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Since, $\mathbb{C} = \mathbb{R}(i)$ where $i = \sqrt{-1}$.
So, $[\mathbb{C} : \mathbb{R}] = [\mathbb{R}(i) : \mathbb{R}] = 2$.
Which $\mathbb{C}$ is a finite field extention of $\mathbb{R}$.

Since, there are infinite number of algebric numbers in $\mathbb{C}$.
So, $[\mathbb{C} : \mathbb{Q}]$ is infinite.
Which means $\mathbb{C}$ is an infinite field extention of $\mathbb{Q}$.

Are they correct ?

xxxxxx
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    There are also an infinite number of algebraic numbers $\mathbb{Q}$ itself, or in $\mathbb{Q}(\sqrt{2}) \setminus \mathbb{Q}$ (e.g. the set ${p \sqrt{2}: p \in \mathbb{Q},, p > 0}$ consists entirely of algebraic irrational numbers, yet lies in a finite extension of $\mathbb{Q}$. I think slightly more precision in language or perhaps a different approach may be helpful. – leslie townes Feb 25 '21 at 07:14

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Remember that if $[F : K]=n\in\mathbb{N}$, $F$ is a vector space over $K$, with dimension $n$. So if $[\mathbb{R} : \mathbb{Q}]=n$, then $\mathbb{R}=\mathbb{Q}\times ... \times\mathbb{Q}$ $n$ times. That would imply that $\mathbb{R}$ is countable.

Thus,$[\mathbb{R} : \mathbb{Q}]=\infty$ and $[\mathbb{C} : \mathbb{Q}]=\infty$.

Lancet S.
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