The ring is algebraic. Rings: definition, properties, examples

Short description

Definition. A ring is an algebra K = ‹K, +, -, ·, 1› of type (2, 1, 2, 0) whose main operations satisfy the following conditions:

Attached files: 1 file

Ring. Definition. Examples. The simplest properties of rings. Homomorphism and isomorphism of rings.

Definition. A ring is an algebra K = ‹K, +, -, ·, 1› of type (2, 1, 2, 0) whose main operations satisfy the following conditions:

1. the algebra ‹K, +, -› is an Abelian group;
2. the algebra ‹K, ·, 1› is a monoid;
3. multiplication is distributive with respect to addition, that is, for any elements a, b, c from K

(a + b) c = a c + b c, c (a + b) = c a + c b.

The base set K of the ring K is also denoted by |K|. The elements of the set K are called the elements of the ring K.

Def. The group ‹K, +, -› is called the additive group of the ring K. The zero of this group, that is, the neutral element with respect to addition, is called the zero of the ring and is denoted by 0 or 0 K.

Def. The monoid ‹K, ·, 1› is called the multiplicative monoid of the ring K. The element 1, also denoted by 1 K, which is neutral under multiplication, is called the identity of the ring K.

A ring K is called commutative if a · b = b · a for any elements a, b of the ring. A ring K is called null if |K| = (0 K).

Def. A ring K is called an integrity domain if it is commutative, 0 K ≠ 1 K, and for any a, b Î K, a b = 0 implies a = 0 or b = 0.

Def. Elements a and b of the ring K are called zero divisors if a ≠ 0, b ≠ 0, or ba = 0. (Any integrity domain has no zero divisors.)

Example. Let K be the set of all real functions defined on the set R of real numbers. sum f + g, product f g, function

f(-1) and the identity function 1 are defined: (f + g) (x) = f (x) + g(x);

(f g)(x) = f(x) g(x); (–f) (x) =–f (x); 1(x) = 1. Direct verification shows that the algebra ‹K, +, -, ·, 1› is a commutative ring.

The simplest properties. Let K be a ring. Since the algebra ‹K, +, -› is an Abelian group, then for any elements a, b, from K, the equation b + x = a has a unique solution a + (-b), which is also denoted by a – b.

1. if a + b = a, then b = 0;
2. if a + b = 0, then b = -a;
3. – (-a) = a;
4. 0 a = a 0 = a;
5. (-a)b = a(-b) = -(ab);
6. (-a)(-b) = a b;
7. (a – b)c = ac – bc and c(a – b) = ca – cb.

Let K = ‹K, +, -, ◦, 1› and K` = ‹K`, +, -, ·, 1`› be rings. It is said that the mapping h of the set K to K` preserves the main operations of the ring K if the following conditions are met:

1. h(a+b)=h(a)+h(b) for any a, b from the ring K;
2. h(-a)=-h(a) for any a from K;
3. h(a b) = h(a)◦h(b) for any a, b from K;
4. h(1) = 1`.

Def. A homomorphism of a ring K into (on) a ring K` is a mapping of a set K into (on) K` that preserves all principal operations of the ring K. A homomorphism of a ring K onto K` is called an epimorphism.

Def. A homomorphism h of a ring K onto a ring K` is called an isomorphism if h is an injective map of the set K onto K`. Rings K and K` are said to be isomorphic if there exists an isomorphism of the ring K onto the ring K`.

Annotation: In this lecture, the concepts of rings are considered. The main definitions and properties of ring elements are given, associative rings are considered. A number of characteristic problems are considered, the main theorems are proved, and problems for independent consideration are given.

Rings

The set R with two binary operations (addition + and multiplication) is called associative ring with unit, If:

If the operation of multiplication is commutative, then the ring is called commutative ring. Commutative rings are one of the main objects of study in commutative algebra and algebraic geometry.

Notes 1.10.1.

Examples 1.10.2 (examples of associative rings).

We have already seen that the group of residues (Z n ,+)=(C 0 ,C 1 ,...,C n-1 ), C k =k+nZ, modulo n with the operation of addition , is a commutative group (see example 1.9.4, 2)).

We define the operation of multiplication by setting . Let's check the correctness of this operation. If C k =C k" , C l =C l" , then k"=k+nu , l"=l+nv , and therefore C k"l" =C kl .

