Q: If we consider $\mathbb{R}$ and $\mathbb{C}$ as $\mathbb{Q}$-vector spaces, then how can we show they are isomorphic?

I know that if two vector spaces have bases with the same cardinality, then they are isomorphic. Also, Zorn's lemma tells us that every vector space has a basis.

In this case, answering my question amounts to showing that any bases of $\mathbb{R}$ and $\mathbb{C}$ over $\mathbb{Q}$ have the same cardinality. In other words, I need to show dim$ \mathbb{R} =$ dim $\mathbb{C}$ over $\mathbb{Q}$, that is, they have bases with the same cardinality. Can anyone help?

Thank you!!

  • $\begingroup$ I am pretty sure that this question has been asked on this site before. Can anyone find the previous question? $\endgroup$ Mar 13 '11 at 21:24
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    $\begingroup$ $\text{dim}_{\mathbb{Q}}(\mathbb{C})=2\text{dim}_{\mathbb{Q}}(\mathbb{R})$ $\endgroup$
    – yoyo
    Mar 14 '11 at 14:59

If $k$ is a field and $V$ is an infinite dimensional $k$-vector space, then $V\cong V\oplus V$. Using this, what you want to show follows from the fact that as $\mathbb Q$-vector spaces, $\mathbb C\cong \mathbb R\oplus \mathbb R$.

Can you see how to prove those two claims?

Later. Ok, apparently not. Let's do it.

(1) If $B$ is a basis for $V$, then the set $$B'=\{(b,0):b\in B\}\cup \{(0,b):b\in B\}$$ is a basis for $V\oplus V$. There is an obvious bijection $B'\cong \{1,2\}\times B$.

Now, if $X$ is an infinite set, then $X$ and $\{1,2\}\times X$ are in bijection. It follows from this that there is a bijection between the basis $B$ of $V$ and the basis $B'$ of $V\oplus V$. As you know, this implies that there is a linear isomorphism between $V$ and $V\oplus V$. This proves my first claim above.

(2) Consider the map $$\phi:a+bi\in\mathbb C\mapsto (a,b)\in\mathbb R\oplus\mathbb R.$$ It is very easy to show that it is an isomorphism of $\mathbb Q$-vector spaces, so that $\mathbb C\cong\mathbb R\oplus\mathbb R$, as my second claim states.

(3) Finally, let's prove your claim that if $\mathbb R$ and $\mathbb C$ are isomorphic $\mathbb Q$-vector spaces: since $\mathbb R$ is a $\mathbb Q$-vector space of infinite dimensions, my first claim tells us that $\mathbb R\cong\mathbb R\oplus\mathbb R$ as $\mathbb Q$-vector spaces. On the other hand, my second claim tells us that $\mathbb R\oplus\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces. Transitivity, then, allows us to conclude that $\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces.

  • $\begingroup$ Thank you for your prompt reply. However, I didn't get the idea. $\endgroup$
    – yaa09d
    Mar 13 '11 at 21:14
  • $\begingroup$ If you prove the first statement, then you can show that $\mathbb{R} \cong \mathbb{R} \oplus \mathbb{R}$ as a vector space over $\mathbb{Q}$. Then you can apply the second claim. $\endgroup$ Mar 13 '11 at 21:44
  • $\begingroup$ @yaa09d: can you be more explicit about what you do not get? I wrote two claims: with which of the two you have problems? $\endgroup$ Mar 13 '11 at 22:00
  • $\begingroup$ I think I didn't understand why the first claim holds. The second claim is what I was trying to show. $\endgroup$
    – yaa09d
    Mar 13 '11 at 22:22
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    $\begingroup$ @yaa09d: no, my second claim is not what you were trying to show: you were trying to show that $\mathbb R\cong\mathbb C$ as $\mathbb Q$-vector spaces. $\endgroup$ Mar 14 '11 at 2:46

Two ways: one using only the fact that $|\mathbb{R}|=|\mathbb{C}|=\mathfrak{c}$; the other using the fact that we can identify the additive structure of $\mathbb{C}$ with the plane.

