Ring of polynomials with integer coefficients. Finite fields based on polynomial rings

The ring of polynomials over a field (as opposed to the case of polynomials over a ring) has a number of specific properties close to the properties of the ring of integers Z. Divisibility of polynomials. The well-known method of division by “angle” for polynomials over the field R uses only arithmetic operations on the coefficients and is therefore applicable to polynomials over any field k. It makes it possible for two non-zero polynomials p,sk[x] to construct polynomials q (incomplete quotient) and r (remainder) such that p = q*s +r, and either r =0 or deg(r)< deg(s). Если r =0 , то говорят, что s делит p (или является делителем p) и обозначают это так: s | p. Будем называть многочлен унитарным (или приведенным), если его старший коэффициент равен 1. Определение. Общим наибольшим делителем ненулевых многочленов p и s называется такой унитарный многочлен ОНД(p, s), что 1. ОНД(p, s) | p; ОНД(p, s) | s. 2. q | p, q | s q | ОНД(p, s). По определению, для ненулевого многочлена р со старшим коэффициентом а ОНД (р, 0) = ОНД (0, р) = р/а; ОНД (0, 0)=0. Аналогично определяется ОНД любого числа многочленов. Единственность ОНД двух многочленов непосредственно вытекает из определения. Существование его следует из следующего утверждения. Основная теорема теории делимости (для многочленов). Для любых двух ненулевых многочленов p и q над полем k можно найти такие многочлены u и v над тем же полем, что ОНД(p, q)= u*p+v*q. Доказательство этой теоремы очень похоже на приведенное в лекции доказательство аналогичной теоремы над Z. Все же наметим основные его шаги. Выберем такие многочлены u и v чтобы сумма w= u*p+v*q имела возможно меньшую степень(но была ненулевой!). Можно при этом считать w унитарным многочленом. Проверим, что w | p. Выполняя деление с остатком, получаем: p= s*w+r. Подставляя это равенство в исходное, находим: r = p - s*w =p - s*(u*p+v*q) = (1-s*u)*p+(-s*v)q = U*p + V*q . Если при этом r 0, то deg(r) Замечание. Используя индукцию, можно доказать, что для любого числа многочленов ОНД для подходящих многочленов. Более того, эта формула сохраняется даже для бесконечного множества многочленов, поскольку их ОНД в действительности является ОНД некоторого их конечного подмножества.

Consequence. Every ideal in the ring of polynomials over a field is principal. In fact, let p be the GND of all polynomials included in the ideal I. Then, where By the definition of an ideal it follows that, and therefore, I =(p). Factorization. Let k be some field, p, q, s be polynomials over k. If p=q*s, and both polynomials q and s have degree less than p, then the polynomial p is called reducible (over the field k). Otherwise p is irreducible. An irreducible polynomial in the ring k[x] is the analogue of a prime number in the ring Z. It is clear that every non-zero polynomial p= can be expanded into a product: p= *, where all polynomials are irreducible over k and have a leading coefficient equal to 1. It can be proven that such an expansion is unique up to the order of the factors. Of course, among these factors there may be identical ones; such factors are called multiples. By combining multiple factors, the same expansion can be written in the form: p= 0. Examples. 1. . Note that polynomials of the first degree are, by definition, irreducible over any field. The factor x is a multiple, the rest are primes. 2. The polynomial is irreducible over the field Q of rational numbers. In fact, if ()=(x-a)*q, then substituting x=a into this equality, we obtain: , which is impossible for any rational number a. We present the same polynomial over the field R of real numbers: , and the second factor has a negative discriminant and therefore cannot be further expanded over R. Finally, over the field C of complex numbers we have: , where = is the cubic root of 1. In this example, we see that the concept of reducibility significantly depends on the field over which the polynomial is considered. Properties of irreducible polynomials. 1.If p is an irreducible polynomial and d = OND(p, q) 1, then p | q. Indeed, p = d*s and if deg(s)>0, then this contradicts the irreducibility of p, and if deg(s)=0, then d | qp | q. 2. If p | and p is irreducible, then either p | or p | . Indeed, otherwise gcd(p,) = gcd(p,) =1 and therefore, by the main theorem of divisibility theory, whence: and that means, that is gcd(p,)=1 and, therefore, deg (p)=0 .

LECTURE7.

Ring of polynomials in one unknown

Definition of a polynomial . From the school course we know the problem of solving a second degree equation of the form

Where
. Solving equation (7.1) means finding such a value of the unknown , which, when substituted into equation ( predicate ) (7.1) turns it into a numerical identity ( into a true statement ).

Example 7.1. Find the truth set of a predicate

.

Solution. Consider the identical transformation of the right side of the specified predicate:

.

