Algebraic independence

Remarks

Equivalently, the evaluation map from the polynomial ring $K[x_s : s \in S]$ to $L$ that sends $x_s$ to $s$ is injective. A one-element set $\set{s}$ is algebraically independent if and only if $s$ is transcendental over $K$. Example: in $K(x, y)$, the set $\set{x, y}$ is independent, since a polynomial $f(x, y)$ that is zero as an element of $K(x, y)$ is the zero polynomial. Example: $\set{\pi}$ is independent over $\Q$, because $\pi$ is a transcendental number. Non-example: $\set{\pi, \pi^2}$ is dependent over $\Q$, since $f(x_1, x_2) = x_2 - x_1^2$ is nonzero and vanishes at $(\pi, \pi^2)$. Algebraic independence implies linear independence over $K$, by taking $f$ linear; the converse fails, as $\set{1, \pi}$ shows.