Algebraic expansion of the field. Forgive expansion of watering. Warehouse expansion of algebra fields

algebraic field extension- — Topic for information protection EN extension field … Dovіdnik technical translation

Field E, which is given field K as a subfield. Tipi extension Extension of algebra extension extension, all elements of such a є algebraic over K, that is, such an element of such a є is the root of a rich term f (x) c ... Wikipedia

Algebraic extension of the field EÉ K, which is normal and separable. For tsikh minds E will mother the largest number of automorphisms over K (as E is unique, then the number of automorphisms is also a significant and more advanced degree of expansion).

Nap_vugroup A nap_vgroup S, scho revenge Av yak p_demigroup. Sound about expanding the names of group A, tying with Atem with other minds. The most advanced theory of ideal R. nap_vgroup (nap_vgroup, what to avenge Av yak ......) Mathematical Encyclopedia

Equal to the mind of the de rich term of the nth stage in the form of one or more of the change. A. in. with one unknown sound. equal to the mind: There is no number, sound. the coefficients are equal and є danimi, hnaz. nevidomim and є… Mathematical Encyclopedia

Fields k algebraic. extension of the field k, which is a closed algebraic field. Such an extension for any field is uniquely assigned up to isomorphism. A. h. fields day numbersє field complex numbers(Div. … … Mathematical Encyclopedia

The normally extended algebraic extension of the field EÉ K for any irreducible rich term f(x) over K, which can have one root E, can be decomposed in E into linear multipliers. Equivalently appointed: Yakscho KÌ EÌ K *, de K * ... ... Wikipedia

A separable extension of an algebraic extension of a field that is composed of separable elements, such that such elements are α, is the minimum annihilator f(x) over K for which there are no multiple roots. Pokhіdna f (x) can buti for vishchevkazanim ... ... Wikipedia

Expanding the field, such that E, is great, over K yak vector space. The expansion of the vector space E over K is called the degree of expansion and is designated. The power of the last expansions In ... ... Wikipedia

The fields are an algebraic extension of the L field K, which satisfies one of the advancing equivalent minds: 1) whether the field L is embedded in the algebraic field. closure of the field є by an automorphism of the field L; 2) L field of arrangement of a given family of polynomials s ... ... Mathematical Encyclopedia

Algebraic expansion of fields

Intro.

Pedagogical universities have launched a program for a unified course in algebra and number theory. The head of the meta-course is the development of the basic systems of algebra and the development of algebraic culture, which is necessary for the future teacher for a deep understanding of goals and the task of the main school course of mathematics, as well as school elective courses.

In our opinion, the most significant introduction to the school curriculum is the elements of contemporary abstract algebra.

The process of algebraicization of mathematics, which originated in the twentieth century, is not accepted, but rather forced to try to understand the basics of algebra in the school mathematical education.

Mathematical depth and superbly wide sphere of fields density will be combined with the simplicity of the basic provisions - to understand fields, a whole number of important theorems can be formulated and brought to light, often appearing in the universe of multiplicity theory. Therefore, field theory is more suitable for showing schoolchildren an insight into modern mathematics.

In addition, the development of elements in the theory of the field is familiar for schoolchildren, sprying to their intellectual growth, which is manifested in the development of those enriched different sides of their minds, qualities and characteristics, as well as the development of scientists, science, and mathematics.

1. A simple extension of the field algebra.

1.1.Simply expand the field.

Let P[x] be a ring of polynomials like x over the field P, where P are subfields of the field F. Let's guess that the element a of the field F is called algebraic over the field P, because a is the root of such a polynomial of positive step P[x].

Appointment. 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] - ring of polynomials in x i

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

so P [a] is impersonal of all in the form a 0 + a 1 a + ... + a n a n de a 0 , a 1, ... a n 0P i n - be a natural number.

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

Theorem 1.1. Let P [x] - a ring of polynomials in x over P and P (a) - a simple extension of the field P. Let y - expand P [x] on P [a] so that y (f) = f (a) for be -th f іz P[x]. Todi:

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

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

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

(e) factor-circle P[x]/Ker y isomorphic to the ring P[a].

Bringing. The assertion (a) and (b) squeal without intermediary from the appointment of y. Introducing y saves the main operations of the ring P[x], so for any f і g з P[x]

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

The firmness (d) blazes forth without a trace from the y.

If the ring y is a homomorphism of the ring P[x] onto P[a], then the factor ring P[x]/Ker y is isomorphic to the ring P[a].

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

Bringing. Looking back at the transcendence of a over P Kery=(0). To that P[x]/(0) - P[a]. In addition, the ring factor P[x] behind the zero ideal is isomorphic to P[x]. Also, P[x] - P[a].

1.2.Minimum polynomial of an algebraic element.

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

Appointment. Let a be an algebraic element over the field P. The minimal polynomial of an element a over P is the valuation polynomial of P [x] of the smallest degree, the root of which is є a. The step of the minimal polynomial is called the step of the element a over P.

It is easy to figure out that for any element a, which is algebraic over P, there is a minimal polynomial.

Proposition 1.3. If a is an element of an algebra over a field P, and g and j are the th minimal polynomial over P, then g = j.

