6.3. QUOTIENT RINGS 285 .a C bx/ ; evaluating at i , we get0 D a C bi , which is possible only if a D b D0 . 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 r0 j 2 B i and r j 2 B j such that r0 j C r j D 1 . Consider the product of all the .r0 j C r j / (in any order). When the product is expanded, all the summands except for one contain at least one factor
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.