"Fields and Galois Theory 001.ps.gz" - читать интересную книгу автораFIELDS AND GALOIS THEORY J.S. MILNE Abstract. These are the notes for the second part of Math 594, University of Michigan, Winter 1994, exactly as they were handed out during the course except for some minor corrections. Please send comments and corrections to me at [email protected] using "Math594" as the subject. v2.01 (August 21, 1996). First version on the web. v2.02 (May 27, 1998). About 40 minor corrections (thanks to Henry Kim). Contents 1. Extensions of Fields 1 1.1. Definitions 1 1.2. The characteristic of a field 1 1.3. The polynomial ring F [X] 2 1.4. Factoring polynomials 2 1.5. Extension fields; degrees 4 1.6. Construction of some extensions 4 1.7. Generators of extension fields 5 1.8. Algebraic and transcendental elements 6 1.9. Transcendental numbers 8 1.10. Constructions with straight-edge and compass. 9 2. Splitting Fields; Algebraic Closures 12 2.1. Maps from simple extensions. 12 2.2. Splitting fields 13 2.3. Algebraic closures 14 3. The Fundamental Theorem of Galois Theory 18 3.1. Multiple roots 18 3.2. Groups of automorphisms of fields 19 3.3. Separable, normal, and Galois extensions 21 3.4. The fundamental theorem of Galois theory 23 3.5. Constructible numbers revisited 26 3.6. Galois group of a polynomial 26 3.7. Solvability of equations 27 Copyright 1996 J.S. Milne. You may make one copy of these notes for your own personal use. i ii J.S. MILNE FIELDS AND GALOIS THEORY 1 1. Extensions of Fields 1.1. Definitions. A field is a set F with two composition laws + and \Delta such that (a) (F; +) is an abelian group; (b) let F \Theta = F \Gamma f0g; then (F \Theta ; \Delta ) is an abelian group; (c) (distributive law) for all a; b; c 2 F , (a + b)c = ac + bc (hence also a(b + c) = ab + ac). Equivalently, a field is a nonzero commutative ring (meaning with 1) such that every nonzero element has an inverse. A field contains at least two distinct elements, 0 and 1. The smallest, and one of the most important, fields is F2 = Z=2Z = f0; 1g. Lemma 1.1. A commutative ring R is a field if and only if it has no ideals other than (0) and R. Proof. Suppose R is a field, and let I be a nonzero ideal in R. If a is a nonzero element of I, then 1 = a\Gamma 1a 2 I, and so I = R. Conversely, suppose R is a commutative ring with no nontrivial ideals; if a 6= 0, then (a) = R, which means that there is a b in F such that ab = 1. Example 1.2. The following are fields: Q , R, C , Fp = Z=pZ: A homomorphism of fields ff : F ! F 0 is simply a homomorphism of rings, i.e., it is a map with the properties ff(a + b) = ff(a) + ff(b); ff(ab) = ff(a)ff(b); ff(1) = 1; all a; b 2 F: Such a homomorphism is always injective, because the kernel is a proper ideal (it doesn't contain 1), which must therefore be zero. |
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