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The p-adics $\mathbb{Q}_p$ are uncountable (because they can be represented by infinite strings of integers in $[0,p-1]$) and hence must be infinite dimensional as a vector space over $\mathbb{Q}$. What is a concrete example of a basis of $\mathbb{Q}_p$ over $\mathbb{Q}$?

usr0192
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1 Answers1

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There is no concrete example for the same reason that there is no concrete example for a Hamel basis for $\Bbb R$ over $\Bbb Q$.

It is consistent with the failure of the axiom of choice that every linear functional from $\Bbb Q_p$ to $\Bbb Q$ is continuous, and therefore is completely decided by values on a countable set (since $\Bbb Q_p$ is separable). There can only be $2^{\aleph_0}$ of these functionals.

But if there is a basis its cardinality must be $2^{\aleph_0}$ and therefore we can generate $2^{2^{\aleph_0}}$ linear functionals. So in models where the above happens, there is no basis for $\Bbb Q_p$ over $\Bbb Q$. And this means that we can't write an explicit basis, and that we must rely on the axiom of choice for proving it exists.

Asaf Karagila
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  • I'm having a bit of trouble parsing your second paragraph. Are you saying that every linear functional form $\mathbb{Q}_p$ to $\mathbb{Q}$ is continuous? What topology are we using for $\mathbb{Q}$ here? – oxeimon Jan 19 '16 at 19:36
  • I believe you, but I don't follow the reasoning of anything you wrote. What is the small argument for how you are concluding $2^{\aleph_0}$ everywhere – usr0192 Jan 19 '16 at 19:41
  • Cardinal arithmetic. It's not hard to show that if $V$ is a vector space of size $2^{\aleph_0}$ over the rationals (like $\Bbb Q_p$) then its basis must have cardinality $2^{\aleph_0}$; in the case of the continuous functions as they are fully decided by a countable dense subset we can restrict to functions from that countable subset to the rationals, and there are exactly $2^{\aleph_0}$ of those; and finally for the $2^{2^{\aleph_0}}$, if you have a basis, every function from the basis to the field extends to a unique linear functional and there are that many of those. – Asaf Karagila Jan 19 '16 at 19:44
  • When it comes to logic, I’m an outsider looking in, but I find this answer very, very pleasing. – Lubin Jan 19 '16 at 19:46
  • @user90219: http://math.stackexchange.com/questions/1267196/cardinality-of-a-basis-of-an-infinite-dimensional-vector-space/1267596#1267596 has the full argument for the least trivial part, the cardinality of the basis. – Asaf Karagila Jan 19 '16 at 19:46