Equating the last expression to zero, we obtain the formula

,

which gives the values ​​of the unknowns that reverse the predicate
into a true statement. Therefore, the truth set predicate
in general it consists of two elements

,

the values ​​of which are calculated through the values ​​of the coefficients of the quadratic trinomial
. Expression
, standing under the square root sign, is called discriminant equations
. Three cases are possible:

1)
– in this case the truth set of the predicate consists of one real number
(quadratic equation
has one real root);

2)
– in this case, the truth set of the predicate consists of two real numbers, which are calculated using the formulas written above (quadratic equation
has two real roots);

3)
– in this case, the truth set of the predicate consists of two complex conjugate numbers:

(the equation
has complex conjugate roots).

In the general case, we arrive at the problem of solving equations - th degree relative to one unknown

odds
which we will consider arbitrary complex numbers , and the leading coefficient
. Solving equation (7.2) means finding such values ​​of the unknown , which, when substituted into equation (7.2), turn it into a numerical identity. The problem of solving equation (7.2) is replaced by a more general problem studying the left side of this equation .

Definition 7.1. Polynomial , orpolynomial of degree from one unknown (or letters )is called a formal expression of the form

, (7.3)

that is, the formal algebraic sum of integer non-negative powers of the unknown , taken with some, generally speaking, complex coefficients , ,
, ,
.

Polynomials are denoted by various letters of the Latin and Greek alphabets, both large and small.

Polynomial degree (7.3) is called the highest degree unknown , at which the coefficient
. Polynomial null degrees is a polynomial consisting of one non-zero complex number. The number zero is also a polynomial the extent of which is undetermined .

Polynomial degree , if necessary, is indicated by a subscript, for example
, or symbol
. Along with writing polynomials in the form (7.3), the form of notation in increasing powers is often used , that is

Equality, sum and product of polynomials . Polynomials can be compared and operations of addition and multiplication can be performed on them.

Definition 7.2. Two polynomials
And
are considered
equal and write
if and only if their coefficients are equal for the same degrees of the unknown
.

No polynomial, at least one coefficient of which is nonzero, can be equal to zero. Therefore, the equal sign in the equation th degree has nothing to do with the equality of polynomials.

In mathematical analysis, equality of polynomials
is considered as the equality of two functions, that is,


.

If polynomials are equal in the sense of Definition 7.2, then they are also equal in the sense of equality of functions. The converse is a consequence of the fundamental theorem of polynomial algebra formulated below.

Let us introduce two algebraic operations on polynomials with complex (in the general case) coefficients: addition And multiplication .

Definition 7.3. Let two polynomials be given

,
,

,
.

For definiteness, let us put
.
Amount of these polynomials is called a polynomial

whose coefficients are equal to the sum of the coefficients for the same degrees of the unknown :


.

Moreover, if
believe
.

Note that the degree of the sum of two polynomials at
equal to , and when
may be less , namely when
.

Definition 7.4. The work polynomials

,
,

,

called a polynomial

whose coefficients are found by the formula


, .
(7.4)

Thus, the coefficient of the product of two polynomials with index
equal to the sum of all possible products of coefficients of polynomials
And
, the sum of the indices is equal to , namely:

,
,
,
.

From the last equality we have
. Hence, the degree of the product of two polynomials is equal to the sum of the degrees of these polynomials:

By definition, it is believed that degree of polynomial

.

We got the following result.

Lemma 7.1. Let
And
– two polynomials. Then their product
.

Example 7.2. Let two polynomials of different degrees be given, for example,

,
.

Then their sum and product are, respectively:

.

So, in the set of polynomials with complex coefficients, two binary algebraic operations are introduced – addition And multiplication . The properties of these operations are established by the following theorem.

Theorem 7.1. The set of all polynomials with complex coefficients is a commutative and associative ring with identity.

The proof of the theorem reduces to checking the axioms of the ring, and we will omit it. Let us only note that the zero for the addition operation is the number (polynomial) , and the unit for the multiplication operation is a number (polynomial) .

The polynomial ring is denoted by
, Where
– symbol of the field over which the polynomial is defined. Thus, Theorem 7.1 states: the set of all polynomials with complex coefficients is a ring
.

Divisibility of polynomials . Polynomial
has an inverse polynomial
, if and only if
– polynomial of zero degree. Indeed, if
, then the inverse polynomial
. If
, then the degree of the left side
provided that
exists, there must be no less
, but the right-hand side of the last equality is a polynomial of degree zero. So, in the polynomial ring
There is no inverse division operation for the multiplication operation
. In the polynomial ring, however, there is division algorithm with remainder .

Theorem 7.2. For any two polynomials
And
there are such polynomials
And
, What

, (7.5)

where, or
. Representation (7.5) is unique
.

Proof. Let
And
. Let's represent polynomials
And
as

If
or
, then we put in (7.5)

,
.

Then, obviously, (7.5) is satisfied. Therefore let's assume that
. Let's put:

. (7.6)

Let us denote the leading coefficient of the polynomial
through . It's obvious that
. If
, then we put:

. (7.7)

The leading coefficient of the polynomial
let's denote . If
, then let's put it again

(7.8)

and so on. Degrees
polynomials
, obviously decrease. After a finite number of steps we get

, (7.9)

where or
, or
. After this, the process stops.

