College Algebra Exam Review 278

# College Algebra Exam Review 278 - 2.7.13 6.3.5 Prove...

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288 6. RINGS Proof. Suppose R is simple and x 2 R is a nonzero element. The ideal Rx is nonzero since x D 1x 2 Rx ; because R is simple, R D Rx . Hence there is a y 2 R such that 1 D yx . Conversely, suppose R is a ﬁeld and M is a nonzero ideal. Since M contains a nonzero element x , it also contains r D rx ± 1 x for any r 2 R ; that is, M D R . n Corollary 6.3.14. If M is a proper ideal in a commutative ring R with 1, then R=M is a ﬁeld if, and only if, M is maximal. Proof. This follows from Propositions 6.3.12 and 6.3.13 . n Exercises 6.3 6.3.1. Work out the rule of computation in the ring R ŒxŁ=.f / , where f.x/ D x 2 ± 1 . Note that the quotient ring consists of elements a C bx C .f / . Compare Example 6.3.3 . 6.3.2. Work out the rule of computation in the ring R ŒxŁ=.f / , where f.x/ D x 3 ± 1 . Note that the quotient ring consists of elements a C bx C cx 2 C .f / . Compare Example 6.3.3 . 6.3.3. Prove Proposition 6.3.7 (the correspondence theorem for rings). 6.3.4. Give another proof of Proposition 6.3.8 , by adapting the proof of Proposition 2.7.13 , rather than appealing to the result of Proposition
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Unformatted text preview: 2.7.13 . 6.3.5. Prove Proposition 6.3.10 (the diamond isomorphism theorem for rings) following the pattern of the proof of Proposition 2.7.18 (the diamond isomorphism theorem for groups). 6.3.6. Prove that an ideal M in R is maximal if, and only if, R=M is simple. 6.3.7. (a) Show that n Z is maximal ideal in Z if, and only if, ˙ n is a prime. (b) Show that .f / D fKŒxŁ is a maximal ideal in KŒxŁ if, and only if, f is irreducible. (c) Conclude that Z n D Z =n Z is a ﬁeld if, and only if, ˙ n is prime, and that KŒxŁ=.f / is a ﬁeld if, and only if, f is irreducible. 6.3.8. If J is an ideal of the ring R , show that JŒxŁ is an ideal in RŒxŁ and furthermore RŒxŁ=JŒxŁ Š .R=J/ŒxŁ . Hint: Find a natural homomorphism from RŒxŁ onto .R=J/ŒxŁ with kernel JŒxŁ ....
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