Questions about right ideals

Question

The question is from the following problem:

Let $R$ be a ring with a multiplicative identity. If $U$ is an additive subgroup of $R$ such that $ur\in U$ for all $u\in U$ and for all $r\in R$, then $U$ is said to be a right ideal of $R$. If $R$ has exactly two right ideals, which of the following must be true?

I. $R$ is commutative.
II. $R$ is a division ring.
III. $R$ is infinite.

I know the definition of every concept here. But I have no idea what is supposed to be tested here.

• Why is the ring $R$ which has exactly two right ideals special?
• What theorem does one need to solve the problem above?

Edit: According to the answers, II must be true. For III, $R$ can be a finite field according to mt_. What is the counterexample for （I） then?

Answer

The trick here is to see that $0$ and $R$ are always right ideals. $R$ is not equal to zero, then there would only be one right ideal, so every ideal must be either $0$ or $R$. So you can prove that every non-zero element has an inverse, since for $a\in R-\{0\}$ we have $aR$ is a right ideal, hence $R$, so there is an $r\in R$ with $ar=1$.

Edit: It is equivalent for a ring to have precisely two right ideals and it being a division ring. Since there exists finite fields and (only infinite non-commutative) division rings, I and III are ruled out. The arguemt why a division ring has exactly two right ideal is the following. (Repeated from a comment below.) Again $0$ and $R$ are right ideals . Assume there is a right ideal $I$ with a non-zero element $a$. Then there is $a'\in R$ with $aa' =1$ (an inverse) therefore $1 \in I$, hence $I=R$.

Check more discussion of this question.