I was looking for a reference that illustrates a $\mathbb{Q}$-vector space basis for the field of p-adic numbers under the following action. Given a rational number $q$. write, $q=\frac{m}{n}$ where $n>0$. Then, for $x\in \mathbb{Q_p}$ $qx=y$ where $y\in \mathbb{Q_p}$ is the unique element s.t. $mx=ny$.
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$\begingroup$ The same question for $\mathbb{R}$ instead of $\mathbb{Q}_p$ has come up before: mathoverflow.net/questions/46063/… $\endgroup$– David LoefflerCommented Jul 6, 2011 at 20:57
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$\begingroup$ Consider the cardinality of a Hamel basis of $\mathbb{Q}_p$.... For $\mathbb{R}$ it is uncountable, and this should give you pause as to how to even describe an 'explicit' basis (as per your comment on Andreas' answer). We need the axiom of choice in this instance, and by definition, you can't write down the result of applying a choice function conjured up by using Choice. $\endgroup$– David Roberts ♦Commented Jul 7, 2011 at 3:17
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$\begingroup$ Why was this downvoted? It's asked in a rather ponderous way (the $\mathbb{Q}$-vector space structure on $\mathbb{Q}_p$ is selfevident) but it's a valid, and interesting, question. $\endgroup$– David LoefflerCommented Jul 7, 2011 at 9:51
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$\begingroup$ Would such a basis have finitely many elements or infinitely many? $\endgroup$– Robert FrostCommented Sep 8, 2023 at 19:28
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2 Answers
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The arguments should mirror those for the reals. For example: If you have a Hamel basis for $\mathbb Q_p$, then you can construct a set that fails the property of Baire, but it is consistent with ZF that every set in a Polish space has the property of Baire.
answered Jul 6, 2011 at 21:30
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I'm not sure what is meant by "illustrating" a basis, but the axiom of choice is needed even to prove the existence of a basis for $\mathbb Q_p$ over $\mathbb Q$.
answered Jul 6, 2011 at 19:16
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1$\begingroup$ I was hoping to see an explicit set of elements of $\mathbb{Q_p}$ which serve as a basis for the above vector space structure. As I understand, the axiom of choice is needed to show that every infinite dimensional vector space has a basis. Yet, we can construct explicit basis in certain cases. For example, the set of polynomials in a variable $x$ over a field forms a vector space over the ground field. The set of power functions $x^n$ forms a basis of this vector space. $\endgroup$ Commented Jul 6, 2011 at 19:57
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$\begingroup$ You need choice to construct a Vitali set but are you entirely sure one is needed to prove its existence? Seems to me we just define the set of equivalence classes say $\Bbb R/\Bbb Q$ and then we know every one of its elements exists, is nonempty, and comprises mutually disjoint subsets of $\Bbb R$ therefore a Vitali set exists, no? $\endgroup$ Commented Sep 13, 2023 at 15:30
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$\begingroup$ @it'sahirecarbaby No. The axiom of choice is exactly what's needed to justify "therefore" at the end of your comment. $\endgroup$ Commented Sep 13, 2023 at 16:09
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$\begingroup$ Thank-you. Not related to this question but I see from your bio you're something of an expert in ultrafilters on the natural numbers, which intersects with an interest of mine as described here: math.stackexchange.com/questions/4763339 The ultrafilter(s) in here are also ultrafilters if you restrict them to the natural numbers. Equating this to the famous open problem is primary research you will not find elsewhere. Any comment or direction how to progres that question would be greatly appreciated. $\endgroup$ Commented Sep 15, 2023 at 15:59