In various branches of mathematics, as well as in the application of mathematics in technology, there is often a situation where algebraic operations are performed not on numbers, but on objects of a different nature. For example, matrix addition, matrix multiplication, vector addition, operations on polynomials, operations on linear transformations, etc.

Definition 1. A ring is a set of mathematical objects in which two actions are defined - "addition" and "multiplication", which compare ordered pairs of elements with their "sum" and "product", which are elements of the same set. These actions meet the following requirements:

1.a+b=b+a(commutativity of addition).

2.(a+b)+c=a+(b+c)(associativity of addition).

3. There is a zero element 0 such that a+0=a, for any a.

4. For anyone a there is an opposite element − a such that a+(−a)=0.

5. (a+b)c=ac+bc(left distributivity).

5".c(a+b)=ca+cb(right distributivity).

Requirements 2, 3, 4 mean that the set of mathematical objects forms a group , and together with item 1 we are dealing with a commutative (Abelian) group with respect to addition.

As can be seen from the definition, in the general definition of a ring, no restrictions are imposed on multiplications, except for distributivity with addition. However, in various situations, it becomes necessary to consider rings with additional requirements.

6. (ab)c=a(bc)(associativity of multiplication).

7.ab=ba(commutativity of multiplication).

8. Existence of the identity element 1, i.e. such a 1=1 a=a, for any element a.

9. For any element of the element a there is an inverse element a−1 such that aa −1 =a −1 a= 1.

In various rings 6, 7, 8, 9 can be performed both separately and in various combinations.

A ring is called associative if condition 6 is satisfied, commutative if condition 7 is satisfied, commutative and associative if conditions 6 and 7 are satisfied. A ring is called a unit ring if condition 8 is satisfied.

Ring examples:

1. Set of square matrices.

Really. The fulfillment of points 1-5, 5 "is obvious. The zero element is the zero matrix. In addition, point 6 (associativity of multiplication), point 8 (the unit element is the identity matrix) are performed. Points 7 and 9 are not performed because in the general case, multiplication of square matrices is non-commutative, and also there is not always an inverse to a square matrix.

2. The set of all complex numbers.

3. The set of all real numbers.

4. The set of all rational numbers.

5. The set of all integers.

Definition 2. Any system of numbers containing the sum, difference and product of any two of its numbers is called number ring.

Examples 2-5 are number rings. Numeric rings are also all even numbers, as well as all integers divisible without remainder by some natural number n. Note that the set of odd numbers is not a ring since the sum of two odd numbers is an even number.

The ring in which the relation "to be greater than zero" is introduced (denoted by a > 0) is called located ring, if two conditions are satisfied for any elements of this ring:

1) one and only one of the conditions is true

a > 0 \/ –a >0 \/ a = 0

2) a > 0 /\ b > 0 => a + b > 0 /\ ab > 0.

A set in which a certain order relation is introduced - non-strict (reflexively, antisymmetric and transitive) or strict (anti-reflexively and transitively) is called orderly. If the law of trichotomy is satisfied, then the set is called linearly orderly. If we consider not an arbitrary set, but some algebraic system, for example, a ring or a field, then for the ordering of such a system, monotonicity requirements are also introduced with respect to the operations introduced in this system (algebraic structure). So ordered ring/field is a non-zero ring/field in which a linear order relation (a > b) is introduced that satisfies two conditions:

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

2) a > b, c > 0 => a c > b c;

Theorem 1. Any located ring is an ordered system (ring).

Indeed, if the relation “to be greater than 0” is introduced in the ring, then it is also possible to introduce a relation greater than for two arbitrary elements, if we assume that

a > b  a - b > 0.

Such a relation is a relation of a strict, linear order.

This “greater than” relation is anti-reflexive, since the condition a > a is equivalent to the condition a - a > 0, the latter contradicts the fact that a - a = 0 (according to the first condition of the located ring, an element cannot be both greater than 0 and equal to 0) . Thus, the statement a > a is false for any element of a, so the relation is anti-reflexive.

Let us prove transitivity: if a > b and b > c, then a > c. By definition, it follows from the conditions of the theorem that a - b > 0 and b - c > 0. Adding these two elements greater than zero, we again obtain an element greater than zero (according to the second condition of the located ring):

a - b + b - c \u003d a - c\u003e 0.

The latter means that a > c. Thus, the introduced relation is a relation of strict order. Moreover, this relation is a relation of linear order, that is, for the set of natural numbers, trichotomy theorem:

For any two natural numbers, one and only one of the following three statements is true:

Indeed (due to the first condition of the located ring) for the number a - b one and only one of the conditions is true:

1) a - b > 0 => a > b

2) - (a - b) = b - a > 0 => b > a

3) a - b = 0 => a = b.

The monotonicity properties also hold for any located ring. Really

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

2) a > b / \ c > 0 => a - b > 0 => (according to the second condition of the located ring) (a - b) c > 0 => ac - bc > 0 => ac > bc.

Thus, we have proved that any located ring is an ordered ring (an ordered system).

For any located ring, the following properties will also be true:

a) a + c > b + c => a > b;

b) a > b /\ c > d => a + c > b + d;

c) a > b / \ c< 0=>ac< bc;

The same properties hold for other signs.<, , .

Let us prove, for example, property (c). By definition, from the condition a > b it follows that a - b > 0, and from the condition c< 0 (0 >c) it follows that 0 - c > 0, and hence the number - c > 0, we multiply two positive numbers (a - b) (-c). The result will also be positive by the second condition of the located ring, i.e.

(a - b)  (-c) > 0 => -ac + bc > 0 => bc - ac > 0 => bc > ac => ac< bc,

Q.E.D.

d) aa = a 2  0;

Proof: According to the first condition of the located ring, either a > 0, or –a > 0, or a = 0. Consider these cases separately:

1) a > 0 => aa > 0 (according to the second condition of the located ring) => a 2 > 0.

