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8.12. Let R be a ring with identity. Suppose that there exist a, b, c e R such that ab = ba = 1 and ac = 0. Show that c = 0.

8.12. Let R be a ring with identity. Suppose that there exist a, b, c e R such that ab = ba = 1 and ac = 0. Show that c = 0.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter6: More On Rings
Section6.2: Ring Homomorphisms
Problem 14E: 14. Let be a ring with unity . Verify that the mapping defined by is a homomorphism.
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Could you explain how to show 8.12 in detail? I also attached definitions and theorems in my textbook.

Theorem 8.2. Let R be a ring. Then the additive identity, 0, is unique. If R has a multiplicative identity 1, then it too is unique. Theorem 8.3. Let R be a ring. If a, b e R, then 1. Оа — а0 3D 0; 2. (-а)b — а(-b) — — (аb); аnd 3. (—а)(-b) — ab. Corollary 8.1. If R is a ring with identity, then (–1)a = -a, for any a e R. Theorem 8.4. Let R be any ring, and a1, a2, .., an € R. Then regardless of how the product a¡a2 · an is bracketed, the result equals (· · · (((a¡a2)az)a4) · · · an-1)aŋ- Example 8.6. In Z6, we have 2 · 3 = 0, but 2 + 0 and 3 # 0. Example 8.7. In M2(R), we have 1 2 -2 4 36 -2 but 1 2 -2 4 36 1 -2 In dealing with groups, we have the cancellation law. We are used to something similar happening in ordinary arithmetic; that is, if ab = ac and a ± 0, then b = c. Again, this does not have to hold in rings. Example 8.8. In Z12, we have 3 · 1 = 3 · 5, but 3 +0 and 1 + 5. Finally, in a group G, we note that if there exists a b e G such that ab = b, then a is the identity. (Just multiply on the right by b-1.) But even if a ring has an identity, the fact that ab = b does not mean that a = 1. Indeed, the previous example points us in the right direction. Example 8.9. In Z12, we have 5 · 3 = 3, but 5 + 1. Thus, to check that a ring element a is the identity, we must make sure that ab = b = ba for every b e R, not just for one such b.
Transcribed Image Text:Theorem 8.2. Let R be a ring. Then the additive identity, 0, is unique. If R has a multiplicative identity 1, then it too is unique. Theorem 8.3. Let R be a ring. If a, b e R, then 1. Оа — а0 3D 0; 2. (-а)b — а(-b) — — (аb); аnd 3. (—а)(-b) — ab. Corollary 8.1. If R is a ring with identity, then (–1)a = -a, for any a e R. Theorem 8.4. Let R be any ring, and a1, a2, .., an € R. Then regardless of how the product a¡a2 · an is bracketed, the result equals (· · · (((a¡a2)az)a4) · · · an-1)aŋ- Example 8.6. In Z6, we have 2 · 3 = 0, but 2 + 0 and 3 # 0. Example 8.7. In M2(R), we have 1 2 -2 4 36 -2 but 1 2 -2 4 36 1 -2 In dealing with groups, we have the cancellation law. We are used to something similar happening in ordinary arithmetic; that is, if ab = ac and a ± 0, then b = c. Again, this does not have to hold in rings. Example 8.8. In Z12, we have 3 · 1 = 3 · 5, but 3 +0 and 1 + 5. Finally, in a group G, we note that if there exists a b e G such that ab = b, then a is the identity. (Just multiply on the right by b-1.) But even if a ring has an identity, the fact that ab = b does not mean that a = 1. Indeed, the previous example points us in the right direction. Example 8.9. In Z12, we have 5 · 3 = 3, but 5 + 1. Thus, to check that a ring element a is the identity, we must make sure that ab = b = ba for every b e R, not just for one such b.
8.12. Let R be a ring with identity. Suppose that there exist a, b, c e R such that ab = ba = 1 and ac = 0. Show that c = 0. 8.16. Let R be a ring in which a? = a for every a e R. 1. Show that a +a = 0 for every a e R. 2. Show that R is commutative.
Transcribed Image Text:8.12. Let R be a ring with identity. Suppose that there exist a, b, c e R such that ab = ba = 1 and ac = 0. Show that c = 0. 8.16. Let R be a ring in which a? = a for every a e R. 1. Show that a +a = 0 for every a e R. 2. Show that R is commutative.
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