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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 9 Rik Sengupta December 10, 2009 1. Let E be the splitting field of x 4 4 x 2 + 2 ∈ Q [ x ] . Describe the automor phisms of E , and find a generator for the group Gal ( E/ Q ) . Solution. Note that if we view the given polynomial as a quadratic in x 2 , then we can solve for x 2 to get x 2 = 4 ± √ 8 2 = 2 ± √ 2 , and so the roots of x 4 4 x 2 + 2 in C are ± p 2 ± √ 2. So we immediately conclude that the roots are of the form a , b , a , b , where a = p 2 + √ 2 and b = p 2 √ 2. We will show first of all that b ∈ Q ( a ). But this follows as soon as we notice that b ∈ Q ( a ) ⇐⇒ ab ∈ Q ( a ). We have ab = q 2 + √ 2 · q 2 √ 2 = p 2 2 2 = √ 2 = ( q 2 + √ 2) 2 2 = a 2 2 , = ⇒ b = a 2 a . So, b is in Q ( a ), and therefore Q ( a ) = Q ( p 2 + √ 2) is the splitting field of the polynomial, and it has degree 4. So the only possibilities are C 4 and C 2 × C 2 . 1 The obvious way of ruling out C 2 × C 2 is by noticing that if the homo morphism σ : a 7→ a 2 a is applied sequentially to p 2 + √ 2, we get the following results. σ ( a ) = b. σ ( b ) = b 2 b = q 2 √ 2 2 p 2 √ 2 = √ 2 p 2 √ 2 = √ 2 · p 2 + √ 2 √ 4 2 = q 2 + √ 2 = a. σ ( a ) = a + 2 a = a 2 a = b. σ ( b ) = σ ( b ) = ( a ) = a. In other words, σ 4 ( b ) = σ 3 ( a ) = σ 2 ( b ) = σ ( a ) = b . So there is no element of order 2, so we can rule out C 2 × C 2 . So the field must be C 4 , as is clear from the cyclic results we have just proven. So, Aut( E ) ’ C 4 , and a generator is σ : a 7→ ( a 2 a ) . 2 2. Find a primitive element (over Q ) for each subfield of Q ( ζ 13 ) , and draw a diagram showing all inclusions among these subfields, and find their relative degrees. Do the same for Q ( ζ 17 ) . What is noteworthy? Solution. Note that definitionally, Q ( ζ 13 ) is the splitting field of Φ 13 ( x ). So, Q ( ζ 13 ) / Q is Galois, and Gal( Q ( ζ 13 ) / Q ) ’ F × 13 , which is a cyclic group generated by σ : Q ( ζ 13 ) → Q ( ζ 13 ) such that σ  Q = Id Q , and σ ( ζ 13 ) = ζ 2 13 , because the powers of 2 reduced modulo 13 are 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 6. 2 Now, Gal( Q ( ζ 13 ) / Q ) is cyclic of order 12, and so it has a subgroup for every divisor of 12, which we write in an upside down lattice as follows immediately after. We claim that, for any subgroup H , the subfield fixed by H is generated by adjoining α df = X σ ∈ H σ ( ζ 13 ) to Q . Clearly, any σ ∈ H fixes α . So now suppose τ / ∈ H , and σ ∈ H . Assume τ ( α ) = α = σ ( α ). This immediately implies that τ ( ζ 13 ) = σ ( ζ 13 ), since ζ 13 ,ζ 2 13 ,...,ζ 12 13 form a basis....
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This note was uploaded on 05/07/2010 for the course MAT 322 at Princeton.
 '09
 CLAUSSORENSEN
 Algebra

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