Algebraic field extensions
Introduction.
In pedagogical universities, a program for a unified course in algebra and number theory has been introduced. The main goal of this course is the study of basic algebraic systems and the education of an algebraic culture, which is necessary for a future teacher for a deep understanding of the goals and objectives of both the main school mathematics course and school elective 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 twentieth century, does not stop, and this causes persistent attempts to introduce basic algebraic concepts into school mathematics education.
The mathematical depth and an unusually wide scope of application of fields are combined with the simplicity of its basic provisions - the concepts of fields; a number of important theorems can be formulated and proved, having initial concepts in the field of set theory. Therefore, field theory is the best suited to show students an example of modern mathematics.
In addition, the study of the elements of field theory is useful for schoolchildren, contributes to their intellectual growth, manifested in the development and enrichment of various aspects of their thinking, qualities and personality traits, as well as fostering students' interest in mathematics and science.
1. Simple algebraic field extension.
1.1 Simple field expansion.
Let P [x] be a ring of polynomials in x over a field P, where P is a subfield of F. Recall that an element a of a field F is called algebraic over a field P if a is a root of some polynomial of positive degree from 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] \u003d (f (a) * f0P [x]),
that is, 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 the symbol P [a].
Theorem 1.1. Let P [x] be a ring of polynomials in x over P and P (a) a simple extension of the field P. Let y be a mapping from P [x] onto P [a] such that y (f) \u003d f (a) for any f from P [x]. Then:
(a) for any a from P y (a) \u003d a;
(c) y is a homomorphism of the ring P [x] onto the ring P [a];
(d) Ker y \u003d (f0P [x] * f (a) \u003d 0);
(f) the quotient ring P [x] / Ker y is isomorphic to the ring P [a].
Evidence. Statements (a) and (b) follow directly from the definition of y. The mapping y preserves the principal operations of the ring P [x], since for any f and g from P [x]
y (f + g) \u003d f (a) + g (a), y (fg) \u003d f (a) g (a), y (1) \u003d 1.
Statement (d) immediately follows from the definition of the mapping 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].
Evidence. Since a is transcendental over P, Kery \u003d (0). Therefore, P [x] / (0) - P [a]. Moreover, the quotient ring of the ring P [x] by the zero ideal is isomorphic to P [x]. Therefore, P [x] - P [a].
1.2 Minimum polynomial of an algebraic element.
Let P [x] be a polynomial ring over a 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 from P [x] of the least degree whose root is a. The degree of the minimal polynomial is called the degree of the element a over P.
It is easy to see that for every element a 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 \u003d j.
Evidence. The degrees of the minimal polynomials g and j coincide. If g ¹ j, then an 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 \u003d j.
Theorem 1.4. Let a be an algebraic element of degree n over the 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) \u003d 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).
Evidence. Suppose that the polynomial g is reducible in the ring P [x], that is, there exist polynomials j and h in P [x] such that
g \u003d jh, 1 £ deg j, deg h Then g (a) \u003d j (a) h (a) \u003d 0. Since P (a) is a field, then j (a) \u003d О or h (a) \u003d 0, which is impossible, since, by hypothesis, the degree of an element a over P is equal to n. Suppose that f 0 P [x] and f (a) \u003d 0. By assumption, g (a) \u003d 0. Therefore, f and g cannot be coprime. Since the polynomial g is irreducible, g divides f. Let j be the homomorphism of the ring P [x] onto the ring P [a] (y (f) \u003d 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, that is, Ker y \u003d (g). Therefore, the quotient ring P \u003d 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 into P. Let f be an element of the coset f. Since f ¹ 0, then f (a) ¹0; therefore 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, in P [x] there exist polynomials u and v such that uf + vg \u003d 1. This implies the equality uf \u003d 1, which shows that the element f is invertible in the ring P. Thus, it has been established that the quotient ring P is a field. By virtue of (c) and (d), P [a] is a field, and therefore P (a) ÌP [a]. Moreover, it is obvious that P [a] ÌP (a). Hence, P [a] \u003d P (a). Consequently, the ring P [a] coincides with the field P (a). 1.3 Construction of a simple algebraic field extension. Theorem 1.5. Let a be an element of positive degree n algebraic over the 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. Evidence. Let b be any element of the field P (a). By Theorem 1.4, P (a) \u003d P [a]; therefore, there exists in P [x] a polynomial f such that Let g be the minimal polynomial for a over P; by the hypothesis of the theorem, its degree is equal to n. By the division theorem with remainder, there exist in P [x] polynomials h and r such that (2) f \u003d gh + r, where r \u003d 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 \u003d c 0 + c 1 a + ... c n -1 a n-1 Let us show that the element b is uniquely representable as a linear combination of elements 1, a, ..., a n-1. Let be (4) b \u003d d 0 + d 1 a +… d n -1 a n-1 (d i 0 P) Any such performance. Consider the 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, since by virtue of (3) and (4) j (a) \u003d 0 and the degree of j is less than the degree of g. The only possible case is when j \u003d 0, that is, with 0 \u003d d 0,. ... ... , with n-1 \u003d d n-1. Therefore, the element b is uniquely representable as a linear combination of elements 1, a,…, a n-1. 1.4 Liberation 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\u003e 1 over the field P; f and h are polynomials from the ring of polynomials P [x] and h (a) ¹0. It is required to represent the element f (a) / h (a) 0P (a) in the form of a linear combination of powers of the element a, that is, in the form j (a), This problem is solved as follows. 