2) –a > 0 => (–a)(–a) > 0, but by the property of the ring (–a)(–a) = aa = a 2 > 0.

3) a \u003d 0 => aa \u003d a 2 \u003d 0.

Thus, in all three cases, a 2 is either greater than zero or equal to 0, which just means that a 2 ≥ 0 and the property is proved (note that we also proved that the square of an element of a located ring is 0 if and only if the element itself is 0).

e) ab = 0  a = 0 \/ b = 0.

Proof: Assume the contrary (ab =0, but neither a nor b are equal to zero). Then only two options are possible for a, either a > 0 or – a > 0 (the option a = 0 is excluded by our assumption). Each of these two cases splits into two more cases depending on b (either b > 0 or – b > 0). Then 4 options are possible:

a > 0, b > 0 => ab > 0;

– a > 0, b > 0 => ab< 0;

a > 0, – b > 0 => ab< 0;

– a > 0 – b > 0 => ab > 0.

As we see, each of these cases contradicts the condition ab = 0. The property is proved.

The last property means that the located ring is an area of ​​integrity, which is also a mandatory property of ordered systems.

Theorem 1 shows that any located ring is an ordered system. The converse is also true - any ordered ring is located. Indeed, if there is a relation a > b in the ring and any two elements of the ring are comparable to each other, then 0 is also comparable to any element a, that is, either a > 0 or a< 0, либо а = 0, что почти точно совпадает с первым условием расположенного кольца, требуется только доказать, что условие а < 0 равносильно в упорядоченном кольце условию –a >0. In order to prove the latter, we apply the monotonicity property of ordered systems: to the right and left sides of the inequality a< 0 прибавим –а, в результате чего получим требуемое.

The second condition of the located ring follows from the properties of monotonicity and transitivity:

a > 0, b > 0 => a + b > 0 + b = b > 0 => a + b > 0,

a > 0, b > 0 => ab > 0b = 0 => ab > 0.

Theorem 2. The ring of integers is an arranged ring (an ordered system).

Proof: Let us use definition 2 of the ring of integers (see 2.1). According to this definition, any integer is either a natural number (the number n is given as [ ], or the opposite of natural (– n corresponds to the class [<1, n / >] , or 0 (class [<1, 1>]). Let's introduce the definition of "be greater than zero" for integers according to the rule:

a > 0  a  N

Then the first condition of the located ring is automatically satisfied for integers: if a is natural, then it is greater than 0, if a is the opposite of natural, then –a is natural, that is, it is also greater than 0, the variant a = 0 is also possible, which also makes true disjunction in the first condition of the located ring. The validity of the second condition of the located ring follows from the fact that the sum and product of two natural numbers (integers greater than zero) is again a natural number, and therefore greater than zero.

Thus, all properties of the arranged rings are automatically transferred to all integers. In addition, for integers (but not for arbitrary arranged rings), the discreteness theorem holds:

Discreteness theorem. No integer can be inserted between two adjacent integers:

( a, x  Z) .

Proof: consider all possible cases for a, and assume the contrary, that is, that there is x such that

A< x < a +1.

1) if a is a natural number, then a + 1 is also a natural number. Then, by the discreteness theorem for natural numbers, no natural number x can be inserted between a and a / = a + 1, that is, x, in any case, cannot be natural. If we assume that x = 0, then our assumption is that

A< x < a +1

will lead us to the condition a< 0 < a + 1 (здесь мы просто подставили х = 0), откуда видим, что а < 0, что противоречит тому, что а – натуральное. Если х не натуральное и не 0, но х – целое, значит –х – натуральное, тогда х < 0. При этом, согласно нашему предположению, а < x, что по свойству транзитивности опять приводит к тому, что а < 0, что невозможно, так как а – натуральное. Таким образом, для натуральных а утверждение а < x < a +1 всегда ложно, и теорема справедлива.

2) a = 0. Then a + 1 = 1. If condition a< x < a + 1, то мы получим 0 < x < 1, то есть с одной стороны х – натуральное (целое, большее нуля), а с другой стороны х меньше 1, что невозможно для натуральных чисел. Таким образом, для нуля наша теорема справедлива.

3) a is negative (–a > 0), then a + 1  0. If a + 1< 0, то умножая условие а < x < a + 1 на –1 получим:

–a–1< – x < –a,

that is, we arrive at the situation considered in the first case (since both -a-1 and -a are natural), whence - x cannot be an integer, and hence x cannot be an integer. The situation when a + 1 = 0 means that a = -1, i.e.

–1 < x < 0.

Multiplying this inequality by (–1), we arrive at case 2. Thus, the theorem is valid in all situations.

Terem of Archimedes. For any integer a and integer b > 0, there exists a positive integer n such that a< bn.

For natural a, the theorem has already been proved, since the condition b > 0 means that the number b is natural. For a  0, the theorem is also obvious, since the right side of bn is a natural number, that is, it is also greater than zero.

In the ring of integers (as in any located ring), we can introduce the concept of a module:

|a| = .

Valid module properties:

1) |a + b|  |a| + |b|;

2) |a – b|  |a| – |b|;

3) |a  b| = |a|  |b|.

Proof: 1) Note that it is obvious from the definition that |a| is a value that is always non-negative (in the first case |a| = a ≥ 0, in the second case |a| = –a, but a< 0, откуда –а >0). The inequalities |a| ≥ a, |a| ≥ –a (the modulus is equal to the corresponding expression if it is non-negative, and greater than it if it is negative). Similar inequalities hold for b: |b| ≥ b, |b| ≥ -b. Adding up the corresponding inequalities and applying property (b) of arranged rings, we obtain

|a| + |b| ≥ a + b |a| + |b| ≥ – a – b.

According to the module definition

|a+b| =
,

but both expressions on the right side of the equality, as shown above, do not exceed |a| + |b|, which proves the first property of modules.

