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F[X] Be an Irreducible Cubic X 3 + Ax 2 + Bx + Cw

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F[X] Be an Irreducible Cubic X 3 + Ax 2 + Bx + Cw Math 404 Assignment 3. Due Friday, May 3, 2013. Cubic equations. Let f(x) F [x] be an irreducible cubic x3 + ax2 + bx + c with roots ∈ x1, x2, x3, and splitting field K/F . Since each element of the Galois group permutes the roots, G(K/F ) is a subgroup of S3, the group of permutations of the three roots, and [K : F ] = G(K/F ) divides (S3) = 6. Since [F (x1): F ] = 3, we see that [K : F ] equals ◦ ◦ 3 or 6, and G(K/F ) is either A3 or S3. If G(K/F ) = S3, then K has a unique subfield of dimension 2, namely, KA3 . We have seen that the determinant J of the Jacobian matrix of the partial derivatives of the system a = (x1 + x2 + x3) − b = x1x2 + x2x3 + x3x1 c = (x1x2x3) − equals (x1 x2)(x2 x3)(x1 x3). − − − Formula. J 2 = a2b2 4a3c 4b3 27c2 + 18abc F . − − − ∈ An odd permutation of the roots takes J =(x1 x2)(x2 x3)(x1 x3) to J and an even permutation of the roots takes J to J. − − − − 1. Let f(x) F [x] be an irreducible cubic polynomial. ∈ (a). Show that, if J is an element of K, then the Galois group G(L/K) is the alternating group A3. Solution. If J F , then every element of G(K/F ) fixes J, and G(K/F ) must be A3, ∈ (b). Show that, if J is not an element of F , then the splitting field K of f(x) F [x] has ∈ Galois group G(K/F ) isomorphic to S3. Solution. If J is not in F , then not every element of G(K/F ) fixes J, and G(K/F ) must be S3. 2. Determine the Galois group of the splitting fields of the following irreducible polynomials.: (a). x3 + 27x 4 Q[x]. − ∈ Solution. By the formula above, J 2 = 4(273) 27( 4)2 < 0. Hence, J is purely imaginary, − − − and so, is not an element of Q. Therefore, by problem 1, the Galois group is S3. 1 (b). x3 27x +4 Q[x]. − ∈ Solution. J 2 = 4( 27)3 27(42) = 4(27)((27)2 4) > 0. However, that number is not − − − − the square of a rational number, and so J is not rational. Hence, by problem 1, the Galois group is S3. (c). x3 + x +1 Q[x]. ∈ 2 Solution. J = 31. As in (a), we conclude that the Galois group is S3. − 3. Let K/Q be the splitting field of the irreducible polynomial x3 + x +1 Q[x]. ∈ Show that there is only one 2-dimensional (over Q) subfield of K and find that field explicitly. Solution. The Galois group of that irreducible polynomial is S3, as we saw in problem 2. A3 Since A3 fixes J, Q(J) K . To show those fields are equal, we show that each is 2- dimensional. Q(J)/Q is 2-dimensional⊂ since J is not an element of K when the Galois group 2 A3 is S3, by problem , but J = 31 is an element of Q. To show that K is 2-dimensional, 3 − consider the ladder Q KA K. The second step is Galois by the numerical theorems, and ⊂ ⊂ its dimension is the order of A3 = 3. Hence, the first step has dimension 2 since the length of the ladder is dimension 6, the order of S3, by the parenthetical remark after theorem 3. By the subfield-subgroup correspondence theorems, and by the nummerical theorems, a field of dimension 2 is the fixed field of a subgroup of order 3 in S3 of which there is only one, A3, the alternating group, i.e., the only subfield of dimension 2 is KA3 = Q(J)= Q(√ 31). − Theorem from Galois Theory. Let F be a field containing Q, and let K/F be a field extension that has a primitive element b K. Let G be a finite subgroup of Gal(K/F ). The G ∈ fixed field K is generated over F by the coefficients of the polynomial f(x) = Πσ∈G(x σ(b)). − 4. For each of the sets of automorphisms of C(y), the field of complex rational functions in a variable y, find the fixed field. (The automorphisms here fix each complex number.) (a). σ(y)= y−1. Solution. The automorphism generates a group G of order 2 since σ has order 2. The fixed field is generated over C by the coefficients of the polynomial f(x)=(x y)(x y−1) = 2 −1 −1 − − x (y + y )x + 1. Hence, C(y)G = C(y + y ), by the result from Galois Theory. − (b). σ(y)= iy. 2 Solution. σ2(y)= y, σ3(y)= iy and σ4(y)= y, i.e., σ4 = I. Let G be the cyclic group I, σ, σ2, σ3 . The irreducible− polynomial− for y in C(y)G[x] is { } (x y)(x iy)(x + y)(x + iy)=(x2 y2)(x2 + y2)= x4 y4. − − − − Hence, C(y)G = C(y4). (c). σ(y)= y and τ(y)= y−1. − Solution. σ and τ are automorphisms of order 2 that commute with each other, since σ τ(y)= y−1 and τ σ(y)=(( 1)y)−1 = 1y−1. ◦ − ◦ − − The group G generated by two distinct commuting elements of order 2 is the Klein group of order 4. The irreducible polynomial for y in C(y)G[x] is (x y)(x + y)(x y−1)(x + y−1)=(x2 y2)(x2 y−2)= x4 (y2 + y−2)x2 +1. − − − − − Hence, C(y)G = C(y2 + y−2). 5. Find the group of automorphisms of C(y) generated by the following pairs. −1 2πi (a). σ(y)= ζy, τ(y)= y , where ζ = e 3 . Solution. σ has order 3 and τ has order 2. Check that τστ −1 = σ−1 and conclude that the group generated is S3. (b). σ(y)= iy and τ(y)= y−1. Solution. σ has order 4 and τ has order 2. Check that τστ −1 = σ−1 and conclude that the group generated is the dihedral group D4. y+i y−1 6. Consider the automorphisms of C(y), σ(y) = y−i , and τ(y) = i y+1 , called Mobius transformations. Show that those two elements generate a group G of automorphisms isomorphic to the alternating group A4. Solution. We find four things for the group generated by σ and τ to permute. It is known that a Mobius transformation takes a circle to either a circle or a line, and a line to a circle or a line. Consider the following four pairs of circle and lines, where each circle passes through three of the finite points in 0, , 1, 1, i, i, and each line passes through the other two points and the point at infinity:∞ pair− 1 consists− of the circle through the points 0, i, 1, − 3 and the line through 1, i, ; pair 2 consists of the circle through the points 0, i, 1, and the line through 1, −i, ;∞ pair 3 consists of the circle through the points 0, 1, i, and the line through i, 1−, −; pair∞ 4 consists of the circle through the points 0, 1, i−, and the line − ∞ − − through 1, i, . Check that σ takes pair 1 to pair 3, takes pair 3 to pair 2, and pair 2 back to pair 1, while∞ it takes pair 4 to itself. Hence, σ gives the permutation (132); in the same way, τ gives the permuation (243). The permutation group of the set of the four pairs that is generated by σ and τ is then the group generated by (132) and (243), which is A4. One can check that the surjective homomorphism from the group G generated by σ and τ to A4 is injective. 4.
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