Algebraic element

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

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

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

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

1. $\sqrt2\in \mathbb{R}$ is algebraic over $\mathbb{Q}$ because it satisfies $x^2-2=0$.
2. $i\in \mathbb{C}$ is algebraic over $\mathbb{R}$ because it satisfies $x^2+1=0$.
3. 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 $x^{p^n}-x\in\mathbb{F}_p[x]$.