Because (C k C l)C m =C (kl)m =C k(lm) =C k (C l C m), C k C l =C kl =C lk =C l C k , C 1 C k =C k =C k C 1 , (C k +C l)C m =C (k+l)m =C km+lm =C k C m +C l C m, then is an associative commutative ring with identity C 1 residue ring modulo n ).

Ring properties (R,+,.)

Lemma 1.10.3 (Newton binomial). Let R be a ring with 1 , , . Then:

Proof.

Definition 1.10.4. A subset S of a ring R is called subring, If:

a) S is a subgroup with respect to addition in the group (R,+) ;

b) for we have ;

c) for a ring R with 1 it is assumed that .

Examples 1.10.5 (examples of subrings).

Task 1.10.6. Describe all subrings in the residue ring Z n modulo n .

Remark 1.10.7. In the ring Z 10 the elements divisible by 5 form a ring with 1 which is not a subring in Z 10 (these rings have different identity elements).

Definition 1.10.8. If R is a ring and , , ab=0 , then element a is called a left zero divisor in R , element b is called a right zero divisor in R .

Remark 1.10.9. In commutative rings, of course, there is no difference between left and right zero divisors.

Example 1.10.10. Z , Q , R have no zero divisors.

Example 1.10.11. The ring of continuous functions C has zero divisors. Indeed, if

then , , fg=0 .

Example 1.10.12. If n=kl , 1

Lemma 1.10.13. If there are no (left) zero divisors in the ring R, then from ab=ac , where , , implies that b=c (i.e., the ability to cancel by a non-zero element on the left if there are no left zero divisors; and on the right if there are no right zero divisors).

Proof. If ab=ac , then a(b-c)=0 . Since a is not a left zero divisor, then b-c=0 , i.e. b=c .

Definition 1.10.14. The element is called nilpotent, if x n =0 for some . The smallest such natural number n is called degree of nilpotency of an element .

It is clear that a nilpotent element is a zero divisor (if n>1, then , ). The converse is not true (there are no nilpotent elements in Z 6, but 2 , 3 , 4 are non-zero zero divisors).

Exercise 1.10.15. The ring Z n contains nilpotent elements if and only if n is divisible by m 2 , where , .

Definition 1.10.16. An element x of the ring R is called idempotent, if x 2 \u003d x. It is clear that 0 2 =0 , 1 2 =1 . If x 2 =x and , , then x(x-1)=x 2 -x=0 , and therefore non-trivial idempotents are zero divisors.

We denote by U(R) the set of invertible elements of the associative ring R , i.e. those for which there is an inverse element s=r -1 (i.e. rr -1 =1=r -1 r ).

Non-empty set TO, on which two binary operations are set - addition (+) and multiplication ( ), satisfying the conditions:

1) regarding the operation of addition TO- commutative group;

2) regarding the operation of multiplication TO- semigroup;

3) the operations of addition and multiplication are related by the law of distributivity, i.e. . (a+b)c=ac+bc, c(a+b)=ca+cb for all a, b, c K, is called ring (K,+, ).

Structure (TO,+) is called additive group rings. If the operation of multiplication is commutative, i.e. ab=ba. for all A, b, then the ring is called commutative.

If with respect to the operation of multiplication there is an identity element, which in the ring is usually denoted by the unit 1,. then they say that TO There is unit ring.

A subset L of a ring is called subring, If L is a subgroup of the additive group of the ring, and L is closed under the operation of multiplication, i.e., for all a, b L is running a+b L And ab L.

The intersection of subrings will be a subring. Then, as in the case of groups, by a subring, generated many S K, is called the intersection of all subrings TO, containing S.

1. The set of integers with respect to the operations of multiplication and addition is a (Z, +, )-commutative ring. Sets nZ whole numbers divisible by P, will be a subring without unity for n>1.

Similarly, the set of rational and real numbers are commutative rings with identity.

2. The set of square matrices of order P with respect to the operations of addition and multiplication of matrices, there is a ring with identity E- identity matrix. At n>1 it is non-commutative.

3. Let K-arbitrary commutative ring. Consider all possible polynomials

with variable X and coefficients a 0, a 1, a 2,..., and n, from TO. With respect to the algebraic operations of addition and multiplication of polynomials, this is a commutative ring. It's called polynomial ring K from a variable X over the ring TO(for example, over the ring of integer, rational, real numbers). The ring of polynomials is defined similarly K from T variables as a polynomial ring in one variable x t over the ring K.

