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Wilkins. Введение в теорию Galois

Wilkins. Introduction to Galois theory(45s).pdf

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Date Jan 14, 2003

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A ring R is said to be unital if it possesses a (necessarily unique) non-zero multiplicative identity element 1 satisfying 1x = x = x1 for all x ∈ R. Deﬁnition A unital commutative ring R is said to be an integral domain if the product of any two non-zero elements of R is itself non-zero. Deﬁnition A ﬁeld consists of a set K on which are deﬁned operations of addition and multiplication satisfying the following axioms: • x+y = y +x for all elements x and y of K (i.e., addition is commutative ); • (x + y ) + z = x + (y + z ) for all elements x, y and z of K (i.e., addition is associative ); • there exists an an element 0 of K known as the zero element with the property that x + 0 = x for all elements x of K ; • given any element x of K , there exists an element −x of K with the property that x + (−x) = 0; • xy = y x for all elements x and y of K (i.e., multiplication is commutative ); • x(y z ) = (xy )z for all elements x, y and z of K (i.e., multiplication is associative ); • there exists a non-zero element 1 of K with the property that 1x = x for all elements x of K ; • given any non-zero element x of K , there exists an element x−1 of K with the property that xx−1 = 1; • x(y + z ) = xy + xz and (x + y )z = xz + y z for all elements x, y and z of K (the Distributive Law ). An examination of the relevant deﬁnitions shows that a unital commutative ring R is a ﬁeld if and only if, given any non-zero element x of R, there exists an element x−1 of R such that xx−1 = 1. Moreover a ring R is a ﬁeld if and only if the set of non-zero elements of R is an Abelian group with respect to the operation of multiplication. Lemma 3.3 A ﬁeld is an integral domain....

Proof Let I be the ideal in K [x] generated by f1 , f2 , . . . , fk . It follows from Lemma 3.11 that the ideal I is generated by some polynomial d. Then d divides all of f1 , f2 , . . . , fk and is therefore a constant polynomial, since these polynomials are coprime. It follows that I = K [x]. The existence of the required polynomials g1 , g2 , . . . , gk then follows using Lemma 3.5. Deﬁnition A non-constant polynomial f with coeﬃcients in a ring K is said to be irreducible over K if there does not exist any non-constant polynomial that divides f whose degree is less than that of f . Prop osition 3.13 Let f , g and h be polynomials with coeﬃcients in some ﬁeld K . Suppose that f is irreducible over K and that f divides the product g h. Then either f divides g or else f divides h. Pro of Suppose that f does not divide g . We must show that f divides h. Now the only polynomials that divide f are constant polynomials and multiples of f . No multiple of f divides g . Therefore the only polynomials that divide both f and g are constant polynomials. Thus f and g are coprime. It follows from Proposition 3.12 that there exist polynomials u and v with coeﬃcients in K such that 1 = ug + v f . Then h = ug h + v f h. But f divides ug h + v f h, since f divides g h. It follows that f divides h, as required. Prop osition 3.14 Let K be a ﬁeld, and let (f ) be the ideal of K [x] generated by an irreducible polynomial f with coeﬃcients in K . Then K [x]/(f ) is a ﬁeld. Pro of Let I = (f ). Then the quotient ring K [x]/I is commutative and has a multiplicative identity element I + 1. Let g ∈ K [x]. Suppose that I + g = I . Now the only factors of f are constant polynomials and constant multiples of f , since f is irreducible. But no constant multiple of f can divide g , since g ∈ I . It follows that the only common factors of f and g are constant polynomials. Thus f and g are coprime. It follows from Proposition 3.12 that there exist polynomials h, k ∈ K [x] such that f h + g k = 1. But then (I + k )(I + g ) = I + 1 in K [x]/I , since f h ∈ I . Thus I + k is the multiplicative inverse of I + g in K [x]/I . We deduce that every non-zero element of K [x]/I is invertible, and thus K [x]/I is a ﬁeld, as required....

i < j and p divides ci for all i < k . But p does not divide bj ck since p does not divide either bj or ck . Therefore p does not divide the coeﬃcient aj +k of g h. This shows that the polynomial g h is primitive, as required. Prop osition 3.16 A polynomial with integer coeﬃcients is irreducible over the ﬁeld Q of rational numbers if and only if it cannot be factored as a product of polynomials of lower degree with integer coeﬃcients. Pro of Let f be a polynomial with integer coeﬃcients. If f is irreducible over Q then f clearly cannot be factored as a product of polynomials of lower degree with integer coeﬃcients. Conversely suppose that f cannot be factored in this way. Let f (x) = g (x)h(x), where g and h are polynomials with rational coeﬃcients. Then there exist positive integers r and s such that the polynomials rg (x) and sh(x) have integer coeﬃcients. Let the positive integers u and v be the highest common factors of the coeﬃcients of the polynomials rg (x) and sh(x) respectively. Then rg (x) = ug∗ (x) and sh(x) = v h∗ (x), where g∗ and h∗ are primitive polynomials with integer coeﬃcients. Then (rs)f (x) = (uv )g∗ (x)h∗ (x). We now show that f (x) = mg∗ (x)h∗ (x) for some integer m. Let l be the smallest divisor of rs such that lf (x) = mg∗ (x)h∗ (x) for some integer m. We show that l = 1. Suppose that it were the case that l > 1. Then there would exist a prime factor p of l. Now p could not divide m, since otherwise (l/p)f (x) = (m/p)g∗ (x)h∗ (x), which contradicts the deﬁnition of l. Theorefore p would have to divide each coeﬃcient of g∗ (x)h∗ (x), which is impossible, since it follows from Gauss’s Lemma (Lemma 3.15) that the product g∗ h∗ of the primitive polynomials g∗ and h∗ is itself a primitive polynomial. Therefore l = 1 and f (x) = mg∗ (x)h∗ (x). Now f does not factor as a product of polynomials of lower degree with integer coeﬃcients. Therefore either deg f = deg g∗ = deg g , or else deg f = deg h∗ = deg h, Thus f is irreducible over Q, as required. 11...

If L: K is a ﬁeld extension then we can regard L as a vector space over the ﬁeld K . If L is a ﬁnite-dimensional vector space over K then we say that the extension L: K is ﬁnite. The degree [L: K ] of a ﬁnite ﬁeld extension L: K is deﬁned to be the dimension of L considered as a vector space over K . Prop osition 3.18 (The Tower Law) Let M : L and L: K be ﬁeld extensions. Then the extension M : K is ﬁnite if and only if M : L and L: K are both ﬁnite, in which case [M : K ] = [M : L][L: K ]. Pro of Suppose that M : K is a ﬁnite ﬁeld extension. Then L, regarded as a vector space over K , is a subspace of the ﬁnite-dimensional vector space M , and therefore L is itself a ﬁnite-dimensional vector space over K . Thus L: K is ﬁnite. Also there exists a ﬁnite subset of M which spans M as a vector space over K , since M : K is ﬁnite, and this ﬁnite subset must also span M over L, and thus M : L must be ﬁnite. Conversely suppose that M : L and L: K are both ﬁnite extensions. Let x1 , x2 , . . . , xm be a basis for L, considered as a vector space over the ﬁeld K , and let y1 , y2 , . . . , yn be a basis for M , considered as a vector space over the ﬁeld L. Note that m = [L: K ] and n = [M : L]. We claim that the set of all products xi yj with i = 1, 2, . . . , m and j = 1, 2, . . . , n is a basis for M , considered as a vector space over K . First we show that the elements xi yj are linearly independent over K . im j n Suppose that λij xi yj = 0, where λij ∈ K for all i and j . Then
=1 =1...