# 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 are
four 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 ${T}_{S}\mathrm{\le}\mathrm{4}\mathrm{-}\mathrm{1}\mathrm{-}\mathrm{1}\mathrm{=}\mathrm{2}$. If
we assume the conjecture that $e$ and $\pi $ are algebraically
independent, then no more dependencies can exist, and we can conclude
that, in fact, ${T}_{S}\mathrm{=}\mathrm{2}$.

Title | transcendence degree |
---|---|

Canonical name | TranscendenceDegree |

Date of creation | 2013-03-22 13:58:11 |

Last modified on | 2013-03-22 13:58:11 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 7 |

Author | mathcam (2727) |

Entry type | Definition |

Classification | msc 12F20 |

Defines | transcendence degree of a set |

Defines | transcendence degree of a field extension |