4. Let X- arbitrary set, TO- arbitrary ring. Consider the set of all functions f: X K, defined on the set X with values ​​in TO We define the sum and product of functions, as usual, by the equalities

(f+g)(x)=f(x)+g(x); (fg)(x)=f(x)g(x),

where + and - operations in the ring TO.

It is easy to check that all the conditions included in the definition of a ring are satisfied, and the constructed ring will be commutative if the original ring is commutative K. It's called function ring on the set X with values ​​in the ring TO.

Many properties of rings are reformulated corresponding properties of groups and semigroups, for example: a m a n = a m + n, (a t) n = a tp for all m, n and all a.

Other specific properties of rings model properties of numbers:

1) for everyone a a 0=0 a=0;

2) .(-a)b=a(-b)=-(ab);

3) - a=(-1)a.

Really:

2) 0=a(similar to (-a)b=-(ab));

3) using the second property, we have- a= (-a)1 =a(-1) = (-1)a.

Field

In the rings of integers, rational and real numbers from the fact that the product ab=0, it follows that either A=0, or b=0. But in the ring of square matrices of order n>1 this property is no longer satisfied, because, for example, = .

If in the ring K ab=0 at A 0, b, That A is called left b- right zero divisor. If in TO there are no zero divisors (except the element 0, which is a trivial zero divisor), then K called a ring no zero divisors.

1. In the ring function f: R R on the set of real numbers R consider the functions f 1 (x)=|x|+x; f 2 (x) =|x|-x. For them f 1 (x)=0 at x And f2(x)=0 at x, and therefore the product f 1 (x) f 2 (x) is a null function though f 1 (x) And f2(x). Therefore, there are zero divisors in this ring.

2. Consider the set of pairs of integers ( a, b) in which the operations of addition and multiplication are given:

(a 1 , b 1)+(a 2 , b 2)=(a 1 +a 2 , b 1 +b 2);

(a 1 , b 1)(a 2 , b 2)= (a 1 a 2 , b 1 b 2).

This set forms a commutative ring with unity (1,1) and zero divisors, since (1,0)(0,1)=(0,0).

If there are no zero divisors in the ring, then the reduction law is satisfied in it, i.e. ab=ac, a=c. Really, ab-ac=0 a(b-c)=0 (b-c)=0 b=c.

Let TO- a ring, with a unit. Element A called reversible if there is such an element a -1 , for which aa -1 =a -1 a=1.

The reversible element cannot be a zero divisor, since. If ab=0 , That a -1 (ab) =0 (a -1 a)b=0 1b=0 b=0(similar ba=0 ).

Theorem. All invertible elements of the ring K with identity form a group with respect to multiplication.

Indeed, the multiplication TO associatively, the unit is contained in the set of invertible elements and the product does not infer from the set of invertible elements, since if A And b reversible, then
(ab) -1 = b -1 a -1 .

An important algebraic structure is formed by the commutative rings TO, in which each nonzero element is invertible, i.e., with respect to the operation of multiplication, the set K\(0) forms a group. Three operations are defined in such rings: addition, multiplication, and division.

commutative ring R with unity 1 0, in which every nonzero element is invertible, is called field.

With respect to multiplication, all non-zero elements of the field form a group called multiplicative group fields.

Work ab -1 is written as a fraction and makes sense only when b 0. The element is the only solution to the equation bx=a. Actions with fractions obey the rules familiar to us:

Let us prove, for example, the second of them. Let x= And y=- solving equations bx=a, dy=c. From these equations it follows dbx=da, bdy=bc bd(x+y)=da+bc t= is the only solution to the equation bdt=da+bc.

1. The ring of integers does not form a field. The field is the set of rational numbers and the set of real numbers.

8.7. Assignments for independent work in chapter 8

8.1. Determine whether the operation of finding the scalar product of vectors in an n-dimensional Euclidean space is commutative and associative. Justify your answer.

8.2. Determine whether the set of square matrices of order n with respect to the operation of matrix multiplication is a group or a monoid.

8.3. Indicate which of the following sets form a group with respect to the operation of multiplication:

a) a set of integers;

b) the set of rational numbers;

c) the set of real numbers other than zero.