2) Let's replace in the first property a with a - b. We get:

|a – b + b| ≤ |a – b| + |b|

|a| ≤ |a – b| + |b|

Move |b| from the right side to the left with the opposite sign

|a| – | b| ≤ |a – b| =>|a-b|  |a| – |b|.

The proof of property 3 is left to the reader.

Task: Solve the equation in integers

2y 2 + 3xy - 2x 2 + x - 2y \u003d 5.

Solution: Factorize the left side. To do this, we represent the term 3xy = – xy + 4xy

2y 2 + 3xy - 2x 2 + x - 2y \u003d 2y 2 - xy + 4xy - 2x 2 + x - 2y \u003d

Y (2y - x) + 2x (2y - x) - (2y - x) \u003d (y + 2x - 1) (2y - x).

Thus, our equation can be rewritten as

(y + 2x - 1) (2y - x) = 5.

Since we need to solve it in integers, x and y must be integers, which means that the factors on the left side of our equation are also integers. The number 5 on the right side of our equation can be represented as a product of integer factors in only 4 ways:

5 = 51 = 15 = –5(–1) = –1(–5). Therefore, the following options are possible:

1)
2)
3)
4)

Among the listed systems, only (4) has an integer solution:

x = 1, y = -2.

Tasks for independent solution

No. 2.4. For elements a, b, c, d of an arbitrary located ring, prove the properties:

a) a + c > b + c => a > b; b) a > b / \ c > d => a + c > b + d.

No. 2.5. Solve the equations in integers:

a) y 2 - 2xy - 2x = 6; b) 2x 2 - 11xy + 12y 2 = 17; c) 35xy + 5x - 7y = 1; d) x 2 - 3xy + 2y 2 = 3; e) ;

f) xy + 3x - 5y + 3 = 0; g) 2xy - 3y 2 - 4y + 2x \u003d 2; h) xy 2 + x = 48; i) 1! +2! + 3! + … + n! = y 2 ; j) x 3 - 2y 3 - 4z 3 \u003d 0

No. 2.6. Find a four-digit number that is an exact square and such that its first two digits are equal to each other and the last two digits are equal to each other.

No. 2.7. Find a two-digit number equal to the sum of its tens and the square of its units.

No. 2.8. Find a two-digit number that is equal to twice the product of its digits.

No. 2.9. Prove that the difference between a three-digit number and a number written in the same digits in reverse order cannot be the square of a natural number.

No. 2.10. Find all natural numbers ending in 91, which, after deleting these digits, decrease by an integer number of times.

No. 2.11. Find a two-digit number equal to the square of its units plus the cube of its tens.

No. 2.12. Find a six-digit number starting with the number 2, which increases by 3 times by rearranging this number to the end of the number.

No. 2.13. More than 40 but less than 48 whole numbers are written on the board. The arithmetic mean of all these numbers is 3, the arithmetic mean of the positive ones is 4, and the arithmetic mean of the negative ones is 8. How many numbers are written on the blackboard? Which number is greater, positive or negative? What is the maximum possible number of positive numbers?

No. 2.14. Can the quotient of a three-digit number and the sum of its digits be 89? Can this quotient be equal to 86? What is the maximum possible value of this quotient?

Federal Agency for Education

State educational institution of higher professional education

Vyatka State University for the Humanities

Faculty of Mathematics

department mathematical analysis and methods
teaching mathematics

Final qualifying work

on the topic: Gauss ring of integers.

Completed:

5th year student

Faculty of Mathematics

Gnusov V.V.

___________________________

Scientific adviser:

senior lecturer of the department

algebra and geometry

Semenov A.N.

___________________________

Reviewer:

Candidate of Physics and Mathematics Sciences, Associate Professor

Department of Algebra and Geometry

Kovyazina E.M.

___________________________

Admitted to defense in the SAC

Head Department ________________ Vechtomov E.M.

« »________________

Dean of the faculty ___________________ Varankina V.I.

Introduction.

Ring of integer complex numbers

was discovered by Carl Gauss and named Gaussian after him.

K. Gauss came to the idea of ​​the possibility and necessity of expanding the concept of an integer in connection with the search for algorithms for solving comparisons of the second degree. He transferred the concept of an integer to numbers of the form

, where are arbitrary integers, and is the root of the equation On this set, K. Gauss was the first to construct a theory of divisibility similar to the theory of divisibility of integers. He substantiated the validity of the basic properties of divisibility; showed that there are only four invertible elements in the ring of complex numbers: ; proved the validity of the theorem on division with a remainder, the theorem on the uniqueness of decomposition into prime factors; showed which prime natural numbers will remain prime in the ring; discovered the nature of simple integer complex numbers.

The theory developed by K. Gauss, described in his work "Arithmetical Investigations", was a fundamental discovery for number theory and algebra.

The following goals were set for the thesis:

1. Develop the theory of divisibility in the ring of Gauss numbers.

2. Find out the nature of simple Gaussian numbers.

3. Show the application of Gaussian numbers in solving ordinary Diophantine problems.

CHAPTER 1. DIVISIBILITY IN THE RING OF GAUSS NUMBERS.

Consider the set of complex numbers. By analogy with the set of real numbers, a subset of integers can be distinguished in it. The set of numbers of the form

, Where will be called complex integers or Gaussian numbers. It is easy to check that the axioms of the ring hold for this set. Thus, this set of complex numbers is a ring and is called ring of Gaussian integers . Let's denote it as , since it is an extension of the ring by the element: .

Since the ring of Gaussian numbers is a subset of complex numbers, then some definitions and properties of complex numbers are valid for it. For example, for each Gaussian number

corresponds to a vector starting at point and ending at . Hence, module Gaussian numbers are . Note that in the set under consideration, the submodule expression is always a non-negative integer. Therefore, in some cases it is more convenient to use the norm , that is, the square of the modulus. Thus . We can distinguish the following properties of the norm. For any Gaussian numbers, the following is true: (1) (2) (3) (4) (5) - the set of natural numbers, that is, positive integers.

