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Subset of the indempotents of a ring

  1. Nov 27, 2011 #1
    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.
  2. jcsd
  3. Nov 27, 2011 #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.
  4. Nov 27, 2011 #3


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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.
  5. Nov 27, 2011 #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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