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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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The same question for $\mathbb{R}$ instead of $\mathbb{Q}_p$ has come up before:… – David Loeffler Jul 6 '11 at 20:57
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. – David Roberts Jul 7 '11 at 3:17
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. – David Loeffler Jul 7 '11 at 9:51

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.

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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 $\mathbb Q$.

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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. – Wilhelm Stevens Jul 6 '11 at 19:57

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