Understanding the connection of the roots of an irreducible polynomial and a basis for field extensions

Question

Let $\alpha,\beta,\gamma \in E$ be the roots of an IRREDUCIBLE polynomial $p(x)\in Q[x]$ (where E/Q is an extension field. Can I use these roots to construct a basis for E over Q? Why?

Answer 1

However for a finite Galois extension with Galois group $G$, you the normal basis theorem: there exists an element $x\in E$ such that its orbit $G\cdot x$ under the action of $G$ is a basis of $E$.

Answer 2

In general, not directly. Just take the example of $x^3 - 2$, whose roots are $\sqrt[3]2$, $\rho \sqrt[3]2$ and $\rho^2 \sqrt[3]2$ where $\rho$ is a third root of unity (i.e. $\rho^3 = 1$ and $\rho \neq 1$). These three are not even linearly independent over $\mathbb Q$ since their sum is zero : $$ \sqrt[3]2 + \rho \sqrt[3]2 + \rho^2 \sqrt[3]2 = (\rho^2 + \rho + 1) \sqrt[3]2 = 0. $$ (You could have also considered any irreducible polynomial of degree $3$ whose quadratic coefficient is zero since that means the sum of the roots is zero.) But since the extension is generated by those roots, there is always a finite sequence of $[E:Q]$ polynomials in the roots which do form a basis for $E/Q$.

Hope that helps,