Isomorphism of Vector spaces over $\mathbb{Q}$ 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!
 A: 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)!. 
A: Two vector spaces whose dimensions over a field are of the same cardinality are easily seen to be isomorphic by mapping one basis onto the other; i.e. if $V$ has basis ${e_i}_{i\in I}$ and $W$ has basis ${f_j}_{j\in J}$ where $I$ and $J$ are of the same cardinality, pick a bijection $\phi:I\rightarrow J$ and define a linear map by sending $e_i\mapsto f_{\phi(i)}$. Since the $e_i$ map bijectively to the $f_j$, this must be an isomorphism.
A: This answer was motivated by some comments of Djaian. As Arturo said, two vector spaces over the same field are isomorphic if and only if they have bases of the same cardinality. The infinite dimensional case is valid in much general context. So the important property of fields is that it is true for finite-dimensional spaces. 
To justify this we'll talk about free modules over a ring with identity. Roughly speaking a module is  a generalization of vector space to arbitrary rings and the word free means that the module has a basis. We have the following result:

If we consider a ring $R$ with identity and a free $R$-module $F$ with an infinite basis $X$, then every basis of $F$ has the same cardinality as $X$.

The proof is based on the Cantor–Schröder-Bernstein theorem and is similar to the one Arturo gave for vector spaces. It needs the Axiom of Choice too. Thus if $F_1$ and $F_2$ are two free modules with infinite bases $X_1$ and $X_2$, respectively, then $F_1 \cong F_2$ if and only if $|X_1| = |X_2|$.
To see how badly it can be, there is a ring $R$ such that, as a free left $R$-module, $R$ has a basis of $n$ elements, for each positive integer $n$. 
We say that a ring with identity has invariant basis number  (IBN) if for every free $R$-module $F$, any two bases of $F$ have the same cardinality. The cardinal number of any basis of $F$ is called the rank (or dimension) of $F$ over $R$. It's easy to see that if $R$ is a ring with IBN and $F_1$ and $F_2$ are two free modules over $R$, then $F_1 \cong F_2$ if and only if they have the same rank.
Any nonzero commutative ring has IBN. Also any division ring has IBN. Other examples include finite rings, left-Noetherian rings, local rings and group rings.
A: All finite dimensional vector spaces of the same dimension over a field $k$ are isomorphic.  Fix a basis of $V$ and $W$; they each have $n$ elements if $V$ and $W$ are $n$-dimensional.  Then the map that sends basis elements $e_i$ of $V$ to basis elements $f_i$ of $W$ is an isomorphism.
We usually say that if $V$ and $W$ are vector spaces over $k$ of the same dimension, they are isomorphic, but not canonically.  This is because to exhibit an isomorphism requires a choice of basis.
