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Unformatted text preview: Math 3124 Tuesday, December 6 Sample Final Solutions 1. Remember that R is an abelian group under addition. Since | R | = 3 and e 6 = 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 6 = /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 \ | = 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 6 = e , consequently a 2 + a + e = 0. 4. We always have two homomorphisms, namely the zero and identity homomorphisms (i.e. x 7→ 0 and x 7→ 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...
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