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College Algebra Exam Review 274

# College Algebra Exam Review 274 - Q W R=I S satisfying Q ı...

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284 6. RINGS r.x/s.x/ C .f / D a.x/ C .f / , where a.x/ is the remainder upon division of r.x/s.x/ by f .x/ . Let’s look at the particular example K D R and f .x/ D x 2 C 1 . Then R OExŁ=.f / consists of cosets a C bx C .f / represented by linear polynomials. Furthermore, we have the computational rule x 2 C .f / D x 2 C 1 1 C .f / D 1 C .f /: Thus .a C bx C .f //.a 0 C b 0 x C .f // D .aa 0 bb 0 / C .ab 0 C a 0 b/x C .f /: All of the homomorphism theorems for groups, which were presented in Section 2.7 , have analogues for rings. The basic homomorphism theo- rem for rings is the following. Theorem 6.3.4. (Homomorphism Theorem for Rings). Let ' W R ! S be a surjective homomorphism of rings with kernel I . Let W R ! R=I be the quotient homomorphism. There is a ring isomorphism Q ' W R=I ! S satisfying Q ' ı D ' . (See the following diagram.) R ' q q S q q q q = Q ' R=I Proof. The homomorphism theorem for groups (Theorem 2.7.6 ) gives us an isomorphism of abelian groups
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Unformatted text preview: Q ' W R=I ! S satisfying Q ' ı ± D ' . We have only to verify that Q ' also respects multiplication. But this follows at once from the deﬁnition of the product on R=I : Q '.a C I/.b C I/ D Q '.ab C I/ D '.ab/ D '.a/'.b/ D Q '.a C I/ Q '.b C I/: n Example 6.3.5. Deﬁne a homomorphism ' W R ŒxŁ ! C by evaluation of polynomials at i 2 C , '.g.x// D g.i/ . For example, '.x 3 ± 1/ D i 3 ± 1 D ± i ± 1 . This homomorphism is surjective because '.a C bx/ D a C bi . The kernel of ' consists of all polynomials g such that g.i/ D . The kernel contains at least the ideal .x 2 C 1/ D .x 2 C 1/ R ŒxŁ because i 2 C 1 D . On the other hand, if g 2 ker .'/ , write g.x/ D .x 2 C 1/q.x/ C...
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