Algebraic independence

In abstract algebra , the algebraic independence is a property of elements of a transcendent field extension , which means that these elements do not satisfy a nontrivial polynomial equation with coefficients in the basic field.

definition

Be a body extension and elements of . Is there a polynomial different from the zero polynomial in variables and coefficients in , i. H. , so that ${\displaystyle L/K}$ ${\displaystyle v_{1},\ldots ,v_{n}}$ ${\displaystyle L}$ ${\displaystyle f}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle f\in K\lbrack X_{1},\ldots ,X_{n}\rbrack \setminus \{0\}}$

${\displaystyle f(v_{1},\ldots ,v_{n})=0}$,

then are called algebraically dependent . If there is no such polynomial, then the elements are called algebraically independent . ${\displaystyle v_{1},\ldots ,v_{n}}$

This concept can be extended to infinite subsets of by calling a set algebraically dependent if it has an algebraically dependent finite subset. ${\displaystyle M}$ ${\displaystyle L}$ ${\displaystyle M}$

Similar to the concept of linear combination (linear homogeneous polynomial) used in vector spaces , which provides the concept of linear independence , one sometimes considers algebraic combinations of transcendent elements in field extensions , i.e. H. any (fractional-rational) polynomials with coefficients in the basic field.

A maximum system of algebraically independent elements is called the basis of transcendence , their power is called the degree of transcendence of expansion.

Relationship with algebraic elements

If there is a field extension, then an element from is algebraically dependent over the field if and only if it is an algebraic element over , because by definition it is the zero of a polynomial with coefficient from . Thus an element from is algebraically independent over if and only if it is a transcendent element over . ${\displaystyle L/K}$ ${\displaystyle L}$ ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle K}$ ${\displaystyle L}$ ${\displaystyle K}$ ${\displaystyle K}$

Examples

• Elements that are inverse to one another with regard to the multiplication are always algebraically dependent, since they are zeros of the polynomial . ${\displaystyle XY-1}$
• The real numbers and (with the circle number pi) are algebraically dependent on the rational numbers , because with and they satisfy the polynomial equation . ${\displaystyle \pi +1}$ ${\displaystyle \pi ^{2}}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle X=\pi +1}$ ${\displaystyle Y=\pi ^{2}}$ ${\displaystyle Y-(X-1)^{2}=0}$
• Likewise, and the imaginary unit are algebraically dependent on , because with and is true . This is of course due to the fact that the amount alone is algebraically dependent. Although and are algebraically dependent, neither belongs to nor to . ${\displaystyle \pi }$ ${\displaystyle i}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle X=\pi }$ ${\displaystyle Y=i}$ ${\displaystyle 0\cdot X+Y^{2}+1=0}$ ${\displaystyle \{i\}}$ ${\displaystyle \pi }$ ${\displaystyle i}$ ${\displaystyle \pi }$ ${\displaystyle \mathbb {Q} (i)}$ ${\displaystyle i}$ ${\displaystyle \mathbb {Q} (\pi )}$

Examples of complex numbers that are over algebraically independent are harder to find, although it has been proven that there are infinitely many (more precisely: continuum - many) over algebraically independent complex numbers. But one suspects that and there are. On the other hand, it is easy to find examples in other bodies: ${\displaystyle \mathbb {Q} }$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle e}$ ${\displaystyle \pi }$

• In the rational function field in two indeterminates and above the rational numbers, the elements and are algebraically independent, because according to the definition of this field, the only polynomial in two variables that equals 0 at this point is the zero polynomial. ${\displaystyle \mathbb {Q} (X,Y)}$ ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle X}$ ${\displaystyle Y}$ ${\displaystyle (X,Y)}$

• A larger example can be found in the function body . Here all elementary symmetric polynomials are algebraically independent. ${\displaystyle K(X_{1},\ldots ,X_{n})}$ ${\displaystyle \sigma _{1},\ldots ,\sigma _{n}}$

Individual evidence

1. ^ Karpfinger, Meyberg: Algebra. 2013, 23.1.1.
2. ^ Karpfinger, Meyberg: Algebra. 2013, February 31

literature

• AB Shidlovskii: Algebraic independence . In: Michiel Hazewinkel (Ed.): Encyclopaedia of Mathematics . Springer-Verlag , Berlin 2002, ISBN 978-1-55608-010-4 (English, online ).
• Christian Karpfinger, Kurt Meyberg: Algebra . Group rings body. 3. Edition. Springer-Verlag, Berlin / Heidelberg 2013, ISBN 978-3-8274-3011-3 , chap. 23 , doi : 10.1007 / 978-3-8274-3012-0 .