8.4. Determine which of the following structures forms a set of square matrices of order n with determinant equal to one: with respect to the usual operations of addition and multiplication of matrices:

a) a group

b) ring;

8.5. Indicate what structure the set of integers forms with respect to the operation of multiplication and addition:

a) non-commutative ring;

b) a commutative ring;

8.6. Which of the following structures forms a set of matrices of the form with real a and b with respect to the usual operations of matrix addition and multiplication:

a) a ring

8.7. Which number must be excluded from the set of real numbers so that the remaining numbers form a group with respect to the usual multiplication operation:

8.8. Find out which of the following structures forms a set consisting of two elements a and e, with a binary operation defined as follows:

ee=e, ea=a, ae=a, aa=e.

a) a group

b) an abelian group.

8.9. Are even numbers a ring with respect to the usual operations of addition and multiplication? Justify your answer.

8.10. Is a ring a set of numbers of the form a+b , where a and b are any rational numbers, with respect to addition and multiplication? Justify the answer.

Definition 2.5. ring called algebra

R = (R, +, ⋅, 0 , 1 ),

whose signature consists of two binary and two null operations, and for any a, b, c ∈ R the equalities hold:

1. a+(b+c) = (a+b)+c;
2. a+b = b+a;
3. a + 0 = a;
4. for every a ∈ R there is an element a" such that a+a" = 0
5. a-(b-c) = (a-b)-c;
6. a ⋅ 1 = 1 ⋅ a = a;
7. a⋅(b + c) = a⋅b + a⋅c, (b + c) ⋅ a = b⋅ a + c⋅a.

The + operation is called ring addition , operation ring multiplication , element 0 - ring zero , element 1 - ring unit .

The equalities 1-7 specified in the definition are called ring axioms . Let us consider these equalities from the point of view of the concept groups And monoid.

Ring axioms 1-4 mean that the algebra (R, +, 0 ), whose signature consists only of the operations of addition of the ring + and zero of the ring 0 , is abelian group. This group is called additive group of the ring R and one also says that, by addition, a ring is a commutative (Abelian) group.

The ring axioms 5 and 6 show that the algebra (R, ⋅, 1), whose signature includes only the multiplication of the ring ⋅ and the identity of the ring 1, is a monoid. This monoid is called multiplicative monoid of the ring R and they say that by multiplication a ring is a monoid.

The connection between the addition of a ring and the multiplication of a ring is established by Axiom 7, according to which the operation of multiplication is distributive with respect to the operation of addition.

Considering the above, we note that a ring is an algebra with two binary and two nullary operations R =(R, +, ⋅, 0 , 1 ) such that:

1. algebra (R, +, 0 ) is a commutative group;
2. algebra (R, ⋅, 1 ) - monoid;
3. the operation ⋅ (multiplication of a ring) is distributive with respect to the operation + (addition of a ring).

Remark 2.2. In the literature there is a different composition of the ring axioms related to multiplication. Thus, axiom 6 may be absent (there are no 1 ) and axiom 5 (multiplication is not associative). In this case, one singles out associative ring-rings (the requirement of associativity of multiplication is added to the ring axioms) and rings with identity. In the latter case, the requirements for the associativity of multiplication and the existence of a unit are added.

Definition 2.6. The ring is called commutative if its multiplication operation is commutative.

Example 2.12. A. The algebra (ℤ, +, ⋅, 0, 1) is a commutative ring. Note that the algebra (ℕ 0 , +, ⋅, 0, 1) will not be a ring, since (ℕ 0 , +) is a commutative monoid, but not a group.

b. Consider the algebra ℤ k = ((0,1,..., k - 1), ⊕ k , ⨀ k , 0,1) (k>1) with the operation ⊕ k of addition modulo l and ⨀ k (multiplication modulo l). The latter is similar to the operation of addition modulo l: m ⨀ k n is equal to the remainder of the division by k of the number m ⋅ n. This algebra is a commutative ring called residue ring modulo k.

V. The algebra (2 A , Δ, ∩, ∅, A) is a commutative ring, which follows from the properties of intersection and the symmetric difference of sets.

G. An example of a non-commutative ring gives the set of all fixed-order square matrices with matrix addition and multiplication operations. The unit of this ring is the identity matrix, and the zero is the zero matrix.

d. Let L- linear space. Consider the set of all linear operators acting in this space.

Recall that sum two linear operators A And IN call the operator A+B, such that ( A + IN) X = Oh +Vx, X∈ L.