The validity of these properties is trivially checked using the module. In passing, we note that (2), (3), (5) are also valid for any complex numbers.

The ring of Gaussian numbers is a commutative ring without divisors 0, since it is a subring of the field of complex numbers. This implies the multiplicative contractibility of the ring

, i.e. (6)

1.1 REVERSIBLE AND ALLOY ELEMENTS.

Let's see which Gaussian numbers will be reversible. Multiplication neutral is

. If a Gaussian number reversible , then, by definition, there exists such that . Passing to the norms, according to property 3, we obtain . But these norms are natural, therefore. Hence, by property 4, . Conversely, all elements of this set are invertible, since . Therefore, numbers with a norm equal to one will be reversible, that is, , .

As you can see, not all Gaussian numbers will be reversible. Therefore, it is interesting to consider the issue of divisibility. As usual, we say that

is divided on if there exists such that . For any Gaussian numbers , as well as invertibles, the properties are true. (7) (8) (9) (10) , where (11) (12)

(8), (9), (11), (12) are easily verified. The validity (7) follows from (2), and (10) follows from (6). Due to property (9), the elements of the set

Examples

a + b i (\displaystyle a+bi) Where a (\displaystyle a) And b (\displaystyle b) rational Numbers, i (\displaystyle i) is the imaginary unit. Such expressions can be added and multiplied according to the usual rules of operations with complex numbers, and each non-zero element has an inverse, as can be seen from the equality (a + b i) (a a 2 + b 2 − b a 2 + b 2 i) = (a + b i) (a − b i) a 2 + b 2 = 1. (\displaystyle (a+bi)\left(( \frac (a)(a^(2)+b^(2)))-(\frac (b)(a^(2)+b^(2)))i\right)=(\frac (( a+bi)(a-bi))(a^(2)+b^(2)))=1.) It follows from this that the rational Gaussian numbers form a field which is a two-dimensional space over (that is, a quadratic field).

• More generally, for any square-free integer d (\displaystyle d) Q (d) (\displaystyle \mathbb (Q) ((\sqrt (d)))) will be a quadratic field expansion Q (\displaystyle \mathbb (Q) ).
• circular field Q (ζ n) (\displaystyle \mathbb (Q) (\zeta _(n))) obtained by adding Q (\displaystyle \mathbb (Q) ) primitive root n th power of unity. The field must also contain all its powers (that is, all roots n th power of unity), its dimension over Q (\displaystyle \mathbb (Q) ) equals the Euler function φ (n) (\displaystyle \varphi (n)).
• Real and complex numbers have infinite power over rational numbers, so they are not number fields. This follows from uncountability: any numeric field is countable.
• Field of all algebraic numbers A (\displaystyle \mathbb (A) ) is not numeric. Although the expansion A ⊃ Q (\displaystyle \mathbb (A) \supset \mathbb (Q) ) algebraically, it is not finite.

Ring of integers numeric field

Since the number field is an algebraic extension of the field Q (\displaystyle \mathbb (Q) ), any of its elements is a root of some polynomial with rational coefficients (that is, it is algebraic). Moreover, each element is a root of a polynomial with integer coefficients, since it is possible to multiply all rational coefficients by the product of the denominators. If a given element is a root of some unitary polynomial with integer coefficients, it is called an integer element (or an algebraic integer). Not all elements of a number field are integers: for example, it is easy to show that the only integer elements Q (\displaystyle \mathbb (Q) ) are regular integers.

It can be proved that the sum and product of two algebraic integers is again an algebraic integer, so the integer elements form a subring of the number field K (\displaystyle K) called whole ring fields K (\displaystyle K) and denoted by . The field does not contain zero divisors and this property is inherited when passing to a subring, so the ring of integers is integral; private ring box O K (\displaystyle (\mathcal (O))_(K)) is the field itself K (\displaystyle K). The ring of integers of any number field has the following three properties: it is integrally closed, Noetherian, and one-dimensional. A commutative ring with these properties is called Dedekind, after Richard Dedekind.

Decomposition into primes and a group of classes

In an arbitrary Dedekind ring, there is a unique decomposition of non-zero ideals into a product of simple ones. However, not every ring of integers satisfies the factorial property: already for the ring of integers, a quadratic field O Q (− 5) = Z [ − 5 ] (\displaystyle (\mathcal (O))_(\mathbb (Q) ((\sqrt (-5))))=\mathbb (Z) [(\sqrt ( -5))]) decomposition is not unique:

6 = 2 ⋅ 3 = (1 + − 5) (1 − − 5) (\displaystyle 6=2\cdot 3=(1+(\sqrt (-5)))(1-(\sqrt (-5) )))

By introducing a norm on this ring, we can show that these expansions are indeed different, that is, one cannot be obtained from the other by multiplying by an invertible element.

The degree of violation of the factorial property is measured using the ideal class group, this group for the ring of integers is always finite and its order is called the number of classes.

Number field bases

whole basis

whole basis number field F degrees n- it's a set

B = {b 1 , …, b n}

from n elements of the ring of integer fields F, such that any element of the ring of integers O F fields F can only be written as Z-linear combination of elements B; that is, for any x from O F there is a unique decomposition

x = m 1 b 1 + … + m n b n,

Where m i are regular integers. In this case, any element F can be written as

m 1 b 1 + … + m n b n,

Where m i are rational numbers. After this the whole elements F are distinguished by the property that these are exactly those elements for which all m i whole.

Using tools such as localization and the Frobenius endomorphism, one can construct such a basis for any number field. Its construction is a built-in feature in many computer algebra systems.

Power basis

Let F- numeric degree field n. Among all possible bases F(How Q-vector space), there are power bases, that is, bases of the form

B x = {1, x, x 2 , …, x n−1 }

for some x ∈ F. According to the primitive element theorem, such x always exists, it is called primitive element this extension.

