Algebraic field extension. Theorem on the structure of a simple algebraic extension

Definition 4.9. Let the field F is an extension of the field R and a e F. We form the set of all elements that are obtained from the elements of the field R and the element a using the operations of addition, subtraction, multiplication and division. Obviously, this set is a field called by a simple extension of the field P by adding the element a and is denoted R(a). If an element a is algebraic over a field P, then P(a) is called simple algebraic expansion, and if a is transcendent over P, then P(a) is called simple transcendent extension fields R.

It is easy to see that P(a) = (--|g(x),h(x)e P[x],/i(a)^0

and P(a) is the minimum field containing the field P and the element a.

Theorem 4.4 (on the structure of a simple algebraic field extension).If a a- algebraic element over a field P of degree n, then:

• 1) P(o0 = (/(a) | /(x) e P[x]);
• 2) P(a) is a vector space over the field P with basis(1 = a 0 , a,..., a "- 1 ), so that |P(a) : P| = P;
• 3) every element(3 e P(a) uniquely represent as a value Yes) some polynomial Dx) with coefficients from the field P of degree not exceeding n - 1.

Proof. 1. It follows from the definition of a simple algebraic extension of the field P that

At the same time, the solution of the problem of getting rid of algebraic irrationality in the denominator of a fraction implies

the ability to represent the relationship -- in the form --= / (a),

hi os) l(a)

where fix) g P[x]. Consequently, P(a) = -(/(a) | Ax) g P[x]).

• 2. Let us prove that the system (1 = a 0 , a, ..., a n_1 ) is a basis of the vector space P(a) over the field P.
• 2.1. We will prove that the system of vectors (1 = a 0, a, ..., a "-1)

linearly independent. Suppose that there are elements a 0 , ..., a n-1 e P such that a 0 1 + a 2 a + ... + a p _ g a p ~ 1 = 0.

Then a turns out to be a root of the polynomial g(x) = a 0 + ajX + ... + + a n _p^- 1g P[x]. If we assume that this polynomial is non-zero, then its degree is less than the degree of the minimal polynomial of the element a, which contradicts the definition of the minimal polynomial. Therefore, the polynomial g(x) is zero, i.e., all its coefficients are equal to zero. But this also means the linear independence of the system of vectors ( 1 = a 0 , a,..., a n_1 ).

• 2.2. Let us prove that any element (3 e P(a) is a linear combination of the vectors of the system (1 = a 0, a, ..., a "-1). In item 1) it is proved that. Divide Dx) by the minimal polynomial φ(x) of an algebraic element a: f(x) = φ(x) q(x) + r(x), where either r(x) = 0 or the degree of the remainder r(x) is strictly less than the degree of the polynomial φ(x) equal to n. In the first case, Δx) = φ(x) q(x), (3 = Da) = φ(a) q(a) = 0 and (3 = 0 is trivially expressed in terms of the elements of this system. In the second case, the remainder has the form r(x) = b 0 + b) x+ ... + b n _ 1 x n_1 . But then (3 = Yes) = f(a) q(a) + r(a) = r(a) = b 0+ bja + ... + b^a* -1 , which was to be proved. So, the system of vectors (1 = a 0 , a, ..., a n-1 ) is the basis.
• 3. The definition of a basis implies the uniqueness of the representation of any element (3 e P(a) in the form of a linear combination of basic elements (1 = a 0, a, ..., a "-1). The theorem is proved.

Consequence. The degree of a simple algebraic extension P(a) coincides with the degree of the minimal polynomial of a.

Consider examples.

• 1. Simple algebraic extensions of the field Q: Q(V2) = (a + bV2 |a,beQ), Q(^3) = (a + b/3 + Cyfi?|a, b, cgQ).
• 2. Consider the field of residue classes modulo 2: Z 2 \u003d (0,1), and the polynomial over this field x 2 + x + 1. By checking, we establish that it has no roots in Z 2, which means that it is irreducible over Z 2. We will assume that in some extension of the field Z 2 this polynomial has a root a. Then we obtain a simple algebraic extension Z 2 (a) = (0,1, a, a +1).

Exercise 4.1. Make tables of addition and multiplication of expansion elements from the previous example 2.

As noted above, a simple algebraic extension F= P(a) fields R is finite, and hence algebraic, i.e. any element (3 e F is a root of some polynomial with coefficients from P. Let us show by an example how to find this polynomial.

Example 4.1

Let F\u003d Q (a) and a is the root of the polynomial / (x) \u003d x 3 + X- 1. Find the minimal polynomial of the element (3 = a 2 + a + 2.

Solution. By condition, a 3 + a - 1 \u003d O, hence a 3 \u003d 1 - a. From the equality given by the condition p \u003d a 2 + a + 2, we find a 2 + a + 2- P \u003d 0. Multiplying this equality by a, we get aP \u003d a 3 + a 2 + 2a \u003d (1-

• - a) + a 2 + 2a \u003d a 2 + a + 1, whence a 2 + (1 - (3) a + 1 \u003d 0. Finally, a 2 p \u003d a 4 + a 3 + 2a 2 \u003d a (1 - a) + (1 - a) + 2a 2 \u003d a 2 + 1, whence (1 -
• - p)a 2 +1 = 0. Thus, we arrive at the system of equalities

Excluding a 2 and a, we arrive at the equality P 3 - 4P 2 + ZP - 1 = 0, which says that P is the root of the polynomial cp (x) \u003d x 3 -

- 4x 2+ Zx- 1. This polynomial, as is easy to see, has no rational roots, and hence is irreducible over the field Q. Consequently, φ(x) is the required minimal polynomial of the element p algebraic over the field Q.

Note that with the exception a from the system of equalities, you can draw knowledge on solving systems linear equations. The system equalities say that the vector (a 2 , a, 1) is a non-zero solution of a homogeneous system of linear equations

Therefore, the determinant of the matrix of the system is equal to zero:

From here we obtain the equality Р 3 - 4Р 2 + ЗР - 1 = 0.

Finally, we note that this exercise can be turned into a school problem: it is known that a 3 + a-1 = 0, P = a 2 + a + 2; find an equation with integer coefficients whose root is p.

Introduction.

A program of a unified course in algebra and number theory has been introduced in pedagogical universities. The main goal of this course is to study the basic algebraic systems and nurture the algebraic culture necessary for a future teacher to deeply understand the goals and objectives of both the main school mathematics course and optional school courses.

In our opinion, the most expedient is the introduction of elements of modern abstract algebra into school teaching.

The process of algebraization of mathematics, which began in the 20th century, does not stop, and this causes persistent attempts to introduce basic algebraic concepts into school mathematical education.

Mathematical depth and an unusually wide scope of fields are combined with the simplicity of its main provisions - the concepts of fields, a number of important theorems can be formulated and proved, having initial ideas in the field of set theory. Therefore, field theory is the best way to show schoolchildren an example of modern mathematics.

In addition, the study of the elements of field theory is useful for schoolchildren, contributes to their intellectual growth, which is manifested in the development and enrichment of various aspects of their thinking, qualities and personality traits, as well as instilling in students an interest in mathematics and science.

1. A simple algebraic field extension.

1.1. Simple field expansion.

Let P[x] be a polynomial ring in x over a field P, where P is a subfield of the field F. Recall that an element a of a field F is called algebraic over a field P if a is a root of some positive degree polynomial in P [x].

Definition. Let P< F и a0F. Простым расширением поля Pс помощью элемента a называется наименьшее подполе поля F, содержащее множество Р и элемент a. Простое расширение Pс помощью a обозначается через P (a), основное множество поля P (a) обозначается через Р(a).

Let a0F, P [x] be the ring of polynomials in x and

P[x]=(f(a)*f0P[x]),

i.e. P [a] is the set of all expressions of the form a 0 + a 1 a+...+ a n a n , where a 0 , a 1, ... a n 0P and n is any natural number.

It is easy to see that the algebra +P[a], +, -, ., 1, - a subring of the field P (a) - is a ring; this ring is denoted by P[a].

