(adsbygoogle = window.adsbygoogle || []).push({}); 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 r_{1}and r_{2}in rR, we can write

r_{1}= rx and r_{2}= ry for some x and y in R.

Hence r_{1}- r_{2}= rx - ry = r(x - y) [tex]\in[/tex] rR.

rR is closed under subraction.

For r_{1}in rR, we can write r_{1}= rx for some x in R.

If y [tex]\in[/tex] R, then r_{1}y = (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?

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# Homework Help: Challenging (star marked) ring theory problem from I N Herstein's Topics in Algebra

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