Prove that $\mathbb{Q}$ has extensions of any finite degree in $\mathbb{C}$

Question

This is a question from a course in Galois Theory and I am quite confused.

In general, the degree of a field extension $E/F$ is the dimension of the vector space $E$.

What would $E$ and $F$ be in these cases? Would $E$ be any field extension, $F$ be $\mathbb{Q}$.. where does $\mathbb{C}$ come into it - does this just mean that all elements in our field extensions over $\mathbb{Q}$ belong to the complex numbers?

Answer 1

Consider the family of polynomials $f_n:=x^n-2$. Can you show that for each $n\geq 2$, $f_n$ is irreducible? What degree extension will adjoining a root of $f_n$ produce?

Answer 2

By Eisenstein's criterion, if $p$ is prime, then the polynomial $f(x):=x^n+px^{n-1}+\cdots+px+p$ is irreducible, and hence the field extension $\mathbb Q[x]/(f)$ over $\mathbb Q$ is of degree $n.$
Hope this helps.

P.S. This extension is the same as adjoining a root of $f$ to $\mathbb Q,$ and hence is a sub-extension of $\mathbb C.$