Bringing. The steps of the minimal polynomials g and j are omitted. If g ¹ j, then the element a (step n over P) will be the root of the polynomial g - j, the step of which is less than the step of the polynomial j (less than n), which is impossible. Later, g = j.

Theorem 1.4. Let a be an algebra element of degree n over the field P (aóP) and g is the th minimal polynomial over P. Then:

(a) the polynomial g is not induced in the circle P [x];

(b) so f (a) = 0, where f 0 P[x], g divide f;

(c) the factor-circle P[x]/(g) isomorphic to the circle P[a];

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

(e) the ring P [a] is matched with the field P (a).

Bringing. Assume that the polynomial g is induced in the circle P [x], then in P [x] such polynomials j and h can be established 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) = Pro or h(a) = 0, which is impossible, shards, behind the mind, steps element a over P is more p.

Assume that f 0 P[x] and f(a) = 0. For the mind, g(a) = 0. Then f and g cannot be mutually forgiven. If the polynomial g is irreducible, then g divide f.

Let j be a homomorphism of the ring P[x] on the ring P[a] (y(f)=f(a) for any f ⊂ P[x]), in view of Theorem 2.1. 3(b) the kernel of the homomorphism y is composed of multiples of the polynomial g, so. Ker y = (g). Also, the ring factor P = P[x]/(g) is isomorphic to the ring P[a].

Oskilki P[a]ÌP(a), then P[a] is the area of ​​validity. Since P @ P [a], then the quotient P is also the domain of integrity. We need to show that any non-zero element f from P can be reduced to P. Let f be an element of the sum class f. Oskіlki f ¹ 0, then f(a)¹0; Therefore, the polynomial g cannot be divided by the polynomial f. Oskіlki polynomial g is irreducible, the stars are clear, but the polynomials f and g are mutually simple. Also, Р[x] establish such polynomials u and v that uf + vg=1. The value uf = 1 shows that the element f is beastly in the P ring.

З (с) і (d) P [a] є field and volume P(a)ÌP[a]. On the other side, obviously, P[a]ÌP(a). Also, P[a] = P(a). Also, the ring P[a] is matched with the field P(a).

1.3. Budov's simple extension of field algebra.

Theorem 1.5. Let a be an algebraic element of positive class n over the field P. Any element of the field P(a) can be uniquely represented by a linear combination of n elements 1, a, ..., a n-1 with coefficients Р.

Bringing. Let the b-be-yakie element of the field P (a). By Theorem 1.4, P(a) = P[a]; also, in P[x] the polynomial f is such that

Let g be the minimal polynomial for a over P; by virtue of the theorem, the first step is more advanced.

(2) f = gh + r, de 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

It is shown that the element is uniquely representable in a linear combination of elements 1, a, ..., a n-1 . Come on

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

Be-yaké such a manifestation. Let's look at polynomial j

j \u003d (s 0 - d 0) + (c 1 - d i .)x + . . . + (З n-1 -d n -1)x n -1

Vipadok, if step j is less than n, impossibly, scalds due to (3) і (4) j(a) = 0 і step j is the smallest type of step g. It is less possible to change, if j \u003d 0, then s 0 \u003d d 0. . . , Z n-1 = d p-1. Also, element b can be uniquely represented as a linear combination of elements 1, a,…,a n-1 .

1.4. Variation in the form of algebraic irrationality in the banner of a fraction.

A task about zvіlnennya in the form of irrationality of algebra in the banner of a fraction in the step. Let a be an algebra element of degree n>1 over the field P; f і h - polynomials from the circle of polynomials P[x] and h(a) ¹0. It is necessary to supply the element f(a)/h(a)0P(a) in the case of a linear combination of steps of the element a, then in the case of j(a),

Tse vdannya virishuєtsya so. Let g be the minimal polynomial for a over P. Oskilki, according to Theorem 1.4, the polynomial is not induced over P і h(a) ¹ 0, then g does not divide h і, also, the polynomials h і g are mutually simple. Therefore, P[x] has such polynomials u and v that

Oskіlki g(a) = 0, іz (1)

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

Also, f(a)/h(a) = f(a)u(a), moreover, f,u 0P[x] and f(a)u(a)0P[a]. Otzhe, we zvіlnilis vіd іrrationalnosti f(a)/h(a) .

Sounds like irrationality at the bannerman

Rich terms p(x) and g(x)=-x 2 +x+1 are mutually simple. Therefore, there are such rich terms j and y that

For vіdshukannya j і y zastosuemo Euclidean algorithm to polynomials p і 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 such a manner,

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

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

Zvіdki know

(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(2/5x2+1/5x+3/5)=1.

in such a manner,

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

Otzhe

.

2. Foldable extension of the field algebra.

2.1. Kіntseve expansion of the field.

Let P be the subfields of the field F. Then we can look at F as a vector space over P, so we can look at the vector space +F, +, (w l ½l 0P),

de w l - the operation of multiplying the elements of F by the scalar l0P.

Appointment. The expansion of the field F is called terminal, like F, as a vector space over P, it is possible to end expansion. Tsya rozmirnіst signified through.