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, then g does not divide h, and hence the polynomials h and g are coprime. Therefore, there exist polynomials u and v in P [x] such that Since g (a) \u003d 0, it follows from (1) that u (a) g (a) \u003d 1, 1 / h (a) \u003d u (a). Therefore, f (a) / h (a) \u003d f (a) u (a), and f, u 0P [x] and f (a) u (a) 0P [a]. So, we got rid of 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) \u003d - x 2 + x + 1 are coprime. Therefore, there exist polynomials j and y such that To find j and y, we apply Euclid's 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 \u003d g (-x-1) + (2x-1), g \u003d (2x-1) (- 1 / 2x + 1/4) +5/4. Where do we find (2x-1) \u003d p + g (x + 1), 5/4 \u003d g- (p + g (x + 1)) (- 1 / 2x + 1/4) p1 / 5 (2x-1) + g (4/5 + 1/5 (2x 2 + x-1)) \u003d 1, p1 / 5 (2x-1) + g (2 / 5x 2 + 1 / 5x + 3/5) \u003d 1. In this way, y (x) \u003d (2 / 5x 2 + 1 / 5x + 3/5). Hence 2. Composite algebraic field extension. 2.1. Ultimate field expansion. Let P be a subfield of F. Then we can regard F as a vector space over P, that is, consider 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 called 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 \u003d n. This proposition follows directly from Theorem 1.5. Definition. An extension F of a field P is said to be 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. Evidence. Let F be n-dimension over P. The theorem is obviously true if n \u003d 0. Suppose that n\u003e 0. Any n + 1 elements of F are linearly dependent over P. In particular, the system of elements 1, a, ..., an is linearly dependent, that is, there exist in P such elements with 0, with 1, ..., cn not all equal zero such that c 0 × 1 + c 1 a +… + cnan \u003d 0. Consequently, the element a is algebraic over P. Note that there are algebraic field extensions that are not finite extensions. 2.2. Composite algebraic field extension. An extension F of a field P is called composite if there is an ascending chain of subfields L i of a field F such that P \u003d L 0 - L 1 -… - L k \u003d F and k\u003e 1. Theorem 2.3. Let F be a finite extension of L and L a finite extension of P. Then F is a finite extension of P and \u003d @ [L: P]. Evidence. Let be (1) a 1,…, a m is a basis of the field L over P (as a vector space) and (2) b 1,…, b n is a basis of the field F over L. Any element d from F can be linearly expressed in terms of the basis: (3) d \u003d l 1 b 1 + ... + l n b n (l k 0L). The coefficients 1 k can be linearly expressed through the basis (1): (4) l k \u003d p 1k a +… + p mk a m \u200b\u200b(p ik 0P). Substituting expressions for the coefficients l k in (3), we obtain d \u003d е 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 \u003d (a i b k1 (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 be (5) еc ik a i b k \u003d 0, where c ik 0 P. Since system (2) is linearly independent over L, it follows from (5) that (6) with 1 k a 1 + ... + with mk a m \u200b\u200b\u003d 0 (k \u003d 1, ..., n). Since the elements a 1, ..., a m are linearly independent over P, it follows from (6) that c 1 k \u003d 0, ..., c mk \u003d 0 (k \u003d 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 \u003d nm \u003d ×. Therefore, F is a finite extension of the field P and formula (I) holds. Definition. An extension F of a field P is called composite algebraic if there is an increasing chain of subfields of the field P P \u003d L 0 - L 1 -… - L k \u003d F and k\u003e 1 (1) such that for i \u003d 1, ..., k the field L i is a simple algebraic extension of the field L i-1. The number k is called the length of the chain (1). Corollary 2.4. A composite algebraic extension F of a field P is a finite extension of P. The proof is easily carried out by induction on the length of the chain (1) based on Theorem 2.3. Theorem 2.5. Let a 1, ..., a k be elements of the field F algebraic over the field P. Then the field P (a 1, ..., a k) is a finite extension of P. L 0 \u003d P, L 1 \u003d P, L 2 \u003d P, ..., L k \u003d P. Then L 1 \u003d 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 \u003d L 0 - L 1 -… - L k \u003d F where L i \u003d L i -1 (a i) for i \u003d 1, ..., k, that is, 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. Hence, by Corollary 2.4, the field F is a finite extension of the field P. Corollary 2.6. A composite algebraic extension of a field is an algebraic extension of this field. 2.3. Simplicity of a composite algebraic field extension. Theorem 2.7. Let a number field F be a composite algebraic extension of the field P. Then F is a simple algebraic extension of P. Evidence. Let P - L - F, and L \u003d P (a), F \u003d L (b) and, therefore, F \u003d P (a, b). Let f and g be minimal polynomials over P for the numbers a and b, respectively, and deg f \u003d m, deg g \u003d n. The polynomials f and g are irreducible over P and, therefore, do not have in the field E complex numbers multiple roots. Let be a \u003d a 1, ..., a m are the roots of the polynomial f in C and b \u003d b 1, ..., b n are the roots of the polynomial g in C. Consider a finite set M: M \u003d ((a i -a) / (b-b k) ½i0 (1,…, m), k0 (2,…, n)). Since P is a numerical set (and, therefore, infinite), then P contains a number c different from the elements of the set M, c0P (M, cóM. Let Then the relations (2) g ¹ a i + cb k \u003d (i0 (1, ..., m), k0 (2, ..., n)). Indeed, in the case of equality a + cb \u003d a i + cb k it would be с \u003d (a i -a) / (b-b k) 0 M which would contradict the choice of the number c. Let F 1 \u003d P (g) and F 1 be a polynomial ring in x. Let h \u003d f (g - cx) be a polynomial from F 1 [x] (g, c0P (g) \u003d 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) \u003d 0, then x-b divides g in E [x]. Further, by virtue of (1) h (b) \u003d f (g-cb) \u003d f (a) \u003d 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 has no roots other than b. Indeed, assume that b k, k0 (2, ..., n) is their common root. Then h (bk) \u003d f (g - cb k) \u003d 0. Therefore, there is an index i0 (1, ..., m) such that g \u003d ai + cb k (k\u003e 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]. therefore (x-b) 0 F 1 [x] and b 0 F 1 \u003d P (g). In addition, a \u003d g - cb 0 F 1. In this way, F \u003d P (a, b) Ì F 1, F 1 ÌF. 2.4. Algebraic number field. 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 algebraic over the 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 \u003d + С, +, -,, 1, complex numbers. The algebra A \u003d + A, +, -,, 1, is a field, a subfield of E. Evidence. 