# Are $C$ and $\bar{Q_p}$ isomorphic?

There is a famous passage on the third page of Deligne's second paper on the Weil conjectures where he expresses his dislike of the axiom of choice, as manifested in the isomorphism between $C$ and $\bar{Q_p}$. The proof of said isomorphism runs as follows. Both $C$ and $\overline{Q_p}$ have transcendence bases, $S$ and $T$. Then $C\simeq \overline{Q(S)}$ and $\overline{Q_p}\simeq \overline{Q(T)}$. But $C$ and $\overline{Q_p}$ have the same cardinality, and hence, so do $S$ and $T$. Therefore, $Q(S)\simeq Q(T)$ and, from there, $C\simeq \overline{Q_p}$.

For myself, this proof is quite convincing. Recently, Torsten Ekedahl expressed his opinion to the contrary, and this led to the following exchange:

Why worry about the axiom of choice?

So I wondered about other expert opinions on this matter. Do you find the isomorphism unbelievable and, if so, why?

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The theorem is true given the axiom of choice. In fact any two fields of a given characteristic that are algebraically closed and share the same cardinality are isomorphic (this is a property of the "theory" - I forget what this property is called, but it has something to do with completeness, which this theory is also. I'll let the model theorists be more specific). Is the problem that you didn't find a proof that was... rigorous enough? It's surely "believable". You can find much less believable iso., like: C is iso. to the ultraproduct of all the F_p bars. –  H. Hasson May 2 '10 at 17:09
Considering injective resolutions, I never really understood AC objections if one uses derived functors in a substantial way. I agree with Minhyong about the way one uses trasncendence bases, so I have no problem with that field isomorphism. However, as a matter of style it shouldn't be invoked if not actually needed. In the case of Weil II, the issue can be avoided at the end of the proof because only countably many complex numbers actually arise (from stalks at closed points for countably many Weil sheaves on finitely many schemes of f.t. over Fbar_p): so only countable AC arises. –  BCnrd May 2 '10 at 17:15
@H.Hasson: the property of ACF_p that you mention is called being "uncountably categorical". (this corrects my previous comment which just said categorical, which is not true.) For the connection to completeness, se the last paragraph of en.wikipedia.org/wiki/Morley's_categoricity_theorem –  Dan Petersen May 2 '10 at 17:34
@Georges: Yes, the correct invariant is the transcendence degree. The theory ACF_p is only uncountably categorical. –  François G. Dorais May 2 '10 at 20:44
H. Hasson, that is not quite said correctly. In the countable case the transcendence degree DOES have to be countable, but the issue is that there are infinitely many different countable cardinalities: all the finite cardinalities, plus the countably infinite cardinalitiy. Each of these occurs as a transcendence degree, leading to non-categoricity. –  Joel David Hamkins May 4 '10 at 10:49

## 4 Answers

First, let me observe that it is consistent with ZF + DC that there is no such isomorphism. (This follows from this answer of mine.) However, as I commented on Torsten's post, the existence of such an isomorphism is a relatively harmless since one can force the existence of such an isomorphism without adding new points to C or Qp. Consequently, any purely field-theoretic fact that can be proved using this generic isomorphism can also be proved without (usually with more work). Since forcing is not widely understood, I will explain this in terms of sheaves instead. (If you're more familiar with forcing and you don't care about sheaves, simply observe that the poset P below is countably closed and ignore the rest of this post.)

Let P be the poset of field isomorphisms p:A→B where A is a countable subfield of C and B is a countable subfield of the algebraic closure of Qp, and p ≤ q iff p ⊇ q (i.e. q is a restriction of p). This ordering is slightly counterintuitive, but it is more convenient than the opposite. The poset P can be viewed as a category where there is one and only one arrow between any two objects p and q iff p ≤ q. The poset P then becomes a Cartesian category where the terminal element is the isomorphism between the two copies of Q in each field, and the product of p and q are is the intersection of the (graphs of) p and q.

There are many Grothendieck topologies that one could define on P. The relevant one for our context is the smallest Grothendieck topology S on P such that, for all x in C and all y in the algebraic closure of Qp, the sieves {q ≤ p : x ∈ dom(q)} and {q ≤ p : y ∈ rng(q)} are both covering sieves at p. (Any larger Grothendieck topology will do; for forcing one uses the double negation topology which includes this one.) Note that the points of (the locale associated to) the site (P, S) are in one-to-one correspondence with isomorphisms between C and the algebraic closure of Qp.

Now, the isomorphisms between C and the algebraic closure of Qp correspond precisely with geometric morphisms Set → Sh(P, S). Whatever is preserved by this geometric morphism can be done equally well on either side. In other words, many things that can be done in Set using such an isomorphism can also be done in Sh(P, S) without this assumption. Of course, this heavily depends on what needs to be done, but there are known ways to carry out this kind of analysis. Since the site (P, S) is relatively nice, this analysis is far from impossible.

It's interesting to see how this formalizes Emerton's view. Objects of Sh(P, S) are functors F:Pop→Set, subject to the usual continuity requirements. One can think of F as a set which evolves along P. This makes sense since we should think of partial isomorphisms p ∈ P as approximations to the desired isomorphism from C onto the algebraic closure of Qp. As more and more information is packed into p, we gain more and more information about the stalk of F at the given point. Although he only considers the first few approximations in his answer, Emerton's view corresponds precisely to working in Set while keeping in mind that the work being done could be done equally well in Sh(P, S) instead.

