Proving R is a Division Ring or Has Prime #Elements

[karthikvs88]

Homework Statement

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 $$\in$$ R.

Homework Equations

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) $$\in$$ rR.
rR is closed under subraction.

For r₁ in rR, we can write r₁ = rx for some x in R.
If y $$\in$$ R, then r₁y = (rx)y = r(xy) $$\in$$ rR.

Hence rR is right ideal.

Now we branch to 2 cases

Case (i): Unit element belongs to the ring R
Let r $$\neq$$ 0.
Then r = r.1 $$\in$$ rR.
Hence rR $$\neq$$ (0). From condition given in the question, we can say rR = R.
As 1 $$\in$$ 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?

 

[karthikvs88]

bump!

 

[AllTheCheese]

I know I'm over a year late to the party, but I figured it was worth posting my solution, since it's a pretty tough problem and I couldn't find the solution elsewhere.

I'm assuming anyone who cares about this problem is working through Herstein, so I'm going to reference some results from the book. In particular, we'll need to make use of

Problem 3.5.01: A ring with a unit element whose only ideals are {0} and the ring itself is a division ring.

Problem 3.4.17: Given a in R, the set {x in R such that ax = 0} is a right ideal.

Problem 3.4.18: L(R) = {x in R such that xa = 0 for all a in R} is a two-sided ideal.

Taking those for granted (they are much easier to demonstrate than this problem), we are now ready to present a solution:

Suppose R is a nontrivial ring. We break the proof into two branches.

Branch 1: For some a in R, we have aR = R.

To prove that R is a division ring, by Problem 3.5.01, it suffices to show that R has a unit element. Problem 3.4.17 tells us that W = {r in R such that ar = 0} is a right ideal. Since a is not in W, it follows from the hypothesis of the problem that W = {0}. Now, since aR=R, there must be e in R satisfying a*e=a. From this, we see that a*e*a=a*a*e, whence

a (e*a - a*e) = 0.

Since W is trivial, we have a*e = e*a.

Because aR=R, we can write any element x as x=a*y for some y. Therefore,

e*x=(e*a)*y=a*y=x.

We would like also for x*e=x, so that e is the unit element we desire. This will be shown to be the case. Suppose first that t is such that t*e=0. Then, for any v, we have

t*v=t*(e*v)=(t*e)*v=0.

Thus, t is an element of the ideal L(R) as described in problem 3.4.18. Note that a is not in L(R) because a*e=e*a=a, which is nonzero. Since L(R) is not equal to R, we have L(R)={0} and t=0.

Now, suppose b is a nonzero element of R and that b*e=c. Since e^2=e, we have b*e=c*e. This gives (b-c)*e=0, whence b=c by the preceding argument. Thus, e is the unit element we sought, and R is a division ring.

Branch 2: For all in R, we have aR =/= R, which, in light of the hypothesis, implies aR = {0} for all a.

This branch is easier. The assumption of the branch already supplies the fact that ab=0 for all a and b. If we suppose a is not equal to 0, and consider the subring <a>={0,a,2a,...}, we notice that <a> is trivially an ideal, since all products equal 0, which is in <a>. By hypothesis, <a>=R. Moreover, <a>=R must have a prime number of elements, because if it did not, there would be some integer n for which <n*a> generates a proper ideal of R, in violation of the hypothesis.

 

[karthikvs88]

There was never, a party, it was just me (and later you).

Anyway, check out Dummit & Foote, it is a better and updated reference for algebra.