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Give a family of elements $\left(v_{1},\ldots,v_{n}\right)$ (where $v_{1},\ldots,v_{n}$ are just some sets), how many vector spaces are there, such that this family is a basis for that vector space ?

For example, if $n=1$ and $\left(v_{1}\right)=(1)$, there are at least two vector spaces, namely $\mathbb{Q}$ and $\mathbb{R}$ over themselves (as fields), such that this is a basis.

Is there even a reasonable way to answer this ?

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This is not really a question in set theory. – Asaf Karagila May 17 '12 at 17:36

You are talking about vector spaces over different fields: $\mathbb{Q}$ is a $\mathbb{Q}$-vector space for which $\{1\}$ is a basis, and $\mathbb{R}$ is an $\mathbb{R}$-vector space for which $\{1\}$ is a basis, but $\mathbb{R}$ is a $\mathbb{Q}$-vector space for which $\{1\}$ is clearly not a basis. Part of the structure of a vector space is the field of scalars, so $\mathbb{R}$-as-an-$\mathbb{R}$-vector space and $\mathbb{R}$-as-a-$\mathbb{Q}$-vector-space are different mathematical objects.

So, if you want to know how many $K$-vector spaces there are having a set $S$ as a basis, where $K$ can be any field, there are "at least as many" as there are fields (so, certainly infinitely many, and in fact there should be "as many" as there are sets, because any set can be chosen to be a transcendence basis over some given field; speaking precisely, they form a proper class).

Even restricting yourself to only a specific field $K$, it is still the case that given a set $S$, there are as many $K$-vector spaces with basis $S$ as there are sets, because (for example) you can always replace the zero element of your vector space with something arbitrary. For example, the $\mathbb{Q}$-vector space $\mathbb{Q}$ has $\{1\}$ as a basis, but so also does the vector space $$\mathbb{Q}\cup\{\star\}\setminus\{0\}$$ (where $\star$ is the zero element of the vector space), and $\star$ can be anything.

But if you're only interested in the number of different $K$-vector spaces with basis $S$ for a given field $K$ up to isomorphism, then there is only one, because any two $K$-vector spaces with equinumerous bases are isomorphic.

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Just to make sure I understood it properly: The section where you explained that there are "as many" vector spaces as there are sets is to be understood as "for every set $S$ there is a field $K$ such that some construction using $K$ yields my vector space" ? I ask that, since $$ $$ 1) I don't know what a transcendence basis is - and I'm also not really interested right know, I just want to grasp the idea $$ $$ 2) in the next section, where $K$ is fixed, we actually need to know, as far as I understood it, that there is at least one (!) field $K$, such that for this field, [...] – temo May 17 '12 at 17:09
[...] the construction from the previous section gives me the suitable vector space. Then we can proceed and modify this vector space by exchanging a finite number of elements (does it necessarily have to be the zero vector?) with arbitrary sets and updating all composition for these new elements to get new vector spaces. $$ $$ Does your argument, that there is only one $K$-vector space up to isomorphism, also hold if $S$ is some (countable/uncountable) infinite set ? – temo May 17 '12 at 17:22
Actually, I take it back, what I wrote in 1): I am interested in what a transcendence basis is. Well not really in this concept for its own sake, but I would like to know how the explicit construction of the vector space looks like, such that $S$ is my basis (as far is I understood, we are viewing the field $K$ as a vector space over itself in that section). Could you maybe give me a reference or outline the construction ? – temo May 17 '12 at 17:27
Bump!............ – temo May 29 '12 at 16:35
@temo: My use of the concept of "transcendence basis" is just for the purposes of saying, informally: there are at least as many fields as there are sets, because given any set I can create a field from it (using it as a transcendence basis). Therefore, if you are interested in how many $K$-vector spaces have $S$ as a basis, where $K$ is allowed to range over all fields, then you have at least as many $K$-vector spaces as there are fields $K$ to choose from, and therefore there are at least as many $K$-vector spaces as there are sets. – Zev Chonoles Jun 2 '12 at 6:23

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