Adding equalities (7.6) – (7.9) we get

Denoting the amount in parentheses
, A
, we obtain (7.5), and either
, or degree
.

Let us prove the uniqueness of (7.5). Let

where or
, or . From (7.5) and (7.11) we have:

The degree of the polynomial on the left side of the last equality is not less than the degree
, and the degree of the polynomial on the right side is either zero or less than the degree
. Therefore, the last equality is satisfied only for the equalities

,
.

Polynomial
in formula (7.5) is called private from dividing a polynomial
to a polynomial
, and the polynomial
called the remainder from this division. If
, then they say that the polynomial
divisible by a polynomial
which is called divisor of a polynomial
. Let's find out when a polynomial
divisible by a polynomial
.

Theorem 7.3. Polynomial
divisible by a polynomial

if and only if such a polynomial exists
, What

. (7.12)

Proof. Indeed, if
divided by
, then as
you should take the quotient of the division
on
. Conversely, let a polynomial satisfying equality (7.12) exist. Then from what was proved in Theorem 7.1. uniqueness of polynomials
And
in representation (7.5) and the conditions that the degree
less degree
, it follows that the quotient of division
on
equals
, and the remainder
.

Corollary to Theorem 7.3.If a polynomial
and its divisor
have rational or real coefficients, then the quotient
will also have rational or real coefficients.

Example 7.3. Perform division with remainder of polynomial

to a polynomial
.

Solution. The division algorithm (7.6) – (7.9) is implemented in the form “ dividing by a corner »:

So, the quotient
, remainder
. Therefore, we have the following representation

which can be verified by direct multiplication.

Definition 7.5. Let
And
– two polynomials. Polynomial
called
greatest common divisor (GCD)of these polynomials if it is their common divisor and is itself divisible by any other common divisor of these polynomials.

GCD of polynomials
And
denoted by . Let us formulate and prove a theorem that gives a constructive algorithm for finding GCD for any two polynomials.

Theorem 7.4 (Euclidean algorithm). For any two polynomials
And
there is a greatest common divisor

Proof. First let's formulate Euclidean algorithm finding
, and then we will prove that the polynomial obtained in the process of implementing this algorithm is the greatest common divisor of the two given polynomials.

First we divide the polynomial
to a polynomial
and in the general case we obtain some remainder
. Next we divide
on
and get the remainder
, divide
on
and get the remainder
and so on. As a result of such successive divisions we arrive at the remainder
, by which the previous remainder is divided
. This remainder will be the greatest common divisor of these polynomials.

To prove it, we write out the chain of divisions sequentially:

The last equality shows that
is a divisor for
. Therefore, both terms on the right side of the penultimate equality are divided by
and therefore on
shares and
. Moving up the chain of divisions, we get that
is a divisor for
,
,
,
. From the second equality of the chain we see that
is a divisor for
and, therefore, based on the first equality – for
. So,
is the common divisor for
And
.

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    Subtitles

Polynomials in one variable over a field

Polynomials

Polynomial from x with coefficients in the field k is an expression of the form

p = p m x m + p m − 1 x m − 1 + ⋯ + p 1 x + p 0 , (\displaystyle p=p_(m)x^(m)+p_(m-1)x^(m-1)+\ cdots +p_(1)x+p_(0),)

Where p 0 , …, p m - elements k, odds p, A x, x 2 , … - formal symbols (“degrees x"). Such expressions can be added and multiplied according to the usual rules for operations with algebraic expressions (commutativity of addition, distributivity, reduction of similar terms, etc.). Members p k xk with zero coefficient p k are usually omitted when recording. Using the sum symbol, polynomials are written in a more compact form:

p = p m x m + p m − 1 x m − 1 + ⋯ + p 1 x + p 0 = ∑ k = 0 m p k x k . (\displaystyle p=p_(m)x^(m)+p_(m-1)x^(m-1)+\cdots +p_(1)x+p_(0)=\sum _(k=0 )^(m)p_(k)x^(k).)

Polynomial ring

It is easy to see that the set of all polynomials with coefficients in k (\displaystyle k) forms a commutative ring, denoted k [ x ] (\displaystyle k[x]) and called ring of polynomials over k (\displaystyle k) . Symbol x (\displaystyle x) usually called a "variable", this terminology arose from consideration polynomial functions above R (\displaystyle \mathbb (R) ) or above C (\displaystyle \mathbb (C) ). However, in general, polynomials and polynomial functions are different things; for example, over a finite field F p (\displaystyle \mathbb (F)_(p)) from a prime number p (\displaystyle p) elements polynomials x 1 (\displaystyle x^(1)) And x p + 1 (\displaystyle x^(p+1)) they define the same function, but these are different polynomials (polynomials are considered equal if and only if all their coefficients coincide). Therefore, the variable x (\displaystyle x) cannot be considered belonging to the field k (\displaystyle k); about the ring k [ x ] (\displaystyle k[x]) you can think like this: we add a new element to the set of field elements x (\displaystyle x) and we only require that the ring axioms be satisfied and that x (\displaystyle x) commuted with field elements.

