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questions :algebraïcally closed field
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[Visitor (111.8.*.*)]answers [Chinese ]Time :2020-04-02
Given a field F, its algebraic closure is equivalent to each of the following properties:

Irreducible polynomial if and only if first degree polynomial

The field F is an algebraic closed field if and only if the irreducible polynomial in the ring F [x] is and can only be a polynomial of the first degree.
The assertion that "a polynomial of the first degree is irreducible" is true for any domain. If F is an algebraic closed field and p (x) is an irreducible polynomial of F [x], then it has some root a, so p (x) is a multiple of x − a. Since p (x) is irreducible, this means that for some k ∈ F \\ {0}, there is p (x) = k (x − a). On the other hand, if F is not an algebraic closed field, then there is some non-constant polynomial p (x) in F [x] that has no root in F. Let q (x) be some irreducible factor of p (x). Since p (x) has no root in F, q (x) has no root in F. Therefore, the degree of q (x) is greater than one, because every degree polynomial has a root in F.

Every polynomial is the product of a polynomial of the first degree
The field F is an algebraic closed field, and if and only if each coefficient is n ≥ 1 in the degree F, the polynomial p (x) can be decomposed into a linear factor. That is, 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 obviously every non-constant polynomial in F [x] has a root in F; that is, F is an algebraic closed field. On the other hand, if F is an algebraic closed field, then according to the previous property, and for any field K, any polynomial in K [x] can be written as the product of an irreducible polynomial, and this property is deduced to be true for F.

Each automorphism of Fn has a eigenvector
The field F is an algebraic closed field. If and only for each natural number n, any linear mapping from F to itself has a certain feature vector.

F ^ n's automorphism has a feature vector if and only if its feature polynomial has a certain root. Therefore, if F is an algebraic closed field, each automorphism of F ^ n has a eigenvector. On the other hand, if every automorphism of F ^ n has a feature vector, let p (x) be an element of F [x]. Dividing by its first term coefficient, we get another polynomial q (x) that has roots if and only if p (x) has roots. But if q (x) = x ^ n an-1x ^ n-1 ··· a0, then q (x) is the characteristic polynomial of the following friend matrix:

0 0 0 …… 0 -a01 0 0 …… 0 -a10 1 0 …… 0 -a2

Decomposition of rational expressions
The field F is a closed algebraic field. If and only if each unary rational function whose coefficients are in F can be written as the sum of a polynomial function and several rational functions of the form a / (x − b) ^ n, where n is Natural numbers, a and b are elements of F.

If F is an algebraic closed field, then since the irreducible polynomials in F [x] are all one degree, according to the theorem of partial fraction decomposition, the above properties hold.

On the other hand, suppose the above properties hold for domain 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 and several rational functions of the form a / (x − b) ^ n. Therefore, rational expressions
Can be written as the quotient of two polynomials, where the denominator is the product of the polynomials of the first degree. Since p (x) is irreducible, it must be able to divide this product, so it must also be a polynomial of the first degree.
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