From this post we see that $\mathbb{R}$ over $\mathbb{Q}$ is infinite dimensional. Similarly $\mathbb{C}$ over $\mathbb{Q}$ is also infinite dimensional, and I rememeber having solved a problem that $\mathbb{R} \cong \mathbb{C}$ when considered as vector spaces over $\mathbb{Q}$. So I would like to extend this question.
Suppose $V$ and $W$ are two vector spaces, over the field $\mathbb{Q}$. Can we say that $V \cong W$ if they have the same dimension over $\mathbb{Q}$.
I really don't know how to prove or disprove this!
Best Answer
It is not hard to prove that for any vector space $\mathbf{V}$, if $\beta$ and $\beta'$ are two bases for $\mathbf{V}$, then the cardinality of $\beta$ equals the cardinality of $\beta'$. For finite dimensional spaces this is standard. For infinite dimensional, for each $x\in\beta$ let $\beta'_x$ be a finite subset of $\beta'$ such that $x\in\mathrm{span}(\beta'_x)$. Then $\cup\beta'_x = \beta'$, so $|\beta'|\leq \aleph_0|\beta|=|\beta|$. Now repeat the process for $y\in\beta'$ to get the other inequality. Edit: Note that this assumes the Axiom of Choice, so that we can talk about the cardinals of $\beta$ and $\beta'$. In the absence of AC, it could be that $\beta$ and $\beta'$ are "incomparable" (no injections going either way), so we cannot conclude that any two bases have the same cardinality without AC.
Fix a field $F$, and let $\mathbf{V}$ and $\mathbf{W}$ be vector spaces over $F$. Assuming the Axiom of Choice we can prove that every vector space has a basis (in fact, "Every vector space over any field has a basis" is equivalent to the Axiom of Choice). So let $\beta=\{v_i\}_{i\in I}$ be a basis for $\mathbf{V}$, and let $\gamma=\{w_j\}_{j\in J}$ be a basis for $\mathbf{W}$.
If $\mathbf{V}$ is isomorphic to $\mathbf{W}$, then let $\varphi\colon\mathbf{V}\to\mathbf{W}$ be an isomorphism. Then $\varphi(\beta)$ is a basis for $\mathbf{W}$, and since $\varphi(\beta)$ and $\gamma$ are both bases of $\mathbf{W}$, and $\varphi$ is a bijection, you have $|\beta|=|\varphi(\beta)|=|\gamma|$. So if $\mathbf{V}$ and $\mathbf{W}$ are isomorphic, then they have bases of the same cardinality.
Conversely, suppose $|\beta|=|\gamma|$. Let $f\colon\beta\to\gamma$ be a bijection, and let $\varphi\colon \mathbf{V}\to\mathbf{W}$ be the unique linear transformation that extends $f$ (that is, extend $f$ linearly to all of $\mathbf{V}$). Since $\varphi(\mathbf{V}) = \varphi(\mathrm{span}(\beta)) = \mathrm{span}(\varphi(\beta)) = \mathrm{span}(f(\beta)) = \mathrm{span}(\gamma)=\mathbf{W}$, then $\varphi$ is onto. It is straightforward to verify that $\varphi$ is one-to-one (it takes a basis to a basis). So $\varphi$ is an isomorphism. Thus, if $\mathbf{V}$ and $\mathbf{W}$ have bases of the same cardinality, then $\mathbf{V}$ is isomorphic to $\mathbf{W}$.
Thus, two vector spaces over the same field are isomorphic if and only if they have bases of the same cardinality, if and only if they have the same dimension (assuming the Axiom of Choice).
In particular, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as vector spaces over $\mathbb{Q}$, since they both have dimension $2^{\aleph_0}$. But you have to specify over what field you are working: $\mathbb{R}$ and $\mathbb{C}$ are not isomorphic as vector spaces over $\mathbb{R}$ (dimensions 1 and 2, respectively)!.