transcendence degree

The transcendence degree of a set $S$ over a field $K$, denoted $T_S$, is the size of the maximal subset $S^{\prime }$ of $S$ such that all the elements of $S^{\prime }$ are algebraically independent.

The transcendence degree of a field extension $L$ over $K$ is the transcendence degree of the minimal subset of $L$ needed to generate $L$ over $K$.

Heuristically speaking, the transcendence degree of a finite set $S$ is obtained by taking the number of elements in the set, subtracting the number of algebraic elements in that set, and then subtracting the number of algebraic relations between distinct pairs of elements in $S$.

Example 1 (Computing the Transcendence Degree).

The set $S\mathrm{=}\mathrm{\{}\sqrt{\mathrm{7}}\mathrm{,}\pi \mathrm{,}{\pi }^{\mathrm{2}}\mathrm{,}e\mathrm{\}}$ has transcendence $T_S\mathrm{\le }\mathrm{2}$ over $\mathrm{Q}$ since there arefour elements, $\sqrt{\mathrm{7}}$ is algebraic, and the polynomial$f\mathit{}\mathrm{(}x\mathrm{,}y\mathrm{)}\mathrm{=}{x}^{\mathrm{2}}\mathrm{-}y$ gives an algebraic dependence between $\pi$ and ${\pi }^{\mathrm{2}}$(i.e. $\mathrm{(}\pi \mathrm{,}{\pi }^{\mathrm{2}}\mathrm{)}$ is a root of $f$), giving $T_S\mathrm{\le }\mathrm{4}\mathrm{-}\mathrm{1}\mathrm{-}\mathrm{1}\mathrm{=}\mathrm{2}$. Ifwe assume the conjecture that $e$ and $\pi$ are algebraicallyindependent, then no more dependencies can exist, and we can concludethat, in fact, $T_S\mathrm{=}\mathrm{2}$.