Since the elements of the polynomial ring can be multiplied by "scalars" from the field k (\displaystyle k), it is actually an associative algebra over the field k (\displaystyle k). If we consider k [ x ] (\displaystyle k[x]) How vector space(that is, “forget” about multiplication), it has an infinite basis of elements 1 = x 0 (\displaystyle 1=x^(0)), x = x 1 (\displaystyle x=x^(1)), x 2 (\displaystyle x^(2)) etc.

Primer decomposition in k[x]

Ring factor k[x]

L ≃ k [ x ] / (p) . (\displaystyle L\simeq k[x]/(p).)

Important special case- when the ring containing k, itself is a field; let's denote it K. Simplicity of the factor module by (p) (\displaystyle (p)) is equivalent to irreducibility p (\displaystyle p). The primitive element theorem states that any finite separable extension can be generated by a single element, and therefore has the form of the quotient of a ring of polynomials over a smaller field by an irreducible polynomial. An example is the field of complex numbers, which is generated over R element i, such that i 2 + 1 = 0. Accordingly, the polynomial x 2 + 1 irreducible over R And

C ≃ R [ x ] / (X 2 + 1) . (\displaystyle \mathbb (C) \simeq \mathbb (R) [x]/(X^(2)+1).)

More generally, for an arbitrary (even non-commutative) ring A containing k and element a rings A, commuting with all elements k, there is a unique homomorphism of rings from k[x] V A, sending x V a:

ϕ : k [ x ] → A , ϕ (x) = a . (\displaystyle \phi:k[x]\to A,\quad \phi (x)=a.)

The existence and uniqueness of such a homomorphism is expressed using a certain universal property of the polynomial ring and explains a certain “uniqueness” of the polynomial ring in various constructions of ring theory and commutative algebra.

Modules

Ring of polynomials in several variables

Definition

Polynomial from n variables X 1 ,…, X n with coefficients in the field K is defined similarly to a polynomial in one variable, but the notation becomes more complex. For any multi-index α = (α 1 ,…, α n), where each α i is a non-zero integer, let

X α = ∏ i = 1 n X i α i = X 1 α 1 … X n α n , p α = p α 1 … α n ∈ K . (\displaystyle X^(\alpha )=\prod _(i=1)^(n)X_(i)^(\alpha _(i))=X_(1)^(\alpha _(1))\ ldots X_(n)^(\alpha _(n)),\quad p_(\alpha )=p_(\alpha _(1)\ldots \alpha _(n))\in \mathbb (K) .\ )

X α called monomial degrees | α | = ∑ i = 1 n α i (\displaystyle |\alpha |=\sum _(i=1)^(n)\alpha _(i)). Polynomial is a finite linear combination of monomials with coefficients in K: ∑ α p α X α (\displaystyle \sum _(\alpha )p_(\alpha )X^(\alpha )).

Polynomials from n variables with coefficients in the field k(with the usual operations of addition and multiplication) form a commutative ring, denoted k[x 1 ,…, x n]. This ring can be obtained by repeated application of the operation “taking a ring of polynomials over a given ring.” For example, k[x 1 , x 2 ] isomorphic k[x 1 ][x 2 ], as well as k[x 2 ][x 1 ]. This ring plays a fundamental role in algebraic geometry. Many results in commutative algebra were achieved through the study of ideals of this ring and modules over it.

Hilbert's zero theorem

Several fundamental results concerning the relationship between ideals of a ring k[x 1 ,…, x n] and algebraic subvarieties k n known collectively as Hilbert's zero theorem.

  • (weak form, algebraically closed field) Let k- algebraically closed field. Then any maximum ideal m rings k[x 1 ,…, x n] has the form
m = (x 1 − a 1 , … , x n − a n) , a = (a 1 , … , a n) ∈ k n . (\displaystyle m=(x_(1)-a_(1),\ldots ,x_(n)-a_(n)),\quad a=(a_(1),\ldots ,a_(n))\in k^(n).)
  • (weak form, any coefficient field) Let k- field, K- algebraically closed field containing k And I- ideal in a ring k[x 1 ,…, x n]. Then I contains 1 if and only if the polynomials from I do not have a common zero in K n .
  • (strong form) Let k- field, K- algebraically closed field containing k, I- ideal in a ring k[x 1 ,…, x n] And V(I) - algebraic subvariety, K n certain I. Let f- a polynomial equal to zero at all points V(I). Then some degree f belongs to the ideal I.
Using the definition of a radical ideal, this theorem states that f belongs to the radical I. An immediate consequence of this form of the theorem is the existence of a bijective correspondence between radical ideals K[x 1 ,…, x n] and algebraic subvarieties n-dimensional affine space K n .

