# Subset of the indempotents of a ring

1. Nov 27, 2011

### 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.

2. Nov 27, 2011

### 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.

3. Nov 27, 2011

### 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.

4. Nov 27, 2011

### 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"