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College Algebra Exam Review 267

# College Algebra Exam Review 267 - R S D h S i C R S Proof...

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6.2. HOMOMORPHISMS AND IDEALS 277 then x 2 I k I n and y 2 I ` I n . Therefore, x C y 2 I n I . If x 2 I and r 2 R , then there exists n 2 N such that x 2 I n . Then rx; xr 2 I n I . Thus I is an ideal. n The analogues of parts (a) and (b) of the Proposition 6.2.21 hold for left and right ideals as well. Proposition 6.2.22. (a) Let I and J be two ideals in a ring R . Then IJ D f a 1 b 1 C a 2 b 2 C C a s b s W s 1; a i 2 I; b i 2 J g is an ideal in R , and IJ I \ J . (b) Let I and J be two ideals in a ring R . Then I C J D f a C b W a 2 I and b 2 J g is an ideal in R . Proof. Exercises 6.2.18 and 6.2.19 . n Ideals generated by subsets Next we investigate ideals, or one–sided ideals, generated by a subset of a ring. Proposition 6.2.23. Let R be a ring and S a subset of R . Let h S i denote the additive subgroup of R generated by S . (a) Define R S D f r 1 s 1 C r 2 s 2 C C r n s n W n 2 N ; r i 2 R; s i 2 S g : Then R S is a left ideal of R . (b) h S i C R S is the smallest left ideal of R containing S , and is equal to the intersection of all left ideals of R containing S . (c) In case R has an identity element,
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Unformatted text preview: R S D h S i C R S . Proof. It is straightforward to check that R S is a left ideal. h S i C R S is a sum of subgroups R , so it is a subgroup. Moreover, for r 2 R , we have r h S i ± R S . It follows from this that h S i C R S is a left ideal. If J is any left ideal of R containing S , then J ´ h S i , because J is a subgroup of R . Since J is a left ideal, J ´ R S as well. Therefore J ´ h S i C R S . This shows that h S i C R S is the smallest left ideal containing S . The intersection of all left ideals of R containing S is also the smallest left ideal of R containing S , so (b) follows. Finally, if R has an identity element, then S ± R S , so h S i ± R S , which implies (c). n...
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