- Algebraically closed field
In

mathematics , a field "F" is said to be**algebraically closed**if everypolynomial in onevariable of degree at least 1, withcoefficient s in "F", has a root in "F".**Examples**As an example, the field of

real number s is not algebraically closed, because the polynomial equation "x"^{2}+ 1 = 0 has no solution in real numbers, even though all its coefficients (1, 0 and 1) are real. The same argument proves that no subfield of the real field is algebraically closed; in particular, the field ofrational number s is not algebraically closed. Also, nofinite field "F" is algebraically closed, because if "a"_{1}, "a"_{2}, …, "a_{n}" are the elements of "F", then the polynomial ("x" − "a"_{1})("x" − "a"_{2}) ··· ("x" − "a"_{"n"}) + 1 has no zero in "F". By contrast, thefundamental theorem of algebra states that the field ofcomplex number s is algebraically closed. Another example of an algebraically closed field is the field of (complex)algebraic number s.**Equivalent properties**Given a field "F", the assertion “"F" is algebraically closed” is equivalent to other assertions:

**The only irreducible polynomials are those of degree one**The field "F" is algebraically closed if and only if the only irreducible polynomials in the ring "F" ["x"] are those of degree one.

The assertion “the polynomials of degree one are irreducible” is trivially true for any field. If "F" is algebraically closed and "p"("x") is an irreducible polynomial of "F" ["x"] , then it has some root "a" and therefore "p"("x") is a multiple of "x" − "a". Since "p"("x") is irreducible, this means that "p"("x") = "k"("x" − "a"), for some "k" ∈ "F" {0}. On the other hand, if "F" is not algebraically closed, then there is some non-constant polynomial "p"("x") in "F" ["x"] without roots in "F". Let "q"("x") be some irreducible factor of "p"("x"). Since "p"("x") has no roots in "F", "q"("x") also has no roots in "F". Therefore, "q"("x") has degree greater than one, since every first degree polynomial has one root in "F".

**Every polynomial is a product of first degree polynomials**The field "F" is algebraically closed if and only if every polynomial "p"("x") of degree "n" ≥ 1, with

coefficient s in "F", splits into linear factors. In other words, there are elements "k", "x"_{1}, "x"_{2}, …, "x_{n}" of the field "F" such that "p"("x") = "k"("x" − "x"_{1})("x" − "x"_{2}) ··· ("x" − "x_{n}").It "F" has this property, then clearly every non-constant polynomial in "F" ["x"] has some root in "F"; in other words, "F" is algebraically closed. On the other hand, that the property stated here holds for "F" if "F" is algebraically closed follows from the previous property together with the fact that, for any field "K", any polynomial in "K" ["x"] can be written as a product of irreducible polynomials.

**The field has no proper algebraic extension**The field "F" is algebraically closed if and only if it has no proper

algebraic extension .If "F" has no proper algebraic extension, let "p"("x") be some irreducible polynomial in "F" ["x"] . Then the quotient of "F" ["x"] modulo the ideal generated by "p"("x") is an algebraic extension of "F" whose degree is equal to the degree of "p"("x"). Since it is not a proper extension, its degree is 1 and therefore the degree of "p"("x") is 1.

On the other hand, if "F" has some proper algebraic extension "K", then the minimal polynomial of an element in "K" "F" is irreducible and its degree is greater than 1.

**The field has no proper finite extension**The field "F" is algebraically closed if and only if it has no finite

algebraic extension because if, within the previous proof, the world “algebraic” is replaced by the word “finite”, then the proof is still valid.**Every endomorphism of "F**^{n}" has some eigenvectorThe field "F" is algebraically closed if and only if, for each natural number "n", every

linear map from "F^{n}" into itself has someeigenvector .An endomorphism of "F

^{n}" has an eigenvector if and only if itscharacteristic polynomial has some root. Therefore, when "F" is algebraically closed, every endomorphism of "F^{n}" has some eigenvector. On the other hand, if every endomorphism of "F^{n}" has an eigenvector, let "p"("x") be an element of "F" ["x"] . Dividing by its leading coefficient, we get another polynomial "q"("x") which has roots if and only if "p"("x") has roots. But if "q"("x") = "x^{n}" + "a"_{"n" − 1}"x"^{"n" − 1}+ ··· + "a"_{0}, then "q"("x") is the characteristic polynomial of thecompanion matrix :$egin\{pmatrix\}00cdots0-a\_0\backslash 10cdots0-a\_1\backslash 01cdots0-a\_2\backslash vdotsvdotsddotsvdotsvdots\backslash 00cdots1-a\_\{n-1\}end\{pmatrix\}.$**Decomposition of rational expressions**The field "F" is algebraically closed if and only if every

rational function in one variable "x", with coefficients in "F", can be written as the sum of a polynomial function with rational functions of the form "a"/("x" − "b")^{n}, where "n" is a natural number, and "a" and "b" are elements of "F".If "F" is algebraically closed then, since the irreducible polynomials in "F" ["x"] are all of degree 1, the property stated above holds by the theorem on partial fraction decomposition.

On the other hand, suppose that the property stated above holds for the field "F". Let "p"("x") be an irreducible element in "F" ["x"] . Then the rational function 1/"p" can be written as the sum of a polynomial function "q" with rational functions of the form "a"/("x" − "b")

^{n}. Therefore, the rational expression:$frac1\{p(x)\}-q(x)=frac\{1-p(x)q(x)\}\{p(x)\}$can be written as a quotient of two polynomials in which the denominator is a product of first degree polynomials. Since "p"("x") is irreducible, it must divide this product and, therefore, it must also be a first degree polynomial.**Other properties**If "F" is an algebraically closed field and "n" is a natural number, then "F" contains all "n"

^{th}roots of unity, because these are (by definition) the "n" (not necessarily distinct) zeroes of the polynomial "x^{n}" − 1. A field extension that is contained in an extension generated by the roots of unity is a "cyclotomic extension", and the extension of a field generated by all roots of unity is sometimes called its "cyclotomic closure". Thus algebraically closed fields are cyclotomically closed. The converse is not true. Even assuming that every polynomial of the form "x^{n}" − "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 of first-order logic is true for an algebraically closed field, then it is true for every algebraically closed field with the same characteristic. Furthermore, if such a proposition is valid for an algebraically closed field with characteristic 0, then not only it is valid for all other algebraically closed field with characteristic 0 as there is some natural number "N" such that the proposition is valid for every algebraically closed field with characteristic "p" when "p" > N. [

*See subsections "Rings and fields" and "Properties of mathematical theories" in §2 of J. Barwise's "An introduction to first-order logic".*]Every field "F" has some extension which is algebraically closed. Among all such extensions there is one and (up to isomorphism) only one which is an

algebraic extension of "F"; [*See Lang's "Algebra", §VII.2 or van der Waerden's "Algebra I", §10.1.*] it is called thealgebraic closure of "F".**Notes****References***

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