Algebraic independence in a field extension

Let $K\mathrm{/}k$ be a field extension and let $S\subseteq K$. The evaluation map ${\mathrm{ev}}_S:k[X_s:s\in S]⟶K$ is the unique ring homomorphism extending the field inclusion $k\hookrightarrow K$ and sending each indeterminate $X_s$ to $s$. The set $S$ is algebraically independent over $k$ when ${\mathrm{ev}}_S$ is injective. Equivalently, no nonzero polynomial involving finitely many indeterminates $X_s$ with $s\in S$ evaluates to zero at the corresponding elements of $S$.

• Empty set: When $S=\mathrm{\varnothing }$, the polynomial ring is $k$ and ${\mathrm{ev}}_S$ is the field inclusion $k\hookrightarrow K$, so the empty set is algebraically independent.
• Zero element: If $0\in S$, then the nonzero polynomial $X_0$ evaluates to zero. Thus any set containing zero is algebraically dependent.
• Singleton: For $S=\{a\}$, the evaluation map is injective exactly when no nonzero polynomial in $k[X]$ vanishes at $a$, that is, exactly when $a$ is transcendental over $k$ (Algebraic and transcendental elements and algebraic extensions).
• Finite tuples: For a finite set $S=\{s_1,\dots ,s_r\}$, the condition uses the polynomial ring in those $r$ variables. It includes $r=0$ and does not require an ordering of $S$; renaming variables preserves injectivity.