Norm and trace

An algebraic number field is a finite-dimensional vector space over Q (\displaystyle \mathbb (Q) )(we denote its dimension as n (\displaystyle n)), and multiplication by an arbitrary element of the field is a linear transformation of this space. Let e 1 , e 2 , … e n (\displaystyle e_(1),e_(2),\ldots e_(n))- any basis F, then the transformation x ↦ α x (\displaystyle x\mapsto \alpha x) corresponds matrix A = (a i j) (\displaystyle A=(a_(ij))), determined by the condition

α e i = ∑ j = 1 n a i j e j , a i j ∈ Q . (\displaystyle \alpha e_(i)=\sum _(j=1)^(n)a_(ij)e_(j),\quad a_(ij)\in \mathbf (Q) .)

The elements of this matrix depend on the choice of the basis, however, all matrix invariants, such as determinant and trace, do not depend on it. In the context of algebraic extensions, the determinant of an element multiplication matrix is ​​called the norm this element (denoted N (x) (\displaystyle N(x))); matrix trace - trace element(denoted Tr (x) (\displaystyle (\text(Tr))(x))).

The trace of an element is a linear functional on F:

Tr (x + y) = Tr (x) + Tr (y) (\displaystyle (\text(Tr))(x+y)=(\text(Tr))(x)+(\text(Tr)) (y)) And Tr (λ x) = λ Tr (x) , λ ∈ Q (\displaystyle (\text(Tr))(\lambda x)=\lambda (\text(Tr))(x),\lambda \in \mathbb (Q) ).

The norm is a multiplicative and homogeneous function:

N (x y) = N (x) ⋅ N (y) (\displaystyle N(xy)=N(x)\cdot N(y)) And N (λ x) = λ n N (x) , λ ∈ Q (\displaystyle N(\lambda x)=\lambda ^(n)N(x),\lambda \in \mathbb (Q) ).

As the initial basis, you can choose an integer basis, multiplication by an integer algebraic number (that is, by an element of the ring of integers) in this basis will correspond to a matrix with integer elements. Therefore, the trace and norm of any element of the ring of integers are integers.

An example of using a norm

Let d (\displaystyle d)- - an integer element, since it is the root of the reduced polynomial x 2 − d (\displaystyle x^(2)-d)). In this basis, multiplication by a + b d (\displaystyle a+b(\sqrt (d))) corresponds matrix

(a d b b a) (\displaystyle (\begin(pmatrix)a&db\\b&a\end(pmatrix)))

Hence, N (a + b d) = a 2 − d b 2 (\displaystyle N(a+b(\sqrt (d)))=a^(2)-db^(2)). On the elements of the ring, this norm takes integer values. The norm is a homomorphism of the multiplicative group Z [ d ] (\displaystyle \mathbb (Z) [(\sqrt (d))]) per multiplicative group Z (\displaystyle \mathbb (Z) ), so the norm of invertible elements of a ring can only be equal to 1 (\displaystyle 1) or − 1 (\displaystyle -1). To solve Pell's equation a 2 − d b 2 = 1 (\displaystyle a^(2)-db^(2)=1), it suffices to find all invertible elements of the ring of integers (also called ring units) and select among them those with the norm 1 (\displaystyle 1). According to Dirichlet's unit theorem, all invertible elements of a given ring are powers of one element (up to multiplication by − 1 (\displaystyle -1)), therefore, to find all solutions of the Pell equation, it suffices to find one fundamental solution.

see also

Literature

• H. Koch. Algebraic Number Theory. - M.: VINITI, 1990. - T. 62. - 301 p. - (Results of science and technology. Series " Contemporary Issues mathematics. Fundamental Directions).
• Chebotarev N.G. Fundamentals of Galois theory. Part 2. - M.: Editorial URSS, 2004.
• Weil G. Algebraic number theory. Per. from English. - M. : Editorial URSS, 2011.
• Serge Lang, Algebraic Number Theory, second edition, Springer, 2000

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Federal Agency for Education

State educational institution of higher professional education

Vyatka State University for the Humanities

Faculty of Mathematics

Department of Mathematical Analysis and Methods
teaching mathematics

Final qualifying work

on the topic: Gauss ring of integers.

Completed:

5th year student

Faculty of Mathematics

Gnusov V.V.

___________________________

Scientific adviser:

senior lecturer of the department

algebra and geometry

Semenov A.N.

___________________________

Reviewer:

Candidate of Physics and Mathematics Sciences, Associate Professor

Department of Algebra and Geometry

Kovyazina E.M.

___________________________

Admitted to defense in the SAC

Head Department ________________ Vechtomov E.M.

« »________________

Dean of the faculty ___________________ Varankina V.I.

« »________________

Kirov 2005

• Introduction. 2
• 3
• 4
• 1.2 DIVISION WITH REMAIN. 5
• 1.3 GCD. EUCLID ALGORITHM. 6
• 9
• 12
• 17
• Conclusion. 23

Introduction.

The ring of complex integers was discovered by Carl Gauss and named Gaussian after him.

K. Gauss came to the idea of ​​the possibility and necessity of expanding the concept of an integer in connection with the search for algorithms for solving comparisons of the second degree. He transferred the concept of an integer to numbers of the form, where are arbitrary integers, and is the root of the equation On a given set, K. Gauss was the first to construct a theory of divisibility, similar to the theory of divisibility of integers. He substantiated the validity of the basic properties of divisibility; showed that there are only four invertible elements in the ring of complex numbers: ; proved the validity of the theorem on division with a remainder, the theorem on the uniqueness of decomposition into prime factors; showed which prime natural numbers will remain prime in the ring; discovered the nature of simple integer complex numbers.

The theory developed by K. Gauss, described in his work "Arithmetical Investigations", was a fundamental discovery for number theory and algebra.