The product of linear operators A And IN is called a linear-linear operator AB, such that ( AB)X = A(Vx) for anyone X ∈ L.

Using the properties of these operations on linear operators, one can show that the set of all linear operators acting in the space L, together with the operations of addition and multiplication of operators, forms a ring. The zero of this ring is null operator, and the unit identity operator.

This ring is called ring of linear operators in linear space L. #

Ring axioms are also called basic ring identities . An identity of a ring is an equality, the validity of which is preserved under the substitution of any elements of the ring instead of the variables appearing in it. Basic identities are postulated, and other identities can then be deduced from them as corollaries. Let's consider some of them.

Recall that the additive group of a ring is commutative and the operation subtraction.

Theorem 2.8. In any ring, the following identities hold:

1. 0 ⋅ a = a ⋅ 0 = 0 ;
2. (-a) ⋅ b = -(a ⋅ b) = a ⋅ (-b);
3. (a-b) ⋅ c = a ⋅ c - b ⋅ c, c ⋅ (a-b) = c ⋅ a - c ⋅ b.

◀Let's prove the identity 0 ⋅ a = 0 . We write for arbitrary a:

a+ 0 ⋅ a = 1 ⋅ a + 0 ⋅ a = ( 1 +0 ) ⋅ a = 1 ⋅ a = a

So, a + 0 ⋅ a = a. The last equality can be considered as an equation in the additive group of the ring with respect to the unknown element 0 ⋅ a. Since in the additive group any equation of the form a + x \u003d b has a unique solution x \u003d b - a, then 0 ⋅ a = a - a = 0 . Identity a⋅ 0 = 0 is proved in a similar way.

We now prove the identity - (a ⋅ b) = a ⋅ (-b). We have

a ⋅ (-b)+a ⋅ b = a ⋅ ((-b) + b) = a ⋅ 0 = 0 ,

whence a ⋅ (-b) = -(a ⋅ b). Similarly, one can prove that (-a) ⋅ b = -(a ⋅ b).

Let us prove the third pair of identities. Let's consider the first of them. In view of what was proved above, we have

a ⋅ (b - c) = a ⋅ (b+(-c)) = a ⋅ b + a ⋅ (-c) = a ⋅ b - a ⋅ c,

those. the identity is true. The second identity of this pair is proved similarly.

Corollary 2.1. In any ring, the identity ( -1 ) ⋅ x = x ⋅ ( -1 ) = -x.

◀This corollary follows from the second identity of Theorem 2.8 for a = 1 and b = x.

The first two identities proved in Theorem 2.8 express a property called annihilating property of zero in the ring. The third pair of identities of this theorem expresses the distributive property of the operation of multiplication of a ring with respect to the operation of subtraction. Thus, when performing calculations in any ring, one can open brackets and change signs in the same way as when adding, subtracting, and multiplying real numbers.

Nonzero elements a and b of the ring R called dividers zero if a ⋅ b = 0 or b ⋅ a = 0 . An example of a zero divisor ring gives any modulo residue ring k if k is a composite number. In this case, the product modulo k of any type that gives a number that is a multiple of k in the usual multiplication will be equal to zero. For example, in the ring of residues modulo 6, the elements 2 and 3 are zero divisors, since 2 ⨀ 6 3 = 0. Another example gives a ring of square matrices of fixed order (at least two). For example, for matrices of the second order we have

For non-zero a and b, the reduced matrices are zero divisors.

By multiplication, a ring is only a monoid. Let us pose the question: in what cases will a multiplication ring be a group? First of all, note that the set of all elements of a ring in which 0 ≠ 1 , cannot form multiplication groups, since zero cannot have an inverse. Indeed, if we assume that such an element 0" exists, then, on the one hand, 0 ⋅ 0" = 0" ⋅ 0 = 1 , and on the other hand, 0 ⋅ 0" = 0" ⋅ 0 = 0 , whence 0 = 1. This contradicts the condition 0 ≠ 1 . Thus, the question posed above can be refined as follows: in what cases does the set of all nonzero elements of a ring form a multiplication group?

If there are zero divisors in the ring, then the subset of all nonzero elements of the ring does not form a multiplication group if only because this subset is not closed under the multiplication operation, i.e. there are non-zero elements whose product is equal to zero.

A ring in which the set of all nonzero elements by multiplication forms a group is called body , commutative body - field , and the group of nonzero elements of the body (field) by multiplication - multiplicative group this body (fields). By definition, a field is a special case of a ring in which operations have additional properties. Let's write down all the properties that are required for field operations. They are also called field axioms .

