all3 - Math 3124 Thursday, November 17 Eighth Homework...

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Unformatted text preview: Math 3124 Thursday, November 17 Eighth Homework Solutions 1. (a) ( AB ) = det ( AB ) = det ( A ) det ( B ) = ( A ) ( B ) , so is a homomorphism. It is onto because given x Z # 5 , we have x 1 = x , and the foregoing matrix is in GL 2 ( Z 5 ) . (b) A ker if and only if det ( A ) = [ 1 ] . (c) Since K = ker and is onto (from above), this follows immediately from the Fundamental Homomorphism Theorem. (d) Since the property of being cyclic is invariant under isomorphism, it will be suffi- cient to determine whether Z # 5 is cyclic. The answer is yes, because Z # 5 = [ 2 ] . 2. Define : G H by ( x ) = x 4 . Since Z # 509 is abelian, we have ( xy ) = ( xy ) 4 = x 4 y 4 = ( x ) ( y ) and so is a homomorphism. It is onto by the definition of H . Therefore by the Fundamental Homomorphism Theorem G / ker = H . We need to determine ker . Now [ 208 ] ker because [ 208 ] 4 = [- 1 ] 2 = [ 1 ] . Also [ 208 ] 2 = [- 1 ] = [ 1 ] , consequently o ([ 208 ]) = 4. Therefore 4 | ker | . Since | G | = 508 = 4 * 127 and 127 is prime, we see from Lagranges theorem that | ker | = 4 or 508. But the latter is not possible because that would imply ker = G , which is clearly not the case. Therefore | ker | = 4 and we deduce that | H | = 127. 3. Problem 24.17 on page 125. Prove that ( a + b ) 2 = a 2 + 2 ab + b 2 if and only if R is commutative. Since ( a + b ) 2 = a 2 + ab + ba + b 2 , we have ( a + b ) 2 = a 2 + 2 ab + b 2 if and only if ab + ba = 2 ab , that is if and only if ba = ab . This last equality is exactly the condition for R to be commutative. 4. Problem 26.14 on page 125. Verify that Q [ 2 ] = { a + b 2 | a , b Q } is a subfield of R . Let F denote the above set. Certainly 1 (the unity of R ) is in F . Now let x , y F . Write x = a + b 2 and y = c + d 2, where a , b , c , d Q . We want to show x + y ,- x and xy F . We have x + y = ( a + b 2 )+( c + d 2 ) = ( a + c )+( b + d ) 2 F , because a + c , b + d Q . Next- x =- a- b 2 F because- a ,- b Q . Finally xy = ( ac + 2 bd )+( ad + bc ) 2 F because ac + 2 bd , ad + bc Q . It remains to show that if x = 0, then x- 1 F . However x- 1 = ( c- d 2 ) / ( c 2- 2 d 2 ) F , because c / ( c 2- 2 d 2 ) and d / ( c 2- 2 d 2 ) are in Q ; we should note that c 2- 2 d 2 = 0. This completes the verification that F is a field. 5. By using the distributive law and the fact that R is commutative, we get ( a + b ) 3 = ( a + b )( a 2 + 2 ab + b 2 ) = a 3 + 2 a 2 b + ab 2 + ba 2 + 2 bab + b 3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 . Now use the fact that R has characteristic 3: we find that 3 a 2 b = 3 ab 2 = 0 and we conclude that ( a + b ) 3 = a 3 + b 3 . Math 3124 Tuesday, December 6 Ninth Homework Solutions 1. Let a , b R . Then ( a + b ) = ( a + b ) p = (by the Freshman calculus rule) a p + b p = ( a ) ( b ) , so respects addition. Also ( ab ) = ( ab...
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This note was uploaded on 01/02/2012 for the course MATH 3124 taught by Professor Parry during the Fall '08 term at Virginia Tech.

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all3 - Math 3124 Thursday, November 17 Eighth Homework...

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