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, -а, ab, 1 are algebraic, that is, they belong to the set A. Thus Thus, the set A is closed with respect to 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 nonzero 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 E. Definition. The field A \u003d + A, +, -,, 1, is called the field of algebraic numbers. Show that the number a \u003d is algebraic. Decision. From a \u003d follows a-. Let's raise both sides of the last equality to the third power: a 3 -3a 2 9a-3 \u003d 2 a 3 + 9a-2 \u003d 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 \u003d 27a 4 + 54a 2 +27 a 6 -9a 4 -4a 3 + 27a 2 -36a-23 \u003d 0. Thus a is a root of the polynomial f (x) \u003d a 6 -9a 4 -4a 3 + 27a 2 -36a-23 \u003d 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 algebraic number field is algebraically closed. Evidence. Let A [x] be a polynomial ring in x over the field A of algebraic numbers. Let be f \u003d a 0 + a 1 x + ... + a n x n (a 0, ..., a n 0 A) Any polynomial of positive degree from 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, that is, there exists a complex number c such that f (c) \u003d 0. Let L \u003d Q (a 0, ..., a n) and L (c) is a simple algebraic extension of the field L using 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 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 from A [x] of positive degree has a root in A, that is, the field A is algebraically closed. 3. Separable and non-separable extensions. Let D be a field. Let us find out whether a polynomial 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 factor, which can be calculated already in D [x]. If the polynomial f (x) is indecomposable, then f (x) cannot have non-constant common factors with any polynomial of lesser degree, therefore, the equality f "(x) \u003d 0 must hold. f (x) \u003d 3a n x n fN (x) \u003d 3na n x n -1 Since fN (x) \u003d 0, each coefficient must vanish: na n \u003d 0 (n \u003d l, 2, ..., n). In the case of characteristic zero, this implies that a n \u003d 0 for all n ¹ 0. Therefore, a non-constant polynomial cannot have multiple roots. In the case of characteristic p, equalities na n \u003d 0 are also possible for n ¹ 0, but then the comparisons must be fulfilled f (x) \u003d a 0 + a p x p + a 2p x 2p +… Conversely: if f (x) has this form, then fN (x) \u003d 0. In this case, we can write: Thus, we have proved the assertion: In the case of characteristic zero, a polynomial f (x) indecomposable in D [x] has only simple roots, in the case of a 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 xp. In the latter case, it may turn out that j (x), in turn, is 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 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 in the form c (x pe + 1), which contradicts the assumption. Therefore, y (y) has only simple roots. We decompose the polynomial y (y) in some extension of the ground field into linear factors: m y (y) \u003d J (y-b i). f (x) \u003d J (x pe -b i) Let a i be some root of the polynomial x pe -b i. Then x i pe \u003d b i, x pe -b i \u003d x pe - a i pe \u003d (x-a i) pe. Therefore, a i is a pe-multiple root of the polynomial x pe -b i and f (x) \u003d J (x -a i) p e. Thus, all roots of the polynomial f (x) 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 of the root a i) over the field D. Between the degree, the reduced degree and the exponent, the relation where m is equal to the number different roots polynomial f (x). If q is a root of a polynomial indecomposable in the ring D [x] and having only simple roots, then q is called a separable element over D or an element of the first kind over D 1). Moreover, 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 elements of which are separable over D, is called 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. We will see later that most of the most important and interesting field extensions are separable and that there are entire classes of fields that do not have inseparable extensions at all (the so-called "perfect fields"). For this reason, in the future, everything related specifically to inseparable extensions is typed in small print. Consider now the algebraic extension S \u003d D (q). When the degree n of the equation f (x) \u003d 0 defining 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]"for which the elements of the subfield D remain fixed and, therefore, S is mapped to an equivalent field S" (isomorphisms of the field S over the field D) and for which the image field S "lies with the field S inside some common field W. these conditions, the following theorem holds: For an appropriate choice of the field W, the extension S \u003d 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. Evidence. Each isomorphism over D must map an element q to its conjugate element q "from W. Choose W so that f (x) decomposes over W into linear factors; then it turns out that q has exactly m conjugates elements q, q", ... Moreover, no matter how the field W is chosen, the element q will have no more than m conjugates in it. Note now that each isomorphism D (q) @D (q") over D is completely determined by specifying the correspondence q® q ". Indeed, if q goes to q" and all elements of 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 \u003d n and, therefore, the number of isomorphisms over the ground field is equal to the degree of the extension. If there is some fixed field containing all the fields under consideration, which contains all the roots of each equation f (x) \u003d 0 (as, for example, in the field of complex numbers), then as W we can take this field once and for all and therefore drop the addition “inside some W” in all isomorphism statements. This is always the case in number field theory. We will see later that such a field W can be constructed for abstract fields as well. The above theorem is generalized by the following statement: If the extension S is obtained from D by successive addition of 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 in no extension, there are no more such isomorphisms of the field S. Evidence. For m \u003d 1, the theorem has already been proved above. Suppose it is valid for the extension S 1 \u003d D (a 1, ..., a m-1): in some suitable extension W 1 is exactly Õ n i ¢ isomorphisms of the field S over D. Let S 1 ®S 1 be one of these Õ n i ¢ isomorphisms. It is asserted that in a suitably chosen field W it can be extended to the isomorphism S \u003d S 1 (a m) @ S \u003d S (a m) in at most n ¢ m ways. The element a m satisfies some equation f 1 (x) \u003d 0 over S 1 with n ¢ m different roots. By means of the isomorphism S 1 ® S 1, the polynomial f 1 (x) is transformed into some polynomial f 1 (x). But then f 1 (x) in a suitable extension has again n ¢ m different roots and at most. Let a m be one of these roots. By the choice of the element a m, the isomorphism S 1 @ S 1 can be extended to an isomorphism S (a m) @ S (a m) with a m ®a m in one and only one way: indeed, this continuation is given by the formula еc k a m \u200b\u200bk ®å c k a m \u200b\u200bk 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 е 1 ® е 1 Since, in turn, this isomorphism can be chosen Õ n "i ways, then everything exists (in the field W, which contains all the roots of all the equations under