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Let $\mathbb Q_p$ be $p$-adic numbers field. I know that the cardinal of $\mathbb Z_p$ (interger $p$-adic numbers) is continuum, and every $p$-adic number $x$ can be in form $x=p^nx^\prime$, where $x^\prime\in\mathbb Z_p$, $n\in\mathbb Z$.

So the cardinal of $\mathbb Q_p$ is continuum or more than that?

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    By your decomposition, we just need to find the cardinality of $\mathbf{Z} \times \mathbf{R}$. It seems like [Cantor-Schroeder-Bernstein](http://en.wikipedia.org/wiki/Cantor%E2%80%93Bernstein%E2%80%93Schroeder_theorem) could be used to show that this has the cardinality of $\mathbf{R}$.2011-11-13

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Well, $\mathbb{Q}_p\subset \mathbb{C}$ ...

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    What is $\mathbb C$?2011-11-13
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    What sort of help is that??? $\{0\}\subseteq\mathbb C$ as well...2011-11-13
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    @Asaf It would show that $\mathbf{Q}_p$ isn't larger than $\mathbf{C}$. And he has the other inequality. What's bogus about this? Besides showing that there is such an embedding :)2011-11-13
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    Sorry for the confusion, I was answering to the part of the question asking "is it bigger than continuum ?" : By $\mathbb{C}$, I indeed meant the fields of complex numbers, but I agree that "$\subset$" is a bit vague. What I have in mind is that $\mathbb{C}$ is fields-isomorphic to $\mathbb{C}_p$, the fields of complex $p$-adic numbers, which contains $\mathbb{Q}_p$ and thus gives an upper bound for its cardinal, but maybe I commit a reasoning mistake !2011-11-13
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    Student, you should aim for a complete and comprehensible answer, not one that is "the first" requires a later addition in the comments.2011-11-13
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The field $\mathbb Q_p$ is the fraction field of $\mathbb Z_p$.

Since you already know that $|\mathbb Z_p|=2^{\aleph_0}$, let us show that this is also the cardinality of $\mathbb Q_p$:

Note that every element of $\mathbb Q_p$ is an equivalence class of pairs in $\mathbb Z_p$, much like the rationals are with respect to the integers.

Since $\mathbb Z_p\times\mathbb Z_p$ is also of cardinality continuum, we have that $\mathbb Q_p$ can be injected into this set either by the axiom of choice, or directly by choosing representatives which are co-prime.

This shows that $\mathbb Q_p$ has at most continuum many elements, since $\mathbb Z_p$ is a subset of its fraction field, then the $p$-adic field has exactly $2^{\aleph_0}$ many elements.

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To complete Student's answer : $\mathbb Z_p \subset \mathbb Q_p \subset \mathbb C_p$, and since $\mathbb C_p$, the algebraic closure of $\mathbb Q_p$, is isomorphic to $\mathbb C$, the field of the complex numbers, we know that $$ |\mathbb R| = |\mathbb Z_p| \le |\mathbb Q_p| \le |\mathbb C_p| = |\mathbb C| = |\mathbb R| \quad \Longrightarrow \quad |\mathbb Q_p| = |\mathbb R|. $$ by the Cantor-Bernstein theorem, which states that if $|A| \le |B|$ and $|B| \le |A|$ then $|A| = |B|$.

EDIT : One REALLY more easy way to see this (in the sense that I can assume less non-trivial things) is that we get an injection from $\mathbb R$ to $\mathbb Q_p$ (and vice-versa) by choosing one representation for each element on each side and associate them in the following manner : $$ \sum_{k=-\infty}^n a_k p^k = \underset{\in \mathbb R}{x} \longleftrightarrow \underset{\in \mathbb Q_p}{y} = \sum_{k=-n}^{\infty} a_{-k} p^k $$ (i.e. you flip the number over) which doesn't give us a bijection but rather two surjections. (If I wanted to have a bijection with this I would have trouble with the possible two representations of a number on each side.) Thus I can apply Cantor-Bernstein.

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    You write "...and since $\mathbb C_p$, the algebraic closure of $\mathbb Q_p$, is isomorphic to $\mathbb C...$" This is far from evident and I am not sure that there is no circularity here. How do you justify this claim?2011-11-13
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    I am not saying it is trivial, I'm just saying it's true, and I don't think that an isomorphism between $\mathbb C_p$ and $\mathbb C$ can possibly assume that $\mathbb Q_p = \mathbb R$... I'll take a look at the actual proof and give you a hint on how to justify this.2011-11-13
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    Whew. It is definitively non-trivial. Quote from Wikipedia, on the article p-adic number r : "The field $\mathbb C_p$ is isomorphic to the field $\mathbb C$ of complex numbers, so we may regard $\mathbb C_p$ as the complex numbers endowed with an exotic metric. It should be noted that the proof of existence of such a field isomorphism relies on the axiom of choice, and does not provide an explicit example of such an isomorphism."2011-11-13
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    I am having trouble making my new solution understandable with the notation ; the idea is just to flip the number over itself.2011-11-13
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    @PatrickDaSilva: You need to check that the map is well defined. It also gives two surjections, not two injections.2013-05-13
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    The answer before the edit is circular. The way one shows that $\Bbb C_p$ is isomorphic to $\Bbb C$ is by using the fact that two algebraically closed fields of the same characteristic and the same cardinality are isomorphic. So one cannot use this isomorphism to determine the cardinality of $\Bbb C_p$.2018-07-23
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    For the two surjections you mentioned it isn't immediately clear why they are surjections. For a real number $x \in \mathbb{R}$, it doesn't have a unique representation as a series in the form you wrote it, as you mentioned. Therefore, for a a choice of such a mapping, we a do not reach all laurent series with coefficients between $0$ and $p-1$, and thus we dont know if your map reaches all p-adic numbers $\mathbb{Q}_p$.2018-10-22
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    Where I come from, $\Bbb C_p$ is standard notation for the *completion of* an algebraic closure of $\Bbb Q_p$, not that algebraic closure itself. They happen to have the same cardinality and are both algebraically closed (which by the way means they are isomorphic as fields as well, which just shows how crazy that whole argument about field isomorphisms with AC is), but one is strictly contained in the other, one is complete and the other is not.2018-12-19