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

This proposition blatantly blazes through Theorem 1.5.

Appointment. An extension F of a field P is called algebraic, since a skin element of F is algebraic over P.

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

Bringing. Let F be n-smooth over P. The theorem is obviously true, since n = 0. Assume that n>0. If n+1 elements of F are linearly fallow over P. Sokrema, a linearly fallow system of elements 1, a, ..., a n , then P such elements of 0 , 1, ..., c n are not all equal to zero , s 0 ×1+ 1 a +…+c n a n = 0.

The element a is also algebraic over P.

It is significant that there are extensions of the field algebra that are not terminal extensions.

2.2. Warehouse expansion of the field of algebra.

The extension F of the field P is called collapsible, as it is

growing lancet subfield L i of field F such that

P = L 0 - L 1 - ... L k = F і k>1.

Theorem 2.3. Let F - end extension of the field L і L - end extension of the field P. Then F - end extension of the field P i

=@[L:P].

Bringing. Come on

(1) a 1 ,…,a m - basis of the field L over P (like 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 through the basis:

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

The coefficient 1 k can be linearly expressed through the basis (1):

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

Substituting the score for the coefficients l k (3), it is acceptable

d = p a a b k .

In this way, the skin element of the field F can be represented as a linear combination of elements of the multiplier B, de

B = (a i b k ½ (1, ..., m), k 0 (l, ..., n)).

It is significant that the multiplier B adds up to nm elements.

We show that F is a basis over P. We need to show that the system of elements of the multiplier B is linearly independent. Come on

(5) åc ik a i b k = 0,

de c ik 0 P. Since the system (2) is linearly independent over L , then (5) follows equality

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

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

c 1 k = 0, ..., c mk = 0 (k = 1, ..., n),

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

Otzhe, inserted, scho = nm = ×. Also F є last extensions of the field P і maє misce formula (I).

Appointment. The extension F of the field P is called foldable algebraic, as it is the growing lance of the subfields of the field P

P \u003d L 0 - L 1 - ... L k \u003d F і k> 1 (1)

such that for i = 1,..., k fields L i є let's just expand the algebra of the field L i-1 . The number k is called dozhina lance (1).

Last 2.4. Warehouse extensions of the algebra F of the field P are terminal extensions of the field P.

The proof can be easily carried out by induction behind the lance (1) on the substantiation of Theorem 2.3.

Theorem 2.5. Let a 1 ,..., ak be algebraic over the field P of elements of the field F . The same field P(a 1 ,..., ak) is the last extension of the field P.

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

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

L 2 = P = (P) = L 1 = L 1 (a 2) etc.

in such a manner,

P = L 0 - L 1 - ... - L k = F

de L i = L i -1 (a i) for i = 1, ..., k, then the skin term of the Lanziuk (2) is a simple extension of the algebra of the forward term of the Lanziuk. Later, the field F is a foldable extension of the algebra of the field P. Again, by Corollary 2.4, the field F is a terminal extension of the field P .

Last 2.6. Warehouse extension of the field algebra є extension of the algebraic field.

2.3. Simplicity of warehouse expansion of field algebra.

Theorem 2.7. Let the number field F be a foldable extension of the field algebra P . Then F є we will simplify the extensions of the algebra of the field P.

Bringing. Let P - L - F, moreover, L = P (a), F = L (b) i, also, F = P (a, b).

Let f and g be minimal polynomials over P, which is valid for numbers a and b and deg f = m, deg g = n. The polynomials f і g cannot be superimposed over P і, therefore, it cannot be in the field E of complex numbers of multiple roots. Come on

a = a 1 ,..., a m - roots of the polynomial f C i

b = b 1 ,..., b n - root of the polynomial g C.

Let's look at the kіtsev bezlіch M:

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

Oskіlki P is a numerical multiplier (i, therefore, not limited), then P is the number c, vidminne in the elements of the multiplier M, c0P (M, cóM. Nehai

Todi vykonuyutsya spіvvіdnoshennia

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

True, in times of equality a + cb = a i + cb k bulo b

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

scho superchilo used the choice of the number c.

Let F 1 = P(g) and F 1 - a ring of polynomials in x. Let h = f(g - cx) be a polynomial from F 1 [x] (g, c0P(g) = F 1). It can be shown that x-b is the largest consonant of the polynomials h and g in the ring F 1 [x]. Scales g(b) = 0, then x-b divide g E[x]. Daly, due to (1)

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

To that x-b divide the polynomial h E[x]. In this order, x-b is a sleeper h and g in the ring E[x].

It is reported that g і h С there are no roots, vіdmіnkh vіd b. Let's just say that b k , k0(2 ,..., n) is its wild root. Then h(b k) = f(g - сb k) = 0. Then, there is such an index i0(1 ,..., m) ). Therefore, it is possible that x-b is the largest sleeper of g and h in E[x]. Oskіlki x - b - normalization polynomial, then the star is clear, scho x - b є the largest hot dilnik g and h y kіltsi F 1 [x]. Tom

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

Moreover, a = g - cb 0 F 1 . in such a manner,

F = P(a, b) Ì F 1 , F 1 ÌF.

