Is Idempotent Equivalence in Rings Transitive?

  • Thread starter Thread starter cogito²
  • Start date Start date
  • Tags Tags
    Equivalence Rings
Click For Summary
The discussion centers on the transitivity of an equivalence relation defined for idempotents in rings, where two idempotents P and Q are equivalent if there exist elements X and Y such that P = XY and Q = YX. The original poster struggles to prove this transitive property but ultimately arrives at a proof by demonstrating that if P~Q and Q~R, then it follows that P~R through a series of algebraic manipulations. The proof involves showing that the relationships between the elements can be combined to establish the equivalence. The poster expresses relief at finally understanding the proof after initial confusion. The conversation highlights the challenges of grasping abstract algebra concepts and the satisfaction of resolving them.
cogito²
Messages
96
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
View full post »

Similar threads

Undergrad Splitting ring of polynomials - why is this result unfindable?
  • · Replies 9 ·
Replies
9
Views
2K
MHB Localization in Commutative Ring Theory
  • · Replies 16 ·
Replies
16
Views
4K
Graduate Structure of modulo-integer matrix group GL(r,Z(n))?
  • · Replies 17 ·
Replies
17
Views
6K
MHB Is There a Typo in the Refinement of Schanuel's Lemma in Passman's Book?
  • · Replies 4 ·
Replies
4
Views
2K
MHB Understanding the Equivalence in Diophantine Relations
  • · Replies 1 ·
Replies
1
Views
1K
Is the Relation Defined by 5 Dividing (2x + 3y) an Equivalence Relation on Z?
  • · Replies 7 ·
Replies
7
Views
2K
Proof of Maximal Left Ideal Generation in the Ring Mn(F)
  • · Replies 1 ·
Replies
1
Views
3K
Finding the Delta Function of a Thin Ring
  • · Replies 11 ·
Replies
11
Views
6K
Proving transitivity in equivalence relation a ~ b iff 2a+3b is div by 5
  • · Replies 1 ·
Replies
1
Views
4K
Are Equivalent Wave Functions Physically Equivalent in Quantum Mechanics?
  • · Replies 5 ·
Replies
5
Views
2K