Let $B$ be a ring with a subring $A$ An element $x \in B$ is algebraic over $A$ if there exist elements $a_1, \dots, a_n \in A$ with $a_n \neq 0$ such that $$ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0. $$ An element $x \in B$ is transcendental over $A$ if it is not algebraic.

The ring $B$ is algebraic over $A$ if every element of $B$ is algebraic over $A$