asf - Math 3124 Tuesday December 6 Sample Final Solutions 1...

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Math 3124 Tuesday, December 6 Sample Final Solutions 1. Remember that R is an abelian group under addition. Since | R | = 3 and e = 0, the order of e is 3, which tells us that e + e = x . Therefore x 2 = ( e + e ) 2 = 4 e = e , the last equality coming from the fact that the order of e under addition is 3. 2. Let a , b C , so ar = ra and br = rb for all r R . We check the conditions required for a subring. (i) C = /0 because 0 C . (ii) ( a + b ) r = ar + br = ra + rb = r ( a + b ) for all r R , so a + b C . (iii) ( - a ) r = - ( ar ) = - ( ra ) = r ( - a ) for all r R , so - a C . (iv) ( ab ) r = abr = arb = r ( ab ) for all r R , so ab C . 3. (a) Since F is a field, the nonzero elements of F form a group under multiplication. Also | F \ 0 | = 3, hence a 3 = e . (b) We have ( a - e )( a 2 + a + e ) = a 3 - e = 0. (c) Since F is a field, we deduce from the above that either a - e = 0 or a 2 + a + e = 0. But a = e , consequently a 2 + a + e = 0. 4. We always have two homomorphisms, namely the zero and identity homomorphisms (i.e. x 0 and x x for all x R ), so we need to find one more. Since F is a field with 9 elements, its characteristic is a prime number which divides 9, so it has characteristic 3.
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