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For any field of p-adic numbers $\mathbb{Q}_p$, one can construct the field $\mathbb{C}_p$, the metric completion of one of its algebraic completions. By the axiom of choice, we can prove this to be isomorphic to the usual field of complex numbers $\mathbb{C}$. Therefore, since $\mathbb{Q}_p$ embeds into $\mathbb{C}_p$, there must be an embedding of $\mathbb{Q}_p$ into $\mathbb{C}$.

Is there any way to explicitly construct such an embedding, so that given an arbitrary p-adic number, we can rewrite it as a complex number to arbitrary precision?

I'm hoping that this sheds some light on what the (algebraic, non-topological) tensor products of things like $\mathbb{Q}_p \otimes_\mathbb{Q} \mathbb{Q}_q$ and $\mathbb{R} \otimes_\mathbb{Q} \mathbb{Q}_p$ and so on might look like.

(The above post was a lot longer, but it was confusing everyone, so I ditched it and wrote my question much more simply.)

Mike Battaglia
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  • Definitely not the complex numbers. For example, the are not algebraically complete, while being topologically complete. So there is no way they can be seen as topological sub-fields of $\mathbb C$. The tensor product of any two pair of rings can only be done relative to some other ring: $\mathbb Q_p\otimes_R \mathbb Q_q$ with some ring $R$ and maps $R\to \mathbb Q_p$ and $R\to\mathbb Q_q$. The most natural ring to take would be $R=\mathbb Q$ if you really want to deal with $\mathbb R\otimes \mathbb Q_p$. – Thomas Andrews Mar 22 '13 at 18:29
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    They might be isomorphic purely algebraically, but the interesting differences are the very different topologies between $\mathbb Q_p$ and $\mathbb C$. For example, in $\mathbb Q_p$ the rational integers are not a discrete topological subspace. That's a pretty big deal. – Thomas Andrews Mar 22 '13 at 18:43
  • Sorry, I think I misunderstood what you wrote - I think you were just referring to the last part I wrote. I mean just the usual algebraic tensor product $\otimes_\mathbb{Z}$, which I was told in a separate post would yield the same field as the tensor product $\otimes_\mathbb{Q}$ in this specific case (at least for finite products of $\mathbb{R} \otimes \mathbb{Q_{p_1}} \otimes \mathbb{Q_{p_2}} \otimes ... \otimes \mathbb{Q_{p_n}}$, since infinite products seem to be less well-behaved). – Mike Battaglia Mar 22 '13 at 19:17
  • You might be interested in the Adele ring: http://en.wikipedia.org/wiki/Adele_ring . It's a product ring rather than a tensor product, so it is, in particular, not a field. – Thomas Andrews Mar 22 '13 at 19:37
  • $\mathbb{Q}_p \otimes \mathbb{Q}_q$ is not a field. – Martin Brandenburg Mar 22 '13 at 19:40
  • Martin: sorry, I should have written "ring" above (too late to edit now), but I'm curious - how do you know immediately that it won't be a field? – Mike Battaglia Mar 22 '13 at 19:48
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    Mike, among other things both $\mathbb{Q}_p$ and $\mathbb{Q}_q$, $p<q$, contain square roots of all the integers congruent to $1\pmod{pq}$ (replace $pq$ with $8q$, if $p=2$). Therefore their tensor product contains too many such square roots to be a field. – Jyrki Lahtonen Mar 22 '13 at 20:20
  • Thanks, Jyrki. I understand your reasoning about it containing square roots of all of those integers, but I don't understand how this would lead to too many square roots for the result to be a field. Is there a reference I could read to understand better why this wouldn't be the case? – Mike Battaglia Mar 22 '13 at 20:28
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    Remember $K[x]/(f) \otimes_K L = L[x]/(f)$. If $f$ splits over $L$, this tensor product is a direct product. – Martin Brandenburg Mar 22 '13 at 23:26

3 Answers3

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Note that every such isomorphism of $\Bbb C_p\to\Bbb C$ is actually a $\Bbb Q$-automorphism of $\Bbb C$.

