Subset of the indempotents of a ring

  • Thread starter pol92
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  • #1
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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.
 

Answers and Replies

  • #2
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.
 
  • #3
Deveno
Science Advisor
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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.
 
  • #4
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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