consideration) Õ n "i × n" m \u003d Õ n "i of isomorphisms of the extension S over the field D, as required. If ni is the full (unreduced) degree of an element ai over D (a 1, ..., a i-1), then ni is equal to the degree of the extension D (a 1, ..., ai) of the field D (a 1, .. ., a i-1); therefore, the degree (S: D) is If we compare this number with the number of isomorphisms The number of isomorphisms of the extension S \u003d D (a 1, ..., am) over D (in some suitable extension W) is equal to degree (S: D) if and only if each element ai is separable over the field D (a 1,. .., a i-1). If at least one element a i is inseparable over the corresponding field, then the number of isomorphisms is less than the degree of the 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 the 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 turns out to be separable if all a i are separable. So: If the elements a i, ..., a n are successively joined to the field D and each element a i turns out to be separable over the field obtained by joining the previous elements a 1, a 2, ..., a i-1, then the extension S \u003d 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 explained by the fact that b satisfies some equation with a finite number of coefficients a 1, ..., am from S and, therefore, is separable over D (a 1, ..., am). Thus, the 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. Endless field expansions. 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, of course, played by the maximal algebraic extensions, that is, those that do not admit further algebraic extensions. 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: \u200b\u200beach polynomial of the ring W [x] can be completely decomposed into linear factors. This condition is also sufficient. Indeed, if every polynomial in W [x] can be decomposed 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], i.e. that is, it coincides with some element a of W. Therefore, we give the following definition: A field W is called algebraically closed if any polynomial in W [x] can be decomposed into linear factors. An equivalent definition is the following: a field W is algebraically closed if every polynomial from W [x] different from a constant has at least one root in W, that is, at least one linear factor in W [x]. Indeed, if such a condition is satisfied and an arbitrary polynomial f (x) is decomposed into indecomposable factors, then all of them must be linear. The "Basic 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, that is, they themselves are algebraic numbers. Here we show how to construct an algebraically closed extension of an arbitrary field P and, moreover, in a purely algebraic way. Steinitz belongs to the following The main theorem. For each field P, there exists 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 a factorization of any polynomial from P [x] in the ring W [x]. Evidence. Let f (x) be an arbitrary polynomial from W [x]. If it does not decompose into linear factors, then we can add some of its roots a and arrive at a proper overfield W. "The element a is algebraic over W, and W is an algebraic extension of the field P; therefore, the element a is algebraic over P. Therefore a root of some polynomial g (x) from P [x]. This polynomial is decomposed in W [x] into linear factors. Therefore, a -root of some linear factor in W [x], that is, 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, so that in this ordering the field P is a segment. Evidence. We define an 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 f (x) is equal to the degree 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 \u003d b i for i a k In this case, an exception is made for the polynomial 0: the degree 0 is assigned to it. Obviously, in this way we obtain some ordering, in the sense of which P [x] is completely ordered. It is shown as follows: in each non-empty set of polynomials there is a non-empty subset of the least degree polynomials; let it be equal to n. This subset contains a non-empty subset of polynomials, the coefficient a 0 of which is the first in the sense of the existing order among the free terms of the polynomials under consideration; in this subset there is, in turn, a subset of polynomials with the first a 1, and so on. The subset with the first a n, which will eventually turn out, can consist only of one single polynomial (since a 0, ..., and n are defined unambiguously due to the consistently fulfilled minimality condition in the choice); this polynomial is the first element in the set. Lemma 3. If the field P is well ordered and a polynomial f (x) of degree n and n symbols a 1 ..., an are given, then the field P (a 1, ..., an), in which f (x) is completely decomposed into linear factors Õ (x-a i), is constructed uniquely and is completely orderly. The field P in the sense of this order is a segment. Evidence. We will attach the roots a 1 ..., a n sequentially, as a result of which the fields P 1, ..., P n will appear sequentially from P \u003d P 0. Suppose that P i-1 \u003d P (a 1 ..., a i-1) is an already constructed field and that P is a segment in P i-1; then Р i will be constructed as follows. First of all, by virtue of Lemma 2, the ring of polynomials Р i-1 [x] is completely ordered. The polynomial f is decomposed in this ring into indecomposable factors, among which x - a 1, ..., x - a i-1 will be in the first place; among the remaining 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 \u003d P i -1 as the collection 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 \u003d P i -1; the symbol 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 the elements of the field are ordered in the same way as the corresponding polynomials are ordered. Obviously, then P i-1 is a segment in P i, and therefore P is a segment in P i. Thus, the fields Р 1, ..., Р n are constructed and are completely ordered. The field Р n is the required uniquely determined 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. Evidence. For any two elements a, b of the union, there are two fields S a, S b, which contain a, and b and one of which precedes the other. The elements a + b and a × b are defined in the enclosing field, 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 \u003d a bg, find among the fields S a, Sb, S g that which contains two other fields (the largest); this field contains a, b and g and the associativity law is satisfied in it. All other calculation rules with union elements are checked in the same way. The proof of the main theorem falls into two parts: the construction of the field W and the proof of uniqueness. Construction of the field W .. Lemma 1 testifies that to construct an algebraically closed extension W of the field P it is sufficient to construct an algebraic extension of the field P such that each polynomial from P [x] decomposes over this extension into linear factors. 