Chapter XI. Polynomials.

Ring of polynomials in one variable over

Associative-commutative ring with identity

Definition 1. Let K- associative-commutative ring with identity. Polynomial over the ring K in the variable x is called an expression of the form , where a iÎ K, and only a finite number of elements a i≠0.

a i called coefficient of the polynomial f(x)at degree i.

The set of all polynomials over the ring K in the variable x denoted by K[x].

Definition 2. Let f(x) And g(x) , Where K is an associative-commutative ring with identity. Polynomials f(x) And g(x) are called equal(algebraically), if their coefficients are respectively equal for the same degrees x.

Definition 3. Zero polynomial is a polynomial whose coefficients are all equal to 0, and is denoted 0=0( x).

Definition 4. Let K- f(x) , f(x)≠0(x). Number n called degree of the polynomial f and is designated deg f =n, if a n≠0 and a i=0 at i>n.

By definition, it is assumed that the degree of the zero polynomial is equal to , i.e. deg 0(x) .

Thus, if , then deg(degℕ {0}).

According to Definition 2, by adding or discarding terms with zero coefficients, we obtain a polynomial equal to the given one. Thus, every polynomial of degree n can be written as

Then a 0 called free or permanent member polynomial f(x), a n - senior coefficient polynomial f(x).

Definition 5. Let K- associative-commutative ring with identity, , , and nm.

Operations of addition and multiplication of polynomials from K[x] are determined by the rules

Theorem 1 . Let K be a nonzero associative-commutative ring with identity. Then K[x]regarding operations according to the rules(1 )And(2 )– is also an associative-commutative ring with identity 1(x)= 1.

Proof. Let's check for K[x] all axioms of an associative-commutative ring with identity.

1. K[x]¹Æ, for example, 0( xK[x], since all its coefficients are equal to 0О K.

2. Operations “+” and “⋅” according to rules (1) and (2) are algebraic on K[x] (i.e. K[x] is closed under these operations). Indeed, let f(x)And g(xK[x], from formulas (1) and (2) it follows that the coefficients of the polynomials f(x)+g(x)And f(x)⋅g(x) are obtained by adding and multiplying the coefficients f(x)And g(x), those. elements from K. Due to the closedness of the ring K regarding addition and multiplication, coefficients of polynomials f(x)+g(x)And f(x)⋅g(x) belong K. That is f(x)+g(xK[x]And f(x)⋅g(xK[x].



3. [ x ], +> is an Abelian group.

a) “+” is associative on K[x]: "f(x),g(x),h(xK[x] (f(x)+g(x))+h(x)=f(x)+(g(x)+h(x))

b) “+” is commutative on K[x]: "f(x),g(xK[x] f(x)+g(x)=g(x)+f(x)

c) There is 0( x)=0+0⋅x+0⋅x 2 +…+0⋅x n+… Î K[x] such that " О K[x] : =

similarly,

d) "О K[x] exists О K[x] such that

= 0+0⋅x+0⋅x 2 +…+0⋅x n = 0(x). Likewise = 0(x).

4. IN K[x]distributive laws are fulfilled:

d) " f(x),g(x),h(xK[x] (f(x)+g(x))⋅h(x)=f(x)⋅h(x)+g(x)⋅h(x)

h(x) ⋅ (f(x)+g(x))=h(x)⋅f(x)+h(x)⋅g(x)

Thus, K[x] - ring.

5. Let's show thatK[x]– associative-commutative ring with 1.

f) “⋅” is associative on K[x]: "f(x),g(x),h(xK[x] (f(x)⋅g(x))⋅h(x)=f(x)⋅(g(x)⋅h(x))

g) “⋅” is commutative on K[x]: "f(x),g(xK[x] f(x)⋅g(x)=g(x)⋅f(x)

h) B K[x]there is a unit polynomial 1( x)= 1+0⋅x+0⋅x 2 +…+0⋅x n +…Î K[x]c coefficients b 0 =1, b i=0 for others i. " Î K[x]

the validity of a), b), e), f), g) follows from the fact that the operations “+” and “⋅” on polynomials are reduced to the corresponding operations on their coefficients - elements from K, and in the ring K“+” and “⋅” are commutative, associative and distributive laws are satisfied.

The theorem has been proven.

Polynomial degree. Properties of the degree of a polynomial

Theorem 2 . Let K be a nonzero associative-commutative ring with identity, , . Then:

1) deg(+ max(deg, deg);

Material from Wikipedia - the free encyclopedia

Polynomials in one variable over a field

Polynomials

Polynomial from x with coefficients in the field k is an expression of the form

p = p_m x^m + p_(m - 1) x^(m - 1) + \cdots + p_1 x + p_0,

Where p 0 , …, p m - elements k, odds p, A x, x 2 , … - formal symbols (“degrees x"). Such expressions can be added and multiplied according to the usual rules for operations with algebraic expressions (commutativity of addition, distributivity, reduction of similar terms, etc.). Members p k xk with zero coefficient p k are usually omitted when recording. Using the sum symbol, polynomials are written in a more compact form:

p = p_m x^m + p_(m - 1) x^(m - 1) + \cdots + p_1 x + p_0 = \sum_(k=0)^m p_k x^k.