The following goals were set for the thesis:

1. Develop the theory of divisibility in the ring of Gauss numbers.

2. Find out the nature of simple Gaussian numbers.

3. Show the application of Gaussian numbers in solving ordinary Diophantine problems.

CHAPTER 1. DIVISIBILITY IN THE RING OF GAUSS NUMBERS.

Consider the set of complex numbers. By analogy with the set of real numbers, a subset of integers can be distinguished in it. A set of numbers of the form where will be called complex integers or Gaussian numbers. It is easy to check that the axioms of the ring hold for this set. Thus, this set of complex numbers is a ring and is called ring of Gaussian integers . Let's denote it as, since it is an extension of the ring by element: .

Since the ring of Gaussian numbers is a subset of complex numbers, then some definitions and properties of complex numbers are valid for it. So, for example, each Gaussian number corresponds to a vector starting at a point and ending at. Hence, module there are Gaussian numbers. Note that in the set under consideration, the submodule expression is always a non-negative integer. Therefore, in some cases it is more convenient to use the norm , that is, the square of the modulus. Thus. We can distinguish the following properties of the norm. For any Gaussian numbers, the following is true:

(1)

(2)

(3)

(4)

(5)

The validity of these properties is trivially checked using the module. In passing, we note that (2), (3), (5) are also valid for any complex numbers.

The ring of Gaussian numbers is a commutative ring without divisors 0, since it is a subring of the field of complex numbers. This implies the multiplicative contractibility of the ring, that is,

1.1 REVERSIBLE AND ALLOY ELEMENTS.

Let's see which Gaussian numbers will be reversible. It is neutral by multiplication. If a Gaussian number reversible , then, by definition, there exists such that Passing to the norms, according to property 3, we get. But these norms are natural, therefore. Hence, by property 4, . Conversely, all elements of a given set are invertible, since. Therefore, numbers with a norm equal to one will be reversible, that is, .

As you can see, not all Gaussian numbers will be reversible. Therefore, it is interesting to consider the issue of divisibility. As usual, we say that is divided on if there exists such that. For any Gaussian numbers, as well as reversible ones, the properties are true.

(7)

(8)

(9)

(10)

, where (11)

(12)

(8), (9), (11), (12) are easily verified. The validity (7) follows from (2), and (10) follows from (6). Due to property (9), the elements of the set behave in the same way with respect to divisibility, and are called allied With. Therefore, it is natural to consider the divisibility of Gaussian numbers up to union. Geometrically, on the complex plane, allied numbers will differ from each other by a multiple angle rotation.

1.2 DIVISION WITH REMAIN.

Let it be necessary to divide by, but it is impossible to make a division entirely. We must receive, and at the same time there must be "little". Then we will show what to take as an incomplete quotient when dividing with a remainder in the set of Gaussian numbers.

Lemma 1. On division with remainder.

In the ring division with a remainder is possible, in which the remainder is less than the divisor in the norm. More precisely, for any And there will be such that . As you can take the closest to the complex number Gaussian number.

Proof.

Divide by in the set of complex numbers. This is possible because the set of complex numbers is a field. Let be. Rounding the real numbers and to integers, we get, respectively, and. Let. Then

.

Multiplying now both parts of the inequality by we get, due to the multiplicativity of the norm of complex numbers, that. Thus, as an incomplete quotient, one can take a Gaussian number, which, as it is easy to see, is the closest to.

C.T.D.

1.3 GCD. EUCLID ALGORITHM.

We use the usual definition of the greatest common divisor for rings. GCD "ohm of two Gaussian numbers is a common divisor that is divisible by any other common divisor.

As in the set of integers, in the set of Gaussian numbers, the Euclidean algorithm is used to find the GCD.

Let and be given Gaussian numbers, moreover. Divide with the remainder by. If the remainder is different from 0, then we will divide by this remainder, and we will continue to sequentially divide the remainders as long as it is possible. We get a chain of equalities:

, Where

, Where

, Where

……………………….

, Where

This chain cannot continue indefinitely, since we have a decreasing sequence of norms, and the norms are non-negative integers.

Theorem 2. On the existence of GCD.

In Euclid's algorithm applied to Gaussian numbers And the last non-zero remainder is gcd( ).

Proof.

Let us prove that in the Euclidean algorithm we indeed obtain a gcd.

1. Consider equalities from bottom to top.

From the last equality it can be seen that. Therefore, as the sum of numbers divisible by. Since and, the next line will give. And so on. Thus, it is clear that i. That is, it is a common divisor of numbers and.

Let us show that this is the greatest common divisor, that is, divisible by any of their other common divisors.

2. Consider equalities from top to bottom.

Let be an arbitrary common divisor of numbers and. Then, as the difference of numbers divisible by, is valid from the first equality. From the second equality we get that. Thus, representing the remainder in each equality as the difference of numbers divisible by, we get from the penultimate equality what is divisible by.

C.T.D.

Lemma 3. On the representation of GCD.

If GCD( , )= , then there are integer Gaussian numbers And , What .

Proof.

Let us consider the chain of equalities obtained in the Euclidean algorithm from bottom to top. Consistently substituting instead of the remainders of their expression through the previous remainders, we express through and.

Gaussian number is called simple , if it cannot be represented as a product of two irreversible factors. The next assertion is obvious.

Statement 4.

Multiplying a simple Gaussian number by an invertible number again results in a simple Gaussian number.

Statement 5.

If we take an irreversible divisor with the smallest norm of a Gaussian number, then it will be a simple Gaussian.

Proof.

Let such a divisor be a composite number. Then, where and are irreversible Gaussian numbers. Let us pass to the norms, and according to (3) we obtain that. Since these norms are natural, we have that, and by virtue of (12), is an irreversible divisor of the given Gaussian number, which contradicts the choice.

Statement 6.

If is not divisible by a prime Gaussian number , then GCD( , )=1.

Proof.

Indeed, a prime number divisible only by allied numbers with 1 or with . Since it is not divisible by , then allied with also not shared. This means that only reversible numbers will be their common divisors.

