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 $$\forall n \in \Z_+: \sum_{k=0}^n a_k \circ \alpha^k = 0_R \Longrightarrow \forall k: 0 \le k \le n: a_k = 0_R$$.

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.