Polynomial ring k[x]

It is easy to see that the set of all polynomials with coefficients in K forms a commutative ring, denoted k[x] and called ring of polynomials over k . Symbol x usually called a "variable", this terminology arose from consideration polynomial functions above R or above C. However, in general, polynomials and polynomial functions are different things; for example, over a finite field \mathbb F_p prime polynomials x And x^p they define the same function, but these are different polynomials (polynomials are considered equal if and only if all their coefficients coincide). Therefore, the variable x cannot be considered belonging to the field k; about the ring k[x] you can think like this: we add a new element to the set of field elements x and we only require that the ring axioms be satisfied and that x commuted with field elements.

Since the elements of the polynomial ring can be multiplied by "scalars" from the field k, it is actually an associative algebra over the field k. If we consider k[x] as a vector space (that is, “forget” about multiplication), it has an infinite basis of elements 1, x, x 2, etc.

Primer decomposition in k[x]

Factor rings k[x]

L\simeq k[x]/(p).

An important special case is when a ring containing k, itself is a field; let's denote it K. Simplicity of the factor module by (p) is equivalent to irreducibility p. The primitive element theorem states that any finite separable extension can be generated by a single element, and hence has the form of the quotient of a polynomial ring over a smaller field by an irreducible polynomial. An example is the field of complex numbers, which is generated over R element i, such that i 2 + 1 = 0. Accordingly, the polynomial x 2 + 1 irreducible over R And

\mathbb(C) \simeq \mathbb(R)[x]/(X^2+1).

More generally, for an arbitrary (even non-commutative) ring A containing k and element a rings A, commuting with all elements k, there is a unique homomorphism of rings from k[x] V A, sending x V a:

\phi: k[x]\to A, \quad \phi(x)=a.

The existence and uniqueness of such a homomorphism is expressed using a certain universal property of the polynomial ring and explains a certain "uniqueness" of the polynomial ring in various constructions of ring theory and commutative algebra.

Modules

Ring of polynomials in several variables

Definition

Polynomial from n variables X 1 ,…, X n with coefficients in the field K is defined similarly to a polynomial in one variable, but the notation becomes more complex. For any multi-index α = (α 1 ,…, α n), where each α i is a non-zero integer, let

X^\alpha = \prod_(i=1)^n X_i^(\alpha_i) =

X_1^(\alpha_1)\ldots X_n^(\alpha_n), \quad p_\alpha = p_(\alpha_1\ldots\alpha_n)\in\mathbb(K).\

X α called monomial degrees |\alpha| = \sum_(i=1)^n \alpha_i. Polynomial is a finite linear combination of monomials with coefficients in K: \sum_\alpha p_\alpha X^\alpha.

Polynomials from n variables with coefficients in the field k(with the usual operations of addition and multiplication) form a commutative ring, denoted k[x 1 ,…, x n]. This ring can be obtained by repeated application of the operation “taking a ring of polynomials over a given ring.” For example, k[x 1 , x 2 ] isomorphic k[x 1 ][x 2 ], as well as k[x 2 ][x 1 ]. This ring plays a fundamental role in algebraic geometry. Many results in commutative algebra were achieved through the study of ideals of this ring and modules over it.

Hilbert's zero theorem

Several fundamental results concerning the relationship between ideals of a ring k[x 1 ,…, x n] and algebraic subvarieties k n known collectively as Hilbert's zero theorem.

  • (weak form, algebraically closed field) Let k is an algebraically closed field. Then any maximal ideal m rings k[x 1 ,…, x n] has the form
m = (x_1-a_1, \ldots, x_n-a_n), \quad a = (a_1, \ldots, a_n) \in k^n.
  • (weak form, any coefficient field) Let k- field, K is an algebraically closed field containing k And I- ideal in a ring k[x 1 ,…, x n]. Then I contains 1 if and only if the polynomials from I do not have a common zero in K n .
  • (strong form) Let k- field, K is an algebraically closed field containing k, I- ideal in a ring k[x 1 ,…, x n] And V(I) - algebraic subvariety, K n certain I. Let f- a polynomial equal to zero at all points V(I). Then some degree f belongs to the ideal I.
Using the definition of an ideal radical, this theorem states that f belongs to the radical I. An immediate consequence of this form of the theorem is the existence of a bijective correspondence between radical ideals K[x 1 ,…, x n] and algebraic subvarieties n-dimensional affine space K n .

