Algebraic element - Imported Knowlpedia Content

Let $E/F$ be a field extension and let $\alpha\in E$ . The element $\alpha$ is algebraic over $F$ if there exists a nonzero polynomial $f(x)\in F[x]$ such that

f(\alpha)=0 \quad \text{in } E.

If no such nonzero polynomial exists, then $\alpha$ is transcendental over F.

Equivalent characterizations

Equivalently, $\alpha$ is algebraic over $F$ iff the evaluation homomorphism

\operatorname{ev}_\alpha: F[x]\to E,\quad f\mapsto f(\alpha),

has nonzero kernel. When $\alpha$ is algebraic, the simple extension $F(\alpha)/F$ (see simple extension) is an algebraic extension and has finite degree.

Examples

$\sqrt2\in \mathbb{R}$ is algebraic over $\mathbb{Q}$ because it satisfies $x^2-2=0$ .
$i\in \mathbb{C}$ is algebraic over $\mathbb{R}$ because it satisfies $x^2+1=0$ .
Every element of $\mathbb{F}_{p^n}$ is algebraic over $\mathbb{F}_p$ : for any $a\in\mathbb{F}_{p^n}$ , one has $a^{p^n}=a$ , so $a$ is a root of $x^{p^n}-x\in\mathbb{F}_p[x]$ .