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This problem is taken from Linear Algebra by Fueng Zhang. R is real numbers, C is complex numbers and Q is rational numbers.

Let $V = \{(x,y) \mid x,y \in \mathbb{C}\}$. Under the standard addition and scalar multiplication for ordered pairs of complex numbers, is $V$ a vector space over $\mathbb{C}$? Over $\mathbb{R}$? Over $\mathbb{Q}$ If so, find the dimension of $V$.

The answer key says:

Yes, over $\mathbb{C}, \mathbb{R}$ and $\mathbb{Q}$. The dimensions are $2,4,\infty$, respectively.

I do understand that the dimension over $\mathbb{C}$ is 2 and over $\mathbb{R}$ is 4, but how come the dimension over $\mathbb{Q}$ is infinite?

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    $\begingroup$ There are numbers like $e$ (transcendental numbers)that are not roots of any polynomial with rational coefficients. Therefore $1,e,e^2,e^3,...$ are linearly independent over $\mathbb{Q}$. If they were dependent then there would be some rational numbers $r_1,...,r_n$, such that $r_1e^{m_1}+...+r_ne^{m_n}=0$, which is a polynomial with rational coefficients of which $e$ is a solution. $\endgroup$ – logarithm May 30 at 14:57
  • $\begingroup$ I think it's also interesting to note that the dimension over $\mathbb{Q}$ is uncountable $\endgroup$ – Qidi May 30 at 15:13
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Hint:

Prove that any finite dimensional vector space over $\mathbb Q$ (or any other countable field) is countable.

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Suppose that $\mathbb{C}^2$ is a finite dimensional $\mathbb{Q}$ vector space, then

$$\mathbb{C}^2 \cong \mathbb{Q}^n$$ for some $n \geq 1$.

In particular, it would follow that $\mathbb{C}^2$ is countable. Contradiction.

The other two are easy: you can just write down bases for them

For example, $\{1,i\}$ is $\mathbb{R}$-linearly independent over $\mathbb{C}$. What is an $\mathbb{R}$-basis for $\mathbb{C}^2$ ?

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