It is consistent that without the axiom of choice there are only two automorphisms of $\Bbb C$, the identity and conjugation. Obviously if $\Bbb C_p$ is a $p$-adic field, such automorphism is neither of the two. Therefore its existence relies on the axiom of choice, and cannot be written explicitly.

Of course if such an embedding of $\Bbb C_p$ does not exist without using the axiom of choice to begin with, then we cannot embed $\Bbb Q_p$ into $\Bbb C$. Otherwise we could have taken the intersection of all algebraically closed subfields of $\Bbb C$ which contain the embedded $\Bbb Q_p$.

Asaf Karagila
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    Asaf, maybe you know the answer to this: I think I have been told that an embedding $\mathbb{Q}_p \to \mathbb{C}$ is necessarily non-measurable, so it seems like it should not be possible to write such a thing down in ZF. – Qiaochu Yuan Mar 22 '13 at 20:57
  • @Qiaochu: If all sets are Lebesgue measurable, then all sets in $\Bbb Q_p$ are Haar measurable (as a question of mine on MO reveals), in which case it is indeed impossible to write such embedding. An argument equally valid and without the need for large cardinals would be Baire measurability, which would probably have the same effects. – Asaf Karagila Mar 22 '13 at 21:00
  • This is a great answer, btw. I guess since my curiosity about this question has led to a foundational dead-end, I should instead ask if the subset of all definable $\mathbb{Q}_p$ also naturally has a field structure, and if so, if there exists any definable embedding of that structure into $\mathbb{C}$. Perhaps looking at the model $V=L$ would help with that. – Mike Battaglia Mar 24 '13 at 10:07
  • It would be quite satisfying to know that my conjecture held true in that model, for then it would make clear that the rest of this stuff about undecidability has to do with the structure of the remaining undefinable "junk" numbers. Then it would be the case that different axioms like AC or AD simply lay out rules for how this "junk" works and what properties the undefinable junk has, which may or may not preclude an embedding into $\mathbb{C}$ based on things like AC. – Mike Battaglia Mar 24 '13 at 10:08
  • @Mike: I'm not sure what it means that the subsets of all definable $\Bbb Q_p$. Every $\Bbb Q_p$ is definable. Metric completions are definable from the metrics, and the $p$-adic metrics on $\Bbb Q$ are definable. Furthermore $V=L$ implies the axiom of choice, so all the "undefinable junk" that follows it exists in $V=L$. – Asaf Karagila Mar 24 '13 at 10:55
  • Asaf, in the first comment I meant "definable" in the sense of being first-order definable. I was curious if either the (countable) subfield of first-order-definable $\mathbb{Q}_p$ might definably embed into $\mathbb{C}$. In the second comment I just meant the "constructible" (uncountable) $\mathbb{Q}_p$ that exists in the universe $L$. That AC holds in $L$ doesn't preclude the being the embedding of $\mathbb{Q}_p \to \mathbb{C}$ from being definable in $L$, even though wouldn't be otherwise. – Mike Battaglia Mar 24 '13 at 17:12
  • To see why, consider that there's a definable well-ordering of the reals in $L$, even though ZFC is consistent with the possibility that no well-ordering exists at all. In that case, there's a certain subset of $\mathbb{R}$ that has a definable well-ordering, and in $L$, you can prove that this subset equals the whole of $\mathbb{R}$. I was curious if a similar thing would apply to $\mathbb{Q}_p$, so that there's always a certain subset which has a definable embedding into $\mathbb{C}$, and in $L$ you can prove that this subset is the entire set $\mathbb{Q}_p$. – Mike Battaglia Mar 24 '13 at 17:17
  • @Mike: First of all, in ZFC there is always a well-ordering of the real numbers. As for the embedding of a subfield, probably you can embed the intersection of the algebraic/real closure of the rationals with the $p$-adic field. And possibly slightly more. – Asaf Karagila Mar 24 '13 at 17:20
  • Sorry, that was a typo; I meant to type ZF, not ZFC. My point is that ZF + "no well-ordering of $\mathbb{R}$ exists" and ZF + "a definable well-ordering of $\mathbb{R}$ exists" are both consistent, and the latter holds in $L$. So I was curious if, likewise, ZF + "no embedding $\mathbb{Q}_p \to \mathbb{C}$ exists" and ZF + "a definable embedding $\mathbb{Q}_p \to \mathbb{C}$ exists" are both consistent. Your last point is a good one to think about. – Mike Battaglia Mar 24 '13 at 17:27
  • @Mike: Definable is a broad word, and has a murky sound to it. In $L$ there is a $\Delta^1_2$ well-ordering of the real numbers, which is fairly close to definable, but still not really definable. It is also consistent that there is no projective ($\Delta^1_n$) well-ordering of the real numbers either. All this in ZFC. But this is not definable in the language of fields, or as a topological construct. This means definable in second-order arithmetic. So what exactly do you mean by definable embedding? – Asaf Karagila Mar 24 '13 at 17:39
  • Hi Asaf, I'm not too knowledgeable about the arithmetic heirarchy (I think that's what the notation $\delta^1_2$ is referring to). I was just intuiting in vague terms, but thanks for pointing me to this. I'll read up on it to get better insight into my question. – Mike Battaglia Mar 24 '13 at 18:18
  • @Mike: If you can talk about sets of natural numbers the. You have a universe of size continuum, and then you can talk about what's definable there. But this is not the same thing as definable in the language of fields, or valued fields. – Asaf Karagila Mar 24 '13 at 18:30
  • @Angel: A topological completion of something exists if that thing exists in the first place. Can you prove the existence of an algebraic closure of $\Bbb Q_p$ without appealing to choice? – Asaf Karagila Feb 02 '16 at 21:12
  • @AsafKaragila, It has already been answered here, and it involves a weak version of AC: http://math.stackexchange.com/questions/1032885/proving-existence-of-overline-bbb-q-p-without-ac – Ángel Valencia Feb 02 '16 at 21:41
  • @Angel: I had a vague memory, but I couldn't find the question. Thanks. In any case, I'd prefer that you leave a comment in the future and let me do the editing. – Asaf Karagila Feb 02 '16 at 21:54
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If you don't care about topology, then any field of characteristic $0$ and cardinality not greater than that of $\Bbb C$ is a subfield of $\Bbb C$.

