Is Idempotent Equivalence in Rings Transitive?

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SUMMARY

The discussion centers on the transitivity of idempotent equivalence in rings, specifically defining two idempotents P and Q as equivalent if there exist elements X and Y such that P = XY and Q = YX. The proof presented demonstrates that if P~Q and Q~R, then it follows that P~R through a series of algebraic manipulations. The proof confirms that the relation is indeed transitive, resolving the initial confusion regarding its complexity.

PREREQUISITES
  • Understanding of ring theory and idempotent elements.
  • Familiarity with equivalence relations in mathematics.
  • Basic knowledge of algebraic manipulation and proof techniques.
  • Experience with mathematical notation and terminology.
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  • Study the properties of idempotent elements in ring theory.
  • Learn about equivalence relations and their applications in abstract algebra.
  • Explore examples of transitive relations in mathematical contexts.
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Mathematicians, students of abstract algebra, and anyone interested in the properties of rings and equivalence relations will benefit from this discussion.

cogito²
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One of my books defines a relation which is "evidently" an equivalence relation. It says that two idempotents in a ring P and Q are said to be equivalent if there exist elements X and Y such that P = XY and Q = YX.

The proof that this relation is transitive eludes me. There is so little information, that I feel like this has to have a really short proof, but I just can't seem to figure it out (or find it on the magical internet). If anyone can can ease my frustration, I would be grateful.
 
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If P~Q and Q~R then, P=XY, Q=YX and Q=VW, R=WV.
Then,
P=P2=XYXY=XQY=(XV)(WY)
R=R2=WVWV=WQV=(WY)(XV)
so P~R.
 
Alright that's about as complicated as I expected it to be...I basically had that written down, but apparently I don't quite have a fully functioning brain and for some reason couldn't see it. Many thanks.
 

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