This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
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.
 Fall '08
 PARRY
 Math, Algebra

Click to edit the document details