College Algebra Exam Review 275

College Algebra Exam Review 275 - 6.3. QUOTIENT RINGS 285...

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6.3. QUOTIENT RINGS 285 .a C bx/ ; evaluating at i , we get 0 D a C bi , which is possible only if a D b D 0 . Thus g is a multiple of x 2 C 1 . That is ker .'/ D .x 2 C 1/ . By the homomorphism theorem for rings, R ŒxŁ=.x 2 C 1/ Š C as rings. In particular, since C is a field, R ŒxŁ=.x 2 C 1/ is a field. Note that we have already calculated explicitly in Example 6.3.3 that multiplication in R ŒxŁ=.x 2 C 1/ satisfies the same rule as multiplication in C . Example 6.3.6. Let R be a ring with identity containing ideals B 1 ; :::;B s . Let B D \ i B i . Suppose that B i C B j D R for all i ¤ j . Then R=B Š R=B 1 ˚ ±±± ˚ R=B s . In fact, ' W r 7! .r C B 1 ;:::;r C B s / is a homomorphism of R into R=B 1 ˚±±±˚ R=B s with kernel B , so R=B Š '.R/ . The problem is to show that ' is surjective. Fix i and for each j ¤ i find r 0 j 2 B i and r j 2 B j such that r 0 j C r j D 1 . Consider the product of all the .r 0 j C r j / (in any order). When the product is expanded, all the summands except for one contain at least one factor
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