Lemma 7. Lemma of Euclid.

If the product of Gaussian numbers is divisible by a prime Gaussian number , then at least one of the factors is divisible by .

Proof.

For the proof, it suffices to consider the case when the product contains only two factors. That is, we show that if is divisible by , then either is divisible by , or divided by .

Let it not be divided into , then GCD(, )=1. Therefore, there are Gaussian numbers and such that. Multiply both sides of the equation by , we get that, it follows that, as the sum of numbers divisible by .

1.4 MAIN THEOREM OF ARITHMETICS.

Any non-zero Gaussian number can be represented as a product of simple Gaussian numbers, and this representation is unique up to union and order of factors.

Remark 1.

A reversible number has zero prime factors in its expansion, that is, it is represented by itself.

Remark 2.

More precisely, uniqueness is formulated as follows. If there are two factorizations into simple Gaussian factors, that is, , That and you can renumber the numbers like this , What will be allied with , for all from 1 to inclusive.

Proof.

We prove it by induction on the norm.

Base. For a number with unit norm, the assertion is obvious.

Let now be a non-zero irreversible Gaussian number, and for all Gaussian numbers with a norm less than the assertion is proved.

Let us show the possibility of decomposition into prime factors. To do this, we denote by the irreversible divisor having the smallest norm. This divisor must be a prime number by Statement 5. Then. Thus, we have and, by the inductive hypothesis, is representable as a product of prime numbers. Hence, decomposes into the product of these simple and.

Let us show the uniqueness of the decomposition into prime factors. To do this, we take two arbitrary such expansions:

According to Euclid's lemma, one of the factors in the product must be divisible by. We can assume that it is divisible by, otherwise we renumber. Since they are simple, where is reversible. Reducing both sides of our equality by, we obtain a prime factorization of a number that is less than in norm.

By the inductive assumption, and it is possible to renumber the numbers in such a way that it will be allied with, with, ..., with. Then for this numbering it is also conjunctive with for all from 1 to inclusive. Hence, the decomposition into prime factors is unique.

An example of a one-generated ring overwithout OTA.

Consider. The elements of this ring are numbers of the form where and are arbitrary integers. Let us show that the fundamental theorem of arithmetic does not hold in it. We define the norm of a number in this ring as follows: . This is indeed the norm, since it is not hard to check that. Let and. Then

Notice, that.

Let us show that the numbers in the ring under consideration are prime. Indeed, let be one of them and. Then we have: Since there are no numbers with norm 2 in this ring, then or. Invertible elements will be numbers with a unit norm and only they. This means that in an arbitrary factorization there is an invertible factor, therefore, it is simple.

CHAPTER 2. GAUSSIAN PRIME NUMBERS.

To understand which Gaussian numbers are prime, consider a number of statements.

Theorem 8.

Every prime Gaussian is a divisor of exactly one prime natural.

Proof.

Let be a simple Gaussian, then. According to the fundamental theorem, the arithmetic of natural numbers is decomposed into a product of prime natural numbers. And by Euclid's lemma, at least one of them is divisible by.

Let us now show that a simple Gaussian cannot divide two distinct prime natural numbers. Indeed, even if there are distinct prime natural numbers divisible by . Since gcd()=1, then, by the theorem on the representation of gcd in integers, there exist and are integers such that. Hence, which is contrary to simplicity.

Thus, decomposing each simple natural into simple Gaussians, we enumerate all simple Gaussians, and without repetitions.

The following theorem shows that each prime natural number "obtains" at most two simple Gaussian ones.

Theorem 9.

If a simple natural factor is decomposed into a product of three simple Gaussian factors, then at least one of the factors is invertible.

Proof.

Let is a simple natural such that . Turning to the rules, we get:

.

From this equality in natural numbers it follows that at least one of the norms is equal to 1. Therefore, at least one of the numbers -- reversible.

Lemma 10.

If a Gaussian number is divisible by a prime number, then u.

Proof.

Let , that is . Then , , that is , .

C.T.D.

Lemma 11.

For a prime natural number of the form, there exists a natural such that.

Proof.

Wilson's Theorem states that an integer is prime if and only if. But from here. Expand and transform the factorial:

Hence we obtain that, i.e. .

Thus, we got that , Where = .

Now we are ready to describe all simple Gaussian numbers.

Theorem 12.

All simple Gaussians can be divided into three groups:

1). Simple natural species are simple Gaussian;

2). Two is allied with the square of a prime Gaussian number;

3). Simple natural types are decomposed into the product of two simple conjugate Gaussian ones.

Proof.

1). We assume that a simple natural kind is not a simple Gaussian. Then , and And . Let's move on to the rules: . Taking these inequalities into account, we obtain , that is is the sum of the squares of two integers. But the sum of the squares of whole numbers cannot give a remainder of 3 when divided by 4.

2). notice, that

.

Number is a simple Gaussian, since otherwise the two would be decomposed into three irreversible factors, which contradicts Theorem 9.

3). Let simple natural look , then by Lemma 11 there exists an integer such that . Let is a simple Gaussian. Because , then by the Euclid lemma on divides at least one of the factors. Let , then there is a Gaussian number such that . Equating the coefficients of the imaginary parts, we get that . Hence, , which contradicts our assumption of simplicity . Means is a composite Gaussian, representable as a product of two simple conjugate Gaussians.

C.T.D.

Statement.

A Gaussian number conjugate to a prime is itself prime.

Proof.

Let the prime number be Gaussian. Assuming that the composite, that is. Then consider the conjugate:, that is, presented as a product of two irreversible factors, which cannot be.

Statement.

A Gaussian number whose norm is a prime natural number is a Gaussian prime number.

Proof.

Let a composite number, then. Let's look at the rules.

That is, we got that the norm is a composite number, and by condition it is a prime number. Therefore, our assumption is not true, and there is a prime number.

Statement.

If a prime natural number is not a simple Gaussian, then it can be represented as the sum of two squares.

