algebraic

Let $B$ be a ring with a subring $A$. An element $x\in B$ is algebraic over $A$ if there exist elements $a_1,\mathrm{\dots },a_n\in A$, with $a_n\ne 0$, such that

$$a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{\cdots }+a_{1}x+a_0=0.$$

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 $A$.