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algebraically dependent (Definition)

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.




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See Also: dependence relation

Also defines:  algebraically independent, algebraic dependence, algebraic independence

Attachments:
Zariski lemma (Derivation) by polarbear
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Cross-references: collection, polynomial, elements, field, field extension
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This is version 5 of algebraically dependent, born on 2003-09-25, modified 2006-02-24.
Object id is 4741, canonical name is AlgebraicallyDependent.
Accessed 7449 times total.

Classification:
AMS MSC12F05 (Field theory and polynomials :: Field extensions :: Algebraic extensions)
 11J85 (Number theory :: Diophantine approximation, transcendental number theory :: Algebraic independence; Gelfond's method)

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elements from algebraic extension always algebraically dependent? by ziegler on 2004-05-14 08:03:08
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

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