see also

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Literature

  • Lam, Tsit-Yuen (2001), A First Course in Noncommutative Rings, Berlin, New York: Springer-Verlag, ISBN 978-0-387-95325-0
  • Lang, Serge(2002), Algebra, Graduate Texts in Mathematics 211 (Revised third ed.), New York: Springer-Verlag - ISBN 978-0-387-95385-4, MR1878556
  • Osborne, M. Scott (2000), Basic homological algebra, vol. 196, Graduate Texts in Mathematics, Berlin, New York: Springer-Verlag, ISBN 978-0-387-98934-1

An excerpt characterizing the Ring of Polynomials

-Where is your head lying? - Nikolai asked, approaching a hundred paces towards the suspicious hunter. But before the hunter had time to answer, the hare, sensing the frost by tomorrow morning, could not stand still and jumped up. A pack of hounds on bows, with a roar, rushed downhill after the hare; from all sides the greyhounds, who were not in the pack, rushed at the hounds and the hare. All these slowly moving hunters are screaming: stop! knocking down the dogs, the greyhounds shout: atu! guiding the dogs, they galloped across the field. Calm Ilagin, Nikolai, Natasha and uncle flew, not knowing how or where, seeing only dogs and a hare, and only fearing to lose sight of the course of the persecution even for a moment. The hare was seasoned and playful. Jumping up, he did not immediately gallop, but moved his ears, listening to the screaming and stomping that suddenly came from all sides. He jumped ten times slowly, allowing the dogs to approach him, and finally, having chosen the direction and realizing the danger, he put his ears to the ground and rushed at full speed. He was lying on the stubble, but in front there were green fields through which it was muddy. The two dogs of the suspicious hunter, who were closest, were the first to look and lay after the hare; but they had not yet moved far towards him, when the Ilaginskaya red-spotted Erza flew out from behind them, approached a dog's distance, with terrible speed attacked, aiming at the hare's tail and thinking that she had grabbed it, rolled head over heels. The hare arched his back and kicked even harder. Wide-bottomed, black-spotted Milka came out from behind Erza and quickly began to sing to the hare.
- Honey! mother! – Nikolai’s triumphant cry was heard. It seemed that Milka would strike and catch the hare, but she caught up and rushed past. The Rusak moved away. The beautiful Erza swooped in again and hung over the hare’s very tail, as if trying to grab him by the back thigh so as not to make a mistake now.
- Erzanka! sister! – Ilagin’s voice was heard crying, not his own. Erza did not heed his pleas. At the very moment when one should have expected her to grab the hare, he whirled and rolled out to the line between the greenery and the stubble. Again Erza and Milka, like a pair of drawbars, aligned themselves and began to sing to the hare; at the turn it was easier for the hare; the dogs did not approach him so quickly.
- Scold! Swearing! Pure march! - shouted at that time another new voice, and Rugai, his uncle’s red, humpbacked dog, stretching out and arching his back, caught up with the first two dogs, moved out from behind them, kicked with terrible selflessness right over the hare, knocked him off the line onto the green, Another time he pushed even harder through the dirty greens, drowning up to his knees, and you could only see how he rolled head over heels, getting his back dirty in the mud, with the hare. The star of dogs surrounded him. A minute later everyone was standing near the crowded dogs. One happy uncle got down and walked away. Shaking the hare so that the blood would drain, he looked around anxiously, running his eyes, unable to find a position for his arms and legs, and spoke, not knowing with whom or what.
“This is a matter of march... here is a dog... here he pulled out everyone, both thousandths and rubles - a pure matter of march!” he said, gasping and looking around angrily, as if scolding someone, as if everyone were his enemies, everyone had offended him, and only now he finally managed to justify himself. “Here are the thousandths for you - a pure march!”
- Scold me, fuck off! - he said, throwing the cut-off paw with the earth stuck on it; – deserved it – pure march!
“She pulled out all the stops, gave three runs on her own,” Nikolai said, also not listening to anyone, and not caring whether they listened to him or not.
- What the hell is this! - said Ilaginsky the stirrup.
“Yes, as soon as she stopped short, every mongrel will catch you from stealing,” said Ilagin at the same time, red-faced, barely catching his breath from the galloping and excitement. At the same time, Natasha, without taking a breath, squealed joyfully and enthusiastically so shrilly that her ears were ringing. With this screech she expressed everything that other hunters also expressed in their one-time conversation. And this squeal was so strange that she herself should have been ashamed of this wild squeal and everyone should have been surprised by it if it had been at another time.
The uncle himself pulled the hare back, deftly and smartly threw him over the back of the horse, as if reproaching everyone with this throwing, and with such an air that he didn’t even want to talk to anyone, sat on his kaurago and rode away. Everyone except him, sad and offended, left and only long after could they return to their former pretense of indifference. For a long time they looked at the red Rugai, who, with his hunchbacked back and dirt stained, rattling his iron, with the calm look of a winner, walked behind the legs of his uncle’s horse.
“Well, I’m the same as everyone else when it comes to bullying. Well, just hang in there!” It seemed to Nikolai that the appearance of this dog spoke.
When, long after, the uncle drove up to Nikolai and spoke to him, Nikolai was flattered that his uncle, after everything that had happened, still deigned to speak with him.

