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ma5c-hw3-soln

# ma5c-hw3-soln - M a 5c HOMEWORK 3 SOLUTION SPRING 09 The...

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Ma 5 c HOMEWORK 3 SOLUTION SPRING 09 The exercises are taken from the text, Abstract Algebra (third edi- tion) by Dummit and Foote, unless stated otherwise. Page 567, 4 . Suppose there is an isomorphism φ : Q ( 2) Q ( 3). Since φ ﬁxes Q and φ is a homomorphism, we must have ( φ ( 2)) 2 = 2, i.e. φ ( 2) must be a root of the polynomial x 2 - 2 = 0. But Q ( 3) has no such root. Therefore the two ﬁelds can’t be isomorphic. Page 567, 10 . Deﬁne f : Aut ( K/F ) Aut ( K 0 /F 0 ) via σ 7→ φσφ - 1 . Since φ and σ are both isomorphisms, f ( σ ) is an automorphism of K 0 . Take x F 0 , then f ( σ )( x ) = φσφ - 1 ( x ) = φφ - 1 ( x ) = x , since φ - 1 ( x ) F , and σ ﬁxes F . Therefore f is well-deﬁned. It’s easy to check that f is a group homomorphism. Also, we can deﬁne g : Aut ( K 0 /F 0 ) Aut ( K/F ) via τ 7→ φ - 1 τφ . Similarly, g is a well- deﬁned group homomorphism. Furthermore, f and g are inverse ho- momorphisms. Therefore f is a group isomorphism. Page 582,

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ma5c-hw3-soln - M a 5c HOMEWORK 3 SOLUTION SPRING 09 The...

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