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Dear Minhyong,

I am quite happy with this isomorphism, but maybe not so much because of the proof using the axiom of choice (although I don't particularly object to AC) but rather because my sense is that, whenever this is used, what is really being used is a choice of isomorphism between the algebraic closure of $\mathbb Q$ in $\mathbb C$ and the algebraic closure of $\mathbb Q$ in $\overline{\mathbb Q}_{\ell}$ (and I have absolutely no objection to identifying these two algebraic closures).

Anytime one uses such an isomorphism in arithmetic, and it isn't ultimately being used to identify algebraic numbers in the two fields, I think it is fairly meaningless. (E.g., for modular forms of wt. $k \geq 1$, I am happy to identify the space over such over $\mathbb C$ with the analogous space over $\mathbb Q_{\ell}$, since the normalized cupsidal eigenforms have algebraic integer coefficients, and so these spaces have a natural underlying $\overline{\mathbb Q}$-structure. But to take non-algebraic Maass eigenforms, and to think of their Fourier coefficients as numbers in $\overline{\mathbb Q}_{\ell}$, while technically possible, is conceptually meaningless.)

In my own papers I often fix such an isomorphism (or even one for each $\ell$), but I don't think of it as having any significance beyond the identification of the two copies of $\overline{\mathbb Q}$.

Added: The comments below have forced me to think a little harder about my position. Here is an attempt to refine it:

Any countably generated extension of $\mathbb Q$ can be embedded into either $\mathbb C$ or $\overline{\mathbb Q}\_{\ell}$, and when I invoke, or seen invoked, an isomorphism between the latter two fields, I think of it as a short-hand for something like the following: in the given proof, a countably generated subfield of $\mathbb C$ will appear (e.g. the field generated by the Hecke eigenvalues of a Maass form). Having fixed the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$, we have in particular fixed an embedding of this field into $\overline{\mathbb Q}\_{\ell}$, and hence have chosen an extension of the $\ell$-adic absolute value to this field. (Of course, one could switch the roles of $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$ here.)

By virtue of fixing the isomorphism between $\mathbb C$ and $\overline{\mathbb Q}\_{\ell}$, one is ensuring that any such extensions are compatible, if along the way we encounter different subfields of $\mathbb C$, and that is one big advantage, when writing an argument, of fixing such an isomorphism once and for all. But in practice I don't know that one encounters anything more serious than one single countably generated subfield that contains all the complex numbers appearing in the proof. And hence one doesn't use anything like the full strength of the isomorphism.

I guess this does put me in Deligne's camp: the isomorphism is convenient, but one could get by with something much weaker, just involving countably generated subfields of $\mathbb C$.

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Hmm. I'm not sure what you mean by meaningless. It sounds like you and a number of other people agree with Deligne, that is, to be happy with the isomorphism in situations where you can do without it. Is this correct? (On the other hand, you say you're fine with AC, in which case I suppose you should be happy with the isomorphism regardless of the situation.) –  Minhyong Kim May 2 '10 at 20:21
Matt, the situation is as you say after Deligne's proof is done, but during the proof we don't know the algebraicity, so the framework of Weil sheaves and their twistings using possibly non-algebraic numbers is important in the method. That's why I pointed out in my comment that only countably many such numbers actually come up in the proof (due to countability of set of closed points), so even for the proof itself only a countably generated subfield of C intervenes. Last year when Akshay and I did a student seminar on Weil II he hoped it could be avoided, but by the end he agreed with me. –  BCnrd May 2 '10 at 20:50

I don't think one should try to determine whether to accept or reject the axiom of choice or any other independent axiom by appealing to "believability" of some consequences of it. With infinite sets, our intuition is just too often misleading. We get used to certain "paradoxa" like Hilbert's hotel because we see them very early in our mathematical life, but nobody should ever claim that he has a complete intuition for set theory.

As for the example, $\bar{\mathbf{Q}}_p$ and $\mathbf{C}$ are isomorphic if the axiom of choice is true, and that's that. Both are constructed using a completion, which makes them topological fields, and they are not homeomorphic or normed isomorphic, that's probably why it feels a bit wrong to some of us.

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I'm afraid this was a bit too long for a comment.

It seems that people don't object too much when we assert that $\mathbb{C}$ and $\overline{\mathbb{Q}_p}$ are isomorphic as sets or as abelian groups. Somehow the use of choice when establishing a ring-theoretic isomorphism bothers mathematicians much more, and I suspect it is because the implications are more at odds with intuition we build up from considering finite extensions of fields, or geometric structure that is normally attached to the fields. When considering purely ring-theoretic maps, we are still forgetting a lot of structural baggage, e.g., an affirmative answer to the similar question about existence of an embedding of fields $\mathbb{C}(t) \hookrightarrow \mathbb{C}$ throws away anything we know about genus zero curves.

I would personally answer your question with "yes" although I would not argue the point with much conviction. I would be interested to know if there were a logical way of chopping up choice so that the isomorphisms of sets and groups were okay but the isomorphisms of rings were not.

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It seems to me that proving $\mathbb{C}$ and $\overline{\mathbb{Q}}_p$ to be bijective as sets is rather straightforward and does not require any axiom of choice at all. This is like proving the unit interval and the Cantor set to be bijective. A little use of the Cantor-Bernstein theorem and you are done. On the other hand, I would be surprised if a proof of group isomorphism were possible without AC. –  Leonid Positselski May 2 '10 at 17:55
You are right. Even group isomorphism of $\mathbb{R}$ and $\mathbb{R}^2$ requires AC. –  Gerald Edgar May 3 '10 at 0:15