Definition:Transcendental (Abstract Algebra)

Rings
Let $\left({R, +, \circ}\right)$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring of $R$.

Let $\alpha \in R$.

Then $\alpha$ is transcendental over $D$ iff:
 * $\displaystyle \forall n \in \Z_+: \sum_{k=0}^n a_k \circ \alpha^k = 0_R \implies \forall k: 0 \le k \le n: a_k = 0_R$

That is, $\alpha$ is transcendental over $D$ iff the only way to express $0_R$ as a polynomial in $\alpha$ over $D$ is by the null polynomial.

If $\alpha \in R$ is not transcendental over $D$ then it is algebraic over $D$.

Fields
Let $E/F$ be a field extension.

Let $\alpha \in E$.

Let $f \left({x}\right)$ be a polynomial in $x$ over $F$.

Then $\alpha$ is transcendental over $F$ if $\nexists ~f \left({x}\right) \in F[x] - \{0\}$ such that $f \left({\alpha}\right) = 0$.

If $\alpha \in E$ is not transcendental over $F$ then it is algebraic over $F$.

Field Extensions
A field extension $E/F$ is said to be transcendental if $\exists ~\alpha \in E: \alpha$ is transcendental over $F$.

That is, a field extension is transcendental if it contains at least one transcendental element.

If no element of $E/F$ is transcendental over $F$, then $E/F$ is algebraic.