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We all know that $ \mathbb{C} $ is naturally a vector space over $ \mathbb{R} $. However, is there some kind of (possibly weird) scalar multiplication law that would make $ \mathbb{R} $ a vector space over $ \mathbb{C} $ instead?

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The best result that I've managed to produce is to show that $ \mathbb{R} $ is a vector space over $ \mathbb{Q} + i \mathbb{Q} $, the field of complex rationals. – Haskell Curry Oct 5 '12 at 19:11
Do you just want a vector space, or an algebra? – Asaf Karagila Oct 5 '12 at 19:13
I only need a vector space. Thanks. – Haskell Curry Oct 5 '12 at 19:14
up vote 10 down vote accepted

Yes. As an additive group, $\mathbb{R}$ is isomorphic to $\mathbb{C}$ (you can see this, for example, from the fact that they're both continuum-dimensional vector spaces over $\mathbb{Q}$). Since the additive group $\mathbb{C}$ can be made into a complex vector space, so can the isomorphic group $\mathbb{R}$.

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Well, note that as abelian groups $(\mathbb C,+)$ and $(\mathbb R,+)$ are isomorphic. Since a vector space is merely an abelian group with scalar multiplication you can just pick a homomorphism between $\mathbb R$ and $\mathbb C$ as additive groups, and use that to define a vector space.

In fact you can do the same trick with any finitely dimensional space over $\mathbb C$.

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Hi Asaf. I believe you meant 'pick an isomorphism between $ \mathbb{C} $ and $ \mathbb{R} $'. That is the only way one can transfer the vector space structure from $ \mathbb{C} $ to $ \mathbb{R} $. – Haskell Curry Oct 7 '12 at 2:45

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