The field is the algebra F = (F, +, ⋅, 0, 1) whose signature consists of two binary and two null operations, and the following identities hold:

1. a+(b+c) = (a+b)+c;
2. a+b = b+a;
3. a+0 = a;
4. for every a ∈ F there is an element -a such that a+ (-a) = 0;
5. a ⋅ (b ⋅ c) = (a ⋅ b) ⋅ c;
6. a ⋅ b = b ⋅ a
7. a ⋅ 1 = 1 ⋅ a = a
8. for every a ∈ F other than 0, there is an element a -1 such that a ⋅ a -1 = 1;
9. a ⋅ (b+c) = a ⋅ b + a ⋅ c.

Example 2.13. A. Algebra (ℚ, +, ⋅, 0, 1) is a field called field of rational numbers .

b. The algebras (ℝ , +, ⋅, 0, 1) and (ℂ, +, ⋅, 0, 1) have fields called fields of real and complex numbers respectively.

V. An example of a body that is not a field is the algebra quaternions . #

So, we see that the axioms of the field correspond to the known laws of addition and multiplication of numbers. Being engaged in numerical calculations, we “work in the fields”, namely, we deal mainly with the fields of rational and real numbers, sometimes we “move” into the field of complex numbers.

The concept of a ring, the simplest properties of rings.

Algebra ( K, +, ∙) is called a ring if the following axioms hold:

1. (K, +) is a commutative group;

2.
a (b+c) = ab+ac (b+c)a = ba+ca;

3. a (bc) = (ab) c.

If the operation of multiplication in a ring is commutative, then the ring is said to be commutative.

Example. Algebras (Z, +, ∙), ( Q, +, ∙), (R, + ,∙) are rings.

The ring has the following properties:

1) a + b = a => b = 0;

2) a+b = 0 => b = - a;

3) – (- a) = a;

4) 0∙a = a∙0 = 0 (0 is the zero of the ring);

5) (-a)∙b = a∙(-b) = -a∙b;

6) (a – b)∙c = a∙c – b∙c, Where a– b = a + (-b).

Let us prove property 6. ( a-b)∙c = (a + (-b))∙c = a∙c+ (-b)∙c = a∙c +(-b∙c)= =a∙c-b∙c.

Let ( K A K is called a subring of the ring ( K,+,∙) if it is a ring with respect to operations in the ring ( K, +, ∙).

Theorem. Let ( K, +, ∙) is a ring. Non-empty subset A K, is a subring of the ring TO if and only if
a- b, a∙b
.

Example. The ring (Q, +, ∙) is a subring of the ring ( A, +, ∙), where A = ={a+ b | a, b Q).

The concept of a field. The simplest field properties.

Definition. commutative ring ( R, +, ∙) with one, where the zero of the ring does not coincide with the unit of the ring, is called a field if
a≠0 there is an inverse element A -1 , A∙ A -1 = e, e is the unit of the ring.

All properties of rings are valid for fields. For field ( R,+,∙) the following properties also hold:

1)
a≠0 equation ah =b has a solution and, moreover, a unique one;

2) ab=e |=> a≠0 b=A -1 ;

3)

c≠0 ac=bc => a=b;

4)ab = 0
a = 0 b = 0;

5) ad = bc (b≠0, d≠0);

6)
;

.

Example. Algebras (Q, +, ∙), ( A, +, ∙), where A = {a+b | a, b Q), ( R, +, ∙) are fields.

Let ( R,+,∙) is a field. Non-empty subset F P, which is a field with respect to the operation on the field ( R,+,∙) is called a subfield of the field R.

Example. The field (Q,+,∙) is a subfield of the field of real numbers (R,+,∙).

Tasks for independent solution

1. Show that the set with respect to the operation of multiplication is an Abelian group.

2. On the set Q\(0) the operation Ab =
. Prove that the algebra (Q\(0),) is a group.

3. A binary algebraic operation is given on the set Z, defined by the rule Ab = a+b – 2. Find out if the algebra (Z,) is a group.

4. On the set A = {(a, b)
) the operation is defined ( A,b) (c, d) = (ac– bd, ad+ bc). Prove that the algebra ( A,) - group.

5. Let T is the set of all mappings
given by rule
, Where A,bQ, a
Prove that T is a group under the composition of mappings.

