Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@QiaochuYuan: What about the question don't you understand? – Thomas Jun 6 '12 at 23:25
@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
up vote 10 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.

share|cite|improve this answer

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.

share|cite|improve this answer
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
The cardinality of the vector space equals the cardinality of the basis times the cardinality of the base field. If $|\Bbb Q|=\aleph_0$ and $|V|=\mathcal{c}$ then... – anon Jun 6 '12 at 23:25
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
@MarkDominus: Without assuming the continuum hypothesis, you don't get that uncountable => at least continuum-sized. – Noah Stein Jun 6 '12 at 23:37
So what? I don't need CH to know that $|{\Bbb R}| > \aleph_0$, and that's all I need here. – MJD Jun 6 '12 at 23:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.