1. Field Р f is the union of the field Р and all fields S g for g 2. The field Р f is completely ordered so that Р and all fields S g for g 3. The field S f is obtained from P f by adding all roots of the polynomial f using the symbols a 1, ..., a n in accordance with Lemma 3. It is necessary to prove that in this way the well-ordered fields Р f, S f are really uniquely determined, if only 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 Р - segment in S h at h S g - segment in S h at g Hence it follows that the field P and the fields S h (h By condition 3, the polynomial f (x) is completely decomposed into linear factors in the field S f. Further, using transfinite induction, it is shown 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; according to Lemma 4, it is a field. This field is algebraic over P, and all polynomials f decompose over it (since every polynomial f is decomposed over S f). Hence, the field W is algebraically closed (Lemma 1). The uniqueness of the field W. Let W and W "be two fields that are algebraic and algebraically closed extensions of the field P. We prove the equivalence of these fields. To this end, we assume that both fields are completely ordered. For each segment J from W (the field W itself is also is considered to be one of such segments) the subset J ¢ in W "and some isomorphism P (Â) @ P ( ¢). The latter must satisfy the following recurrence relations. 1. The isomorphism P (J) @ P (J) must leave each element of the field P in place. 2. The isomorphism P (J) @ P (J,) for ÁÌ J must be an extension of the isomorphism P (Á) @ P ("). 3. If J has the last element a, so that J \u003d ÁÈ (a), and if a is a root of a polynomial f (x) indecomposable in P (), then the element a "must be the first root of the corresponding by virtue of P (Á) @ Р (Á "), the polynomial f ¢ (x) in a well-ordered field W". It is necessary to show that these three requirements really define the isomorphism P (J) @ P (J,), if only it is already defined for all previous segments ÁÌ J. Two cases must be distinguished here. First case. The set  has no last element. Then each element a belongs to some previous segment Á; therefore J is the union of the segments Á, and therefore P (J) is the union of the fields P (Á) for ÁÌ J. Since each of the isomorphisms P (Á) @ P (") is a continuation of all the previous ones, each element a for all these isomorphisms is associated with only one element a". Therefore, there is one and only one mapping P (J) → P (J ¢), which continues all the previous isomorphisms P (Á) → P ("), namely the -map a®a". Obviously, it is an isomorphism and satisfies requirements 1 and 2. Second case. The set  has the last element a; therefore,  \u003d ÁÈ (a). Due to Requirement 3, the element a ", assigned to an element a, is uniquely determined. Since a" over the field P (") (in the sense of the isomorphism under consideration) satisfies the" same "indecomposable equation as a over P (), then the isomorphism P (Á) → P (") (and in the case when is empty, that is, the identity isomorphism P®P) extends to an isomorphism P (Á, a) ®P (Á", a), for where a goes into a ". This isomorphism is uniquely defined by each of the above requirements, because each rational function j (a) with coefficients from J must go over to the function j "(a") with the corresponding coefficients from Á ". The isomorphism P (J) ® P (J ¢) satisfies requirements 1 and 2, obviously. This completes the construction of the isomorphism P (J) → P (J). We denote by W "the union of all the fields P (J ¢); then there exists an isomorphism P (W) ®W" or W®W ", which leaves in place each element of the field P. Since the field W is algebraically closed, the same must be W ", and therefore W" coincides with the entire field W. Hence 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 P and S is an arbitrary algebraic extension of P, then inside W there exists an extension S 0 equivalent to S. Evidence. We extend S to some algebraically closed algebraic extension W ". It will also be algebraic over P, and therefore equivalent to the extension W. For some isomorphism taking W" to W and keeping each element of P fixed, the field S goes into some equivalent subfield S 0 in W. 4.2. Simple transcendental extensions. Each 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 rational functions Theorem. Every element h of degree n different from a constant is transcendental over D and the field D (x) is an algebraic extension of the field D (h) of degree n. Evidence. The representation h \u003d f (x) / g (x) will be considered irreducible. Then the element x satisfies the equation g (x) × h - f (x) \u003d 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 nonzero coefficient of the polynomial g (x) for the same degree x, and b k a nonzero coefficient of the polynomial f (x), then the equality whence h \u003d b k / a k \u003d const, which contradicts the assumption. Consequently, the element x is algebraic over D (h). If h were algebraic over D, then x would also be algebraic over D, which, however, is not the case. Consequently, 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 decomposable in n 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. Consequently, the element x is algebraic of degree n over the field D (h). This implies the statement that (D (x): D (h)) \u003d n For what follows, note that the polynomial has no factors depending only on z (ie, 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), thereby 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 one or another choice of the generating element x. 2. The equality D (h) \u003d D (x) holds if and only if h has degree 1, that is, is a linear fractional function. This means: the generating element of the field, in addition to 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 in place each element of the field D, must take the element x to some generating element of the field. Conversely, if x is mapped to some generating element x \u003d (ax + b) / (cx + d) and each function j (x) is to a function j (x), then an automorphism is obtained in which all elements from D remain in place. Hence, All automorphisms of the field D (x) over the field D are linear fractional permutations x \u003d (ax + b) / (cx + d), ad - bc ¹ 0. Important for some geometric studies is Luroth's theorem. Each intermediate field S for which DÌSÍD (x) is a simple transcendental extension: S \u003d D (q). Evidence. 