Take the 2-minute tour ×  I'm not sure how to go about finding
the solution to this question.  Let R be a ring with identity
such that $x^2 = 1_R$ for all $0_R \neq x\in R$. How many
elements are in $R$?  I've just been playing around with
squaring elements, like $$(x+1_R)^2 = x^2+2.x+1_R =2.x+2.1_R =
1_R.$$  But I'm not sure where to go with this. Any help?
share|improve this question Either $1_R+1_R=0$ or
$(1_R+1_R)^2=1_R$ so $3\cdot 1_R=0$. –  Davide Giraudo Mar 5 '12
at 20:15 add comment  2 Answers  One possibility, of course, is
$R=\{0\}$. Assume $1_R\neq 0$.  $R$ has no zero divisors: if
$xy=0$ and $x\neq 0$, then $y = 1_Ry = xxy = x(0)=0$.  $R$ is
commutative: if $x$ and $y$ are nonzero, then so is $xy$ by the
above; hence $(xy)^2 = x^2y^2$; canceling from $xyxy=xxyy$ we
get $yx=xy$.  (Of course, the ring satisfies $x^3=x$ for all
$x$, so by a famous theorem of Jacobson, the ring is necessarily
commutative; but we don't need to call in that heavy cannon to
the fray).  Since every nonzero element is in