Galois Theory and Noether's Problem Meredith Blue ... - MAA Sections

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3. Symmetric Group: Let X be a set with n objects X = {x 1 , . . . , x n }.Let G be the set of all one-to-one onto maps σ : X → X. G is a groupwith binary operation, ◦, composition of functions. This group is calledthe Symmetric Group on n letters denoted S n .4. C = {f : R → R | f is increasing and continuous} with binary operationcomposition of functions.If G is a finite group, we let |G| represent the number of elements of G.Next, we define fields:Definition 2 A field F is a set of elements and 2 binary operations: + and× such that F is a group under + with identity 0, and F − {0} is a groupunder multiplication. Moreover:∀x, y ∈ F, x + y = y + x and x × y = y × xFor simplicity we write x × y as xy.EXAMPLES:1. Q2. R3. C4. Z 7 = {0, 1, 2, 3, 4, 5, 6} with addition and multiplication modulo 7.Definition 3 A field F is a subfield of a field K if F ⊂ K.Definition 4 If f ⊂ K are fields, then we say K is an extension field of F .This is denoted K/F .Let α ∈ C. Let Q(α) be the smallest field containing Q and α; in particularany field K that contains Q and α contains Q(α). The field Q(α) caneither be an algebraic or a transcendental extension. We give an example ofeach and express some important properties of each type of extension.We first consider Q(π). π is transcendental over Q: i.e. there is nopolynomial with rational coefficients that has π as a root. Q(π) is isomorphicto the field of rational functions in x over Q. In addition, we may add anothertranscendental element e. Q(π, e) ∼ = Q(x, y) the field of rational functions2
in 2 variables over Q. These are infinite extensions because Q(x) is infinitedimensional as a vector space over Q. For example, {x i | i ∈ Z} is an infinitelinearly independent set in Q(x) when viewed as a vector space over Q.We next consider Q( √ 2). √ 2 is a root of the polynomial x 2 −2. Thus, wesay √ 2 is algebraic over Q, and Q( √ 2) is an algebraic extension field of Q.An easy exercise shows that any element in Q( √ 2) can be written as a+b √ 2with a and b rational numbers. Thus, Q( √ 2) is a vector space of dimension2 over Q. This is a finite extension.If an extension field L ⊃ Q can be obtained by adding finitely manyroots of polynomials, we say L is a finite algebraic extension of Q. The nexttheorem states that any finite algebraic extension can be obtained by addinga single element to Q.Theorem 5 (Primitive Element Theorem) Any finite algebraic extension,L, of Q, has the form:L = Q(α) for some α ∈ CLet us return to our original interpretation in terms of polynomials.Definition 6 Let α be algebraic over Q. f(x) is the minimal polynomialfor α (denoted min Q (α)) if:1. f(x) has leading coefficient 1.2. f(α) = 0 and3. If g(x) is any polynomial with rational coefficients such that g(α) = 0then the degree of g(x) < the degree of f(x).A polynomial p(x) is irreducible if whenever p(x) = h(x)k(x) with h(x)and k(x) polynomials, then either h(x) or k(x) is a constant (i.e. rationalnumber.)Theorem 7 min Q (α) is irreducible.Proof: Let f(x) = min Q (α). Suppose f(x) = p(x)q(x). Since α is a rootof f(x) ∈ C, f(α) = 0 = p(α)q(α). p(α) and q(α) are just complex numberswhose product is 0. Thus, either p(α) = 0 or q(α) = 0. Without loss ofgenerality, assume p(α) = 0. Since f(x) is the polynomial with minimumdegree with α as a root, the degree of p(x) = degree f(x). Thus, the degreeof q(x) is 0. This shows f(x) is irreducible.q.e.d.3
Page 1: Galois Theory and Noether’s Probl
Page 5 and 6: Theorem 11 If L/Q is a field extens
Page 8 and 9: andσ(x 0 x 1 ) = σ(x 0 )σ(x 1 )

Galois Theory and Noether's Problem Meredith Blue ... - MAA Sections

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