2.4. Field algebraic numbers.

The class of subfields of the field of complex numbers is one of the most important - the field of algebraic numbers.

Appointment. An algebraic number is called a complex number, which is the root of a polynomial of positive degree with rational coefficients.

It is significant that the number of an algebra, be it a complex number, be algebraic over the field Q. Sokrema, be it a rational number, be algebraic.

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

Bringing. Let a and b be elements of A. For last 2.6, the field Q(a, b) is algebraic over Q. Therefore, the numbers a + b, -a, ab, 1 are algebraic, so that the multiples of A lie. , impersonal A is closed according to the head operations of the cycle E. Therefore, the algebra A is a subcycle of the cycle E - is a cycle.

In addition, since a is a non-zero element in A, a -1 0 Q (a, b) and that a -1 lie in A. Again, the algebra A is a field, subfields of the field E.

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

Show that the number a = algebraic.

Solution. Z a \u003d screaming a-.

Zvedomly insulting parts of the remaining equivalence in the third step:

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

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

Now the offending parts of jealousy are brought to another level:

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.

In this rank a є the root of a rich term

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

from rational coefficients. Ce means that a is an algebraic number.

2.5. Algebraic closure of the field of numbers of algebra.

Theorem 2.9. The number field of an algebra is algebraically closed.

Bringing. Let A [x] be a ring of polynomials in x over the field A of algebraic numbers. Come on

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

Be some polynomial of positive step A[x]. We need to prove that f can be rooted in A. If f0C[x] and the field E is algebraically closed, then f can be rooted in E so that it has such a complex number s, that f (c) = 0. Let L = Q (a 0 , ..., and n) and L(c) is a simple extension of the algebra of the field L beyond the help of c. Then Q - L - L (c) is a terminal extension of the algebra of the field L. By Theorem 2.2, L is a terminal extension of the field Q. By virtue of Theorem 2.3, L (c) is a terminal extension of the field Q. the field L (c) is an extension of the algebra of the field Q i, hence, c0A. Thus, if there is any polynomial in A[x] of positive step A can have a root, then the field A is algebraically closed.

3. Separable and inseparable extensions.

Come on D - field.

Surely, how can a non-decomposable D[x] polynomial be a mother of multiple roots?

In order for f(x) to be multiple roots, the rich terms f(x) and fN(x) are due to the mother's common double constant multiplier, which can be calculated already in D[x]. Even though the polynomial f(x) is indecomposable, then with any rich term of the lower degree f(x) it cannot be the mother of incomprehensible global multipliers, also, there can be equality f "(x) = 0.

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

So fN(x) = O, the skin coefficient is guilty of zero:

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

What is important is the characteristic zero of the star, that a n = 0 to all n ¹ 0. Also, an inconsistent polynomial can be the mother of multiple roots. At the time of the characteristics p_evenness na n \u003d 0 it is possible to have n ¹ 0, but it can also be equal

f(x) = a 0 + a p x p + a 2p x 2p +…

Back: if f(x) can look like this, then fN(x)=0.

With this vipadka we can write:

Tim himself brought the assertion: In the case of the characteristic zero, the rich term f (x) is not divisible in D [x], it can only be a simple root, in the case of the characteristic p, the polynomial f (x) (which is also the same as the constant) can be a multiple of the root, if it is possible to show it as a polynomial j vіd x p.

At times, it is possible that j(x) is a polynomial in its own way x p . Then f(x) is a polynomial like x p 2 . Let f(x) - rich term like xpe

ale є polynomial vіd x pe +1 . Understandably, the polynomial y(y) is indecomposable. Dali, y¢(y) ¹ 0, because otherwise y(y) would look like c(y p) i, then f(x) would look like c(x pe + 1), which would supersede the omission. Otzhe, y (y) can only be a simple root.

Let's expand the polynomial y in order to expand the main field on linear factors: m

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

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

Let a i be the root of the polynomial x pe - bi. Then x i pe \u003d b i,

x pe - bi = x pe - a i pe = (x-a i) pe .

Also, a i є r e -multiple root of the polynomial x pe - b i

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

The mustache of the root of the polynomial f(x) can, in this way, have the same multiplicity of p e.

The step m of the polynomial y is called the reduction step 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.

de m more expensive number of different roots of the polynomial f(x).

If q is the root of a polynomial that is not decomposable in the ring D[x], which can only be simple roots, then q is called a separable element over D or an element of the first kind over D 1). With this, an inextricable rich term, all the roots of which are separable, are called separable. Otherwise, the algebraic element q and the indecomposable rich term f(x) are called inseparable or an element (like a rich term) of a different kind. Now, an extension of the algebra S, all elements of which are separable over D, are called separable over D, and any other extension of the algebra is called inseparable.

In times of characteristic zero, it is said that the skin is not an indecomposable rich term (and therefore the skin extension of algebra) is separable. We would like to know that most of the most important and most important extensions of fields are separable and that we know the quality of the class of fields, so that inseparable extensions (the so-called “finished the field”) are not possible. Z tsієї causa all pov'yazane specially with inseparable extensions typed in a different font.

