# Algebraic degree

A field is called algebraically closed if every non-constant polynomial with coefficients in has a zero in . A field is an algebraic closure of if it is algebraically closed and is an algebraic extension field of . Since an algebraic closure is unambiguous except for isomorphism , one often speaks of the algebraic closure. Finding the zeros of polynomials is an important mathematical task; their existence can at least be ensured in an algebraic conclusion. In fact, one can show that there is an algebraic closure for every field. ${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle L}$${\ displaystyle K}$${\ displaystyle K}$

## Definitions

${\ displaystyle K [x]}$as usual, denote the polynomial ring over . ${\ displaystyle K}$

In general, a field is called algebraically closed if one of the following equivalent statements holds: ${\ displaystyle K}$

• Every polynomial from has a zero in .${\ displaystyle K [x] \ setminus K}$${\ displaystyle K}$
• Each polynomial from is broken down into linear factors , i.e. polynomials of degree 1.${\ displaystyle K [x] \ setminus K}$
• ${\ displaystyle K}$ has no real algebraic extensions.
• Every irreducible polynomial in has degree 1.${\ displaystyle K [x]}$

An algebraic closure of a field can now be defined in two ways: ${\ displaystyle L}$${\ displaystyle K}$

• ${\ displaystyle L}$is an algebraic expansion field of , in which every polynomial from has a root.${\ displaystyle K}$${\ displaystyle K [x] \ setminus K}$
• ${\ displaystyle L}$is an algebraic expansion field of , in which every polynomial from has a root.${\ displaystyle K}$${\ displaystyle L [x] \ setminus L}$

The second condition is a seemingly stronger statement, but it turns out to be equivalent to the first.

## existence

For a single polynomial one can easily find an algebraic extension in which the polynomial has a zero. With Zorn's lemma one can find an algebraic extension in which all non-constant polynomials from have a zero. According to the remark above, this is then an algebraic closure of . ${\ displaystyle K [x] \ setminus K}$${\ displaystyle L}$${\ displaystyle K [x]}$${\ displaystyle K}$

Ernst Steinitz was the first to succeed in showing in 1910 that every body has an algebraically closed upper body and thus an algebraic closure. Steinitz uses the axiom of choice , which is equivalent to the above mentioned lemma of Zorn. The proof of existence necessarily requires transfinite methods such as the axiom of choice: If the axioms of set theory are consistent, then the axioms of set theory (without the axiom of choice) are also together with the sentence "There is a field that has no algebraic closure." consistent.

## Uniqueness

Also connected to the Zorn's Lemma can show you that two algebraic statements to each other are -isomorph, that is, for algebraic statements of there is a Körperisomorphismus , of limited to the identity is. However, there is no canonical , i.e. no excellent, isomorphism, but in general a large number of equals. So being an algebraic degree is not a universal property . ${\ displaystyle K}$${\ displaystyle L, L '}$${\ displaystyle K}$${\ displaystyle \ varphi \ colon L \ to L '}$${\ displaystyle K}$

The algebraic closure of has the same power as if is infinite, and is countable if is finite. An algebraically closed body, on the other hand, cannot be finite: If the body is finite with elements and the product of all linear factors, the polynomial has no zero. ${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle K}$${\ displaystyle q}$${\ displaystyle a_ {1}, \ dotsc, a_ {q}}$${\ textstyle f: = \ prod _ {k = 1} ^ {q} (x-a_ {k})}$${\ displaystyle f + 1}$

## Examples

• The fundamental theorem of algebra says that the field of complex numbers is algebraically closed and thus an algebraic closure of real numbers . If there is another algebraic closure of and are and the solutions of in , then there are two -isomorphisms of after . Either is shown on or on . Both options are equal.${\ displaystyle \ mathbb {C}}$ ${\ displaystyle \ mathbb {R}}$${\ displaystyle L}$${\ displaystyle \ mathbb {R}}$${\ displaystyle j_ {1}}$${\ displaystyle j_ {2} = - j_ {1}}$${\ displaystyle x ^ {2} = - 1}$${\ displaystyle L}$${\ displaystyle \ mathbb {R}}$${\ displaystyle L}$${\ displaystyle \ mathbb {C}}$${\ displaystyle j_ {1}}$${\ displaystyle i}$${\ displaystyle -i}$
• An algebraic closure of the rational numbers is the field of the algebraic numbers .${\ displaystyle \ mathbb {Q}}$ ${\ displaystyle \ mathbb {A}}$
• There are many countable algebraically closed real torsos of algebraic numbers in . They are algebraic closings of transcendent extensions of .${\ displaystyle \ mathbb {C}}$${\ displaystyle \ mathbb {Q}}$
• For a finite field of the prime number order , the algebraic closure is a countably infinite field of the characteristic , and for every natural number contains a subfield of the order , it even consists of the union of these subfields.${\ displaystyle \ mathbb {F} _ {p}}$${\ displaystyle p}$ ${\ displaystyle p}$${\ displaystyle n}$${\ displaystyle p ^ {n}}$

## meaning

The importance of the algebraic closure is to find the zeros of polynomials. In the algebraic conclusion, every polynomial -th degree has exactly zeros, which are to be counted with multiples. However, nothing is said about how these can be found specifically, see the article zero . ${\ displaystyle n}$${\ displaystyle n}$

## Individual evidence

1. ^ Kurt Meyberg, Algebra II , Carl Hanser Verlag (1976), sentence 6.10.6
2. Ernst Steinitz: Algebraic theory of bodies . In: Journal for Pure and Applied Mathematics , Volume 137, 1910, pp. 167-309
3. ^ Thomas Jech : The Axiom of Choice . North Holland, 1973, ISBN 0-7204-2275-2 , pp. 147 .