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Inmathematics, afieldF isalgebraically closed if everynon-constant polynomial inF[x] (the univariatepolynomial ring with coefficients inF) has aroot inF. In other words, a field is algebraically closed if thefundamental theorem of algebra holds for it.

Every field${\displaystyle K}$ is contained in an algebraically closed field${\displaystyle C,}$ and the roots in${\displaystyle C}$ of the polynomials with coefficients in${\displaystyle K}$ form an algebraically closed field called analgebraic closure of${\displaystyle K.}$ Given two algebraic closures of${\displaystyle K}$ there are isomorphisms between them that fix the elements of${\displaystyle K.}$

Algebraically closed fields appear in the following chain ofclass inclusions:

rngs ⊃rings ⊃commutative rings ⊃integral domains ⊃integrally closed domains ⊃GCD domains ⊃unique factorization domains ⊃principal ideal domains ⊃euclidean domains ⊃fields ⊃algebraically closed fields

As an example, the field ofreal numbers is not algebraically closed, because the polynomial equation${\displaystyle x^2+1=0}$ has no solution in real numbers, even though all its coefficients (1 and 0) are real. The same argument proves that no subfield of the real field is algebraically closed; in particular, the field ofrational numbers is not algebraically closed. By contrast, thefundamental theorem of algebra states that the field ofcomplex numbers is algebraically closed. Another example of an algebraically closed field is the field of (complex)algebraic numbers.

Nofinite fieldF is algebraically closed, because ifa₁,a₂, ...,a_n are the elements ofF, then the polynomial (x − a₁)(x − a₂) ⋯ (x − a_n) + 1has no zero inF. However, the union of all finite fields of a fixed characteristicp (p prime) is an algebraically closed field, which is, in fact, thealgebraic closure of the field${\displaystyle {\mathbb{F}}_p}$ withp elements.

The field${\displaystyle \mathbb{C}(x)}$ of rational functions with complex coefficients is not closed; for example, the polynomial${\displaystyle y^2-x}$ has roots${\displaystyle \pm \sqrt{x}}$, which are not elements of${\displaystyle \mathbb{C}(x)}$.

Given a fieldF, the assertion "F is algebraically closed" is equivalent to other assertions:

The fieldF is algebraically closed if and only if the onlyirreducible polynomials in thepolynomial ringF[x] are those of degree one.

The assertion "the polynomials of degree one are irreducible" is trivially true for any field. IfF is algebraically closed andp(x) is an irreducible polynomial ofF[x], then it has some roota and thereforep(x) is a multiple ofx −a. Sincep(x) is irreducible, this means thatp(x) =k(x −a), for somek ∈F \ {0}. On the other hand, ifF is not algebraically closed, then there is some non-constant polynomialp(x) inF[x] without roots inF. Letq(x) be some irreducible factor ofp(x). Sincep(x) has no roots inF,q(x) also has no roots inF. Therefore,q(x) has degree greater than one, since every first degree polynomial has one root inF.

The fieldF is algebraically closed if and only if every polynomialp(x) of degreen ≥ 1, withcoefficients inF,splits into linear factors. In other words, there are elementsk, x₁, x₂, ..., x_n of the fieldF such thatp(x) = k(x − x₁)(x − x₂) ⋯ (x − x_n).

IfF has this property, then clearly every non-constant polynomial inF[x] has some root inF; in other words,F is algebraically closed. On the other hand, that the property stated here holds forF ifF is algebraically closed follows from the previous property together with the fact that, for any fieldK, any polynomial inK[x] can be written as a product of irreducible polynomials.

If every polynomial overF of prime degree has a root inF, then every non-constant polynomial has a root inF.^[1] It follows that a field is algebraically closed if and only if every polynomial overF of prime degree has a root inF.

The fieldF is algebraically closed if and only if it has no properalgebraic extension.

IfF has no proper algebraic extension, letp(x) be some irreducible polynomial inF[x]. Then thequotient ofF[x] modulo theideal generated byp(x) is an algebraic extension ofF whosedegree is equal to the degree ofp(x). Since it is not a proper extension, its degree is 1 and therefore the degree ofp(x) is 1.

