Subset of the indempotents of a ring

[pol92]

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={x\in A , x^2=x} . 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.

 

[COURAGE_WOLF]

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^{n}_{2}, where n is the dimension. So the ring must have cardinality 2ⁿ.

 

[Deveno]

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.

 

[COURAGE_WOLF]

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"