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 has been shown, the element x is algebraic over D (h) and, all the more, 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. (1) Let us clarify the structure of this polynomial. Elements a i are rational functions of x. By multiplying them by a common denominator, they can be made entire rational functions and, in addition, a polynomial with respect to x with content 1 can be obtained: f (x, z) \u003d b 0 (x) z n + b 1 (x) z n-1 +… + b n (x). The degree of this polynomial in x is denoted by m, and in z by n. The coefficients a i \u003d b i / b 0 from (1) cannot all be independent of x, since otherwise x would be an algebraic element over D; so one of them, say q \u003d a i \u003d b i (x) / b 0 (x), should actually depend on x; we 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) \u003d g (z) - (g (x) / h (x)) h (z) (which is not the identical zero) has a root z \u003d 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 polynomials that are integral in x with content 1, then the divisibility relation is preserved, and we get h (x) g (z) -g (x) h (z) \u003d q (x, z) f (x, z). The left-hand side of this equality has degree in x not exceeding m. But on the right, the polynomial f already has degree m; therefore, the degree of the left-hand side is exactly m and q (x, z) is independent of x. However, a factor depending only on z cannot divide the left-hand side (see above); therefore q (x, z) is constant: h (x) g (z) -g (x) h (z) \u003d qf (x, z). Since the presence of the constant q does not matter, the structure of the polynomial f (x, z) is described in full. The degree of the polynomial f (x, z) in x is m therefore (for reasons of symmetry), and the degree in z is equal to m, so that m \u003d 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 degree m in x. Thus, since on the one hand the equality (D (x): D (q)) \u003d t, and on the other - equality then, since S contains D (q), Conclusion. In the course work, the following types of extensions of the number field P were considered: Simple algebraic field extension. Composite algebraic field extension. Separable and non-separable extensions. Endless field expansions. Analyzing the work, we can draw some conclusions. Of the extensions discussed in the first two parts, such as: simple algebraic extensions; final extensions; compound algebraic extensions. It follows that all these types of extensions coincide and, in particular, are exhausted by simple algebraic extensions of P. List of references 1. L. Ya. Kulikov. Algebra and Number Theory.- Moscow: 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 were used materials from the site 10. Theorem on the structure of a simple algebraic extension ten . The concept of a minimal polynomial. Let a be an algebraic number over the field k, that is, the 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 the minimal polynomial of a number a if the following conditions are satisfied: 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) \u003d 0, i.e. a is the root of the polynomial m (x). 20 . Basic properties of minimal polynomials. 1. If f (x) Î k [x] and f (a) \u003d 0, then f (x) is divisible by the minimal polynomial m (x) of a. Evidence. Indeed, assuming that f is not divisible by m, we write f \u003d mg + r, deg r< deg m based on the remainder division theorem. Whence r (a) \u003d 0. Since the polynomials r and m are coprime, they cannot have common roots - a contradiction. 2. Suppose that a is an algebraic number, and g (x) is a normalized polynomial of the least positive degree such that g (x) Î k [x] and g (a) \u003d 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 determined. To prove it, it suffices to apply property 2. Definition. The degree of the minimal polynomial of the number a is called the degree of the number a; notation deg k a. 4.a Î k Û deg k a \u003d 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, that is, ("c 0, c 1, ..., c n-1 Îk) c 0 + c 1 a + ... + c n-1 an -1 \u003d 0 is possible only if c 0 \u003d c 1 \u003d.. . \u003d c n-1 \u003d 0. Evidence. Indeed, if the indicated powers of a are linearly dependent, then this number is a root of some polynomial over k of degree less than m. 6. Let a be an algebraic number, f (x) Î k [x], and f (a) ¹ 0. Then the fraction can be represented as \u003d g (a) for some g (x) Î k [x]. Evidence. Indeed, the polynomials f and m are coprime (otherwise f would be divisible by m), hence, by the theorem on the linear representation of gcd: for some polynomials g and h over k, the equality Whence f (a) g (a) \u003d 1, as required. thirty . The structure of simple algebraic extensions. Definition. Let k be a subfield in L; a Î L. The smallest subfield in L containing a number a and a subfield k, denoted by k (a), is called a simple extension of the field k (they also say that k (a) is obtained by joining the number a to the field k). It is easy to deduce a theorem from these 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 \u003d deg k a. Evidence. It is easy to see 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] is the ring of values \u200b\u200bof polynomials at point a, i.e. k [a] \u003d (f (a) ½f (x) Î k [x]). Property 6 implies the equality k (a) \u003d k [a]. From the theorem on division with remainder it follows that the value of an arbitrary polynomial over a field k at a point a is a linear combination over the field k of the powers of the element a specified in the theorem. Finally, property 5 implies linear independence over the field k of these degrees. ÿ 4 0. Liberation from irrationality in the denominator of the fraction. Let us examine various ways to solve the problem of freeing from irrationality in the denominator of a fraction. The possibility of its solution in principle follows from the theorem on the structure of a simple algebraic extension. Example 1. Get rid of irrationality in the denominator of a fraction: Decision. Let us denote a number through c, and use the well-known formula for the sum of members of a geometric progression: 1+ c + c 2 + c 3 + c 4 \u003d (c 5 - 1) / (c- 1) \u003d 1 / (c- 1), hence, Example 2. Get rid of irrationality in the denominator of a fraction: Decision. Let c denote a number, and first write the fraction as the sum of the simplest: Now, using Horner's scheme, each of the indicated fractions can be replaced by a polynomial with respect to c. First, divide c 5 - 2 by c + 1: hence, 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) \u003d 31s 4 - 40s 3 + 22s 2 - 10s - 14, Example 3. Get rid of irrationality in the denominator of a fraction: Decision. Let's denote by c a number. Find a linear representation of the GCD of the polynomials f (x) \u003d x 3 - 2 and g (x) \u003d 1 + 2x - x 2: f (x) \u003d - g (x) × (x + 2) + r (x), where r (x) \u003d 5x 5g (x) \u003d r (x) × (x - 2) - 