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Let $p$ be a fixed prime, $v:\mathbb{Q}\rightarrow\mathbb{Z}$ be the $p$-adic valuation on $\mathbb{Q}$ and $\mathbb{Q}^h$ the Henselization of $\mathbb{Q}$ with respect to $v$. I want to show that for each integer $n$ there are only a finite number of algebraic extensions of $\mathbb{Q}^h$ of degree $n$.

I know that the unramified extensions of $Q^h$ correspond to extensions by roots of $X^q-X$ for appropriate $q$.

I know that the totally ramified extensions of $Q^h$ correspond to extensions by roots of Eisenstein polynomials.

I also think I should use Krasner's lemma (for Henselian valuations) at some point. (I have seen a proof where $Q^h$ is replaced by $Q_p$ but without the completeness I can't make it work.)

Does anyone have any clues or references? Thanks, Conrad.

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up vote 2 down vote accepted

Yes, Krasner's Lemma holds for Henselian (rank one) valuations. The statement and proof of KL in $\S 3.5$ of these notes is given in that context. The key point is that a valuation $v$ on $K$ is Henselian iff it extends uniquely to each finite degree field extension $L/K$.

To establish the basic finiteness result however, I think one should first prove it over $\mathbb{Q}_p$ by a compactness argument as in Theorem 14 here and then use the fact that the Henselization $\mathbb{Q}^h$ of $\mathbb{Q}$ with respect to the $p$-adic valuation is the algebraic closure of $\mathbb{Q}$ in $\mathbb{Q}_p$. It follows that if $K_1$ and $K_2$ are two number fields such that $K_1 \otimes_{\mathbb{Q}} \mathbb{Q}_p$ and $K_2 \otimes_{\mathbb{Q}} \mathbb{Q}_p$ are isomorphic $p$-adic fields, then already $K_1 \otimes_{\mathbb{Q}} \mathbb{Q}^h \cong K_2 \otimes_{\mathbb{Q}} \mathbb{Q}^h$.

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Thanks for the answer. Unfortuately I still have a problem. When $K=\mathbb{Q}^h$, the valuation ring is not compact. So I don't see why the Eisenstein locus is compact and thus don't get finitely many discs. I think I'm really missing something. – Conrad Sep 5 '11 at 15:04
I'm concocting a slightly bizarre proof to transfer the proof from $\mathbb{Q}_p$ to $\mathbb{Q}^h$ using model completeness of $p$-adically closed fields of $p$-rank $d$ and the fact that $\mathbb{Q}^h$ is dense in $\mathbb{Q}_p$. Btw, thanks for making me think about it a bit more and for the notes on compact fields. My background is not in number theory, so this is probably my best bet! – Conrad Sep 5 '11 at 16:37
@Conrad: yes, I agree that the part of the argument which uses the compactness of the Eisenstein locus cannot be carried over directly to the Henselian case. Originally I read your question as asking for a version of KL for Henselian valuations, but rereading it this may not be the case. Anyway, yes, I think the idea of establishing the the complete case by compactness and then the Henselian case by an approximation argument is a good way to go. You can express this model theoretically, but it should also be possible to see it more directly. Let me know if you want to hear more about that. – Pete L. Clark Sep 5 '11 at 20:22
Yeah. I just figured out how to get it from $\mathbb{Q}_p$. I took the centres of the discs in theorem 14 to have rational coordinates (which can be done since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$) and then used Krasner's lemma over $\mathbb{Q}^h$. I hoped something like the tensor product thing in your answer was true but I didn't know. I'll look it up, should come in handy in the future. Model completeness doesn't help! Thanks for the answer. – Conrad Sep 5 '11 at 20:45

To elaborate on Pete L. Clark's answer: a lot of results stated in algebraic number theory books for complete valued fields (with, say, a discrete valuation) work just as fine for henselian fields (such as the algebraic numbers in $\mathbb{Q}_p$).

Why is this? In this particular example, the claim is that if $K$ is a henselian field (w/ a discrete valuation, say) and $L$ a finite separable extension, then the valuation on $K$ extends uniquely to $L$. The usual proof of this fact in the complete case -- basically, some argument based on the equivalence of all norms on a finite-dimensional space over a complete field -- doesn't (unless I'm missing something?) work here without completeness hypotheses.

But the result is still true. Namely, the point is that if $O$ is the ring of integers in $K$, then $O$ is a henselian ring. One way to state this is that Hensel's lemma holds for $O$, but a more transparent one is that any finite $O$-algebra splits as a direct product of local $O$-algebras. (You might find helpful Raynaud's book "Anneaux locaux henseliens," or sec. 4 of, here for the equivalence of the two claims.)

Now if $O'$ is the integral closure of $O$ in $L$, then we know $O$ is a finite $O'$-module, so in particular it splits as a product of local rings. However, it's a domain! It follows that $O'$ is itself local, and consequently a discrete valuation ring itself. In other words, the maximal ideal of $O$ extends in precisely one way to $O'$ (this is equivalent to $O'$ being local). But these extensions of maximal ideals correspond precisely to the extensions of valuations to $L$, so we get the claim.

The moral of the story is that much true for complete valued fields is true for henselian fields, because this (the uniqueness of extensions) is a key property of complete fields for statements in algebraic number theory.

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I'm not sure I understand. My definition of a henselian valued field $(K,\mathcal{O})$ is exactly that $\mathcal{O}$ has a unique prolongation to every algebraic extension. Also the version of Kranser's lemma I know is exactly for Henselian valued fields (in fact it is equivalent to being Henselian). I don't understand from your answer why I can transfer $\mathbb{Q}_p$ having finitely many extensions of degree $n$ to $\mathbb{Q}^h$ having finitely many extensions of degree $n$. Am I missing something? – Conrad Sep 5 '11 at 15:45
@Conrad: OK. I had assumed you were using the definition of a henselian field as one where Hensel's lemma is true. Certainly the definition you are using is equally valid, though in that case my answer is probably unhelpful. Sorry! – Akhil Mathew Sep 5 '11 at 19:43
No worries. I'm more familiar with valuation theory than number theory but I guess it's the other way round for most people. – Conrad Sep 5 '11 at 20:47

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