Transcendental element

Let $E/F$ be a field extension and let $\alpha\in E$ . The element $\alpha$ is transcendental over $F$ if there is no nonzero polynomial $f(x)\in F[x]$ such that $f(\alpha)=0$ . Equivalently, $\alpha$ is transcendental over $F$ iff $\alpha$ is not algebraic over F.

Equivalent characterizations

A useful equivalent condition is: $\alpha$ is transcendental over $F$ iff the evaluation map

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

is injective. When $\alpha$ is transcendental, the simple extension $F(\alpha)/F$ is a transcendental extension and has infinite degree.

Remarks

Transcendence depends on the base field: an element may be transcendental over $F$ but algebraic over a larger intermediate field $K$ with $F\subseteq K\subseteq E$ (see intermediate field).

Examples

If $t$ is an indeterminate, then $t$ is transcendental over $\mathbb{Q}$ inside $\mathbb{Q}(t)$ : no nonzero polynomial in $\mathbb{Q}[x]$ vanishes at $t$ .
The classical constants $\pi$ and $e$ are transcendental over $\mathbb{Q}$ (deep theorems, e.g. Lindemann–Weierstrass).
Let $t$ be an indeterminate. Then $t$ is transcendental over $\mathbb{Q}$ , but $t$ is algebraic over the intermediate field $\mathbb{Q}(t^2)\subseteq \mathbb{Q}(t)$ , because $t$ satisfies the polynomial $x^2-t^2=0$ with coefficients in $\mathbb{Q}(t^2)[x]$ .