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Here is a problem so beautiful that I had to share it. I found it in Paul Halmos's autobiography. Everyone knows that $\mathbb{C}$ is a vector space over $\mathbb{R}$, but what about the other way around?

Problem: Prove or disprove: $\mathbb{R}$ can be written as vector space over $\mathbb{C}$

Of course, we would like for $\mathbb{R}$ to retain its structure as an additive group.

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Depends on what structures on $\mathbb{R}$ you want to preserve. I don't think this is a well-specified question. –  Qiaochu Yuan Jun 6 '12 at 23:08
    
All you want to preserve is the structure as an additive group. –  Potato Jun 6 '12 at 23:12
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@QiaochuYuan: What about the question don't you understand? –  Thomas Jun 6 '12 at 23:25
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@Thomas: there's nothing I don't understand. The question as it is written just doesn't specify what structure on $\mathbb{R}$ is supposed to be retained. For example, by restriction of scalars, every complex vector space is also a real vector space. Is the induced real vector space structure on $\mathbb{R}$ supposed to have any relationship to its usual real vector space structure? Halmos doesn't specify this. –  Qiaochu Yuan Jun 6 '12 at 23:36
    
The structure was specified when I first posted the problem. It's in the last line. –  Potato Jun 6 '12 at 23:38

2 Answers 2

up vote 9 down vote accepted

If you want the additive vector space structure to be that of $\mathbb{R}$, and you want the scalar multiplication, when restricted to $\mathbb{R}$, to agree with multiplication of real numbers, then you cannot.

That is, suppose you take $\mathbb{R}$ as an abelian group, and you want to specify a "scalar multiplication" on $\mathbb{C}\times\mathbb{R}\to\mathbb{R}$ that makes it into a vector space, and in such a way that if $\alpha\in\mathbb{R}$ is viewed as an element of $\mathbb{C}$, then $\alpha\cdot v = \alpha v$, where the left hand side is the scalar product we are defining, and the right hand side is the usual multiplication of real numbers.

If such a thing existed, then the vector space structure would be completely determined by the value of $i\cdot 1$: because for every nonzero real number $\alpha$ and every complex number $a+bi$, we would have $$(a+bi)\cdot\alpha = a\cdot \alpha +b\cdot(i\cdot \alpha) = a\alpha + b(i\cdot(\alpha\cdot 1)) = a\alpha + b\alpha(i\cdot 1).$$ But say $i\cdot 1 = r$. Then $(r-i)\cdot 1 = 0$, which is contradicts the properties of a vector space, since $r-i\neq 0$ and $1\neq \mathbf{0}$. So there is no such vector space structure.

But if you are willing to make the scalar multiplication when restricted to $\mathbb{R}\times\mathbb{R}$ to have nothing to do with the usual multiplication of real numbers, then you can indeed do it by transport of structure, as indicated by Chris Eagle.

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As an additive group, $\mathbb{R}$ is isomorphic to $\mathbb{C}$ (they're both continuum-dimensional rational vector spaces). Clearly $\mathbb{C}$ can be given the structure of a $\mathbb{C}$-vector space, thus $\mathbb{R}$ can too.

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This is correct. If you could expand on why they are isomorphic, that would great. –  Potato Jun 6 '12 at 23:15
    
The expansion is in the parenthesis: They are both continuum-dimensional rational vector spaces. What more is there to say? Things that are isomorphic as vector spaces are ipso facto also isomorphic as additive groups. –  Henning Makholm Jun 6 '12 at 23:16
    
Perhaps I'm being dense, but why is a Hamel basis continuum-dimensional? –  Potato Jun 6 '12 at 23:19
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Wouldn't the existence of a countable basis for $\Bbb R$ over $\Bbb Q$ immediately imply that $\Bbb R$ was a countable union of countable sets and therefore countable? –  MJD Jun 6 '12 at 23:27
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@MarkDominus: Without assuming the continuum hypothesis, you don't get that uncountable => at least continuum-sized. –  Noah Stein Jun 6 '12 at 23:37

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