Is $\mathbb{R}$ a free $\mathbb{Q}$-module? $\mathbb{R}$ is the real numbers and $\mathbb{Q}$ is the rational numbers. I seem to get that the algebraic closure on $\mathbb{Q}$ is a free $\mathbb{Q}$-module, by adding linearly independent elements and using Zorn's lemma. For the transcendental elements, I'm not sure how to prove it.
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14Every vector space over a field is free. – Randall May 30 '24 at 14:01
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Oh, I get it. As a vector space, it's trivial. Thank you. – cute dunkey May 30 '24 at 14:15
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5@cutedunkey Not trivially, you still need Zorn's lemma, fundamentally. Since the proof has to pass through Zorn or some other assumption as strong as it, and you don't see how to compose a proof with it, just saying "it is a vector space" does not constitute a proof or any better understanding. – ameg May 30 '24 at 14:29
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@ameg So with the help of Zorn's lemma, I can get bases for any vector space. And using it, We can find bases for $\mathbb{R}$ over $\mathbb{Q}$. – cute dunkey May 30 '24 at 15:09
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This is surprisingly delicate, as indicated in the comments, so worth clarifying a bit. The axiom of choice / Zorn's lemma can be used to prove that every vector space has a basis and hence that every module over a field is free; this statement is in fact equivalent to AC, and so is independent of ZF, and hence can be false in models of ZF where AC is false.
In fact even the existence of a basis for $\mathbb{R}$ over $\mathbb{Q}$ is independent of ZF. This is because such a basis allows you to construct a Vitali set (or something like it, anyway), which is non-measurable, and Solovay constructed a model of ZF in which all subsets of $\mathbb{R}$ are measurable.
With a little more work you can show that it's independent of ZF whether there exists even a single nonzero $\mathbb{Q}$-linear functional $\mathbb{R} \to \mathbb{Q}$. In other words, it's consistent with ZF that infinite-dimensional vector spaces can have trivial dual.
Qiaochu Yuan
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