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Can we make the real numbers into a $\mathbb{C}$-vector space?

We should take the real numbers as a group and the complex numbers as our ring. Please tell me what is the operation?

user26857
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kpax
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    Hi. Welcome to Math.SE. It will be easier to help you if you can write more in your question describing what you've thought of or tried. – Sammy Black Sep 27 '13 at 08:14
  • Suppose that you had some map $\mathbb{C} \times \mathbb{R} \to \mathbb{R}$. What two properties would it have to satisfy in order to define $\mathbb{R}$ as a $\mathbb{C}$-module? – Sammy Black Sep 27 '13 at 08:16
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    If it is possible, I think the multiplication will not be continuous. – Karl Kroningfeld Sep 27 '13 at 12:03
  • @KarlKronenfeld yes you are right, thank you I removed my comment. – Math137 Sep 27 '13 at 12:13
  • If you want the endomorphisms of $\mathbb R$ be continuous then they are determined by where they send $1 \in \mathbb R$ and so the endomorphism ring is isomorphic to $\mathbb R$. A $\mathbb C$-module would then be a ring homomorphism $\mathbb C \to \mathbb R$ which is impossible since $\mathbb C$ is a field. If you allow discontinuous endomorphisms then perhaps the answer might be positive. – Marek Sep 27 '13 at 12:48

2 Answers2

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Yes, we can!

$\mathbb R$ is a $\mathbb C$-module iff there exists a ring homomorphism $\mathbb C\to\operatorname{End}(\mathbb R)$.

A group endomorphism of $\mathbb R$ is a $\mathbb Q$-linear map. Now let's define such a linear map which satisfies the condition $(f\circ f)(x)=-x$ for all $x\in\mathbb R$. Let $(x_i)_{i\in I}$ be a $\mathbb Q$-basis of $\mathbb R$ and write $I$ as a disjoint union of subsets made of two elements. For such a subset, say $\{u,v\}$, define $f(x_u)=x_v$ and $f(x_v)=-x_u$. Now check that $(f\circ f)(x_i)=-x_i$ for all $i\in I$.

Then define a ring homomorphism $\phi:\mathbb C\to\operatorname{End}(\mathbb R)$ by $\phi(a+bi)=\phi_a+\phi_b\circ f$, where $\phi_a(x)=ax$, respectively $\phi_b(x)=bx$ for all $x\in\mathbb R$.

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    $\phi$ defined here is not multiplicative. If it were, we would have $\phi_a\circ f = \phi(ai) = \phi(ia) = \phi(i)\circ\phi(a) = f\circ \phi_a$, so $f$ would be $\mathbb R$-linear. Then, $f$ would necessarily be of the form $f(x) = cx$ for some real $c$, so $f(f(x)) = c^2x$, but $c^2 \neq -1$. Of course, the problem is that $\mathrm{End}(\mathbb R)$ is not a commutative ring or $\mathbb R$-algebra, so you can't simply define $\phi(1)$ and $\phi(i)$ and extend by linearity. – Ennar Jul 24 '23 at 01:56
  • In spite of its 8 upvotes and accept, this answer is wrong. – user26857 Jul 26 '23 at 07:30
  • For a correct answer see here. – user26857 Jul 26 '23 at 07:54
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As an abelian group, $\mathbb{R}$ is uniquely divisible (that is, for each $x$ and nonzero integer $n$, there is a unique $y$ so that $x = n \cdot y$), so $\mathbb{R}$ is isomorphic to a vector space over $\mathbb{Q}$. (there's a more obvious proof that $\mathbb{R}$ is a rational vector space -- the point is that "uniquely divisible abelian group" means the same thing as "rational vector space", so I will talk about the latter instead)

In fact, (I'm surely assuming axiom of choice), $\mathbb{R}$ is a $\mathfrak{c}$-dimensional rational vector space, where $\mathfrak{c}$ is the cardinality of $\mathbb{R}$.

$\mathbb{C}$ is also a $\mathfrak{c}$-dimensional rational vector space. Therefore, $\mathbb{C} \cong \mathbb{R}$ as rational vector spaces, and thus as abelian groups.

Thus, your question is equivalent to the following:

Can we make the complex numbers into a $\mathbb{C}$-vector space?

We should take the complex numbers as a group and the complex numbers as our ring. Please tell me what is the operation?

user26857
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  • While the "question is equivalent to the following" seems to be obvious, the details are missing. We have to show how to transport the $\mathbb C$-vector space structure from $\mathbb C$ to $\mathbb R$ via an isomorphism of addive groups. – user26857 Jul 26 '23 at 07:39