On the other hand, ifF has some proper algebraic extensionK, then theminimal polynomial of an element inK \ F is irreducible and its degree is greater than 1.

The fieldF is algebraically closed if and only if it has no properfinite extension because if, within theprevious proof, the term "algebraic extension" is replaced by the term "finite extension", then the proof is still valid. (Finite extensions are necessarily algebraic.)

The fieldF is algebraically closed if and only if, for each natural numbern, everylinear map fromFⁿ into itself has someeigenvector.

Anendomorphism ofFⁿ has an eigenvector if and only if itscharacteristic polynomial has some root. Therefore, whenF is algebraically closed, every endomorphism ofFⁿ has some eigenvector. On the other hand, if every endomorphism ofFⁿ has an eigenvector, letp(x) be an element ofF[x]. Dividing by its leading coefficient, we get another polynomialq(x) which has roots if and only ifp(x) has roots. But ifq(x) =xⁿ +a_n − 1 x^n − 1 + ⋯ +a₀, thenq(x) is the characteristic polynomial of then×ncompanion matrix

The fieldF is algebraically closed if and only if everyrational function in one variablex, with coefficients inF, can be written as the sum of a polynomial function with rational functions of the forma/(x − b)ⁿ, wheren is a natural number, anda andb are elements ofF.

IfF is algebraically closed then, since the irreducible polynomials inF[x] are all of degree 1, the property stated above holds by thetheorem on partial fraction decomposition.

On the other hand, suppose that the property stated above holds for the fieldF. Letp(x) be an irreducible element inF[x]. Then the rational function 1/p can be written as the sum of a polynomial functionq with rational functions of the forma/(x – b)ⁿ. Therefore, the rational expression

${\displaystyle \frac{1}{p(x)}-q(x)=\frac{1-p(x)q(x)}{p(x)}}$

can be written as a quotient of two polynomials in which the denominator is a product of first degree polynomials. Sincep(x) is irreducible, it must divide this product and, therefore, it must also be a first degree polynomial.

For any fieldF, if two polynomialsp(x),q(x) ∈F[x] arerelatively prime then they do not have a common root, for ifa ∈F was a common root, then p(x) and q(x) would both be multiples ofx −a and therefore they would not be relatively prime. The fields for which the reverse implication holds (that is, the fields such that whenever two polynomials have no common root then they are relatively prime) are precisely the algebraically closed fields.

If the fieldF is algebraically closed, letp(x) andq(x) be two polynomials which are not relatively prime and letr(x) be theirgreatest common divisor. Then, sincer(x) is not constant, it will have some roota, which will be then a common root ofp(x) andq(x).

IfF is not algebraically closed, letp(x) be a polynomial whose degree is at least 1 without roots. Thenp(x) andp(x) are not relatively prime, but they have no common roots (since none of them has roots).

IfF is an algebraically closed field andn is a natural number, thenF contains allnth roots of unity, because these are (by definition) then (not necessarily distinct) zeroes of the polynomialxⁿ − 1. A field extension that is contained in an extension generated by the roots of unity is acyclotomic extension, and the extension of a field generated by all roots of unity is sometimes called itscyclotomic closure. Thus algebraically closed fields are cyclotomically closed. The converse is not true. Even assuming that every polynomial of the formxⁿ − a splits into linear factors is not enough to assure that the field is algebraically closed.

If a proposition which can be expressed in the language offirst-order logic is true for an algebraically closed field, then it is true for every algebraically closed field with the samecharacteristic. Furthermore, if such a proposition is valid for an algebraically closed field with characteristic 0, then not only is it valid for all other algebraically closed fields with characteristic 0, but there is some natural numberN such that the proposition is valid for every algebraically closed field with characteristic p whenp > N.^[2]

Every fieldF has some extension which is algebraically closed. Such an extension is called analgebraically closed extension. Among all such extensions there is one and only one (up to isomorphism, but notunique isomorphism) which is analgebraic extension ofF;^[3] it is called thealgebraic closure ofF.

The theory of algebraically closed fields hasquantifier elimination.

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