What is $\mathbb{Q}_p \cap \overline{\mathbb{Q}}$ ?

For instance, we know that $\mathbb{Q}_p$ contains the $p-1$st roots of unity, so we might say that $\mathbb{Q}(\zeta) \subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$, where $\zeta$ is a primitive $p-1$st root.

As a more specific example, $x^2 - 6$ has 2 solutions in $\mathbb{Q}_5$, so we could also say that $\mathbb{Q}(\sqrt{6},\sqrt{-1})\subset \mathbb{Q}_p \cap \overline{\mathbb{Q}}$.

Edit: I removed the motivation for this question (which I think stands by itself), as it will be better as a separate question once I think it through a bit better.

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    $\begingroup$ Note that, as an abstract field, $\overline{\mathbb{Q}_p}$ is isomorphic to $\mathbb{C}$: both are algebraically closed fields of characteristic zero and with transcendence degree $2^{\aleph_0}$ over $\mathbb{Q}$. I saw a talk recently on $p$-adic L-functions where this isomorphism was used (somewhat apologetically) in a critical way. $\endgroup$ Mar 4, 2010 at 0:59
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    $\begingroup$ (I think from the phrasing of your question that you knew this already, so my comment is more for the benefit of the other readers.) $\endgroup$ Mar 4, 2010 at 1:00
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    $\begingroup$ Victor: Pete was only saying the two fields were abstractly isomorphic -- he did not claim the isomorphism preserved any topology. I've also used that isomorphism before. I would describe its use as "violent." $\endgroup$ Mar 4, 2010 at 2:35
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    $\begingroup$ @Pete: if you ever talk about "the" p-adic Galois representation attached to a modular form then you might be implicitly fixing an iso Q_p-bar=C. But in fact one can get away with less, and I wonder whether the p-adic L-function talker could have got away with less: usually it suffices to fix embeddings from Q-bar into Q_p-bar and C (so any algebraic number in C can be interpreted as an element of Q_p-bar). I would be interested to know if the p-adic L-function talker really needed more than that. $\endgroup$ Mar 4, 2010 at 11:40
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    $\begingroup$ I would go so far as to state that such an isomorphism could never be needed for any investigation related to $p$-adic $L$-functions, $p$-adic Galois reps. attached to modular forms, and so on. Although choosing it will save later circumlocutions, it is always just shorthand for fixing embeddings of $\bar{\mathbb Q}$ into $\bar{\mathbb Q}_p$ and $\mathbb C$. $\endgroup$
    – Emerton
    Mar 4, 2010 at 15:53

2 Answers 2


The field $K_p = \mathbb{Q}_p \cap \overline{\mathbb{Q}}$ is a very natural and well-studied one. I can throw some terminology at you, but I'm not sure exactly what you want to know about it.

1) It is often called the field of "$p$-adic algebraic numbers". This comes up in model theory: it is a $p$-adically closed field, which is the $p$-adic analogue of a real-closed field. In particular, it is elementarily equivalent to $\mathbb{Q}_p$.

2) It is the Henselization of $\mathbb{Q}$ with respect to the $p$-adic valuation, or the fraction field of the Henselization of the ring $\mathbb{Z}_{(p)}$ -- i.e., $\mathbb{Z}$ localized at the prime ideal $p$.

The idea is that this field is not complete but is Henselian -- it satisfies the conclusion of Hensel's Lemma. Alternately and somewhat more gracefully, Henselian valued fields are characterized by the fact that the valuation extends uniquely to any algebraic field extension.

Roughly speaking, Henselian fields are as good as complete fields for algebraic constructions but are not "big enough" to have the same topological properties. For instance, note that $K_p$ cannot possibly be complete with respect to the $p$-adic valuation, because it is countably infinite and without isolated points: apply the Baire Category Theorem.

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    $\begingroup$ Pete, are you sure about the "totally" terminology in (1)? The real numbers which are algebraic over Q are not just the totally real numbers (those who min. poly. over Q splits completely over R) but includes a whole lot more. I'm not quibbling with the usage of the term "p-adically closed". (I accidentally posted this comment as a full reply before. If someone can delete that one, please do.) $\endgroup$
    – KConrad
    Mar 4, 2010 at 1:52
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    $\begingroup$ I agree with KConrad on this one. The 2nd option is right, and for the first it's just a suitably directed union of number fields equipped with a p-adic place that has its own e=f=1. That's a down to earth way of getting at description #2. Rather interesting, and less evident, is that something similar is true for completions of higher-dimensional normal local noetherian domains, and more specifically the excellent ones. This comes out from the Artin approximation theorem (in the general form of Popescu, for excellent rings). $\endgroup$
    – BCnrd
    Mar 4, 2010 at 3:41
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    $\begingroup$ @Conrads: I agree; the "totally" should not be there. I removed it. $\endgroup$ Mar 4, 2010 at 4:24

There's a slightly subtle point near here of which some people are not aware: that it is dangerous (perhaps even nonsensical) to compare algebraic numbers under various different completions. So, to talk about $Q_p\cap \bar Q$, you should be talking about a completion of $Q$ containing $Q_p$, not, e.g., a completion of $Q$ lying inside $C$. I don't think this is what is happening here, but some people may find this interesting.

Now, there are lots of isomorphisms floating around, so usually everything turns out just fine, but sometimes not. Here are two examples.

(1) The following fallacious argument that $e$ is transcendental is from a talk by Gouvêa, "Hensel's p-adic Numbers: early history" (originally due to Hensel himself).

The series expansion of $e^p$ converges in $Q_p$, thus $e$ is a solution to the equation $X^p=1+p\epsilon$, where $\epsilon$ is a $p$-adic unit. So $[Q_p(e):Q_p]=p$ (of course you need to argue that the polynomial is irreducible), and so $[Q(e):Q]\ge p$. Since $p$ was arbitrary, $e$ must be transcendental over $Q$.

The fallacy is that even though the series for $e$ (and $e^p$) converges in $R$ and $Q_p$, the numbers they converge to are not the same.

(2) The following is from Koblitz's $p$-adic book, page 83 (with an example and some other fallacious arguments).

It is not true that if an infinite sum of rational numbers (a) converges $p$-adically to a rational number for some $p$ and (b) converges in the real topology to a rational number, then the rational numbers the two series converge to are the same!

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    $\begingroup$ I'm pretty the original poster simply meant "what are the numbers in Q_p which are algebraic over Q?" But your examples are worth pointing out to any readers who are not sensitive to the danger of talking about intersections of fields which are not lying in some common field. (For a similar reason, the notion of composite of two fields can be dangerous if the fields are not already given inside a common field.) Your reference to a series in (2) is not necessary: a sequence in Q can converge to different rational numbers in different topologies (real and p-adic or p-adic and q-adic, say). $\endgroup$
    – KConrad
    Mar 9, 2010 at 21:50

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