Subset of the indempotents of a ring

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Discussion Overview

The discussion revolves around a problem concerning the subset of idempotents in a ring. Participants are exploring the implications of the cardinality of this subset, specifically focusing on proving that if the subset of idempotents is finite, it must have an even cardinality. The scope includes mathematical reasoning and proof techniques related to ring theory.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes that to show the subset of idempotents has even cardinality, an involution without fixed points could be utilized.
  • Another participant suggests that the proof relates to finite Boolean rings and mentions the additive group of the ring as a vector space over F2, leading to the conclusion that the cardinality must be 2n.
  • A third participant points out that the original poster did not specify that addition is idempotent, questioning whether S forms a ring.
  • The second participant acknowledges the oversight and suggests that the proof can still be valid by omitting the reference to the "additive group."

Areas of Agreement / Disagreement

Participants express differing views on the validity of the proof approach, particularly regarding the assumptions about the structure of S and whether it forms a ring. There is no consensus on the correctness of the initial claims or the implications of the proof.

Contextual Notes

There are limitations regarding the assumptions made about the operations within the subset S and the nature of the ring A. The discussion highlights the need for clarity on whether S forms a ring and the implications of that on the proof.

pol92
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Hello,
This is my first post on this forum, and I'm not used to the english mathematical vocabulary, I'll try my best to explain what is my problem.

Let (A,+,x) be a ring, ans S be the subset of the indempotents of A, i.e S={[itex]x\in A , x^2=x[/itex]} . I must show that if S is a finite set, then S has an even cardinality.

My idea was to find an involution of S without any fixed point (since an involution of a set with odd cardinality has always a fixed point), but all my trials had failed.
Could you help in solving this problem? Thank you.
 
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Interesting. So the proof is to show that every finite Boolean ring has even cardinality. Well the additive group of the ring can be thought of as a vector space over F2, a field consisting of two elements. If we consider only a finite case, then it must be a finite space over this field. It shouldn't be too much of a surprise to now notice that it's isomorphic to Z[itex]^{n}_{2}[/itex], where n is the dimension. So the ring must have cardinality 2n.
 
he did not say that the addition was idempotent, only the multiplication on S. there is nothing given in the problem that S even forms a ring.
 
Yeah you're right and I'm not sure why I decided to say that. Oh well. The proof still works, just take out the phrase "additive group"
 

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