When Ilagin said goodbye to Nikolai in the evening, Nikolai found himself at such a far distance from home that he accepted his uncle’s offer to leave the hunt to spend the night with him (with his uncle), in his village of Mikhailovka.
- And if they came to see me, it would be a pure march! - said the uncle, even better; you see, the weather is wet, the uncle said, if we could rest, the countess would be taken in a droshky. “Uncle’s proposal was accepted, a hunter was sent to Otradnoe to pick up the droshky; and Nikolai, Natasha and Petya went to see their uncle.
About five people, large and small, courtyard men ran out onto the front porch to meet the master. Dozens of women, old, big and small, leaned out from the back porch to watch the approaching hunters. The presence of Natasha, a woman, a lady on horseback, brought the curiosity of the uncle's servants to such limits that many, not embarrassed by her presence, came up to her, looked into her eyes and in her presence made their comments about her, as if about a miracle being shown, which is not a person, and cannot hear or understand what is said about him.
- Arinka, look, she’s sitting on her side! She sits herself, and the hem dangles... Look at the horn!
- Father of the world, that knife...
- Look, Tatar!
- How come you didn’t somersault? – said the bravest one, directly addressing Natasha.
The uncle got off his horse at the porch of his wooden house overgrown with a garden and, looking around at his household, shouted imperiously that the extra ones should leave and that everything necessary for receiving guests and hunting would be done.
Everything ran away. Uncle took Natasha off the horse and led her by the hand along the shaky plank steps of the porch. The house, unplastered, with log walls, was not very clean - it was not clear that the purpose of the people living was to keep it stain-free, but there was no noticeable neglect.
The hallway smelled of fresh apples, and there were wolf and fox skins hanging. Through the front hall the uncle led his guests into a small hall with a folding table and red chairs, then into a living room with a birch round table and a sofa, then into an office with a torn sofa, a worn carpet and with portraits of Suvorov, the owner’s father and mother and himself in a military uniform. There was a strong smell of tobacco and dogs in the office. In the office, the uncle asked the guests to sit down and make themselves at home, and he himself left. Scolding, his back not cleaned, entered the office and lay down on the sofa, cleaning himself with his tongue and teeth. From the office there was a corridor in which screens with torn curtains could be seen. Women's laughter and whispers could be heard from behind the screens. Natasha, Nikolai and Petya undressed and sat on the sofa. Petya leaned on his arm and immediately fell asleep; Natasha and Nikolai sat in silence. Their faces were burning, they were very hungry and very cheerful. They looked at each other (after the hunt, in the room, Nikolai no longer considered it necessary to show his male superiority in front of his sister); Natasha winked at her brother, and both did not hold back for long and laughed loudly, not yet having time to think of an excuse for their laughter.
A little later, the uncle came in wearing a Cossack jacket, blue trousers and small boots. And Natasha felt that this very suit, in which she saw her uncle with surprise and mockery in Otradnoye, was a real suit, which was no worse than frock coats and tails. Uncle was also cheerful; Not only was he not offended by the laughter of his brother and sister (it could not enter his head that they could laugh at his life), but he himself joined in their causeless laughter.
- That’s how the young countess is - a pure march - I’ve never seen another like it! - he said, handing one pipe with a long shank to Rostov, and placing the other short, cut shank with the usual gesture between three fingers.
“I left for the day, at least on time for the man and as if nothing had happened!”
Soon after the uncle, the door opened, obviously a barefoot girl by the sound of her feet, and a fat, ruddy, beautiful woman of about 40, with a double chin, and full, ruddy lips, entered the door with a large tray in her hands. She, with hospitable presence and attractiveness in her eyes and every movement, looked around at the guests and bowed respectfully to them with a gentle smile. Despite her greater-than-usual thickness, which forced her to stick her chest and stomach forward and hold her head back, this woman (the uncle’s housekeeper) walked extremely lightly. She walked up to the table, put down the tray and deftly with her white, plump hands removed and placed bottles, snacks and treats on the table. Having finished this, she walked away and stood at the door with a smile on her face. - “Here I am!” Do you understand uncle now?” her appearance told Rostov. How not to understand: not only Rostov, but also Natasha understood her uncle and the meaning of the frowning eyebrows, and the happy, self-satisfied smile that slightly wrinkled his lips as Anisya Fedorovna entered. On the tray were herbalist, liqueurs, mushrooms, cakes of black flour on yuraga, comb honey, boiled and sparkling honey, apples, raw and roasted nuts and nuts in honey. Then Anisya Fedorovna brought jam with honey and sugar, and ham, and freshly fried chicken.

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