Let us now look at the extension of the algebra S = D (q). If the steps n are equal f(x) = 0, which signifies a larger, more advanced step (S: D), the reduction of the steps m is equal to the number of isomorphisms of the field S in the advancing sense: we can only look at these isomorphisms [email protected]", for any elements of the subfield D are filled with non-violent i, then, S is transferred to the equivalent field S" (isomorphism of the field S over the field D) and for any field-image S "to lie together with the field S in the middle of the field W. tsikh umovah maє mistse theorem:

With an appropriate choice of the field W, the extension S=D(q) can have exactly m isomorphisms over D, and for any choice of the field W, the field S cannot have more than m such isomorphisms.

Bringing. The skin isomorphism over D is responsible for translating the element q into its associations with the element q" from W. Choose W so that f(x) expands over W into linear multipliers; then it appears that the element q can have exactly m occurrences elements q,q If so, as bi, the field W was not chosen, the element q is not matima in any more than m cases. It is respectable now that the skin isomorphism D(q)@D(q") over D is fully dependent on the given identity of q® q". Obviously, if q goes over to q "and all elements from D are left on the spot, then the element

3a k q k (yak 0D)

guilty go to

and cym stands for isomorphism.

Sokrema, since q is a separable element, then m = n і, hence, the number of isomorphisms over the main field is more evenly extended.

If so, if the field is fixed, which can cover all the fields that are looked at, in which all roots of the skin equalization f (x) = 0 can be located (like, for example, in the field of complex numbers), then in the capacity of W you can take the field i once and for all To this, add the addition of “in the middle of the deaky W” in all statements about isomorphism. So start repairing theoretically numerical fields. We would like to remind you that for abstract fields you can also use the W field.

The cited theorem is the following statement:

How to expand S to exit from D to subsequent arrivals m

algebraic elements a 1 , ..., a m , moreover, skin behind a i , є root

non-expandable over D(a 1 , ..., a i-1) is equal to the reduced stage n" i , then

expansion of S can be exactly ?n i ¢ isomorphisms over D i in the same way

no extensions greater number such isomorphisms of the field S.

Bringing. For m = 1, the theorem has been further developed. Assume її valid for the extension S 1 = D(a 1 , ..., a m-1):

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

Let S 1 ®S 1 be one of the Õ n i ¢ isomorphisms. It is argued that in the reverse order of the reversed field W vin can be continued to isomorphism S = S 1 (am) @ S = S (am) no more than n_zh n m ways.

The element a m satisfies the equation f 1 (x) = 0 over S 1 with n¢ m different roots. After the additional isomorphism S 1 ® S 1, the rich term f 1 (x) can be translated into another rich term f 1 (x). Ale todі f 1 (x) in a broadly expanded way, but n m different roots and no more. Let a m - one of these roots. Looking at the choice of the element a m, the isomorphism S 1 @S 1 is three to the isomorphism S (a m) @ S (am) for a m ®a m in one and only one way: effectively, the continuation is given by the formula

åc k a m ​​k ®å c k a m ​​k

The selections for the element a m can be defined in n" m ways, using n" m to continue such a sort for the reverse isomorphism å 1 ®å 1

Oskіlki have their own line, and this isomorphism can be converted

Х n" i ways,

then everything is true (that field W, in which all the roots of all equals, which are looked at) are located)

Õ n" i ×n" m = Õ n" i

isomorphisms of the extension of S over the field D, which was necessary to bring.

If n i is a real (unreduced) step of the element a i over D (a 1 ,...,a i-1), then n i more steps of the extension D (a 1 , ... , a i) of the field D(a 1 , .. ., a i-1);

otzhe, steps (S: D) more

How to match the number with the number of isomorphisms

The number of isomorphisms of the extension S = D(a 1 , ... , a m) over D (for any given extension W) an additional step (S: D) even and only once, if the skin element a i is separable over the field D(a 1 , . .. , a i-1). If we want one element a i to be inseparable over a separate field, then the number of isomorphisms is less than the degree of expansion.

From the point of the theorem, a few important remarks will immediately appear. For us, the theorem states that the power of the skin element a i is separable over the front field, and the power of the extension S itself is independent of the choice of elements that generate a i . Since an additional element of the field can be taken as the first generation, element b appears to be separable, as all a i are so. Father:

Elements a i , ... ,a n i are added sequentially to the field D, the skin element a i appears separable over the field, we remove the adjoining front elements a 1, a 2 ,...,a i-1 expansion

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

separable over D.

Zokrema, suma, retail, tvir that privately separatized elements are separable.

Further, since 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 the final number of coefficients a 1 , ... , a m з S i, again, is separable over D (a 1, ..., a m). Tim itself separable extension

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

Nareshti, may the same place be given: the number of isomorphisms of a terminal separable extension S over a field D to a higher degree of extension (S: D).

4. Unlimited expansion of irrigation.

The skin field emerges from its simple sub-field for the help of the final chi of the inexhaustible expansion. In this division, innumerable expansions of fields are seen, first of all algebraic, and then transcendental.

4.1. Algebraically closed fields

Among the expansion of the algebra of a given field, an important role is played, especially, by the maximum expansion of the algebra, so as not to allow a further expansion of the algebra. The reason for such extensions will be brought to this paragraph.

