Is $\mathbb{R}$ a vector space over $\mathbb{C}$? 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.
 A: 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.
A: 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.