In fact, using the Axiom of Choice, for each cardinal $\kappa > \aleph_0$, there is "only one" algebraically closed field of cardinal $\kappa$. (while there are many countable algebrically closed fields of characteristic $0$ : $\overline{\Bbb Q},\overline{\Bbb Q(X)},\overline{\Bbb Q(X,Y)}$ and so on, which can also be realised as subfields of $\Bbb C$).

So if $|K| \le |\Bbb C|$ then $K$ is a subfield of $\overline{K}$, which is either isomorphic to $\Bbb C$ or to one of the countable algebraically closed fields.

mercio
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    I believe the OP was asking whether one could construct an explicit injection from $\mathbb{Q}_p\hookrightarrow\mathbb{C}$ so that given a $p$-adic number, one could compute to arbitrary precision the corresponding complex number. –  Mar 22 '13 at 18:48
  • Right, what Jason said. I'll edit my post to make that clearer. – Mike Battaglia Mar 22 '13 at 19:09
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    The formulation "is a subfield of $\mathbb{C}$" might be misleading for some beginners. I would choose "embeds into $\mathbb{C}$" or something like that. After all, there are lots of embeddings, no canonical one, and often they only exist due to AC. – Martin Brandenburg Mar 22 '13 at 19:30
  • Good point. I think at this point that the way I wrote this post was a total disaster, so I rewrote the whole thing much more simply. Note to self: time for some coffee. – Mike Battaglia Mar 22 '13 at 19:42
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A little off topic but thinking of C af "the" extension of reals may be an issue. In quaternions, otherwise, 2-adic numbers have their explicit representation. Or may be I misunderstood something?

Ernesto Iglesias
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