algebraically closed field
https://en.wikipedia.org/wiki/Algebraically_closed_field
As an example, the field of real numbers is not algebraically closed, because the polynomial equation x2 + 1 = 0 has no solution in real numbers, even though all its coefficients (1 and 0) are real.
Examples
As an example, the field of real numbers is not algebraically closed, because the polynomial equation x2 + 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 of rational numbers is not algebraically closed. Also, no finite field F is algebraically closed, because if a1, a2, ..., an are the elements of F, then the polynomial (x − a1)(x − a2) ··· (x − an) + 1 has no zero in F. By contrast, the fundamental theorem of algebra states that the field of complex numbers is algebraically closed. Another example of an algebraically closed field is the field of (complex) algebraic numbers.
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 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 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 coefficients in F, splits into linear factors. In other words, there are elements k, x1, x2, ..., xn of the field F such that p(x) = k(x − x1)(x − x2) ··· (x − xn).
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
J. Shipman showed in 2007 that if every polynomial over F of prime degree has a root in F, then every non-constant polynomial has a root in F, thus F is algebraically closed.
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 word "algebraic" is replaced by the word "finite", then the proof is still valid.
Every endomorphism of Fn has some eigenvector
The field F is algebraically closed if and only if, for each natural number n, every linear map from Fn into itself has some eigenvector.
An endomorphism of Fn has an eigenvector if and only if its characteristic polynomial has some root. Therefore, when F is algebraically closed, every endomorphism of Fn has some eigenvector. On the other hand, if every endomorphism of Fn 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) = xn + an − 1xn − 1+ ··· + a0, then q(x) is the characteristic polynomial of the n×n companion matrix
- {\displaystyle {\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}}.}
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 qwith rational functions of the form a/(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. 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 p(x),q(x) ∈ F[x] are relatively prime then they do not have a common root, for if a ∈ F was a common root, then p(x) and q(x) would both be multiples of x − 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).
Other properties
If F is an algebraically closed field and n is a natural number, then F contains all nth roots of unity, because these are (by definition) the n (not necessarily distinct) zeroes of the polynomial xn − 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 xn − 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 is it valid for all other algebraically closed fields with characteristic 0, but there is some natural number N such that the proposition is valid for every algebraically closed field with characteristic p when p > N.[1]
Every field F has some extension which is algebraically closed. Such an extension is called an algebraically closed extension. Among all such extensions there is one and only one (up to isomorphism, but not unique isomorphism) which is an algebraic extension of F;[2] it is called the algebraic closure of F.
The theory of algebraically closed fields has quantifier elimination.
https://zh.wikipedia.org/wiki/代數閉域
例子
举例明之，实数域并非代数闭域，因为下列实系数多项式无实根：
- {\displaystyle x^{2}+1=0}
同理可证有理数域非代数闭域。此外，有限域也不是代数闭域，因为若{\displaystyle a_{1},\ldots ,a_{n}}
列出{\displaystyle F}
的所有元素，则下列多项式在{\displaystyle F}中没有根：
- {\displaystyle (x-a_{1})(x-a_{2})\cdots (x-a_{n})+1\,}
反之，复数域则是代数闭域；这是代数基本定理的内容。另一个代数闭域之例子是代数数域。
等价的刻划[编辑]
给定一个域{\displaystyle F}，其代数封闭性与下列每一个性质等价：
不可约多项式当且仅当一次多项式[编辑]
域F是代数闭域，当且仅当环F[x]中的不可约多项式是而且只能是一次多项式。
“一次多项式是不可约的”的断言对于任何域都是正确的。如果F是代数闭域，p(x)是F[x]的一个不可约多项式，那么它有某个根a，因此p(x)是x − a的一个倍数。由于p(x)是不可约的，这意味着对于某个k ∈ F \ {0}，有p(x) = k(x − a)。另一方面，如果F不是代数闭域，那么存在F[x]内的某个非常数多项式p(x)在F内没有根。设q(x)为p(x)的某个不可约因子。由于p(x)在F内没有根，因此q(x)在F内也没有根。所以，q(x)的次数大于一，因为每一个一次多项式在F内都有一个根。
每一个多项式都是一次多项式的乘积[编辑]
域F是代数闭域，当且仅当每一个系数位于次数F内的n ≥ 1的多项式p(x)都可以分解成线性因子。也就是说，存在域F的元素k， x1， x2， ……， xn，使得p(x) = k(x − x1)(x − x2) ··· (x − xn)。
如果F具有这个性质，那么显然F[x]内的每一个非常数多项式在F内都有根；也就是说，F是代数闭域。另一方面，如果F是代数闭域，那么根据前一个性质，以及对于任何域K，任何K[x]内的多项式都可以写成不可约多项式的乘积，推出这个性质对F成立。
Fn的每一个自同态都有特征向量[编辑]
域F是代数闭域，当且仅当对于每一个自然数n，任何从Fn到它本身的线性映射都有某个特征向量。
Fn的自同态具有特征向量，当且仅当它的特征多项式具有某个根。因此，如果F是代数闭域，每一个Fn的自同态都有特征向量。另一方面，如果每一个Fn的自同态都有特征向量，设p(x)为F[x]的一个元素。除以它的首项系数，我们便得到了另外一个多项式q(x)，它有根当且仅当p(x)有根。但如果q(x) = xn + an − 1xn − 1+ ··· + a0，那么q(x)是以下友矩阵的特征多项式：
- {\displaystyle {\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}}.}
有理表达式的分解
域F是代数闭域，当且仅当每一个系数位于F内的一元有理函数都可以写成一个多项式函数与若干个形为a/(x − b)n的有理函数之和，其中n是自然数，a和b是F的元素。
如果F是代数闭域，那么由于F[x]内的不可约多项式都是一次的，根据部分分式分解的定理，以上的性质成立。
而另一方面，假设以上的性质对于域F成立。设p(x)为F[x]内的一个不可约元素。那么有理函数1/p可以写成多项式函数q与若干个形为a/(x − b)n的有理函数之和。因此，有理表达式
可以写成两个多项式的商，其中分母是一次多项式的乘积。由于p(x)是不可约的，它一定能整除这个乘积，因此它也一定是一个一次多项式。