Algebraic expansion

In algebra , a field extension is called algebraic if every element of algebraic is over , i.e. H. when each element is the zero of a polynomial with coefficients in . Body extensions that are not algebraic, that is, contain transcendent elements , are called transcendent. ${\displaystyle L/K}$ ${\displaystyle L}$ ${\displaystyle K}$ ${\displaystyle L}$ ${\displaystyle K}$

For example, the extensions and are algebraic while is transcendent. ${\displaystyle \mathbb {C} /\mathbb {R} }$ ${\displaystyle \mathbb {Q} ({\sqrt {2}})/\mathbb {Q} }$ ${\displaystyle \mathbb {R} /\mathbb {Q} }$

If an upper body is of then one can understand it as - vector space and determine its dimension . This vector space dimension is called the degree of body expansion. Depending on whether this degree is finite or infinite, the expansion of the body is also called finite or infinite. Every transcendent expansion is infinite. It follows that every finite expansion is algebraic. ${\displaystyle L}$ ${\displaystyle K,}$ ${\displaystyle L}$ ${\displaystyle K}$

But there are also infinite algebraic extensions, for example the algebraic numbers form an infinite extension of ${\displaystyle \mathbb {Q} .}$

Is algebraic over then the ring of all polynomials in over is even a field. is a finite algebraic extension of Such extensions that arise from the adjunction of a single element are called simple extensions. ${\displaystyle a}$ ${\displaystyle K,}$ ${\displaystyle K[a]}$ ${\displaystyle a}$ ${\displaystyle K}$ ${\displaystyle K[a]}$ ${\displaystyle K.}$

A field that has no real algebraic extension is algebraically closed .

If and are body extensions, the following statements are equivalent: ${\displaystyle M/L}$ ${\displaystyle L/K}$

• ${\displaystyle M/K}$ is algebraic.
• ${\displaystyle M/L}$and are algebraic. ${\displaystyle L/K}$

example

With is an algebraic field extension over , because with ${\displaystyle a={\sqrt {2}}+{\sqrt {3}}}$ ${\displaystyle \mathbb {Q} (a)}$ ${\displaystyle \mathbb {Q} }$

${\displaystyle a^{2}=5+2{\sqrt {6}}}$, and ${\displaystyle a^{3}=11{\sqrt {2}}+9{\sqrt {3}}}$ ${\displaystyle a^{4}=49+20{\sqrt {6}}}$

is the zero of the polynomial and thus algebraically over . Since it is an irreducible polynomial of the fourth degree, the degree of body expansion is also greater than four. As for every algebraic element, there is a basis of as a vector space over . A simpler one, however, is the basis . ${\displaystyle a}$ ${\displaystyle x^{4}-10x^{2}+1}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle \mathbb {Q} (a)}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle \{1,a,a^{2},a^{3}\}}$ ${\displaystyle \mathbb {Q} (a)}$ ${\displaystyle \mathbb {Q} }$ ${\displaystyle \{1,{\sqrt {2}},{\sqrt {3}},{\sqrt {6}}\}}$

literature

Kurt Meyberg: Algebra II. Carl Hanser Verlag, 1976, ISBN 3-446-12172-2 , Chapter 6.3 Algebraic field extensions .