1. The problem statement, all variables and given/known data Problem 3.5.2 Let R be a ring such that the only right ideals of R are (0) and R. Prove that either R is a division ring or that R is ring with a prime number of elements in which ab = 0 for every a, b [tex]\in[/tex] R. 2. Relevant equations 3. The attempt at a solution First, prove that for all r in R, rR is a right ideal. For r1 and r2 in rR, we can write r1 = rx and r2 = ry for some x and y in R. Hence r1 - r2 = rx - ry = r(x - y) [tex]\in[/tex] rR. rR is closed under subraction. For r1 in rR, we can write r1 = rx for some x in R. If y [tex]\in[/tex] R, then r1y = (rx)y = r(xy) [tex]\in[/tex] rR. Hence rR is right ideal. Now we branch to 2 cases Case (i): Unit element belongs to the ring R Let r [tex]\neq[/tex] 0. Then r = r.1 [tex]\in[/tex] rR. Hence rR [tex]\neq[/tex] (0). From condition given in the question, we can say rR = R. As 1 [tex]\in[/tex] R, there exists r-1 such that rr-1 = 1. This is true for every non zero r. Hence every non zero element is invertible. Ring may not be commutative, we conclude R is a division ring. Case (i): Unit element does not belong to R rR = (0) or rR = R. This is the case where I am stuck. I am guessing that we should somehow prove rR = (0) and then prove that R has prime number of elements. How do we proceed?