The field extension $\mathbb{C}/\mathbb{Q}$ is infinite

Question

I am reading the following proof about the field extension $\mathbb{C}/\mathbb{Q}$, that this extension is infinite.

It is sufficient to show that, for no matter how big natural number $n$, there are $n$ complex numbers, that are $\mathbb{Q}$-linearly independent.

For that, we consider the polynomial $f(x)=x^n-2$ and let $z\in \mathbb{C}$ one of its roots, e.g., $z=\sqrt[n]{2}\left (\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}\right )$.

We claim that the complex numbers $1, z, z^2, \dots , z^{n-1}$ are $\mathbb{Q}$-linearly independent. Indeed, in other case, there would be rationals $c_0, c_1, \dots , c_{n-1}$, not all zero, such that $c_0+c_1z+\dots +c_{n-1}z^{n-1}=0$, i.e., $g(z)=0$, where $g(x)=c_0+c_1x+\dots +c_{n-1}x^{n-1}$. Therefore, the non-zero polynomials $f(x)$ and $g(x)$ would have $z$ as a common root, so we would conclude that $f(x)$ and $g(x)$ are not coprime. But $f(x)$ is irreducible, so we would have to conclude that $f(x)\mid g(x)$. That is not possible, because the non-zero polynomial $g(x)$ has degree smaller than the degree of $g(x)$.

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We have that $f(x)$ and $g(x)$ have a common root, $z$. Why do we conclude that $f(x)\mid g(x)$ ?

How have we shown in that way that the field extension is infinite? Is it because we take $n\in \mathbb{N}$ ?

Answer 1

A faster way to show that $\mathbb{C}$ is an infinite extension of $\mathbb{Q}$ is to observe that $\mathbb{C}$ is uncountable, while any finite extension of $\mathbb{Q}$ is countable.

A more interesting question is showing that $\overline{\mathbb{Q}}$ is an infinite extension of $\mathbb{Q}$, which your argument in fact shows. What you've shown is that $\overline{\mathbb{Q}}$ contains a degree $n$ extension of $\mathbb{Q}$ for all $n\in\mathbb{N}$ (the reason that $f|g$ is that $f$ and $g$ are not coprime because they have a common root, and $f$ is irreducible). Would this be possible if $\overline{\mathbb{Q}}$ were a finite extension of $\mathbb{Q}$?

Answer 2

If $f$ and $g$ are not coprime, they have a non constant common divisor, say $d$. So $d\mid g$ and $d \mid f$. But since $f$ is irreducible, $d=f$, so $f\mid g$.

Answer 3

The gcd of two polynomials $f(x), g(x) \in F[x]$ is the same as their gcd in $K[x]$, where $K/F$ is a field extension. The only divisors of $f(x)$ in $\Bbb{Q}[x]$ are 1 and itself. Sincd $f$ and $g$ have a common root in some extension of $F$ their gcd is not 1, so it must be $f$. Since $f$ has degree greater than $g$ it cannot divide $g$, so this is a contradiction and $g$ cannot exist.

We have shown that the complex numbers contain arbitrarily large finite sets of elements that are linearly independent over the rationals. Therefore the complex numbers cannot have a finite basis over the rationals, since any set with more elements than that basis would be dependent. We conclude that the degree of $\Bbb{C} / \Bbb{Q}$ is infinite.