5. From these equalities, we obtain a linear representation of the gcd f (x) and g (x): f (x) × (x - 2) + g (x) × (x 2 + 1) \u003d 5. Substituting the number c instead of x in the last equality, we obtain therefore, \u003d. Example 4. Get rid of irrationality in the denominator of a fraction: Decision. Let's denote a number by c and apply the method of undefined coefficients. By the theorem on the structure of a simple algebraic extension, there exist rational numbers x, y, z such that Xc 2 + yc + z or 89 \u003d (c 2 + 16c - 11) (xc 2 + yc + z). Expanding the brackets and using the equality c 3 \u003d 2, we get: 89 \u003d (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 \u003d 89, 2x - 11y + 16z \u003d 0, 11x + 16y + z \u003d 0. The solution to the last system is the set of numbers (3, 2, 1). So, we get the answer: Let the field P contained in field T and a- element T not owned P... Consider the smallest field P(a) containing all elements from P and a... All view elements belong to P(a). Let's consider two cases. End fields. Theorem 4.2. The number of elements in a finite field is p n, where p is a prime number. Evidence... Since the field P is finite, its characteristic is nonzero. Let p be its characteristic. The field P can be viewed as a vector space over Z p. Let v 1,…, v n denote a basis P. Any element of the field P is uniquely characterized by coordinates (x 1,…, x n) in this basis. Each coordinate takes p values, therefore, the number of different sets of coordinates, and hence the elements of the field P, is equal to p n. Lemma 4.1 In the characteristic field p Evidence. Theorem 4.3. For any natural n and prime p, there is a field of order p n. We extend Z p so that the resulting field contains all the roots of the polynomial. The polynomial has no multiple roots, since its derivative is –1. Let M denote the set of roots of the polynomial. It is easy to check that M is a field and the number of its elements is equal to p n Theorem 4.4. The order field is unique up to isomorphism. Evidence. Since the number of elements of the field, then its characteristic is equal. Therefore, any field P order can be viewed as an extension of the residue ring. The multiplicative group of the field () has order, and, therefore, for any it is true. Thus, all elements of the field are the roots of the equation over. Theorem 4.5. Multiplicative root group n-th degree of 1 in P is cyclic. Evidence. Let be p field characteristic P... If, then 1. n - Prime number. Then the root group is of order n, and hence cyclic 2. - power of a prime number. Find the root of the equation, which is not the root of the equation. Element order is a divisor of group order n and is not a divisor. Hence the order is n and the group is cyclic. 3. Let. Let us denote by the generator of the cyclic group of roots of degree 1. Let's put. Induction on k we will show that the order is equal. When k\u003d 1 the statement is obvious. Let it be proven for k-1. The element order is. Greatest common divisor t and is equal to 1, and, therefore, there are numbers u and v, what . Since and, the order of the element is divisible by t and on. Further, from equality, it follows that the order of the element is a divisor. The theorem is proved. Galois theory Field T is called finite field extension P, if a T is a finite-dimensional linear space over P... The dimension of space is called the degree of expansion. Any algebraic field extension P is finite. Its degree is equal to the degree of the irreducible polynomial. Theorem 5.1. Final expansion U fields Twhich is the final extension of the field P, is the final extension P... Moreover, the degree of expansion U over P is equal to the product of the degrees of expansion. Evidence... It's almost obvious. Field element T is called algebraic over Pif it is a root of some polynomial over P. All elements of the final extension P algebraic over P. Any finite extension can be obtained by adding a finite number of algebraic extensions. Theorem 5.2. Any finite field expansion P characteristics 0 is a simple extension. Evidence not obvious. Final expansion T called normal expansion Pif from the fact that an irreducible polynomial over P has in T root, it follows its decomposability into linear factors. It is clear that the normal extension of a field of characteristic 0 is a polynomial decomposition field. The converse is also true. The decomposition field of a polynomial is a normal extension. An automorphism of a field is an isomorphic mapping onto itself. Galois group of normal extension T fields P is the group of automorphisms of the field Tpreserving field elements P. Theorem 5.3. To every intermediate field U, there corresponds some subgroup of the Galois group, namely, the collection of those automorphisms that do not change the elements. The field is uniquely determined by the subgroup. Introduction. In pedagogical universities, a program for a unified course in algebra and number theory has been introduced. The main goal of this course is the study of basic algebraic systems and the education of an algebraic culture, which is necessary for a future teacher for a deep understanding of the goals and objectives of both the main school mathematics course and school elective 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 twentieth century, does not stop, and this causes persistent attempts to introduce basic algebraic concepts into school mathematics education. The mathematical depth and an unusually wide scope of application of fields are combined with the simplicity of its basic provisions - the concepts of fields; a number of important theorems can be formulated and proved, having initial concepts in the field of set theory. Therefore, field theory is the best suited to show students an example of modern mathematics. In addition, the study of the elements of field theory is useful for schoolchildren, contributes to their intellectual growth, manifested in the development and enrichment of various aspects of their thinking, qualities and personality traits, as well as fostering students' interest in mathematics and science. 1. Simple algebraic field extension. 