\begin{definition}[Definition:Algebraically Independent]
Let $L / K$ be a field extension.
Let $A \subseteq L$ be a subset of $L$.
Let $\map K {\set {X_\alpha}_{\alpha \mathop \in A} }$ be the Field of Rational Functions in the indeterminates $\set {X_\alpha: \alpha \mathop \in A}$.
Then $A$ is '''algebraically independent''' over $K$ if there exists a homomorphism:
:$\phi: \map K {\set {X_\alpha}_{\alpha \mathop \in A} } \to L$
such that, for all $\alpha \in A$:
:$\map \phi {X_\alpha} = \alpha$
\end{definition}