Proof.

Let a prime natural number and not be a simple Gaussian. Then. Since the numbers are equal, their norms are also equal. That is, from here we get.

Two cases are possible:

1). , that is, presented as the sum of two squares.

2). , that is, it means a reversible number, which cannot be, so this case does not satisfy us.

CHAPTER 3. APPLICATION OF GAUSS NUMBERS.

Statement.

The product of numbers representable as a sum of two squares is also representable as a sum of two squares.

Proof.

Let's prove this fact in two ways, using Gaussian numbers, and without using Gaussian numbers.

1. Let, be natural numbers representable as the sum of two squares. Then, and. Consider the product, that is, presented as a product of two conjugate Gaussian numbers, which is represented as the sum of two squares of natural numbers.

2. Let . Then

Statement.

If, where is a simple natural of the form, then and.

Proof.

It follows from the condition that in this case, too, is a simple Gaussian. Then, by Euclid's lemma, one of the factors is divisible by. Suppose then, by Lemma 10, we have that and.

Let us describe the general form of natural numbers representable as the sum of two squares.

Fermat's Christmas theorem or Fermat's theorem--Euler.

A nonzero natural number can be represented as a sum of two squares if and only if in the canonical decomposition all prime factors of the form are in even powers.

Proof.

Note that 2 and all prime numbers of the form can be represented as the sum of two squares. Let there be prime factors of the form in the canonical decomposition of a number that occur in an odd degree. We put in brackets all the factors representable as the sum of two squares, then the factors of the form will remain, and all in the first degree. Let us show that the product of such factors cannot be represented as a sum of two squares. Indeed, if we assume that, then we have that one of the factors or should divide, but if one of these Gaussian numbers divides, then it must also divide the other, as conjugate to it. That is, and, but then it should be in the second degree, and it in the first. Therefore, the product of any number of prime factors of the form of the first degree cannot be represented as a sum of two squares. This means that our assumption is not true and all prime factors of the form in the canonical decomposition of the number enter in even powers.

Task 1.

Let's see the application of this theory on the example of solving the Diaphantian equation.

Solve in integers.

Note that the right side can be represented as a product of conjugate Gaussian numbers.

That is. Let it be divisible by some simple Gaussian number, and the conjugate is also divisible by it, that is. If we consider the difference of these Gaussian numbers, which should be divisible by, we get that it should divide 4. But, that is, allied with.

All prime factors in the decomposition of the number are included in the power of a multiple of three, and factors of the form, in the power of a multiple of six, since a simple Gaussian number is obtained from the decomposition into simple Gaussian 2, but, therefore. How many times it occurs in the decomposition into prime factors of a number, the same number of times it occurs in the decomposition into prime factors of a number. Because it is divisible by if and only if it is divisible by. But allied with That is, they will be distributed equally, which means that they will be included in the expansions of these numbers in powers of a multiple of three. All other prime factors included in the decomposition of a number will only enter either in the decomposition of a number or a number. This means that in the expansion into simple Gaussian factors of a number, all factors will be included in a power of a multiple of three. Therefore, the number is a cube. Thus we have that. From here we get that, that is, it must be a divisor of 2. Hence, or. From where we get four options that satisfy us.

1. , . Where do we find that, .

2. , . Hence, .

3. , . Hence, .

4. , . Hence, .

Task 2.

Solve in integers.

Let's represent the left side as a product of two Gaussian numbers, that is. Let us decompose each of the numbers into simple Gaussian factors. Among the simple ones there will be those that are in the expansion of and. We group all such factors and denote the resulting product. Then only those factors that are not in the expansion will remain in the expansion. All the simple Gaussian factors in the expansion enter to an even degree. Those that are not included in will be present either only in or in. So the number is a square. That is. Equating the real and imaginary parts, we get that, .

Task 3.

The number of representations of a natural number as a sum of two squares.

The problem is equivalent to the problem of representing a given natural number as the norm of some Gaussian number. Let be a Gaussian number whose norm is equal to. Let us decompose into simple natural factors.

Where are prime numbers of the form and are prime numbers of the form. Then, in order to be representable as a sum of two squares, it is necessary that all be even. We decompose the number into simple Gaussian factors, then

where are the simple Gaussian numbers into which they are decomposed.

Comparison of a norm with a number leads to the following relations, which are necessary and sufficient in order to:

The number of views is calculated from the total number of options for selecting indicators. For indicators, there is an opportunity, since the number can be divided into two non-negative terms in the following way:

For a couple of indicators, there is an option, and so on. Combining in all possible ways the allowable values ​​for the indicators, we will get a total of different values ​​for the product of simple Gaussian numbers, with a norm of the form or 2. The indicators are chosen uniquely. Finally, four meanings can be given to the reversible: Thus, there are all possibilities for a number, and therefore, a number in the form of the norm of a Gaussian number, that is, in the form can be represented in ways.

In this calculation, all solutions of the equation are considered different. However, some solutions can be seen as defining the same representation as the sum of two squares. So, if -- solutions to the equation, then you can specify seven more solutions that determine the same representation of the number as the sum of two squares: .

Obviously, out of eight solutions corresponding to one representation, only four different ones can remain if and only if or, or. Such representations are possible if a full square or a doubled full square, and moreover, there can be only one such representation: .

Thus, we have the following formulas:

If not all are even and

If all are even.

Conclusion.

In this paper, we studied the theory of divisibility in the ring of Gaussian integers, as well as the nature of Gaussian prime numbers. These questions are covered in the first two chapters.

The third chapter considers the application of Gauss numbers to the solution of well-known classical problems, such as:

· The question of the possibility of representing a natural number as a sum of two squares;

· The problem of finding the number of representations of a natural number as a sum of two squares;

· Finding general solutions of the indefinite Pythagorean equation;

and also to the solution of the Diaphantine equation.

I also note that the work was performed without the use of additional literature.

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