## You are here

Homealgebraically dependent

## Primary tabs

# algebraically dependent

Let $L$ be a field extension of a field $K$. Two elements $\alpha,\beta$ of $L$ are *algebraically dependent* if there exists a non-zero polynomial $f(x,y)\in K[x,y]$ such that $f(\alpha,\beta)=0$. If no such polynomial exists, $\alpha$ and $\beta$ are said to be *algebraically independent*.

More generally, elements $\alpha_{1},\ldots,\alpha_{n}\in L$ are said to be algebraically dependent if there exists a non-zero polynomial $f(x_{1},\ldots,x_{n})\in K[x_{1},\ldots,x_{n}]$ such that $f(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})=0$. If no such polynomial exists, the collection of $\alpha$’s are said to be algebraically independent.

Defines:

algebraically independent, algebraic dependence, algebraic independence

Related:

DependenceRelation

Type of Math Object:

Definition

Major Section:

Reference

Groups audience:

## Mathematics Subject Classification

12F05*no label found*11J85

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Comments

## elements from algebraic extension always algebraically depen...

Hi, maybe I'm confusing things but are't elements of an ALGEBRAIC field extension *always* algebraically dependent? Perhaps, L should rather be an *arbitrary* field extension of K...

Thanks