Solving Rings and Idempotent Problems - RK

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SUMMARY

The discussion centers on solving rings and idempotent problems, specifically addressing multiplicative inverses in algebra. The user, RK, presents two attempts at a solution, highlighting a concern about proving the existence of multiplicative inverses after establishing the necessity of a multiplicative identity. RK's first attempt is deemed unsatisfactory due to the lack of proof for inverses, while the second attempt, although more complex and unfinished, is considered a better approach. The conclusion emphasizes the importance of proving that each element in the ring is its own inverse.

PREREQUISITES
  • Understanding of ring theory and its properties
  • Familiarity with multiplicative identities and inverses
  • Knowledge of algebraic structures and their operations
  • Basic proof techniques in abstract algebra
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  • Study the properties of rings in abstract algebra
  • Learn about the existence of multiplicative inverses in algebraic structures
  • Explore the concept of idempotent elements in rings
  • Review proof strategies for algebraic identities and their implications
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Mathematicians, students of abstract algebra, and anyone interested in understanding the properties of rings and multiplicative inverses.

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I have a tablet so I have made a PDF of all my work and the problem. the file is attached to this post. please let me know if i am on the right track or give me a hint. I am currently stuck in attempt 2 and don't like my solution in attempt 1.

attempt 1: at the very last step I am using multiplicative inverses and I haven't proved that they must exist. but since I have shown that a multiplicative identity is required, it shouldn't be hard to prove that inverses exist also but i don't know if it will be required..?

attempt 2: took a different approach at the problem, and while it's a bit more complicated in the end and is unfinished (this is where i am stuck), I think it's the better attempt.



thank you,
RK
 

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a^3 = a implies that aaa = a.
Multiply by "a inverse" to obtain aa=1.
Multiply by "a inverse" again to obtain a=a^-1.
So each element is it's own inverse.
 

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