6. Let A={1,2,…,n). One-to-one mapping f:
called a substitution n- oh degree. substitution n- oh degree, it is convenient to write in the form of a table
, where The product of two substitutions
sets A is defined as a composition of mappings. A-priory

Prove that the set of all permutations n- oh degree is a group under the product of permutations.

7. Find out if the ring forms with respect to addition, multiplication:

a) N; b) the set of all odd integers; c) the set of all even integers; d) a set of numbers of the form
Where A,b

8. Is a set a ring? TO={A+b
) with respect to the operations of addition and multiplication.

9. Show that the set A={a+b) with respect to the operations of addition and multiplication is a ring.

10. On the set Z two operations are defined: ab=a+b+1, ab= ab+ a+ b. Prove that algebra

11. On the set of residue classes modulo m two binary operations are given: Prove that the algebra
commutative ring with identity.

12 . Describe all subrings of the ring
.

13. Find out which of the following sets of real numbers are fields with respect to addition and multiplication operations:

a) rational numbers with odd denominators;

b) numbers of the form
with rational A,b;

c) numbers of the form
with rational A, b;

d) numbers of the form
with rational a, b, c.

§5. The field of complex numbers. Operations on complex

numbers in algebraic form

Field of complex numbers.

Let two algebras ( A,+,∙), (Ā , , ◦). Display f: A in (on) >Ā , satisfying the conditions:
f(a+b) = f(a) f(b) f(a◦b) = f(a) ◦ f(b), is called a homomorphism of the algebra ( A, +, ∙) into (on) the algebra ( Ā , , ◦).

Definition. Homomorphic mapping f algebras ( A, +, ∙) onto the algebra ( Ā , , ◦) is called an isomorphic mapping if the mapping f sets A on Ā injectively. From the point of view of the algebra, isomorphic algebras are indistinguishable, that is, have the same properties.

Above the field R equation of the form x 2 +1 = 0 has no solutions. Let us construct a field that contains a subfield isomorphic to the field ( R,+,∙), and in which the equation of the form x 2 +1 = 0 has a solution.

On the set C = R× R = {(a, b) | a, b R) we introduce the operations of addition and multiplication as follows: ( a, b) (c, d) = (a+ c, b+ d), (a, b) ◦ (c, d) = (ac-bd, ad+bc). It is easy to prove that the algebra (C, ,◦) is a commutative ring with unit. The pair (0,0) is the zero of the ring, (1,0) is the unit of the ring. Let us show that the ring ( WITH, ,◦) is a field. Let ( a, b) C, ( a, b) ≠ (0,0) and ( x,y) C is a pair of numbers such that ( a, b)◦(x, y) = (1,0). (a, b)◦(x, y) = (1,0) (ax– by, ay+ bx) = (1,0)

(1)

From (1) =>
,
(a, b) -1 =
. Hence (С, +, ∙) is a field. Consider the set R 0 = {(a,0) | a R). Because ( a,0) (b,0) = (a- b,0)R 0 , (a,0)◦(b,0) = (ab,0) R 0 ,
(a,0) ≠ (0,0) (a,0) -1 = (,0) R 0 , then the algebra ( R 0, ,◦) is a field.

Let's build a mapping f: R
R 0 , defined by the condition f(a)=(a,0) . Because f is a bijective mapping and f(a+ b)= (a+ b,0) = =(a,0)(b,0) = f(a)f(b), f(a∙b) = (a∙ b,0) = (a,0)◦(b,0) =f(a)◦f(b), That f is an isomorphic mapping. Hence, ( R , +,∙)
(R 0, ,◦). (R 0, ,◦) is the field of real numbers.

Let us show that an equation of the form X 2 +1 = 0 in the field (C , , ◦) has solutions. ( x,y) 2 + (1,0) = (0,0) (x 2 - y 2 +1, 2xy) = (0,0)

(2)

(0,1), (0, -1) are solutions of system (2).

The constructed field (C , ,◦) is called the field of complex numbers, and its elements are called complex numbers.

Algebraic form of a complex number. Operations on complex numbers in algebraic form.

