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Wilkins. Введение в теорию Galois

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

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The veriﬁcation of the following result is a straightforward exercise. Prop osition 3.7 Let ϕ: R → S be a homomorphism from a ring R to a ring S , and let I be an ideal of R satisfying I ⊂ ker ϕ. Then there exists a unique homomorphism ϕ: R/I → S such that ϕ(I + x) = ϕ(x) for al l x ∈ R. Moreover ϕ: R/I → S is injective if and only if I = ker ϕ. Corollary 3.8 Let ϕ: R → S be ring homomorphism. Then ϕ(R) is isomorphic to R/ ker ϕ....

bj c0 is not divisible by p, since p is prime and neither bj nor c0 is divisible by p. Therefore aj is not divisible by p, and hence j = n and deg g ≥ n = deg f . Thus deg g = deg f and deg h = 0. Thus the polynomial f does not factor as a product of polynomials of lower degree with integer coeﬃcients, and therefore f is irreducible over Q (Proposition 3.16)....

xi yj span M as a vector space over K , and thus {xi yj : 1 ≤ i ≤ m and 1 ≤ j ≤ n}...

Let g and h be polynomials with coeﬃcients in K1 . Now g (α) = h(α) if and only if m divides g − h. Similarly σ∗ (g )(β ) = σ∗ (h)(β ) if and only if σ∗ (m) divides σ∗ (g ) − σ∗ (h). Therefore σ∗ (g )(β ) = σ∗ (h)(β ) if and only if g (α) = h(α), and thus there is a well-deﬁned isomorphism ϕ: K1 (α) → K2 (β ) which sends g (α) to σ∗ (g )(β ) for any polynomial g with coeﬃcients in K . Now L1 and L2 are splitting ﬁelds for the polynomials f and σ∗ (f ) over the ﬁelds K1 (α) and K2 (β ) respectively, and [L1 : K1 (α)] < [L1 : K1 ]. The induction hypothesis therefore ensures the existence of an isomorphism τ : L1 → L2 extending ϕ: K1 (α) → K2 (β ). Then τ : L1 → L2 is the required extension of σ : K 1 → K2 . Corollary 3.31 Let L: K be a splitting ﬁeld extension, and let α and β be elements of L. Then there exists a K -automorphism of L sending α to β if and only if α and β have the same minimum polynomial over K . Pro of Suppose that there exists a K -automorphism σ of L which sends α to β . Then h(β ) = σ (h(α)) for all polynomials h ∈ K [x] with coeﬃcients in K . Therefore h(α) = 0 if and only if h(β ) = 0. It follows that α and β must have the same minimum polynomial over K . Conversely suppose that α and β are elements of L that have the same minimum polynomial m over K . Let h1 and h2 be polynomials with coefﬁcients in K . Now h1 (α) = h2 (α) if and only if h1 − h2 is divisible by the minimum polynomial m. It follows that h1 (α) = h2 (α) if and only if h1 (β ) = h2 (β ). Therefore there is a well-deﬁned K -isomorphism ϕ: K (α) → K (β ) that sends h(α) to h(β ) for all polynomials h with coeﬃcients in K . Then ϕ(α) = β . Now L is the splitting ﬁeld over K for some polynomial f with coeﬃcients in K . The ﬁeld L is then a splitting ﬁeld for f over both K (α) and K (β ). It follows from Theorem 3.30 that the K -isomorphism ϕ: K (α) → K (β ) extends to a K -automorphism τ of L that sends α to β , as required....

Corollary 3.34 Let K be a ﬁeld. An irreducible polynomial f is inseparable if and only if Df = 0. Pro of Let f ∈ K [x] be an irreducible polynomial. Suppose that f is inseparable. Then f has a repeated root in a splitting ﬁeld, and it follows from Proposition 3.33 that there exists a non-constant polynomial g in K [x] dividing both f and its formal derivative Df . But then g = cf for some non-zero element c of K , since f is irreducible, and thus f divides Df . But if Df were non-zero then deg Df < deg f , and thus f would not divide Df . Thus Df = 0. Conversely if Df = 0 then f divides both f and Df . It follows from Proposition 3.33 that f has a repeated root in a splitting ﬁeld, and is thus inseparable. Deﬁnition An algebraic ﬁeld extension L: K is said to be separable over K if the minimum polynomial of each element of L is separable over K . Suppose that K is a ﬁeld of characteristic zero. Then n.k = 0 for all n ∈ Z and k ∈ K satisfying n = 0 and k = 0. It follows from the deﬁnition of the formal derivative that Df = 0 if and only if f ∈ K [x] is a constant polynomial. The following result therefore follows immediately from Corollary 3.34. Corollary 3.35 Suppose that K is a ﬁeld of characteristic zero. Then every polynomial with coeﬃcients in K is separable over K , and thus every ﬁeld extension L: K of K is separable...

or all j satisfying 0 < j < p. But px = 0 for all x ∈ K , j since charK = p. Therefore (x + y )p = xp + y p for all x, y ∈ K . The identity (xy )p = xp y p is immediate from the commutativity of K . that p divides Let K be a ﬁeld of characteristic p, where p > 0. The monomorphism x → xp is referred to as the Frobenius monomorphism of K . If K is ﬁnite then this monomorphism is an automorphism of K , since any injection mapping a ﬁnite set into itself must be a bijection. Theorem 3.37 A ﬁeld K has pn elements if and only if it is a splitting ﬁeld n for the polynomial xp − x over its prime subﬁeld Fp , where Fp ∼ Z/pZ. = Pro of Suppose that K has q elements, where q = pn . If α ∈ K \ {0} then αq−1 = 1, since the set of non-zero elements of K is a group of order q − 1 with respect to multiplication. It follows that αq = α for all α ∈ K . Thus all elements of K are roots of the polynomial xq − x. This polynomial must therefore split over K , since its degree is q and K has q elements. Moreover the polynomial cannot split over any proper subﬁeld of K . Thus K is a splitting ﬁeld for this polynomial. Conversely suppose that K is a splitting ﬁeld for the polynomial f over Fp , where f (x) = xq − x and q = pn . Let σ (α) = αq for all α ∈ K . Then σ : K → K is a monomorphism, being the composition of n successive applications of the Frobenius monomorphism of K . Moreover an element α of K is a root of f if and only if σ (α) = α. It follows from this that the roots of f constitute a subﬁeld of K . This subﬁeld is the whole of K , since K is a splitting ﬁeld. Thus K consists of the roots of f . Now Df (x) = q xq−1 − 1 = −1, since q is divisible by the characteristic p of Fp . It follows from Proposition 3.33 that the roots of f are distinct. Therefore f has q roots, and thus K has q elements, as required. Let K be a ﬁnite ﬁeld of characteristic p. Then K has pn elements, where n = [K : Fp ], since any vector space of dimension n over a ﬁeld of order p must have exactly pn elements. The following result is now a consequence of the existence of splitting ﬁelds (Corollary 3.29) and the uniqueness of splitting ﬁelds up to isomorphism (Theorem 3.30) Corollary 3.38 There exists a ﬁnite ﬁeld GF(pn ) of order pn for each prime number p and positive integer n. Two ﬁnite ﬁelds are isomorphic if and only if they have the same number of elements....

of ϕ(d) taken over all divisors of a positive integer n is equal to n. Lemma 3.39 Let n be a positive integer. Then d ϕ(d) = n....