$ is unramified, i.e. the residue field extension $\lambda|\kappa$ is separable and $[L:K] = [\lambda:\kappa]$.
  • Every subextension is unramified.
  • $L$ is a directed union of unramified subextensions.
  • $L$ is a union of unramified subextensions.
  • $L$ is a compositum of unramified subextensions.

Only one implication is nontrivial: If $L$ is a compositum of unramified subextensions, then finitely many of them suffice and $L$ is unramified by 7.3.

Once one has the lemma, one can make the definition for not necessarily finite extensions by calling $L|K$ unramified if every finite subextension of $L$ is unramified, or equivalently if $L$ is a directed union (resp. a union, resp. a compositum) of finite unramified subextensions, all these conditions being equivalent.

I, too, do not see how $L = \bigcup L_j$ would imply $LK' = \bigcup L_j K'$, unless the union is directed. However, we can use the definition with $L$ being a compositum $L = \prod L_j$ of finite unramified subextensions, then we do have $LK' = \prod L_jK'$ and the reduction to the finite case in the proof of 7.2 is justified.

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  • $\begingroup$ That's very nice. I also suspected that $L$ needs be defined as a compositum of finite unramified subextensions but I was somehow hoping that the result of corollary 7.3 would not be needed as the reduction problem occurs at the start of the proof of proposition 7.2. $\endgroup$
    – eddie
    May 22, 2016 at 20:26

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