Algebraic element explained

In mathematics, if is an extension field of, then an element of is called an algebraic element over, or just algebraic over, if there exists some non-zero polynomial with coefficients in such that . Elements of that are not algebraic over are called transcendental over .

These notions generalize the algebraic numbers and the transcendental numbers (where the field extension is, with being the field of complex numbers and being the field of rational numbers).

Examples

Properties

The following conditions are equivalent for an element

a

of

L

:

a

is algebraic over

K

,

K(a)/K

is algebraic, i.e. every element of

K(a)

is algebraic over

K

(here

K(a)

denotes the smallest subfield of

L

containing

K

and

a

),

K(a)/K

has finite degree, i.e. the dimension of

K(a)

as a

K

-vector space is finite,

K[a]=K(a)

, where

K[a]

is the set of all elements of

L

that can be written in the form

g(a)

with a polynomial

g

whose coefficients lie in

K

.

\varepsilona:K[X]K(a),P\mapstoP(a)

. This is a homomorphism and its kernel is

\{P\inK[X]\midP(a)=0\}

. If

a

is algebraic, this ideal contains non-zero polynomials, but as

K[X]

is a euclidean domain, it contains a unique polynomial

p

with minimal degree and leading coefficient

1

, which then also generates the ideal and must be irreducible. The polynomial

p

is called the minimal polynomial of

a

and it encodes many important properties of

a

. Hence the ring isomorphism

K[X]/(p)im(\varepsilona)

obtained by the homomorphism theorem is an isomorphism of fields, where we can then observe that

im(\varepsilona)=K(a)

. Otherwise,

\varepsilona

is injective and hence we obtain a field isomorphism

K(X)K(a)

, where

K(X)

is the field of fractions of

K[X]

, i.e. the field of rational functions on

K

, by the universal property of the field of fractions. We can conclude that in any case, we find an isomorphism

K(a)\congK[X]/(p)

or

K(a)\congK(X)

. Investigating this construction yields the desired results.

This characterization can be used to show that the sum, difference, product and quotient of algebraic elements over

K

are again algebraic over

K

. For if

a

and

b

are both algebraic, then

(K(a))(b)

is finite. As it contains the aforementioned combinations of

a

and

b

, adjoining one of them to

K

also yields a finite extension, and therefore these elements are algebraic as well. Thus set of all elements of

L

that are algebraic over

K

is a field that sits in between

L

and

K

.

Fields that do not allow any algebraic elements over them (except their own elements) are called algebraically closed. The field of complex numbers is an example. If

L

is algebraically closed, then the field of algebraic elements of

L

over

K

is algebraically closed, which can again be directly shown using the characterisation of simple algebraic extensions above. An example for this is the field of algebraic numbers.

See also