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.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F8a8aa7db-742d-448d-afe5-4e8d014934ea%2Flkepk5p_processed.png&w=3840&q=75)
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2F8a8aa7db-742d-448d-afe5-4e8d014934ea%2Fsmg4i5_processed.png&w=3840&q=75)
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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