1.1 Simple field expansion. Let P [x] be a ring of polynomials in x over a field P, where P is a subfield of F. Recall that an element a of a field F is called algebraic over a field P if a is a root of some polynomial of positive degree from 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] \u003d (f (a) * f0P [x]), that is, 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 the symbol P [a]. Theorem 1.1. Let P [x] be a ring of polynomials in x over P and P (a) a simple extension of P. Let y be a mapping from P [x] onto P [a] such that y (f) \u003d f (a) for any f from P [x]. Then: (a) for any a from P y (a) \u003d a; (c) y is a homomorphism of the ring P [x] onto the ring P [a]; (d) Kery \u003d (f0P [x] * f (a) \u003d 0); (f) the quotient ring P [x] / Ker y is isomorphic to the ring P [a]. Evidence. Statements (a) and (b) follow directly from the definition of y. The mapping y preserves the principal operations of the ring P [x], since for any f and g from P [x] y (f + g) \u003d f (a) + g (a), y (fg) \u003d f (a) g (a), y (1) \u003d 1. Statement (d) immediately follows from the definition of the mapping 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]. Evidence. Due to the transcendence of a over PKery \u003d (0). Therefore, P [x] / (0) –P [a]. Moreover, the quotient ring of the ring P [x] by the zero ideal is isomorphic to P [x]. Therefore, P [x] –P [a]. 1.2 Minimum polynomial of an algebraic element. Let P [x] be a polynomial ring over a 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 from P [x] of the least degree whose root is a. The degree of the minimal polynomial is called the degree of the element a over P. It is easy to see that for every element a 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 \u003d j. Evidence. The degrees of the minimal polynomials g and j coincide. If g¹j, then an element a (of degree n over P) will be a root of the polynomial g - j whose degree is less than the degree of the polynomial j (less than n), which is impossible. Therefore, g \u003d j. Theorem 1.4. Let a be an algebraic element of degree n over the 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) \u003d 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). Evidence. Suppose that the polynomial g is reducible in the ring P [x], that is, there exist polynomials j and h in P [x] such that g \u003d jh, 1 £ deg j, deg h Then g (a) \u003d j (a) h (a) \u003d 0. Since P (a) is a field, then j (a) \u003d О or h (a) \u003d 0, which is impossible, since, by hypothesis, the degree of an element a over P is equal to n. Suppose that f0 P [x] and f (a) \u003d 0. By assumption, g (a) \u003d 0. Therefore, f and g cannot be coprime. Since the polynomial g is irreducible, g divides f. Let j be the homomorphism of the ring P [x] onto the ring P [a] (y (f) \u003d 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, that is, Ker y \u003d (g). Therefore, the quotient ring P \u003d 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 into P. Let f be an element of the coset f. Since f¹ 0, then f (a) ¹0; therefore 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, in P [x] there exist polynomials u and v such that uf + vg \u003d 1. This implies the equality uf \u003d 1, which shows that the element f is invertible in the ring P. Thus, it has been established that the quotient ring P is a field. By virtue of (c) and (d), P [a] is a field, and therefore P (a) ÌP [a]. Moreover, it is obvious that P [a] ÌP (a). Hence, P [a] \u003d P (a). Consequently, the ring P [a] coincides with the field P (a). 1.3 Construction of a simple algebraic field extension. Theorem 1.5. Let a be an algebraic element of positive degree n over the 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. Evidence. Let b be any element of the field P (a). By Theorem 1.4, P (a) \u003d P [a]; therefore, there exists in P [x] a polynomial f such that Let g be the minimal polynomial for a over P; by the hypothesis of the theorem, its degree is equal to n. By the division theorem with remainder, there exist polynomials h and r in P [x] such that (2) f \u003d gh + r, where r \u003d 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 \u003d c 0 + c 1 a + ... c n -1 a n-1 Let us show that the element b is uniquely representable as a linear combination of elements 1, a, ..., a n-1. Let be (4) b \u003d d 0 + d 1 a +… d n -1 a n-1 (d i 0P) Any such performance. Consider the 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, since by virtue of (3) and (4) j (a) \u003d 0 and the degree of j is less than the degree g. The only possible case is when j \u003d 0, that is, with 0 \u003d d 0,. ... ... , with n-1 \u003d d n-1. Therefore, the element b is uniquely representable as a linear combination of elements 1, a,…, a n-1. 1.4 Liberation 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\u003e 1 over the field P; f and h are polynomials from the ring of polynomials P [x] and h (a) ¹0. It is required to represent the element f (a) / h (a) 0P (a) in the form of a linear combination of the powers of the element a, that is, in the form j (a), This problem is solved as follows. 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, then g does not divide h, and hence the polynomials h and g are coprime. Therefore, there exist polynomials u and v in P [x] such that Since g (a) \u003d 0, it follows from (1) that u (a) g (a) \u003d 1, 1 / h (a) \u003d u (a). Therefore, f (a) / h (a) \u003d f (a) u (a), and f, u0P [x] and f (a) u (a) 0P [a]. So, we got rid of irrationality in the denominator of the fraction f (a) / h (a). Get rid of irrationality in the denominator of a fraction Decision. In our case, a \u003d The polynomials p (x) and g (x) \u003d - x 2 + x + 1 are coprime. Therefore, there exist polynomials j and y such that To find j and y, we apply Euclid's 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 \u003d g (-x-1) + (2x-1), g \u003d (2x-1) (- 1 / 2x + 1/4) +5/4. Where do we find (2x-1) \u003d p + g (x + 1), 5/4 \u003d g- (p + g (x + 1)) (- 1 / 2x + 1/4) p1 / 5 (2x-1) + g (4/5 + 1/5 (2x 2 + x-1)) \u003d 1, p1 / 5 (2x-1) + g (2 / 5x 2 + 1 / 5x + 3/5) \u003d 1. In this way, y (x) \u003d (2 / 5x 2 + 1 / 5x + 3/5). Hence 2. Composite algebraic field extension. 2.1. Ultimate field expansion. Let P be a subfield of F. Then we can regard F as a vector space over P, that is, 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 called 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 \u003d n. This proposition follows directly from Theorem 1.5. Definition. An extension F of a field P is said to be algebraic if every element of F is algebraic over P.
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, where is the multiplicity of the occurrence of the element. The quantity is not divisible by p only in case i \u003d0;p. Because pe \u003d 0then .
, and, therefore, the set of roots of the equation coincides with the set of roots of degree. Without breaking the generality, you can consider. It suffices to carry out the proof for the case when all roots n-th power of 1 are contained in the field P... Otherwise, we will extend the field and use the fact that any subgroup of a cyclic group is cyclic. Insofar as 
has only a single root equal to zero, then the number of roots n-th power of 1 equals n... Consider three cases: