Questions about right ideals 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?
 A: The trick here is to see that $0$ and $R$ are always right ideals. $R$ is not equal to zero, as otherwise 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 $aR = R$ is a right ideal, 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 argument 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 $I=R$.
A: The concept that needs to be understood here is that if a ring has EXACTLY two right ideals, those ideals must be the trivial zero ideal and the whole ring. We further know two facts:


*

*There exists a division ring (i.e. a field) that is finite (i.e. $\mathbb Z_3$, the integers mod 3). This is a finite field (and thus a division ring). So, we have a ring that is a commutative ring and is finite. So we can satisfy II and III

*There exists division rings that are non commutative, but infinite. The real quaternions are an example of this of course. So we can satisfy I and II
In total, we have found two rings that satisfy two of three properties. But, the important thing to note is that both 1 and 2 show that I and III are not ALWAYS satisfied. Thus II is the only sure thing.
If you want to find a ring that satisfies ONLY II, you could be looking a long time. (It could exist, but I cannot think of an example). But, finding two rings that satisfy only two of the three properties (I, II, III) is possible. The only property these have in common however is II. Thus B is the answer. Good luck figuring that out in about 3 minutes on the exam, though.
A: The proof that $R$ with only two right ideals and an identity must be a division ring gets only half the mark because you have shown that every non-zero element has a right inverse, not a two sided inverse.
A: A ring with exactly two right ideals is special because any right ideal is equal either to the zero ideal or the whole ring.  For example, any field has this property (giving you a hint for III).  II is the statement that is true: suppose $R$ has exactly two right ideals, and let $0 \neq r \in R$, to show $R$ is a division ring you must prove $r$ has a multiplicative inverse.  So consider the right ideal generated by $r$, written $rR = \{ rs: s \in R\}$.  This must be either the zero ideal or the whole ring...
A: 2nd option must be true.... first is incorrect as every division ring which is not commutative has exactely two ideals and hence two right ideals,  3rd is wrong as take any finite field which is simple and hence only two ideal and hence only two right ideal.. so 2nd must be true