  1. If $\mathbf{F}$ is a field and $\mathbf{V}$ is a vector space over $\mathbf{V}$, what is the cardinality of $\mathbf{V}$? If $\beta$ is a basis for $\mathbf{V}$, then every vector of $\mathbf{V}$ can be written uniquely as an $\mathbf{F}$-linear combination of vectors in $\beta$. Therefore, there is a bijection between the elements of $\mathbf{V}$ and the set $$\bigl\{ f\colon\beta\to \mathbf{F}\bigm| f(\mathbf{b})=\mathbf{0}\text{ for almost all }\mathbf{b}\in\beta\bigr\}.$$ That is, the set of functions of finite support from $\beta$ to $\mathbf{F}$, $\mathbf{F}^{(\beta)}$. If $\mathbf{F}$ is infinite, then this cardinality is equal to $|\mathbf{F}||\beta|=\max\{|\mathbf{F}|,|\beta|\}$. Here, $\mathbf{F}=\mathbb{Q}$. So if $\beta$ is a basis for $\mathbb{R}$ over $\mathbb{Q}$, we need $|\mathbb{R}| = \aleph_0|\beta|$. What is $|\beta|$? For $\mathbb{C}$, we need $|\mathbb{C}| = \aleph_0|\gamma|$, where $\gamma$ is a basis for $\mathbb{C}$ over $\mathbb{Q}$. What is $|\gamma|$?

  2. Since the additive structure of $\mathbb{C}$ is just the same as the additive structure of $\mathbb{R}\oplus\mathbb{R}$, then they are isomorphic as $\mathbb{Q}$-vector spaces. So if $\{\mathbf{v}_{b}\}_{b\in\beta}$ is a basis for $\mathbb{R}$ over $\mathbb{Q}$, then $$\bigl\{\mathbf{v}_{b}\bigr\}_{b\in\beta}\cup \bigl\{i\mathbf{v}_{b}\bigr\}_{b\in\beta}$$ is a basis for $\mathbb{C}$ over $\mathbb{Q}$ (prove it). So $\dim_{\mathbb{Q}}(\mathbb{C}) = 2\dim_{\mathbb{Q}}(\mathbb{R})$. If the dimension of $\mathbb{R}$ over $\mathbb{Q}$ were finite, this would show they are not isomorphic. But is the dimension finite or infinite? And what does that tell you about the dimensions?

  • $\begingroup$ I think I understood your proof. Thank you very much. $\endgroup$
    – yaa09d
    Mar 13 '11 at 21:55
  • $\begingroup$ What you mentioned in 1 will imply that $|\beta|=|\mathbb{R}|$ and $|\gamma|=|\mathbb{C}|$. I am ok with that, but basically why did you say that $|\mathbb{R}| = \aleph_0|\beta|$? $\endgroup$
    – yaa09d
    Mar 13 '11 at 22:08
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    $\begingroup$ @yaa09d: As I say earlier, if $\mathbf{V}$ is a vector space with basis $\beta$ over the infinite field $\mathbf{F}$, then $|\mathbf{V}|=|\mathbf{F}||\beta|$. Here, $\mathbf{F}=\mathbb{Q}$, so simply substituting into that equation, we have:$$|\mathbb{R}|=|\mathbb{Q}||\beta|=\aleph_0|\beta|.$$ $\endgroup$ Mar 13 '11 at 22:11
  • $\begingroup$ I see. I think I got it. I appreciate your help. Thank you. $\endgroup$
    – yaa09d
    Mar 13 '11 at 22:17
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    $\begingroup$ @yaa09d: The functions from $\beta$ to $\mathbf{F}$ that have finite support have that cardinality when $\mathbf{F}$ is infinite. For finite $\beta$, this is immediate (cardinality of $\mathbf{V}$ is just the cardinality of $\mathbf{F}$); for infinite $\beta$, you can decompose the functions by the cardinality of the support, and there are $|\beta|$ finite subsets of $\beta$; so you get a sum of $|\beta|$ terms, each of cardinality $|\mathbf{F}|$. $\endgroup$ Mar 14 '11 at 13:14

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