Let (C, +, ∙) be the field of complex numbers,
c,
=(a, b). Because ( R 0 ,+, ∙) (R, +, ∙), then any pair ( a,0) is identified with the real number a. Denote by ί = (0,1). Because ί 2 = (0.1)∙(0.1) = (-1.0) = -1, then ί is called the imaginary unit. Imagine a complex number
=(a,b) in the form: =( a,b)=(a,0) +(b,0) ◦(0,1)=a+b∙ί. Representation of a complex number in the form, = A + bί is called the algebraic form of the number. a is called the real part of a complex number and is denoted by Re, b is the imaginary part of a complex number and is denoted by Im.

Addition of complex numbers:

α = a+bί, β = c+dί , α +β = (A,b) + (c, d) = (a+ c, b+ d) = a+ c+ (b+ d)ί.

Multiplication of complex numbers:

α∙β = (a, b)(c, d) = (a∙ c– b∙ d, a∙ d+ b∙ c) = a∙ c - b∙ d + (a∙ d + b∙ c)ί.

To find the product of complex numbers a+bί And c+dί , you need to multiply a+bί on c+dί as a binomial by a binomial, given that ί 2 = -1.

quotient from division by β , β ≠ 0 is a complex number γ such that = γ∙ β .

= γ∙ β => γ = ∙ β -1 . Because
, then = ∙β -1 = =(a, b)∙
Thus

This formula can be obtained if the numerator and denominator of a fraction are multiplied by the complex number conjugate to the denominator, i.e. on

With -dί.

Example. Find the sum, product, quotient of complex numbers

2+ 3ί , β = 3 - 4ί .

Solution. + β =(2 + 3ί ) + (3 – 4ί ) =5– ί, ∙β = (2 + 3ί) (3– 4ί ) = 6 –8ί + 9ί – –12ί 2 = 18 + ί .

§6. root extractionnth power of a complex number in trigonometric form

Trigonometric form of a complex number.

On a plane in a rectangular coordinate system, a complex number

z = a + bί will be represented by a dot A(A,b) or radius vector
.

Draw a complex number z = 2 – 3ί .

Definition. Number
is called the modulus of a complex number z = a + bί and is denoted by | z |.

The angle formed between the positive direction of the O axis X and a radius vector representing a complex number z= a+ bί, is called the number argument z and denoted Argz.

Argz defined up to the term 2π k, .

Complex number argument z, satisfying the condition 0≤< 2π , называется главным значением аргумента комплексного числа z and denoted arg z.

From OAA 1 a=
cos ,b= sin
. Complex number representation z= a+ bί as z= r(cos + ί sin) is called the trigonometric form of the number z (r=). To write a complex number z = a + bί in trigonometric form, you need to know | z| And Arg z, which are determined from the formulas
, cos =
sin =

Let z 1 = r 1 (cos φ 1 + ί sin φ 1), z 2 = r 2 (cos φ 2 + ί sin φ 2). Then z 1∙ z 2 = =r 1∙ r 2 [(cos φ 1 ∙cos φ 2-sin φ 1 sin φ 2)+i]= r 1∙ r 2 [(cos (φ 1+ φ 2) + i sin( φ 1+ φ 2)]. It follows from this that | z 1 z 2 | = |z 1 | |z 2 |, Arg z 1 ∙z 2 = Arg z 1 + Arg z 2 .

Arg
Arg – Arg .

root extractionn– th power of a complex number in trigonometric form.

Let zC, nN. n – th power of a complex number z the work is called
it is denoted z n. Let m=- n. By definition, let's say that
z≠0, z0 = 1, z m = . If z =r(cos φ + ί sin φ ) , That z n =

= r n(cos nφ + ί sin nφ). At r = 1 we have z n = cos nφ + ί sin nφ - Moivre's formula. De Moivre's formula holds
.

root n z such a complex number is called ω , What ω n = z. Fair assertion.

Theorem. Exists n different root values n-th degree from a complex number z = r(cos φ + ί sin φ ) . All of them are obtained from the formula for k = 0, 1, … , n-1. In this formula
is an arithmetic root.

Let's denote by ω 0 , ω 1 ,…, ω n-1 - root values n th degree of z, which are obtained with k = 0, 1, ... , n-1. Since | ω 0 | = |ω 1 | = |ω 2 |= … =|ω n -1 |,

arg ω 0 = , ω 1 = arg ω 0 +
, … , arg ω n -1 = arg ω n - 2 + , then the complex numbers ω 0 , ω 1 ,…, ω n-1 on the plane are represented by points of a circle with a radius equal to
and divide this circle into n equal parts.