In order for the field W to be the maximum extension of the algebra, it is necessary to advance the mind: the skin polynomial of the circle W[x] can be decomposed into linear multipliers. Tsya mind is sufficient. Indeed, since a skin polynomial in W[x] is decomposed into linear multipliers, then all simple polynomials in W[x] are linear and skin elements of any extension of the algebra W" of the field W appear to be the root of any linear rich term x - a in W[x] , i.e. it works with the actual element a of the field W.

To that damo is the same fate:

The field W is called a closure of the algebra, because any polynomial in W [x] can be decomposed into linear factors.

Equally important is the following: the field W is algebraically closed, so that the polynomial in W[x] can be a distinct polynomial in W[x] with a single root, that is, with a single linear multiplier in W[x].

Indeed, as such a clever vikonan and quite a lot of takings, the polynomial f (x) is decomposed into factors that are not decomposed, then all the stench is to blame but linear.

The "Basic Theorem of Algebra" states that the field of complex numbers is algebraically closed. An approaching butt of an algebraically closed field can be the field of all complex algebraic numbers, so that impersonal complex numbers, like being satisfied with any kind of equality with rational coefficients. The complex root is equal to the coefficients of algebra є and really algebraic not only over the field of algebraic numbers, but also over the field rational numbers, i.e., themselves are algebraic numbers.

Here we will show how to induce a closed algebraic extension of a sufficiently given field P and in a purely algebraic way. Steinitz to lie down like that

Main theorem. For the skin field P, a closed algebraic extension of the algebra W. Exactly up to equivalence, the extension is uniquely defined: whether there are two algebraically closed algebraic extensions W, W "of the field P are equivalent.

The proof of these theorems is due to the surplus of the lem:

Lemma 1. Let W, be an extension of the field algebra P. Sufficient mind in order for W to be a closure of the algebra, є expansion into linear factors of any polynomial in P[x] in the ring W[x].

Bringing. Let f(x) be an additional polynomial from W[x]. If vin is not decomposed into linear multipliers, then one can take a th root a i to come to the upper superfield W. The element a is algebraic over W, and W is an extension of the algebra of the field P; the root of the next polynomial g(x) in P[x]

Lemma 2. If the field P is holistically ordered, then the ring of polynomials P[x] can be holistically ordered and up to the extent that this ordered field P will be triple.

Bringing. Significantly change the order between the polynomials f(x) in P[x] as follows: let f(x)

1) step f(x) is a smaller type of step g(x);

2) step f(x) more step g(x) and more n, then.

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

i for the next index k:

and i = b i for i

a k

If so, for the polynomial 0, blame is given: it is assigned a step 0. It is obvious that such a way to come out in order is, for the sense of which P [x] is completely ordered. It will be shown as follows: in the skin non-empty plural of rich segments, there is a non-empty sub-multiple of rich segments of the smallest degree; don’t hai such a good number. In each submultiple, there is a non-empty submultiple of rich-terms, the coefficient is 0, which is the first in the sense of the main order of the middle of the large parts of the rich-terms, which are looked at; at the appointed submultiple є have its own line submultiplier of rich terms with the first a 1 and so on. minimnosti, which are consecutively victorious, at choice); this polynomial is the first element of the given multiplier.

Lemma 3. If the field P is ordered as a whole, the rich term f(x) of the stage n і n symbols a 1 ..., a n then the field P (a 1 ,..., a n), in which f(x) will be expanded on linear multipliers

Õ(x-a i), will be a single rank and a whole

order. Field P in sensi tsiy є vіdrіzkom.

Bringing. We add the root a 1 ..., a n successively, after which P = P 0 successively win the fields Р 1 , ..., Р n . Let's assume that R i-1 = P(a 1 ..., a i-1) - the field has already been induced and that P is a contract with R i-1; then R i will be so.

Before the problem 2, the ring of polynomials Р i-1 [x] is ordered in a whole. The polynomial f is decomposed at each kіltsi into inextricable factors, the middle of which is the first place x - a 1 ,..., x - a i-1 ; among the other plurals, let f i (x) be the first in the sense of the clear order. Together with the symbol a i, which designates the root of the rich term f i (x), we signify the field P i = P i -1 as the totality of the sums

de h is the step of the rich term f i (x). If f i (x) is linear, then, of course, we respect P i = P i -1; the character a i is not needed. Encourage the field as a whole to be ordered for additional offensive intelligence: the skin element of the field

perhaps a rich member

And the elements of the field are ordered in the same way, as the ordering of their rich terms.

Obviously, the same Р i-1 is in relation to Р i, and to that і P - in relation to Р i.

Tim the fields P 1 ,..., P n themselves are motivated by a whole ordering. The field Р n is uniquely searchable by the first field P(a 1 ,..., a n).

Lemma4

Bringing. For any two elements a, b, combine two fields S a , S b , so as to replace a, b and from any one before the other. In the hoarse field, the elements a + b and a × b are assigned to the elements in the skin field, so that a and b can be avenged, since two such fields are one ahead of the other and yogo subfield. For example, to bring the law of associativity

ab g = a bg,

we know the middle fields S a , Sb, S g those that cover two other fields (the largest); in which field there is a, b and g i in the new law of associativity vikonano. In the same way, the reshta rules for calculating the elements of the association are revised.

