Editing Algebraically closed field (section)

==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 polynomial]]s in the [[polynomial 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 {{math|''x'' &minus; ''a''}}. Since ''p''(''x'') is irreducible, this means that {{math|1=''p''(''x'') = ''k''(''x'' &minus; ''a'')}}, for some {{math|''k'' ∈ ''F'' \&thinsp;{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''&nbsp;≥&nbsp;1, with [[coefficient]]s in ''F'', [[factorization|splits into linear factors]]. In other words, there are elements ''k'',&nbsp;''x''<sub>1</sub>,&nbsp;''x''<sub>2</sub>,&nbsp;...,&nbsp;''x<sub>n</sub>'' of the field ''F'' such that ''p''(''x'')&nbsp;=&nbsp;''k''(''x''&nbsp;&minus;&nbsp;''x''<sub>1</sub>)(''x''&nbsp;&minus;&nbsp;''x''<sub>2</sub>)&nbsp;⋯&nbsp;(''x''&nbsp;&minus;&nbsp;''x<sub>n</sub>'').

If ''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.

===Polynomials of prime degree have roots===
If every polynomial over ''F'' of prime degree has a root in ''F'', then every non-constant polynomial has a root in ''F''.<ref>Shipman, J. [http://www.jon-arny.com/httpdocs/Gauss/Shipman%20Intellig%20Mod%20p%20FTA.pdf Improving the Fundamental Theorem of Algebra]  ''The Mathematical Intelligencer'', Volume 29 (2007), Number 4. pp. 9–14</ref> It follows that a field is algebraically closed if and only if every polynomial over ''F'' of prime degree has a root in ''F''.

===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 ring|quotient]] of ''F''[''x''] modulo the [[ideal (ring theory)|ideal]] generated by ''p''(''x'') is an algebraic extension of ''F'' whose [[degree of a field extension|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&nbsp;1.

On the other hand, if ''F'' has some proper algebraic extension ''K'', then the [[Minimal polynomial (field theory)|minimal polynomial]] of an element in ''K''&nbsp;\&nbsp;''F'' is irreducible and its degree is greater than&nbsp;1.

===The field has no proper finite extension===
The field ''F'' is algebraically closed if and only if it has no proper [[finite extension]] because if, within the [[#The field has no proper algebraic extension|previous proof]], the term "algebraic extension" is replaced by the term "finite extension", then the proof is still valid. (Finite extensions are necessarily algebraic.)

===Every endomorphism of ''F<sup>n</sup>'' has some eigenvector===
The field ''F'' is algebraically closed if and only if, for each natural number ''n'', every [[linear map]] from ''F<sup>n</sup>'' into itself has some [[eigenvector]].

An [[endomorphism]] of ''F<sup>n</sup>'' has an eigenvector if and only if its [[characteristic polynomial]] has some root. Therefore, when ''F'' is algebraically closed, every endomorphism of ''F<sup>n</sup>'' has some eigenvector. On the other hand, if every endomorphism of ''F<sup>n</sup>'' 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 {{math|1=''q''(''x'') = ''x<sup>n</sup>'' + ''a''<sub>''n''&thinsp;&minus;&thinsp;1</sub>&thinsp;''x''<sup>''n''&thinsp;&minus;&thinsp;1</sup>&thinsp;+ ⋯ + ''a''<sub>0</sub>}}, then ''q''(''x'') is the characteristic polynomial of the ''n×n'' [[companion matrix]]
:<math>\begin{pmatrix}
  0 & 0 & \cdots & 0 & -a_0\\
  1 & 0 & \cdots & 0 & -a_1\\
  0 & 1 & \cdots & 0 & -a_2\\
  \vdots & \vdots & \ddots & \vdots & \vdots\\
  0 & 0 & \cdots & 1 & -a_{n-1}
\end{pmatrix}.</math>

===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''&nbsp;−&nbsp;''b'')<sup>''n''</sup>, 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 [[Partial fraction decomposition#Statement of theorem|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''&nbsp;–&nbsp;''b'')<sup>''n''</sup>. Therefore, the rational expression
:<math>\frac1{p(x)}-q(x)=\frac{1-p(x)q(x)}{p(x)}</math>
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.

===Relatively prime polynomials and roots===
For any field ''F'', if two polynomials {{math|''p''(''x''), ''q''(''x'') ∈ ''F''[''x'']}} are [[coprime|relatively prime]] then they do not have a common root, for if {{math|''a'' ∈ ''F''}} was a common root, then&nbsp;''p''(''x'') and&nbsp;''q''(''x'') would both be multiples of {{math|''x'' &minus; ''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 field ''F'' is algebraically closed, let ''p''(''x'') and ''q''(''x'') be two polynomials which are not relatively prime and let ''r''(''x'') be their [[greatest common divisor]]. Then, since ''r''(''x'') is not constant, it will have some root ''a'', which will be then a common root of ''p''(''x'') and ''q''(''x'').

If ''F'' is not algebraically closed, let ''p''(''x'') be a polynomial whose degree is at least 1 without roots. Then ''p''(''x'') and ''p''(''x'') are not relatively prime, but they have no common roots (since none of them has roots).