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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!

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In the question do you mean "having solved a problem that $\mathbb{R} \cong \mathbb{C}$"? –  Willie Wong Nov 4 '10 at 14:07
    
If I wanted to disprove this, I would take $2^{mathbb{R}}$, see if I can express this set as a vector space over $\mathbb{Q}$ and then prove no isomorphism is possible with $\mathbb{R}$ thanks to the diagonal argument from Cantor. This makes me think your assertion is not true for infinite dimensions, but I didn't actually checked the details. –  Djaian Nov 4 '10 at 14:11
    
I think the problem comes from what you exactly mean by having "the same dimension". If you mean the same cardinality then an isomorphism exists, like some people answered already. But, if you just say that 2 vector spaces with infinite dimension have the same dimension, then we could have a problem. For example, like I said you cannot find an isomorphism from 2^$\mathbb{R}$ and $\mathbb{R}$ considered as vector spaces over $\mathbb{Q}$. –  Djaian Nov 4 '10 at 21:59
    
@Djaian: Yes, that is the issue. Generally, "infinite" is not considered to be the dimension of a vector space. "Infinite dimensional" is the term to mean "does not have a finite basis", but this is not taken to mean "the dimension is equal to infinity." When talking about the dimension of a vector space, it almost invariably refers to the cardinality of a basis; i.e., a cardinal number, finite or infinite. –  Arturo Magidin Nov 4 '10 at 23:44
    
To prove that if two spaces have the same dimension (i.e., bases of the same cardinality) then they are isomorphic is almost trivial, and the answer by user1306 gives the proof. The somewhat harder part is to show that dimension is always well defined (as a cardinal number) in the first place, namely that one space cannot have two bases of different cardinalities; the accepte answer deals with that question (but not with the one stated). –  Marc van Leeuwen Feb 13 at 11:17

4 Answers 4

up vote 6 down vote accepted

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)!.

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@Aruturo: Yes, i was speaking over $\mathbb{Q}$ only. –  anonymous Nov 4 '10 at 15:05
    
@Arturo: As $\mathbb{R} \cong \mathbb{C}$ over $\mathbb{Q}$, what would be an example of a bijection between these two! –  anonymous Nov 4 '10 at 15:10
    
What happens when we don't assume Axiom of Choice? –  Djaian Nov 4 '10 at 15:49
    
@Chandru1: We cannot write down a basis (it requires the Axiom of Choice to even prove there is one), so we cannot write down a vector space isomorphism. But if we assume the Axiom of Choice, then there has to be one. –  Arturo Magidin Nov 4 '10 at 15:53
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@Chandru1: No: you asked "what would be an example of a bijection between these two!" (confusing the exclamation point with the question mark, as is your wont). The comment was a response to that. –  Arturo Magidin Nov 4 '10 at 15:56

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.

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What is the basis of $\mathbb{R}$ over $\mathbb{Q}$ ? –  Djaian Nov 4 '10 at 14:15
    
Couldn't say, but I didn't realize that was the question. Isn't it "Suppose V and W are two vector spaces, over the field Q. Can we say that V≅W if they have the same dimension over Q?" –  user1306 Nov 4 '10 at 14:17
    
@user1306 : I think Chandru (who asked the question) knows that vector spaces with same finite dimensions are isomorphic. As I understand his question, he wants to know if this is true also for infinite dimensions. –  Djaian Nov 4 '10 at 14:25
    
This onswer does not assume anything about the dimension; it is perfectly OK (except that it should not talk about the cardinality of the dimensions; dimensions are cardinalities). –  Marc van Leeuwen Feb 13 at 11:13

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.

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@Nuno: Cantor-Schröder-Bernstein by itself does not require the Axiom of Choice (by C-S-B, I mean "if there is an injection $f\colon A\to B$ and an injection $g\colon B\to A$, then there is a bijection $h\colon A\to B$"; the precise statement is important, as, e.g., if you change 'injection' to 'surjection', then AC becomes relevant). But the proof that any two infinite bases have the same cardinality may require choice when you are showing that $X$ injects into $X'$ and vice-versa. If you can establish the injections without AC, then equality follows without AC as well. –  Arturo Magidin Nov 4 '10 at 19:08
    
@Arturo: I see. I checked in wikipedia before posting and there says the same thing about Cantor-Schröder-Bernstein theorem. But what about cardinal arithmetic? As I remember to prove that $\alpha \beta$ = $\beta$, if $0 \ne \alpha \le \beta$ and $\beta$ is infinite, requires Zorn's lemma. Here we only need to prove that $\aleph_0 \beta = \beta$. Do we need the axiom of choice? P.S. I'm considering $\alpha$,$\beta$ cardinals. –  Nuno Nov 4 '10 at 19:49
    
@Nuno: yes, the problem here is whether you can show that there are injections going each way without AC; but if you can construct the injections then the bijection will follow without AC. I think you can prove the formula for the product/sum of infinite cardinals without AC; you need to consider the canonical ordering on $\alpha\times\alpha$ ($\alpha$ an ordinal) to show $\aleph_{\alpha}\aleph_{\alpha}=\aleph_{\alpha}$. The sum/product formula follows by C-S-B. (Theorem 8, p 25, of Jech's Set Theory). –  Arturo Magidin Nov 4 '10 at 20:00
    
@Nuno: However, the formula for cardinal arithmetic is not enough, since in the absence of AC, your sets $X$ and $X'$ may fail to be bijectable with a cardinal in the first place. –  Arturo Magidin Nov 4 '10 at 20:01
    
@Arturo: You are right. Thanks for helping me with these things. I've also edited the question clarifying that passage. –  Nuno Nov 4 '10 at 20:25

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.

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The important word in your answer is "finite". For infinite dimensional vector spaces, this is not so easy. And I think that is exactly the point of the question. –  Djaian Nov 4 '10 at 14:08
    
@Djaian: The only difficulty in infinite dimension is with the existence of bases; that any two bases for the same space have the same cardinality is not hard; that any two vector spaces with bases of the same cardinality are isomorphic works the same as in the finite dimensional case; and that if two vector spaces are isomorphic and one has a basis, then they both do and the bases have the same cardinality is easy. So the only issue is whether any infinite dimensional vector space has a basis or not. –  Arturo Magidin Nov 4 '10 at 18:11
    
@Arturo: sorry but I strongly disagree. Two vector spaces might have infinite dimension but basis of different cardinality. If $V$ has an $\aleph_0$ basis and $W$ an $\aleph_1$ basis, they cannot be isomorphic. –  Djaian Nov 4 '10 at 21:54
    
@Djaian: You are not disagreeing with me: any two infinite dimensional vector spaces will be isomorphic if and only if they have the same dimension. Here, "dimension" does not mean "finite or infinite", it means "cardinality of a basis". That is, "dimension" is not defined to be either "infinite" or a finite number, but rather "infinite dimensional" means "no finite basis", but the dimension of an infinite dimensional vector space is the cardinality of any basis (if there is one). –  Arturo Magidin Nov 4 '10 at 23:35
    
@Djaian: (cont) So the proposition "isomorphic if and only if they have the same dimension" follows immediately from the existence of a basis in any dimension, finite or infinite. But if there are vector spaces that don't have bases, then it is possible that two vector spaces are isomorphic and neither has a "dimension". And it is also possible, if not every vector space has a basis, to have to vector spaces that don't have "dimension" but that are not isomorphic. That's why I say that the only difficulty is with the existence of bases; if bases always exist, \problem is solved like in fin.dim –  Arturo Magidin Nov 4 '10 at 23:37

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