The proof of the main theorem is divided into parts: the subfield W and the proof of unity.

Pobudov fields W. Lemma 1 prove that for a seemingly algebraically closed extension W of the field P it is enough to induce such an extension of the algebra of the field P, so that the polynomial in P[x] can be expanded over these extensions into linear multipliers.

1. The field P f є ob'ednannyam field P і all fields S g for g

2. The field P f is ordered in such a way that P and all fields S g with g

3. The field S f comes from R f to the given roots of the rich term f after the additional symbols a 1 ,..., a n is valid up to lemi 3.

It is necessary to state that in this way the whole ordering of the fields Р f , S f can be explicitly assigned by the whole ordering field, as well as all the forward ones Р g , S g are already assigned more often.

Yakshcho vikonano 3, then nasampered P f - vіdrіzok S f . Z ogo i vimogi 2 we see that the field P i skin field S g (g

Р - vіdrіzok S h at h

S g - double S h at g

Sounds like P i fields S h (h b, yak can be saved in Pf. The same order is one and the same in all fields P abo S g yak yak yak a, so ib, to that all ts field є v_drіzkami one of one. Otzhe, setting up to order is appointed. Those who are completely ordered impersonal, obviously, so as the skin is not empty impersonal x in Р f to avenge at least one element of the work of the deyakogo field S g, and that is the first element of x x Ç Work x Ç S g. This element is one hour є i the first element x.

Looking at your mind 3, the polynomial f(x) is again decomposed into linear factors in the field S f . Further, after the help of transfinite induction, it is shown that S f is algebraic over P. Indeed, it is assumed that all fields S g (g

Now we store the pool W of all fields Sf; zgіdno z lemoy 4 won є field. The entire field is algebraically over P and all rich terms f are expanded over it (small skin polynomials f are already expanded over S f). Also, the field W is algebraically closed (Lema 1).

The unity of the field W. Let W and W" be two fields that are algebraic and closed algebraic extensions of the field P. Let us bring the equivalence of these fields. is also considered by one of these arguments) submultiple ¢ in W "and some isomorphism

P(Â) @ P(¢).

The rest of May will be satisfied with the upcoming recurring spiving.

1. The isomorphism P(Â) @ P(¢) is due to the depletion of the skin element of the field P on the field.

2. The isomorphism P(Â) @ P(¢) with ÁÌ Â can be an extension of the isomorphism P(Â) @ P(Á").

3. If  is the remaining element a, so that  = ÁÈ(a), and if a is the root of the rich term f(x) that cannot be decomposed in P (Á), then the element a" is to blame for the first root of the genus P(Á) @ P(I"), a polynomial f¢(x) in a well-ordered field W".

It is necessary to show that the isomorphism P(Â) @ P(¢) is effectively assigned in the same way, even though the wines are already assignments for all the forward edges of ÁÌ Â. Here it is necessary to distinguish two points.

First drop. Impersonal  can't have the rest of the element. The same leather element should lie on the singing front breech Á; to that  є to the combined waterings of Á, to that P(Â) - to the cumulative fields P(Á) for ÁÌ Â. If the skin elements from the isomorphisms P(Á) @P(Á") proceed from the previous ones, then the skin element a with all these isomorphisms is given only one element a". Therefore, there is one and more than one inflection P(Â) → P(¢), which continues all forward isomorphisms P(Á) → P(Á"), and the inflection itself a®a". It is obvious that it is an isomorphism and a combination of 1 and 2.

Another drop. Anonymous maє remaining element a; also,  = ÁÈ(a). Finally, the element a" associated with the element a is uniquely assigned. Since a" over the field P(I") (in the sense of the analyzed isomorphism) satisfies "the same" inconsistently equal that i a over P(I), then the isomorphism P(I) → P(I") (in the case, if I is empty, then the same isomorphism P®P) goes up to the isomorphism P(I, a) ®P(I", a¢), when a passes at a". The dermal isomorphism was unambiguously identified by the suggestion of the skin, so the rational cutaneous function j(a) with the coefficients of the general language passes to the function j "(a") with the equivalent coefficients of the Á". ) ® P(¢) obviously matches 1 and 2.

Thus, the substitution of the isomorphism P(Â)→P(¢) is completed. Significantly through W" the generalization of all fields P(В¢); then there is an isomorphism P(W)®W" or W®W", which adds an element of the field P to the space of the skin. Since the field W is algebraically closed, so can Buti і W ", and to that W" is matched with the required field W¢.

The meaning of an algebraically closed extension of a given field is the same in that, up to the point of equivalence, it is possible to overcome the possible extensions of the algebraic field. More precisely:

If W is an algebraically closed extension of the algebra of the field P and S is a fairly algebraic extension of the field P, then in the middle of W there is a general extension of S 0 , which is equivalent to an extension of S.

Bringing. We can extend S to a certain closed algebraic extension W". It will be algebraic and over P, and therefore equivalent to an extension W. Under any isomorphism, in order to translate W" into W, taking the unbreakable skin element of P, the field S passes into a deak equivalent to yoma subfield S 0W.

