Basis of Vector space $\Bbb C$ over rational numbers. What will be the basis of vector space $\Bbb C$ over a field of rational numbers $\Bbb Q$?
I think it will be an infinite basis! I think it will be $B=\{r_1+r_2i \mid  r_1,  r_2 \in \Bbb Q^{c}\}\cup\{1,i\}$. But this generator is an uncountable set. Can a basis of a vector space be that big? If it is true does it mean that $\Bbb Q$-module (Vector space) $\Bbb C$ is free?
 A: Note first that for any ring $R$ and any $R$-module $M$, if the cardinality $\lvert M\rvert$ is infinite and greater than $\lvert R \rvert$, then any generating set of $M$ (in particular, any basis) must have cardinality equal to $\lvert M\rvert$ (this is simple combinatorics).
In particular, any basis of $\bf C$ over $\bf Q$ must have cardinality of the continuum, which is rather uncountable no matter how you look at it.
Secondly, any module over a field (i.e. any vector space) is free (it's a basic theorem in linear algebra that every vector space has a basis, assuming axiom of choice).
On the other hand, in ZF alone (without axiom of choice), it is consistent that $\bf C$ has no basis over $\bf Q$ (because such a basis can't be measurable, and there are models of ZF where every set is measurable). Therefore, loosely speaking, you can't write down such a basis. A little more precisely, it's impossible to write a formula which will provably define a basis of $\bf C$ over $\bf Q$ without some strong set-theoretic assumptions like $V=L$.
The set you've written down certainly generates $\bf C$ as a vector space, but it's not hard to see that it is not linearly independent.
A: The straightforward answer for "what is a basis for $\mathbb{C}/\mathbb{Q}$" is that we don't know.  The sneaky answer is that we do know there is one, because any maximal linearly independent set is a basis, and exists by Zorn's Lemma.  This shows that $\mathbb{C}$ is a free $\mathbb{Q}$-module, since that concept is literally equivalent to the existence of a basis, and the same argument in fact proves that any vector space is a free module over its field.  As noted in the comments, for finite-dimensional vector spaces one could deduce this from the fact that fields are PIDs and their modules are torsion-free, but that uses a very fancy theorem of ring theory that, moreover, reduces to computing a basis!  So it's enough just to say that we could, in principle, find one.
As for how you would find one, the answer is not satisfying.  One way is by ordinal induction: for every ordinal $\alpha$ less than the cardinality of $\mathbb{C}$, suppose we have found a $\mathbb{Q}$-linearly independent set in $\mathbb{C}$ of cardinality that of $\alpha$.  If it doesn't span $\mathbb{C}$, then it can be extended by some new element and that gives such a set for $\alpha + 1$.  For any limit ordinal $\beta$, construct the union of the linearly independent sets for lesser $\alpha$, which is linearly independent since they are, by construction, nested.  Then by the time we pass through all ordinals less than or equal to the cardinality of $\mathbb{C}$, we must have found a basis, since we will have run out of complex numbers!
I do not encourage you to actually implement this construction.
A: Any basis for $\Bbb C$ as a $\Bbb Q$ vector space must be infinite, since any finite dimensional $\Bbb Q$ vector space is countable, but $\Bbb C$ is not. Constructing such a basis requires the axiom of choice.
