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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 could be used to show that this has the cardinality of $\mathbf{R}$. –  Dylan Moreland Nov 13 '11 at 21:38

3 Answers 3

up vote 8 down vote accepted

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 injections. (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? –  Georges Elencwajg Nov 13 '11 at 22:55
    
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. –  Patrick Da Silva Nov 13 '11 at 23:05
    
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." –  Patrick Da Silva Nov 13 '11 at 23:08
    
I am having trouble making my new solution understandable with the notation ; the idea is just to flip the number over itself. –  Patrick Da Silva Nov 13 '11 at 23:17
    
@PatrickDaSilva: You need to check that the map is well defined. It also gives two surjections, not two injections. –  Eric Naslund May 13 '13 at 14:44

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

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What is $\mathbb C$? –  Hai Minh Nov 13 '11 at 21:31
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What sort of help is that??? $\{0\}\subseteq\mathbb C$ as well... –  Asaf Karagila Nov 13 '11 at 21:34
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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 :) –  Dylan Moreland Nov 13 '11 at 21:44
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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 ! –  Student Nov 13 '11 at 21:48
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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. –  Asaf Karagila Nov 13 '11 at 22:05

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