4.2. Forgive the transcendent expansion.

The skin is simply a transcendental extension of the field D, apparently equivalent to the field of private D(x) of the ring of polynomials D[x]. To that mi vivchimo tse private field

The elements of the field W are rational functions

Theorem. The transcendental element h of step n is transcendental over D і the field D(x) is an extension of the algebra of the field D(h) of step n.

Bringing. The submission h = f(x)/g(x) is not short-lived. Same element x satisfied

g(x)×h - f(x)=0

with coefficients D(h). Numbers of coefficients cannot be equal to zero. Indeed, if all stinks were equal to zero and ak letter bi in the same world x be a non-zero coefficient of the polynomial g (x), and b k - a non-zero coefficient of the polynomial f (x), then it would not be enough for the mother to be equal

stars h = b k / ak = const, which is a superstition. Again, the element x is algebraic over D(h).

If the element h is albeit algebraic over D, then x is albeit bi algebraic over D, which, however, is not so. Again, the element h is transcendental over D.

The element x is the root of the rich term of step n

in the ring D(h)(z). This polynomial is indecomposable in D(h)[z], the shards are also vin bouv bi can be decomposed n in the kіlci D, і, the shards of vin are linear in h, one of the multiples of maw bi is not possible to deposit h, or less z. But such a multiplier cannot be, because g(z) and f(z) are mutually simple.

Also, the element x is a step of the algebra n over the field D(h). The stars are solid, so (D(x) : D(h)) = n

For a meaner, it is significant that a rich member

there are no multiples that can lie only near z (to lie near D[z]). Tse solidification is overridden, if h is replaced by its values ​​f (x) / g (x) and multiplied by the banner g (x), we ourselves are a polynomial.

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

kіltsya D there are no multipliers, fall only in vіd z.

From the theorems brought above, there are three remarks.

1. The function step h - f(х)/g(х) should be deposited only in the fields D(h) and D(x), and not in the choice of another element that generates x.

2. Rivnist D(h) = D(x) is less than the same, if h is less than 1, then it is a shot-linear function. Tse means: the parent element of the field, the crim of the element x, can be a fractional-linear function like x and only such a function.

3. Any automorphism of the field D(x), which leaves an element of the field D on the canvas, is guilty of translating the element x into any element of the field. Back, if x is translated into a parent element x = (ax + b) / (cx + d) and skin function j (x) - y function j (x), then an automorphism comes out, when all elements D are left on the target. Otzhe,

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

x = (ax+b)/(cx+d), ad – bc ¹ 0.

Important for some geometrical achievements

Lurot's theorem. The skin-intermediate field S, for which DÌSID(x) is simple transcendental extensions: S = D(q).

Bringing. The element x is guilty of being algebraic over S, because if h - if any element of S does not belong to the field D, then, as was shown, the element x is algebraic over D (h) and even more algebraic over S. S [z] rich term with senior coefficient 1 and root x may look

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

Z'yasuєmo Budov's rich member.

Elements a i є rational functions x. For the help of multiplying by a sleeping banner of їх, you can use it with many rational functions and, moreover, take a rich term like x іz instead of 1:

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

The steps of the polynomial are significant in terms of m, and in z - in terms of n.

The coefficients a i \u003d b i / b 0 z (1) cannot be independent in x, so that x would otherwise appear as an algebraic element over D; so one of them, say,

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

is actually guilty of deposition vіd x; Let's write down yoga in a short look:

Steps of polynomials g(x) and h(x) do not exceed m. Polynomial

g(z) - qh(z) = g(z) – (g(x)/h(x))h(z)

(which is not the same zero) if root z = x, then vin is divisible by f 0 (z) in the ring S[z]. If you want to go from thir rational in terms of x rich terms to tsilih in x rich terms with zmist 1, then you should save your divisibility, and we will take it

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

The left part of this equanimity has steps along x, but it doesn’t move t. Ale on the right is already a rich member of f stupіn t; otzhe, the steps of the left part are exactly exactly old and q(x, z) do not lie in x. However, it is impossible to deposit less than a z multiplier to divide the left part (div. more); to that 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 Budov polynomial f(x, z) is described completely. The steps of the polynomial f(x, z) in x are more advanced (with the symmetry of symmetry), and the steps in z are more advanced, so m = n. m, later, i function q is due to the mother of steps m x.

Tim ourselves, shards from one side are set equal

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

and for the rest - jealousy

those shards to avenge D(q),

Visnovok.

The robots looked like this, see the expansion of the numeric field P:

A simple extension of the field algebra.

Warehouse expansion of the field of algebra.

Separable and inseparable extensions.

Unlimited expansion of watering.

Analyzing the work, you can create deaky visnovki.

Z looked at the first two parts of the expansion, such as:

simple expansion of algebra;

end expansion;

warehouse expansion of algebra.

Next, if you see the extensions zbіgayutsya і, zokrema, are drawn by simple algebraic extensions of the field P.

List of references

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

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

3. E.F. Shmigiryov, S.V. Ignatovich. Theory of rich terms. - Mosir 2002.

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