Theorem 1.1. Let P[x] be a polynomial ring in x over P and P(a) be a simple extension of the field P. Let y be a mapping from P[x] to P[a] such that y(f)=f(a) for any f from P[x]. Then:

(a) for any a in P y (a) = a;

(c) y is a homomorphism of the ring P [x] onto the ring P [a];

(d) Kery =(f0P[x]*f(a)=0);

(f) the quotient ring P[x]/Ker y is isomorphic to the ring P[a].

Proof. Statements (a) and (b) follow directly from the definition of y. The mapping y preserves the main operations of the ring P[x], since for any f and g from P[x]

y(f + g)=f(a)+g(a), y(fg)= f(a)g(a), y(1)=1.

Assertion (d) follows directly from the definition of the map y.

Since y is a homomorphism of the ring P[x] onto P[a], the quotient ring P[x]/Ker y is isomorphic to the ring P[a].

Corollary 1.2. Let a be a transcendental element over a field P. Then the polynomial ring P[x] is isomorphic to the ring P[a].

Proof. Due to the transcendence of a over PKery=(0). Therefore P[x]/(0)P [a]. Moreover, the quotient ring of the ring P [x] is isomorphic to P [x] by the zero ideal. Hence P[x]P[a].

1.2 Minimal polynomial of an algebraic element.

Let P [x] be the ring of polynomials over the field P.

Definition. Let a be an algebraic element over a field P. The minimal polynomial of an element a over P is a normalized polynomial in P[x] of the least degree whose root is a. The degree of a minimal polynomial is called the degree of an element a over P.

It is easy to see that for any element a that is algebraic over P , there exists a minimal polynomial.

Proposition 1.3. If a is an algebraic element over a field P, and g and j are its minimal polynomials over P, then g=j.

Proof. The degrees of the minimal polynomials g and j coincide. If g¹j, then the element a (of degree n over P) will be a root of the polynomial g - j, the degree of which is less than the degree of the polynomial j (less than n), which is impossible. Therefore, g=j.

Theorem 1.4. Let a be an algebraic element of degree n over a field P (aóP) and g be its minimal polynomial over P. Then:

(a) the polynomial g is irreducible in the ring P [x];

(b) if f (a) = 0, where f0P[x], then g divides f;

(c) the quotient ring P [x]/(g) is isomorphic to the ring P [a];

(d) P[x]/(g) is a field;

(f) the ring P[a] coincides with the field P(a).

Proof. Let us assume that the polynomial g is reducible in the ring P [x], i.e., there are polynomials j and h in P[x] such that

g = jh, 1£deg j, deg h

Then g(a) = j(a)h(a) = 0. Since P (a) is a field, then j(a) = 0 or h(a) = 0, which is impossible because, by assumption, the degree element a over P is equal to n.

Assume that f0 P[x] and f(a) = 0. By assumption, g(a) = 0. Therefore, f and g cannot be coprime. Since the polynomial g is irreducible, then g divides f.

Let j be the homomorphism of the ring P [x] onto the ring P [a] (y(f)=f(a) for any f from P[x]) considered in Theorem 2.1. By virtue of (b), the kernel of the homomorphism y consists of multiples of the polynomial g, i.e., Ker y = (g). Therefore, the quotient ring P = P [x]/(g) is isomorphic to the ring P [a].

Since P[a]ÌP(a), then P [a] is a domain of integrity. Because [email protected][a], then the quotient ring P is also a domain of integrity. We need to show that any nonzero element f from P is invertible to P. Let f be an element of the coset f. Since f¹ 0, then f(a)¹0; so the polynomial g does not divide the polynomial f. Since the polynomial g is irreducible, it follows that the polynomials f and g are coprime. Therefore, there are polynomials u and v in Р[x] such that uf + vg=1. This implies the equality uf = 1, which shows that the element f is invertible in the ring P. Thus, we have established that the quotient ring P is a field.

By virtue of (c) and (d), P[a] is a field, and so P(a)ÌP[a]. Moreover, obviously, P[a]ÌP(a). Hence P[a] = P(a). Therefore, the ring P[a] coincides with the field P(a).

1.3 Structure of a simple algebraic field extension.

Theorem 1.5. Let a be an algebraic element over the field P of positive degree n. Then any element of the field P(a) can be uniquely represented as a linear combination of n elements 1, a, ..., a n-1 with coefficients from P.

Proof. Let b be any element of the field P (a). By Theorem 1.4, P(a) = P[a]; hence there exists a polynomial f in P[x] such that

Let g be the minimal polynomial for a over P; by virtue of the hypothesis of the theorem, its degree is equal to n. By the division theorem with remainder, there are polynomials h and r in P[x] such that

(2) f = gh + r, where r = 0 or derr< derg = n , т. е. r=c 0 +c 1 x +…c n -1 x n -1 (c i 0P). Полагая в (2) x = а и учитывая равенство (1), имеем

(3) b = c 0 +c 1 a +…c n -1 a n-1

Let's show that the element b is uniquely representable as a linear combination of elements 1, a, ..., a n-1 . Let

(4) b = d 0 +d 1 a +…d n -1 a n-1 (d i 0P)

Any such performance. Consider a polynomial j

j \u003d (c 0 - d 0) + (c 1 - d i .)x + . . . + (with n-1 –d n -1)x n -1

The case when the degree of j is less than n is impossible because, by virtue of (3) and (4), j(a) = 0 and the degree of j is less than the degree of g. The only possible case is when j = 0, i.e. with 0 = d 0 , . . . , with n-1 = d n-1. Therefore, the element b is uniquely representable as a linear combination of elements 1, a,…,a n-1 .

1.4. Exemption from algebraic irrationality in the denominator of a fraction.

The problem of getting rid of algebraic irrationality in the denominator of a fraction is as follows. Let a be an algebraic element of degree n>1 over a field P; f and h are polynomials from the polynomial ring P [x] and h(a) ¹0. It is required to represent the element f(a)/h(a)0P(a) as a linear combination of powers of the element a, i.e., as j(a),

This problem is solved in the following way. Let g be the minimal polynomial for a over P. Since, by Theorem 1.4, the polynomial is irreducible over P and h(a) ¹ 0, it follows that g does not divide h and hence the polynomials h and g are coprime. Therefore, there are polynomials u and v in P[x] such that

Since g(a) = 0, it follows from (1) that

u(a)g(a) = 1, 1/h(a) = u(a).

Therefore, f(a)/h(a) = f(a)u(a), where f,u 0P[x] and f(a)u(a)0P[a]. So, we got rid of the irrationality in the denominator of the fraction f(a)/h(a) .

Get rid of irrationality in the denominator of a fraction

Solution. In our case a=

The polynomials p(x) and g(x)=-x 2 +x+1 are coprime. Therefore, there are polynomials j and y such that

To find j and y, we apply the Euclid algorithm to the polynomials p and g:

X 3 -2 -x 2 +x+1 -x 2 +x+1 2x-1

x 3 -x 2 -x -x-1 -x 2 +1/2x -1/2x+1/4

x 2 -x-1 1/2x-1/4

In this way,

p=g(-x-1)+(2x-1),

g=(2x-1)(-1/2x+1/4)+5/4.

Where do we find

(2x-1)=p+g(x+1),

5/4=g-(p+g(x+1))(-1/2x+1/4)

p1/5(2x-1)+g(4/5+1/5(2x 2 +x-1))=1,

p1/5(2x-1)+g(2/5x 2 +1/5x+3/5)=1.

In this way,

y(x)= (2/5x 2 +1/5x+3/5).

Consequently

2.Composite algebraic field extension.

2.1. Finite field extension.

Let P be a subfield of the field F. Then we can consider F as vector space over P, i.e., consider the vector space +F, +, (w l ½l0P),

where w l is the operation of multiplying elements from F by the scalar l0P.

Definition. An extension F of a field P is said to be finite if F, as a vector space over P, has finite dimension. This dimension is denoted by .

Proposition 2.1. If a is an algebraic element of degree n over P, then =n.

This proposition follows directly from Theorem 1.5.

Definition. An extension F of a field P is called algebraic if every element of F is algebraic over P.

10. Theorem on the structure of a simple algebraic extension

ten . The concept of a minimal polynomial.

Let a be an algebraic number over a field k, i.e. root of a nonzero polynomial with coefficients from the field k.

Definition. A normalized polynomial m(a, k, x) over a field k is called a minimal polynomial of a if the following conditions are met:

a) m(x) is irreducible over the field k, i.e. does not decompose into a product of polynomials of positive degree with coefficients from k;

b) m(a) = 0, i.e. a is the root of the polynomial m(x).

twenty . Basic properties of minimal polynomials.

1. If f(x) н k[x] and f(a) = 0, then f(x) is divisible by the minimal polynomial m(x) of a.

Proof. Indeed, assuming that f is not divisible by m, we write

f = mg + r, deg r< deg m

based on the division with remainder theorem. Whence r(a)=0. Since the polynomials r and m are coprime, they cannot have common roots - a contradiction.

2. Let a be an algebraic number and g(x) be a normalized polynomial of least positive degree such that g(x) н k[x] and g(a) = 0. Then g(x) is the minimal polynomial of the number a.

The proof follows immediately from property 1.

3. The minimal polynomial of an algebraic number a over a given field is uniquely defined.

To prove it, it suffices to apply property 2.

Definition. The degree of the minimal polynomial of a is called the degree of a; notation deg k a.

4. a н k w deg k a = 1.

The proof follows immediately from the definitions.

5. If a is an algebraic number of degree n, then 1, a, a 2 , ..., a n -1 are linearly independent over the field k, i.e. ("c 0 , c 1 , ..., c n-1 нk) c 0 + c 1 a + ... + c n-1 a n -1 = 0 is possible only in the case c 0 = c 1 = . . .=cn-1=0.

Proof. Indeed, if the indicated powers of the number a are linearly dependent, then this number is the root of some polynomial over k, of degree less than m.

6. Let a be an algebraic number, f(x) н k[x] and f(a) ¹ 0. Then the fraction can be represented as = g(a) for some g(x) н k[x].

Proof. Indeed, the polynomials f and m are coprime (otherwise f would be divisible by m), which means that by the linear representation theorem gcd: for some polynomials g and h over k the equality

Whence f(a) g(a) = 1, as required.

thirty . The structure of simple algebraic extensions.

Definition. Let k be a subfield in L; a н L. The smallest subfield in L containing the number a and the subfield k, denoted by k(a), is called a simple extension of the field k (it is also said that k(a) is obtained by adding the number a to the field k).

It is easy to deduce a theorem from the above properties.

Theorem (on the structure of a simple algebraic extension).

For any algebraic number a over a field k, the linear space k(a) has a basis of elements of the form

1, a, a 2 , . . . , a n -1 , where n = deg k a.

Proof. It is easy to understand that k(a) consists of fractions f(a)/g(a), where f(x), g(x) are polynomials over the field k and g(a) ¹ 0. Denote by k[a] - the ring of polynomial values ​​at the point a, i.e. k[a] = ( f(a)½f(x)н k[x]).

Property 6 implies the equality k(a) = k[a]. It follows from the division theorem with remainder that the value of an arbitrary polynomial over the field k at the point a is a linear combination over the field k of the powers of the element a indicated in the theorem. Finally, it follows from property 5 that these powers are linearly independent over the field k. ÿ

40 . Exemption from irrationality in the denominator of a fraction.

Let us analyze various ways of solving the problem of liberation from irrationality in the denominator of a fraction. The fundamental possibility of its solution follows from the theorem on the structure of a simple algebraic extension.

Example 1. Get rid of irrationality in the denominator of a fraction:

Solution. Denote by c the number , and use the well-known formula for the sum of the terms of a geometric progression:

1+ c + c 2 + c 3 + c 4 = (c 5 - 1)/(c- 1) = 1/(c- 1),

Consequently, .

Example 2. Get rid of irrationality in the denominator of a fraction:

Solution. Denote by c the number , and first write down the fraction

as a simple sum:

.

Now, using Horner's scheme, each of the indicated fractions can be replaced by a polynomial in c. First we divide c 5 - 2 by c + 1:

Consequently,

C 4 - 2c 3 + 4c 2 - 8c + 16.

Then we get

34(c 4 - c 3 + c 2 - c + 1) - 3(c 4 - 2c 3 + 4c 2 - 8c + 16) =

31c 4 - 40c 3 + 22c 2 - 10c - 14,

Example 3. Get rid of irrationality in the denominator of a fraction:

Solution. Denote by c the number . Find a linear representation of the gcd of the polynomials f(x) = x 3 - 2 and g(x) = 1 + 2x - x 2:

f(x) = - g(x)×(x + 2) + r(x), where r(x) = 5x

5g(x) = r(x)×(x - 2) - 5.

From these equalities, we obtain a linear representation of gcd f(x) and g(x):

f(x)×(x - 2) + g(x)×(x 2 + 1) = 5.

Substituting the number c instead of x into the last equality, we get

hence =.

Example 4. Get rid of irrationality in the denominator of a fraction:

.

Solution. Denote by c the number and apply the method of indeterminate coefficients. By the theorem on the structure of a simple algebraic extension, there are rational numbers x, y, z such that

Xc 2 + yc + z or 89 = (c 2 + 16c - 11)(xc 2 + yc + z).

Expanding the brackets and using the equality c 3 = 2, we get:

89 = (32x + 2y - 11z) + (2x - 11y + 16z)c + (-11x + 16y + z)c 2 .

Since the numbers 1, c, c 2 are linearly independent over Q, we have

32x + 2y - 11z = 89, 2x - 11y + 16z = 0,

11x + 16y + z = 0.

The solution of the last system is the set of numbers (3, 2, 1). So we get the answer: .

Algebraic field extensions

Introduction.

A program of a unified course in algebra and number theory has been introduced in pedagogical universities. The main goal of this course is to study the basic algebraic systems and nurture the algebraic culture necessary for a future teacher to deeply understand the goals and objectives of both the main school mathematics course and optional school courses.

In our opinion, the most expedient is the introduction of elements of modern abstract algebra into school teaching.

The process of algebraization of mathematics, which began in the 20th century, does not stop, and this causes persistent attempts to introduce basic algebraic concepts into school mathematical education.

Mathematical depth and an unusually wide scope of fields are combined with the simplicity of its main provisions - the concepts of fields, a number of important theorems can be formulated and proved, having initial ideas in the field of set theory. Therefore, field theory is the best way to show schoolchildren an example of modern mathematics.

In addition, the study of the elements of field theory is useful for schoolchildren, contributes to their intellectual growth, which is manifested in the development and enrichment of various aspects of their thinking, qualities and personality traits, as well as instilling in students an interest in mathematics and science.

1. A simple algebraic field extension.

1.1. Simple field expansion.

Let P[x] be a polynomial ring in x over a field P, where P is a subfield of the field F. Recall that an element a of a field F is called algebraic over a field P if a is a root of some positive degree polynomial in P [x].

Definition. Let P< F и a0F. Простым расширением поля P с помощью элемента a называется наименьшее подполе поля F, содержащее множество Р и элемент a. Простое расширение P с помощью a обозначается через P (a), основное множество поля P (a) обозначается через Р(a).

Let a0F, P [x] be the ring of polynomials in x and

P[x]=(f(a)*f0P[x]),

i.e. P [a] is the set of all expressions of the form a 0 + a 1 a+...+ a n a n , where a 0 , a 1, ... a n 0P and n is any natural number.

It is easy to see that the algebra +P[a], +, -, ., 1, - a subring of the field P (a) - is a ring; this ring is denoted by P[a].

Theorem 1.1. Let P[x] be a polynomial ring in x over P and P(a) be a simple extension of the field P. Let y be a mapping from P[x] to P[a] such that y(f)=f(a) for any f from P[x]. Then:

(a) for any a in P y (a) = a;

(c) y is a homomorphism of the ring P [x] onto the ring P [a];

(d) Ker y =(f0P[x]*f(a)=0);

(f) the quotient ring P[x]/Ker y is isomorphic to the ring P[a].

Proof. Statements (a) and (b) follow directly from the definition of y. The mapping y preserves the main operations of the ring P[x], since for any f and g from P[x]

y(f + g)=f(a)+g(a), y(fg)= f(a)g(a), y(1)=1.

Assertion (d) follows directly from the definition of the map y.

Since y is a homomorphism of the ring P[x] onto P[a], the quotient ring P[x]/Ker y is isomorphic to the ring P[a].

Corollary 1.2. Let a be a transcendental element over a field P. Then the polynomial ring P[x] is isomorphic to the ring P[a].

Proof. Due to the transcendence of a over P Kery=(0). Therefore P[x]/(0) P [a]. Moreover, the quotient ring of the ring P [x] is isomorphic to P [x] by the zero ideal. Hence P [x] P [a].

1.2 Minimal polynomial of an algebraic element.

Let P [x] be the ring of polynomials over the field P.

Definition. Let a be an algebraic element over a field P. The minimal polynomial of an element a over P is a normalized polynomial in P[x] of the least degree whose root is a. The degree of a minimal polynomial is called the degree of an element a over P.

It is easy to see that for any element a that is algebraic over P , there exists a minimal polynomial.

Proposition 1.3. If a is an algebraic element over a field P, and g and j are its minimal polynomials over P, then g=j.

Proof. The degrees of the minimal polynomials g and j coincide. If g ¹ j, then the element a (of degree n over P) will be the root of the polynomial g - j, the degree of which is less than the degree of the polynomial j (less than n), which is impossible. Therefore, g=j.

Theorem 1.4. Let a be an algebraic element of degree n over a field P (aóP) and g be its minimal polynomial over P. Then:

(a) the polynomial g is irreducible in the ring P [x];

(b) if f (a) = 0, where f 0 P[x], then g divides f;

(c) the quotient ring P [x]/(g) is isomorphic to the ring P [a];

(d) P[x]/(g) is a field;

(f) the ring P[a] coincides with the field P(a).

Proof. Let us assume that the polynomial g is reducible in the ring P [x], i.e., there are polynomials j and h in P[x] such that

g = jh, 1£deg j, deg h

Then g(a) = j(a)h(a) = 0. Since P (a) is a field, then j(a) = 0 or h(a) = 0, which is impossible because, by assumption, the degree element a over P is equal to n.

Assume that f 0 P[x] and f(a) = 0. By assumption, g(a) = 0. Therefore, f and g cannot be coprime. Since the polynomial g is irreducible, then g divides f.

Let j be the homomorphism of the ring P [x] onto the ring P [a] (y(f)=f(a) for any f from P[x]) considered in Theorem 2.1. By virtue of (b), the kernel of the homomorphism y consists of multiples of the polynomial g, i.e., Ker y = (g). Therefore, the quotient ring P = P [x]/(g) is isomorphic to the ring P [a].

Since P[a]ÌP(a), then P [a] is a domain of integrity. Since P @ P[a], the quotient ring P is also a domain of integrity. We need to show that any nonzero element f from P is invertible to P. Let f be an element of the coset f. Since f ¹ 0, then f(a)¹0; so the polynomial g does not divide the polynomial f. Since the polynomial g is irreducible, it follows that the polynomials f and g are coprime. Therefore, there are polynomials u and v in Р[x] such that uf + vg=1. This implies the equality uf = 1, which shows that the element f is invertible in the ring P. Thus, we have established that the quotient ring P is a field.

By virtue of (c) and (d), P[a] is a field, and so P(a)ÌP[a]. Moreover, obviously, P[a]ÌP(a). Hence P[a] = P(a). Therefore, the ring P[a] coincides with the field P(a).

1.3 Structure of a simple algebraic field extension.

Theorem 1.5. Let a be an algebraic element of positive degree n over a field P. Then any element of the field P(a) can be uniquely represented as a linear combination of n elements 1, a, ..., a n-1 with coefficients from P.

Proof. Let b be any element of the field P (a). By Theorem 1.4, P(a) = P[a]; hence there exists a polynomial f in P[x] such that

Let g be the minimal polynomial for a over P; by virtue of the hypothesis of the theorem, its degree is equal to n. By the division theorem with remainder, there are polynomials h and r in P[x] such that

(2) f = gh + r, where r = 0 or der r< der g = n , т. е. r=c 0 +c 1 x +…c n -1 x n -1 (c i 0P). Полагая в (2) x = а и учитывая равенство (1), имеем

(3) b = c 0 +c 1 a +…c n -1 a n-1

Let's show that the element b is uniquely representable as a linear combination of elements 1, a, ..., a n-1 . Let

(4) b = d 0 +d 1 a +…d n -1 a n-1 (d i 0 P)

Any such performance. Consider a polynomial j

j \u003d (c 0 - d 0) + (c 1 - d i .)x + . . . + (with n-1 –d n -1)x n -1

The case when the degree of j is less than n is impossible because, by virtue of (3) and (4), j(a) = 0 and the degree of j is less than the degree of g. The only possible case is when j = 0, i.e. with 0 = d 0 , . . . , with n-1 = d n-1. Therefore, the element b is uniquely representable as a linear combination of elements 1, a,…,a n-1 .

1.4. Exemption from algebraic irrationality in the denominator of a fraction.

The problem of getting rid of algebraic irrationality in the denominator of a fraction is as follows. Let a be an algebraic element of degree n>1 over a field P; f and h are polynomials from the polynomial ring P [x] and h(a) ¹0. It is required to represent the element f(a)/h(a)0P(a) as a linear combination of powers of the element a, i.e., as j(a),

This problem is solved in the following way. Let g be the minimal polynomial for a over P. Since, by Theorem 1.4, the polynomial is irreducible over P and h(a) ¹ 0, it follows that g does not divide h and hence the polynomials h and g are coprime. Therefore, there are polynomials u and v in P[x] such that

Since g(a) = 0, it follows from (1) that

u(a)g(a) = 1, 1/h(a) = u(a).

Therefore, f(a)/h(a) = f(a)u(a), and f,u 0P[x] and f(a)u(a)0P[a]. So, we got rid of the irrationality in the denominator of the fraction f(a)/h(a) .

Get rid of irrationality in the denominator of a fraction

The polynomials p(x) and g(x)=-x 2 +x+1 are coprime. Therefore, there are polynomials j and y such that

To find j and y, we apply the Euclid algorithm to the polynomials p and g:

X 3 -2 -x 2 +x+1 -x 2 +x+1 2x-1

x 3 -x 2 -x -x-1 -x 2 +1/2x -1/2x+1/4

x 2 -x-1 1/2x-1/4

In this way,

p=g(-x-1)+(2x-1),

g=(2x-1)(-1/2x+1/4)+5/4.

Where do we find

(2x-1)=p+g(x+1),

5/4=g-(p+g(x+1))(-1/2x+1/4)

p1/5(2x-1)+g(4/5+1/5(2x 2 +x-1))=1,

p1/5(2x-1)+g(2/5x 2 +1/5x+3/5)=1.

In this way,

y(x)= (2/5x 2 +1/5x+3/5).

Consequently

.

2.Composite algebraic field extension.

2.1. Finite field extension.

Let P be a subfield of the field F. Then we can consider F as a vector space over P, i.e., consider the vector space +F, +, (w l ½l 0P),

where w l is the operation of multiplying elements from F by the scalar l0P.

Definition. An extension F of a field P is said to be finite if F, as a vector space over P, has finite dimension. This dimension is denoted by .

Proposition 2.1. If a is an algebraic element of degree n over P, then =n.

This proposition follows directly from Theorem 1.5.

Definition. An extension F of a field P is called algebraic if every element of F is algebraic over P.

Theorem 2.2. Any finite extension F of a field P is algebraic over P.

Proof. Let F be an n-dimension over P. The theorem is obviously true if n = 0. Suppose that n>0. Any n+1 elements from F are linearly dependent over P. In particular, the system of elements 1, a, ..., a n is linearly dependent, i.e., there exist in P such elements with 0 , with 1,…, c n not all equal zero, which is c 0 ×1+ c 1 a +…+c n a n = 0.

Therefore, the element a is algebraic over P.

Note that there are algebraic extensions of the field that are not finite extensions.

2.2. Composite algebraic field extension.

An extension F of a field P is said to be composite if there exists

an increasing chain of subfields L i of the field F such that

P = L 0 L 1 … L k = F and k>1.

Theorem 2.3. Let F be a finite extension of the field L and L a finite extension of the field P. Then F is a finite extension of the field P and

= @[L:P].

Proof. Let

(1) a 1 ,…,a m is a basis of the field L over P (as a vector space) and

(2) b 1 ,…,b n - basis of the field F over L . Any element d from F can be linearly expressed in terms of the basis:

(3) d = l 1 b 1 +...+l n b n (l k 0L).

The coefficients 1 k can be linearly expressed in terms of the basis (1):

(4) l k = p 1k a +…+ p mk a m ​​(p ik 0P).

Substituting the expressions for the coefficients l k in (3), we obtain

d = å p ik a i b k .

Thus, each element of the field F can be represented as a linear combination of elements of the set B, where

B = ( a i b k 1(1,..., m), k 0 (l,..., n)).

Note that the set B consists of nm elements.

Let us show that B is a basis of F over the field P. We need to show that the system of elements of the set B is linearly independent. Let

(5) åc ik a i b k = 0,

where c ik 0 P. Since system (2) is linearly independent over L , then (5) implies the equalities

(6) c 1 k a 1 +...+c mk a m ​​= 0 (k = 1,..., n).

Since the elements a 1 , ..., a m are linearly independent over P, then (6) implies the equalities

c 1 k = 0,…,c mk = 0 (k = 1, ..., n),

showing that all coefficients in (5) are equal to zero. Thus, the system of elements B is linearly independent and is a basis of F over P.

So it is established that = nm = ×. Therefore, F is a finite extension of the field P and formula (I) holds.

Definition. An extension F of a field P is said to be composite algebraic if there exists an increasing chain of subfields of the field P

P = L 0 L 1 … L k = F and k>1 (1)

such that for i = 1,..., k the field L i is a simple algebraic extension of the field L i-1 . The number k is called the chain length (1).

Corollary 2.4. A composite algebraic extension F of a field P is a finite extension of the field P.

The proof is easily carried out by induction on the length of the chain (1) on the basis of Theorem 2.3.

Theorem 2.5. Let a 1 ,..., a k be algebraic over the field P elements of the field F . Then the field P(a 1 ,..., a k) is a finite extension of the field P.

L 0 = P, L 1 = P , L 2 = P ,..., L k = P .

Then L 1 = P is a simple algebraic extension of the field L 0 ; L 2 is a simple algebraic extension of the field L 1 since

L 2 \u003d P \u003d (P) \u003d L 1 \u003d L 1 (a 2), etc.

In this way,

P = L 0 L 1 … L k = F

where L i = L i -1 (a i) for i = 1, ..., k, i.e., each member of the chain (2) is a simple algebraic extension of the previous member of the chain. Thus, the field F is a composite algebraic extension of the field P. Therefore, by Corollary 2.4, the field F is a finite extension of the field P .

Corollary 2.6. A compound algebraic extension of a field is an algebraic extension of that field.

2.3. Simplicity of compound algebraic field extension.

Theorem 2.7. Let the number field F be a composite algebraic extension of the field P . Then F is a simple algebraic extension of the field P.

Proof. Let P L F , and let L = P(a), F = L(b), and hence F = P(a, b).

Let f and g be minimal polynomials over P for numbers a and b, respectively, and deg f = m, deg g = n. The polynomials f and g are irreducible over P and, therefore, have no complex numbers of multiple roots in the field E. Let

a = a 1 ,..., a m are the roots of the polynomial f in C and

b = b 1 ,..., b n are the roots of the polynomial g in C.

Consider a finite set M:

M = ((a i -a)/(b-b k)½i0(1,…,m), k0(2,…,n)).

Since P is a numerical set (and therefore infinite), then in P there exists a number c distinct from the elements of the set M, c0P(M, cóM. Let

Then the relations

(2) g ¹ a i + cb k = (i0(1,..., m), k0(2, ..., n)).

Indeed, in the case of equality a + cb = a i + cb k it would be

c \u003d (a i -a) / (b-b k) 0 M

which would contradict the choice of c.

Let F 1 = P (g) and F 1 be a polynomial ring in x. Let h = f(g - cx) be a polynomial in F 1 [x] (g, c0P(g) = F 1). Let us show that x-b is the greatest common divisor of the polynomials h and g in the ring F 1 [x]. Since g(b) = 0, x-b divides g in E[x]. Further, due to (1)

h(b) = f(g-cb) = f(a) = 0.

Therefore x-b divides the polynomial h in E[x]. Thus x-b is a common divisor of h and g in the ring E[x].

Let us prove that g and h in C have no roots other than b. Indeed, suppose that b k , k0(2 ,..., n), is their common root. Then h(b k) = f(g - cb k) = 0. Therefore, there is an index i0(1 ,..., m) such that g = a i + cb k (k>1), which contradicts (2 ). Based on this, we conclude that x-b is the greatest common divisor of g and h in E[x]. Since x - b is a normalized polynomial, it follows that x - b is the greatest common divisor of g and h in the ring F 1 [x]. That's why

(x-b) 0 F 1 [x] and b 0 F 1 = P(g).

In addition, a = g - cb 0 F 1 . In this way,

F = P(a, b)Ì F 1 , F 1 ÌF.

2.4. Field of algebraic numbers.

In the class of subfields of the field of complex numbers, one of the most important is the field of algebraic numbers.

Definition. An algebraic number is a complex number that is a root of a polynomial of positive degree with rational coefficients.

Note that an algebraic number is any complex number that is algebraic over a field Q. In particular, any rational number is algebraic.

Theorem 2.8. The set A of all algebraic numbers is closed in the ring E = +C, +, -, , 1, of complex numbers. The algebra A = +A, +, -, , 1, is a field, a subfield of the field E.

Proof. Let a and b be any elements of A. By Corollary 2.6, the field Q(a, b) is algebraic over Q. Therefore, the numbers a + b, -a, ab, 1 are algebraic, that is, they belong to the set A. Thus Thus, the set A is closed under the principal operations of the ring E. Therefore, the algebra A, a subring of the ring E, is a ring.

In addition, if a is a non-zero element of A, then a -1 0 Q (a, b) and therefore a -1 belongs to A. Therefore, the algebra A is a field, a subfield of the field E.

Definition. The field A = +A, +, -, , 1, is called the field of algebraic numbers.

Show that the number a= is algebraic.

Solution. From a= follows a-.

Let's raise both parts of the last equality to the third power:

a 3 -3a 2 9a-3=2

a 3 +9a-2=3(a 2 +1).

Now we raise both sides of the equality to the second power:

a 6 +18a 4 +81a 2 -4a 3 -36a+4=27a 4 +54a 2 +27

a 6 -9a 4 -4a 3 +27a 2 -36a-23=0.

Thus a is the root of the polynomial

f(x)= a 6 -9a 4 -4a 3 +27a 2 -36a-23=0

with rational coefficients. This means that a is an algebraic number.

2.5. Algebraic closedness of the field of algebraic numbers.

Theorem 2.9. The field of algebraic numbers is algebraically closed.

Proof. Let A [x] be a polynomial ring in x over a field A of algebraic numbers. Let

f = a 0 + a 1 x+... + a n x n (a 0 ,…, a n 0 A)

Any positive degree polynomial in A[x]. We need to prove that f has a root in A. Since f0C[x] and the field E is algebraically closed, then f has a root in E, i.e., there exists a complex number c such that f(c) = 0. Let L= Q(a 0 , ..., a n) and L (c) is a simple algebraic extension of the field L with the help of c. Then Q L L (c) is a finite algebraic extension of the field L. By Theorem 2.2, L is a finite extension of the field Q. By virtue of Theorem 2.3, L (c) is a finite extension of the field Q. Hence, by Theorem 2.2, it follows that the field L(c) is an algebraic extension of the field Q and hence c0A. Thus, any polynomial in A[x] of positive degree has a root in A, i.e., the field A is algebraically closed.

3. Separable and non-separable extensions.

Let D be a field.

Let us find out whether a polynomial that is indecomposable in D[x] can have multiple roots?

In order for f(x) to have multiple roots, the polynomials f(x) and fN(x) must have a common non-constant multiplier that can be computed already in D[x]. If the polynomial f (x) is indecomposable, then with any polynomial of lesser degree f (x) cannot have non-constant common factors, therefore, the equality f "(x) \u003d 0 must hold.

f(x) =3a n x n fN(x) =3na n x n -1

Since fN(x) = 0, each coefficient must go to zero:

na n = 0 (n = l, 2, ..., n).

In the case of characteristic zero, this implies that a n = 0 for all n ¹ 0. Therefore, a non-constant polynomial cannot have multiple roots. In the case of characteristic p, the equalities na n = 0 are also possible for n ¹ 0, but then the comparisons

f(x) = a 0 + a p x p + a 2p x 2p +…

Conversely, if f(x) has this form, then fN(x)=0.

In this case, we can write:

This proves the statement: In the case of characteristic zero, the polynomial f (x) indecomposable in D [x] has only simple roots; in the case of characteristic p, the polynomial f(x) (if it is different from a constant) has multiple roots if and only if when it can be represented as a polynomial j in x p .

In the latter case, it may turn out that j(x) is in turn a polynomial in x p . Then f(x) is a polynomial in x p 2 . Let f(x) be a polynomial in x pe

but is not a polynomial in x pe +1 . Of course, the polynomial y(y) is indecomposable. Further, y¢(y) ¹ 0, because otherwise y(y) would have the form c(y p) and, therefore, f(x) would be represented by c(x pe+1), which contradicts the assumption. Therefore, y(y) has only simple roots.

Let us decompose the polynomial y(y) in some extension of the main field into linear factors: m

y(y) = J(y-b i).

f(x) = J(x pe -b i)

Let a i be some root of the polynomial x pe -b i . Then x i pe = b i ,

x pe -b i \u003d x pe - a i pe \u003d (x-a i) pe.

Therefore, a i is a p e -fold root of the polynomial x pe -b i and

f(x) = J(x -a i) p e.

All roots of the polynomial f(x) thus have the same multiplicity p e.

The degree m of the polynomial y is called the reduced degree of the polynomial f(x) (or the root a i); the number e is called the exponent of the polynomial f (x) (or the root a i) over the field D. The relation between the degree, the reduced degree and the exponent is

where m is equal to the number of different roots of the polynomial f(x).

If q is a root of a polynomial that is indecomposable in the ring D[x] and has only simple roots, then q is called a separable element over D or an element of the first kind over D 1). In this case, an indecomposable polynomial, all of whose roots are separable, is called separable. Otherwise, the algebraic element q and the indecomposable polynomial f(x) are called inseparable or an element (respectively, a polynomial) of the second kind. Finally, an algebraic extension S all of whose elements are separable over D is said to be separable over D, and any other algebraic extension is said to be inseparable.

In the case of characteristic zero, according to what was said above, every indecomposable polynomial (and therefore every algebraic extension) is separable. Later we will see that most of the most important and interesting field extensions are separable and that there are whole classes of fields that do not have non-separable extensions at all (the so-called "perfect fields"). For this reason, in the following, everything related specifically to non-separable extensions is typed in small print.

Consider now the algebraic extension S = D (q). When the degree n of the equation f(x) = 0 that defines this extension is equal to the degree (S: D), the reduced degree m turns out to be equal to the number of isomorphisms of the field S in the following sense: consider only such isomorphisms [email protected]", under which the elements of the subfield D remain fixed and, therefore, S is transferred to the equivalent field S" (isomorphisms of the field S over the field D) and under which the image field S" lies together with the field S inside some common field W. In under these conditions, the following theorem holds:

For a suitable choice of the field W, the extension S=D(q) has exactly m isomorphisms over D, and for any choice of the field W, the field S cannot have more than m such isomorphisms.

Proof. Every isomorphism over D must take the element q to its conjugate element q" from W. We choose W so that f(x) decomposes over W into linear factors; then it turns out that the element q has exactly m conjugates elements q,q", ... Moreover, no matter how the field W is chosen, the element q will not have more than m conjugates in it. We now note that each isomorphism D(q)@D(q") over D is completely determined by specifying the correspondence q® q". Indeed, if q passes into q" and all elements from D remain in place, then the element

3a k q k (a k 0D)

should go to

and this defines an isomorphism.

In particular, if q is a separable element, then m = n, and hence the number of isomorphisms over the ground field is equal to the degree of extension.

If there is some fixed field containing all the fields under consideration, which contains all the roots of each equation f(x) = 0 (as, for example, in the field of complex numbers), then this field can be taken as W once and for all and therefore drop the addition "inside some W" in all isomorphism sentences. This is always done in number field theory. Later we will see that such a field W can also be constructed for abstract fields.

The generalization of the above theorem is the following statement:

If an extension S is obtained from D by successively adding m

algebraic elements a 1 , ..., a m , and each of a i ,- is a root

indecomposable over D(a 1 , ..., a i-1) equation of reduced degree n" i , then

extension S has exactly Õn i ¢ isomorphisms over D and none

extension, there is no greater number of such isomorphisms of the field S.

Proof. For m = 1 the theorem has already been proved above. Suppose it is valid for the extension S 1 = D(a 1 , ..., a m-1): in some suitable extension

W 1 are exactly q n i ¢ isomorphisms of the field S over D.

Let S 1 ®S 1 be one of these q n i ¢ isomorphisms. It is claimed that in a suitably chosen field W it can be extended to the isomorphism S = S 1 (a m) @ S= S(a m) in no more than n¢ m ways.

The element a m satisfies some equation f 1 (x) = 0 over S 1 with n¢ m different roots. Using the isomorphism S 1 ®S 1, the polynomial f 1 (x) is taken to some polynomial f 1 (x). But then f 1 (x) in a suitable extension has again n¢ m different roots and no more. Let a m be one of these roots. Due to the choice of the element a m, the isomorphism S 1 @S 1 extends to the isomorphism S (a m) @ S (a m) with a m ®a m in one and only one way: indeed, this extension is given by the formula

åc k a m ​​k ®å c k a m ​​k

Since the choice of an element a m can be done in n" m ways, there are n" m extensions of this kind for the chosen isomorphism å 1 ®å 1

Since, in turn, this isomorphism can be chosen

Х n" i ways,

then everything exists (in the field W, which contains all the roots of all considered equations)

Õ n" i ×n" m = Õ n" i

isomorphisms of the extension S over the field D, which was to be proved.

If n i is the full (non-reduced) degree of an element a i over D (a 1 ,...,a i-1), then n i is equal to the degree of the extension D (a 1 , ... , a i) of the field D(a 1 , .. . , a i-1);

hence the degree of (S: D) is

If we compare this number with the number of isomorphisms

The number of isomorphisms of the extension S = D(a 1 , ... , a m) over D (in some suitable extension W) is equal to the degree (S: D) if and only if each element a i is separable over the field D(a 1 , . .. , a i-1). If at least one element a i is not separable over the corresponding field, then the number of isomorphisms is less than the degree of extension.

Several important consequences immediately follow from this theorem. First of all, the theorem states that the property of each element a i to be separable over the previous field is a property of the extension S itself, regardless of the choice of generating elements a i . Since an arbitrary element b of the field can be taken as the first generator, the element b is separable if all a i are separable. So:

If elements a i , ... ,a n are sequentially added to the field D and each element a i turns out to be separable over the field obtained by adding the previous elements a 1, a 2 ,…,a i-1 then the extension

S = D(a 1 , ... ,a n)

separable over D.

In particular, the sum, difference, product, and quotient of separable elements are separable.

Further, if b is separable over S and the field S is separable over D, then the element b is separable over D. This is because b satisfies some equation with a finite number of coefficients a 1 , ... ,a m from S and, therefore, is separable over D (a 1 , ... ,a m). Thereby separable and extension

D (a 1 ,..., a m , b).

Finally, the following proposition holds: the number of isomorphisms of a finite separable extension S over a field D is equal to the degree of the extension (S: D).

4. Infinite field extensions.

Each field is obtained from its simple subfield using a finite or infinite extension. This chapter deals with infinite field extensions, first algebraic and then transcendental.

4.1. Algebraically closed fields

Among the algebraic extensions of a given field, an important role is played, of course, by maximal algebraic extensions, i.e., those that do not admit a further algebraic extension. The existence of such extensions will be proved in this section.

For the field W to be a maximal algebraic extension, the following condition is necessary: ​​each polynomial of the ring W[x] can be completely factorized into linear factors. This condition is also sufficient. Indeed, if every polynomial in W[x] decomposes into linear factors, then all simple polynomials in W[x] are linear and every element of any algebraic extension W" of the field W turns out to be a root of some linear polynomial x - a in W[x], m i.e., coincides with some element a of the field W.

Therefore, we give the following definition:

A field W is called algebraically closed if any polynomial in W[x] decomposes into linear factors.

A definition equivalent to this is the following: a field W is algebraically closed if every non-constant polynomial in W[x] has at least one root in W, i.e. at least one linear factor in W[x].

Indeed, if such a condition is satisfied and an arbitrarily taken polynomial f(x) decomposes into indecomposable factors, then all of them must be linear.

The "Fundamental Theorem of Algebra" states that the field of complex numbers is algebraically closed. The next example of an algebraically closed field is the field of all complex algebraic numbers, that is, the set of those complex numbers that satisfy some equation with rational coefficients. The complex roots of an equation with algebraic coefficients are indeed algebraic not only over the field of algebraic numbers, but also over the field of rational numbers, i.e., they themselves are algebraic numbers.

Here we show how to construct an algebraically closed extension of an arbitrarily given field P and, moreover, in a purely algebraic way. Steinitz owns the following

Main theorem. For every field P, there is an algebraically closed algebraic extension W. Up to equivalence, this extension is uniquely defined: any two algebraically closed algebraic extensions W, W" of the field P are equivalent.

We must preface the proof of this theorem with several lemmas:

Lemma 1. Let W, be an algebraic extension of the field P. A sufficient condition for W to be algebraically closed is the factorization of any polynomial from P[x] in the ring W[x].

Proof. Let f(x) be an arbitrary polynomial in W[x]. If it does not decompose into linear factors, then we can add some of its root a and arrive at a proper superfield W". is a root of some polynomial g(x) in P[x].This polynomial decomposes into linear factors in W[x], Therefore, a is a root of some linear factor in W[x], i.e., belongs to the field W, which contradicts the assumption .

Lemma 2. If the field P is well ordered, then the polynomial ring P[x] can be well ordered and, moreover, in such an order that the field P is a segment in this ordering.

Proof. We define the order relation between polynomials f(x) from P[x] as follows: let f(x)

1) the degree f(x) is less than the degree g(x);

2) the degree of f(x) is equal to the degree of g(x) and is equal to n, i.e.

f (x) \u003d a 0 x n + ... + a n, g (x) \u003d b 0 x n + ... + b n

and for some index k:

and i = b i for i

a k

In this case, an exception is made for the polynomial 0: it is assigned the degree 0. It is obvious that in this way some ordering is obtained, in the sense of which P[x] is completely ordered. It is shown as follows: in every non-empty set of polynomials there is a non-empty subset of polynomials of the least degree; let this be equal to n. In this subset there is a non-empty subset of polynomials whose coefficient a 0 is the first in the sense of the available order among the free terms of the considered polynomials; in the specified subset, there is in turn a subset of polynomials with the first a 1, etc. The subset with the first a n that eventually turns out can consist of only one single polynomial (since a 0 , ..., and n are determined by uniquely due to the consistently satisfied minimality condition in the choice); this polynomial is the first element in the given set.

Lemma 3. If the field P is well ordered and a polynomial f(x) of degree n and n symbols a 1 ..., a n are given, then the field P (a 1 ,..., a n) in which f(x) is completely decomposed into linear factors

q(x-a i), is constructed in a unique way and is completely

orderly. The field P in the sense of this order is a segment.

Proof. We will add the roots a 1 ..., a n sequentially, as a result of which the fields P 1 , ..., P n will appear sequentially from P = Р 0 . Assume that P i-1 = P(a 1 ..., a i-1) is a field already constructed and that P is a segment in P i-1 ; then R i will be constructed as follows.

First of all, by virtue of Lemma 2, the polynomial ring P i-1 [x] is completely ordered. The polynomial f decomposes in this ring into indecomposable factors, among which x - a 1 ,..., x - a i-1 will take the first place; among other factors, let f i (x) be the first in the sense of the existing order. Together with the symbol a i denoting the root of the polynomial f i (x), we define the field Р i = P i -1 as the set of all sums

where h is the degree of the polynomial f i (x). If f i (x) is linear, then of course we set P i = P i -1 ; the character a i is not needed in this case. The constructed field is completely ordered using the following condition: each element of the field

compare the polynomial

and we order the elements of the field in exactly the same way as the corresponding polynomials are ordered.

Obviously, then Р i-1 is a segment in Р i , and therefore P is a segment in Р i .

Thus the fields Р 1 ,..., Р n are constructed and well ordered. The field P n is the desired uniquely defined field P(a 1 ,..., a n).

Lemma 4. If in an ordered set of fields each preceding field is a subfield of the next one, then the union of these fields is a field.

Proof. For any two elements a, b of a union, there are two fields S a , S b that contain a and b, and one of which precedes the other. The enclosing field defines the elements a + b and a×b, and this is how these elements are defined in each of the fields containing a and b, because of any two such fields, one precedes the other and is its subfield. For example, to prove the law of associativity

ab g = a bg,

find among the fields S a , Sb, S g the one that contains the other two fields (the largest); this field contains a, b and g and the law of associativity is satisfied in it. All other calculation rules with union elements are checked in the same way.

The proof of the main theorem is divided into two parts: the construction of the field W and the proof of uniqueness.

Construction of the field W. Lemma 1 shows that in order to construct an algebraically closed extension W of the field P, it suffices to construct an algebraic extension of the field P such that each polynomial from P[x] decomposes over this extension into linear factors.

1. The field P f is the union of the field P and all fields S g for g

2. The field Р f is completely ordered so that Р and all fields S g with g

3. The field S f is obtained from P f by adding all the roots of the polynomial f using the symbols a 1 ,..., a n in accordance with Lemma 3.

It is necessary to prove that well-ordered fields Р f , S f are indeed uniquely determined in this way, if all the previous Р g , S g are already defined by the requirements listed above.

If requirement 3 is satisfied, then, first of all, Р f is a segment in S f . It follows from this and Requirement 2 that the field P and each field S g (g

P - segment in S h at h

S g - segment in S h at g

It follows from this that the field P and the fields S h (h b, which must be stored in P f . This order relation is the same in all fields P or S g that contain both a and b, because all these fields are segments of each other. So the order relation is defined. That it defines a well-ordered set is obvious, since every non-empty set x in P f contains at least one element from P or from some field S g , and therefore also the first element from x Ç P or from x Ç S g . This element is also the first element in x.

By condition 3, the polynomial f(x) is completely decomposed into linear factors in the field S f . Next, we show by transfinite induction that S f is algebraic over P. Indeed, suppose that all fields S g (g

Let us now compose the union W of all fields S f ; by Lemma 4 it is a field. This field is algebraic over P, and all polynomials f decompose over it (since every polynomial f decomposes already over S f). Therefore, the field W is algebraically closed (Lemma 1).

Uniqueness of the field W. Let W and W" be two fields that are algebraic and algebraically closed extensions of the field P. Let us prove the equivalence of these fields. is considered to be one of such segments) a subset ¢ in W" and some isomorphism

P(Â) @ P(¢).

The latter must satisfy the following recurrence relations.

1. The isomorphism P(Â) @ P(¢) must leave each element of the field P in place.

2. The isomorphism P(Â) @ P(¢) under IÌ Â must be an extension of the isomorphism P(I) @ P(I").

3. If  has a last element a, so that  = ÁÈ(a), and if a is a root of a polynomial f(x) that is indecomposable in P(I), then the element a" must be the first root of the corresponding due to P(I) @P(I"), a polynomial f¢(x) in a well-ordered field W".

It must be shown that these three requirements really define the isomorphism P(Â) @ P(¢), if only it is already defined for all previous segments ÁÌ Â. Here it is necessary to distinguish between two cases.

First case. The set  does not have a last element. Then each element a belongs to some previous segment I; therefore,  is the union of segments Á, and therefore P(Â) is the union of the fields P(I) for ÁÌ Â. Since each of the isomorphisms P(I) @P(I") is an extension of all the previous ones, each element a under all these isomorphisms corresponds to only one element a". Therefore, there is one and only one mapping P(I) → P(I¢) that extends all previous isomorphisms P(I) → P(I"), namely, the -map a®a". Obviously, it is an isomorphism and satisfies requirements 1 and 2.

Second case. The set  has a last element a; hence  = ÁÈ(a). By virtue of requirement 3, the element a" assigned to the element a is uniquely determined. Since a" over the field P(I") (in the sense of the considered isomorphism) satisfies the "same" indecomposable equation as a over P(I"), the isomorphism P(I)→P(I") (and in the case where I is empty, i.e., the identity isomorphism P®P) extends to an isomorphism P(I, a) ®P(I", a¢), for in which a goes to a. This isomorphism is uniquely determined by each of the above requirements, because every rational function j(a) with coefficients from  necessarily goes to the function j"(a") with the corresponding coefficients from Á". That the so-defined isomorphism P(Â) ® Р(В¢) obviously satisfies requirements 1 and 2.

Thus, the construction of the isomorphism P(Â)→P(¢) is completed. Denote by W" the union of all fields P(¢); then there exists an isomorphism P(W)®W" or W®W", leaving each element of the field P in place. Since the field W is algebraically closed, so must W ", and therefore W" coincides with the entire field W¢. This implies the equivalence of the fields W and W¢.

The meaning of an algebraically closed extension of a given field is that, up to equivalence, it contains all possible algebraic extensions of this field. More precisely:

If W is an algebraically closed algebraic extension of the field P and S is an arbitrary algebraic extension of the field P, then inside W there exists an extension S 0 equivalent to the extension S.

Proof. Let us extend S to some algebraically closed algebraic extension W". It will also be algebraic over P, and therefore equivalent to an extension W. subfield S 0 in W.

4.2. Simple transcendental extensions.

Every simple transcendental extension of the field D, as we know, is equivalent to the field of quotients D(x) of the polynomial ring D[x]. Therefore, we will study this field of quotients

The elements of the field W are the rational functions

Theorem. Every non-constant element h of degree n is transcendental over D, and the field D(x) is an algebraic extension of the field D(h) of degree n.

Proof. The representation h = f(x)/g(x) will be considered irreducible. Then the element x satisfies the equation

g(x)×h - f(x)=0

with coefficients from D(h). These coefficients cannot all be zero. Indeed, if all of them were equal to zero and a k would be any non-zero coefficient of the polynomial g(x) with the same degree x, and b k - a non-zero coefficient of the polynomial f(x), then the equality would have to take place

whence h = b k /a k = const, which contradicts the assumption. Hence the element x is algebraic over D(h).

If the element h were algebraic over D, then x would also be algebraic over D, which, however, is not the case. Hence the element h is transcendental over D.

The element x is the root of a polynomial of degree n

in the ring D(h)(z). This polynomial is indecomposable in D(h)[z], because otherwise it would be n-decomposable in the ring D, and since it is linear in h, one of the factors would have to depend not on h, but only on z. But there cannot be such a factor, because g(z) and f(z) are coprime.

Therefore, the element x is algebraic of degree n over the field D(h). This implies the assertion that (D(x) : D(h)) = n

For what follows, we note that the polynomial

has no factors depending only on z (i.e., lying in D[z]). This statement remains true when h is replaced by its value f(x)/g(x) and multiplied by the denominator g(x), thus the polynomial

g(z)f(x) - f(z)g(x)

the ring D has no factors depending only on z.

Three corollaries follow from the proved theorem.

1. The degree of the function h - f(x)/g(x) depends only on the fields D(h) and D(x), and not on this or that choice of the generating element x.

2. The equality D(h) = D(x) holds if and only if h has degree 1, i.e., is a linear-fractional function. This means that the generating element of the field, except for the element x, can be any linear-fractional function of x and only such a function.

3. Any automorphism of the field D(x) leaving every element of the field D in place must map the element x to some generating element of the field. Conversely, if x is translated into some generating element x = (ax + b) / (cx + d) and each function j (x) - into a function j (x), then an automorphism is obtained, in which all elements from D remain in place. Consequently,

All automorphisms of the field D(x) over the field D are fractional linear substitutions

x = (ax+b)/(cx+d), ad – bc ¹ 0.

Important for some geometric studies is

Lurot's theorem. Every intermediate field S for which DÌSÍD(x) is a simple transcendental extension: S = D(q).

Proof. An element x must be algebraic over S, because if h is any element of S that does not belong to the field D, then, as was shown, the element x is algebraic over D(h) and even more so algebraic over S. Let S be indecomposable in the polynomial ring [z] polynomial with leading coefficient 1 and root x has the form

f 0 (z) \u003d z n + a 1 z n -1 + ... + a n. (one)

Let us find out the structure of this polynomial.

The elements a i are rational functions of x. By multiplying by a common denominator, they can be made rational entire functions and, in addition, a polynomial in x with content 1 can be obtained:

f(x, z) =b 0 (x)z n +b 1 (x)z n-1 +…+b n (x).

The degree of this polynomial with respect to x will be denoted by m, and with respect to z by n.

The coefficients a i = b i / b 0 from (1) cannot all be independent of x, since otherwise x would turn out to be an algebraic element over D; so one of them, let's say

q = a i = b i (x)/ b 0 (x),

must actually depend on x; Let's write it in an irreducible form:

The degrees of the polynomials g(x) and h(x) do not exceed m. The polynomial

g(z) - qh(z) = g(z) – (g(x)/h(x))h(z)

(which is not identically zero) has a root z = x, and therefore it is divisible by f 0 (z) in the ring S[z]. If we pass from these polynomials rational in x to integer polynomials in x with content 1, then the divisibility relation is preserved, and we get

h(x)g(z)-g(x)h(z) = q(x, z)f(x, z).

The left side in this equality has a degree in x not exceeding m. But the polynomial f already has degree m on the right; hence the degree of the left-hand side is exactly m and q(x, z) does not depend on x. However, a factor that depends only on z cannot divide the left side (see above); so q(x, z) is a constant:

h(x)g(z)-g(x)h(z) = qf(x, z).

Since the presence of the constant q does not play a role, the structure of the polynomial f(x, z) is described in full. The degree of the polynomial f(x, z) in x is therefore (for symmetry reasons) and the degree in z is m, so m = n. At least one of the degrees of the polynomials g(x) and h(x) must actually reach the value m, therefore, the function q must also have a degree m in x.

Thus, since on the one hand the equality

(D(x):D(q)) = m,

on the other hand, equality

then, since S contains D(q),

Conclusion.

In the course work, the following types of extensions of the numeric field P were considered:

A simple algebraic field extension.

Composite algebraic field extension.

Separable and inseparable extensions.

Infinite field extensions.

Analyzing the work, some conclusions can be drawn.

Of the extensions discussed in the first two parts, such as:

simple algebraic extensions;

end extensions;

compound algebraic extensions.

It follows that all these types of extensions coincide and, in particular, are exhausted by simple algebraic extensions of the field P.

Bibliography

1. L.Ya. Kulikov. Algebra and number theory.- M.: Vyssh. School, 1979.-528-538s.

2. B.L. Van der Waerden. Algebra. - M., 1976 - 138-151s., 158-167s., 244-253s.

3. E.F. Shmigirev, S.V. Ignatovich. Theory of polynomials. - Mozyr 2002.

For the preparation of this work, materials from the site were used.