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# HW 09 - MAT 322 Algebra with Galois Theory Professor Claus...

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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

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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. Then, σ - 1 τ ( ζ 13 ) = ζ 13 . So, this implies σ - 1 τ = Id, which necessarily implies σ = τ , a contradiction. So, if τ / H , α is not fixed by τ . So, Gal( Q ( ζ 13 ) / Q ( α )) = H . e h σ 12 i 2 < h σ 3 i 3 > h σ 8 i 2 h σ 4 i 2 h σ i 2 < 3 > Hence, by Galois correspondence, the lattice for the intermediate fields with relative degrees is the following. Note that each intermediate field is generated by a primitive element α = σ H σ ( ζ 13 ).

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