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1) what is an explicit basis for $\mathbb R$ as a $\mathbb Q$-vector space?

2) what is a basis for $\mathbb C$ as a $\mathbb C$ vector space? i think you will say $\{1\}$ is a basis since $\forall z \in \mathbb C$, $z=z*1$, but why can't we say the following: $\{1,i\}$ is a basis for $\mathbb C$ as a $\mathbb C$ vector space since $\forall z \in \mathbb C$, $z=x*1+y*i$ for some $x,y\in \mathbb R \subset \mathbb C$, i know this gives that $\{1,i\}$ is a basis for $\mathbb C$ as a $\mathbb R$ vector space but why it is not valid as a $\mathbb C$ vector space knowing that $\mathbb R \subset \mathbb C$?

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See math.stackexchange.com/questions/73400/… –  lhf Oct 27 '11 at 9:56
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2 Answers

up vote 3 down vote accepted

1) Read about Hamel basis. I have a small hope though that there an explicit form for it.

2) The basis should be linear independent, but $i = z\cdot 1$ for $z = i\in \mathbb C$. In fact, any non-zero $z\in \mathbb C$ forms a one-component basis $\{z\}$ of the vector space $\mathbb C$ over a field $\mathbb C$.

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1) One can’t be given: the existence of such a basis requires the axiom of choice.

2) $\{1,i\}$ isn’t a basis for $\mathbb{C}$ over $\mathbb{C}$ because it isn’t linearly independent: $1\cdot 1+i\cdot i=0$ even though the coefficients $1$ and $i$ are non-zero.

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