Algebraic Henselian extensions of the rationals

Question

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.

Answer 1

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 $K_1 \otimes_{\mathbb{Q}} \mathbb{Q}^h \cong K_2 \otimes_{\mathbb{Q}} \mathbb{Q}^h$.

Answer 2

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 http://people.fas.harvard.edu/~amathew/chcompletion.pdf, 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.