Primitive element theorem

Let $L/K$ be a finite field extension .

Theorem (Primitive element theorem). If $L/K$ is a finite separable extension , then there exists an element $\alpha\in L$ such that

i.e. $L/K$ is a simple extension .

A useful strengthening (often used in proofs) is: if $L=K(\beta,\gamma)$ is finite and separable and $K$ is infinite, then for all but finitely many $c\in K$, the element $\alpha=\beta+c\gamma$ satisfies $L=K(\alpha)$.

Examples

1. $\mathbb{Q}(\sqrt2,\sqrt3)/\mathbb{Q}$ is separable because $\operatorname{char}(\mathbb{Q})=0$ (see characteristic 0 or prime ).
   A primitive element is $\alpha=\sqrt2+\sqrt3$. One checks $\mathbb{Q}(\alpha)$ contains both $\sqrt2$ and $\sqrt3$, hence equals $\mathbb{Q}(\sqrt2,\sqrt3)$.

2. Any finite field extension $\mathbb{F}_{p^n}/\mathbb{F}_p$ is separable (finite fields are perfect ).
   Therefore $\mathbb{F}_{p^n}=\mathbb{F}_p(\alpha)$ for some $\alpha$. Concretely, choosing $\alpha$ to generate the cyclic group $\mathbb{F}_{p^n}^\times$ from cyclicity of the multiplicative group guarantees $\mathbb{F}_p(\alpha)=\mathbb{F}_{p^n}$.

3. If $f\in K[x]$ is separable and $L$ is its splitting field over $K$, then $L/K$ is finite and separable; hence $L=K(\alpha)$ for some $\alpha\in L$. For instance, the splitting field of $x^3-2$ over $\mathbb{Q}$ can be generated by a single element $\alpha\in \mathbb{Q}(\sqrt[3]{2},\zeta_3)$.

(For contrast: without separability, the conclusion can fail; there exist finite inseparable extensions that are not simple.)