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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
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.

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online