Finite field extension over the rationals need only one generator?

507 Views Asked by At

Some book stated (without proof) that in every finite dimensional field over the rationals of dimension $n$, there is an element of degree $n$ (i.e. any field $Q[\alpha_1, ..., \alpha_n]$ is of the form $Q[\theta]$).

Can someone give a confirmation, reference, proof-sketch, hint?

And is there a similar result for $Z$ ?

1

There are 1 best solutions below

5
On BEST ANSWER

You are looking for the primitive element theorem. A quick Google search will give you several different sets of self-contained notes that include the proof.

If your second question is asking whether given a collection of algebraic numbers $\alpha_1, \dots, \alpha_n$, there exists a $\theta$ such that $\mathbf{Z}[\theta] = \mathbf{Z}[\alpha_1, \dots, \alpha_n]$, then the answer is no.

For a concrete example, let $\alpha$ be a root of $x^3 - x^2 - 2x - 8$. Then for any $\theta$ (necessarily in $\mathbf{Q}(\alpha)$), we have $\mathbf{Z}[\theta] \neq \mathbf{Z} \left[\alpha, \frac{\alpha^2 + \alpha}{2} \right]$, the latter of which is the ring of integers in $\mathbf{Q}(\alpha)$. All of this can be proved via methods from basic algebraic number theory.