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This is question #66 from http://www.ets.org/s/gre/pdf/practice_book_math.pdf

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 (that is, all elements except the additive identity have multiplicative inverses).
III. $R$ is infinite.

Here is my reasoning:
Because $R$ is a ring, $R$ is also a additive group with some identity element $0$. We have a theorem that says $0r = 0 $ in any ring, so $\{0\}$ is a right ideal of $R$. Also, $R$ is a right ideal of $R$. Now I have found two different right ideals and there mustn't be any more.

Edit: As mentioned in the comments, the example below is not a ring, so it is not applicable to the problem. I could not fix it by taking additive closure because that introduced more than two ideals.

A possible candidate for $R$ could be the set of $2\times2$ matrices $\{0,I,-I,a,-a\}$ where $a = [[^1_0] ,[^0_0]]$. The only right ideals are $R$ and $\{0\}$. This ring satisfies only property I, but the answer key says that II is the correct answer.

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2  
Your candidate $R$ is not a ring. –  Michael Joyce Mar 18 '13 at 18:58
    
$R$ is not a ring since for example $I+I=2I\notin R$. –  Sami Ben Romdhane Mar 18 '13 at 19:00
    
$I-a\notin R$ (even if we had the entries from the field of two elements, when $I+I=0$ would be in there). –  Jyrki Lahtonen Mar 18 '13 at 19:01
    
My intuition about rings is very lacking. I guess fixing my counterexample may also lead to the correct answer. –  Mark Mar 18 '13 at 19:01
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If $u$ is a non-zero quaternion, then you get all the quaternions $q$ as products $ur=q$, because you can select $r=u^{-1}q$. –  Jyrki Lahtonen Mar 18 '13 at 20:58

2 Answers 2

up vote 4 down vote accepted

Well since $R$ has exactly two right ideals we know a priori that the two ideals are just $\{0\},R$ since these two are always ideals. Therefore if $I$ is a non zero ideal of $R$ then $I=R$.

Let $a\in R, \ a\neq0$. We will show that $a$ is invertible and therefore $R$ is a division ring. Consider the ideal $\langle a\rangle=\{ar:r\in R\}$. Since $\langle a\rangle\neq\{0\} \Longrightarrow \langle a\rangle=R \Longrightarrow 1\in \langle a\rangle\Longrightarrow 1=ar$ for some $r\in R.$ Now by considering the ideal $\langle r\rangle$ there is some $s\in R$ such that $rs=1$. It remains to show that $a=s$. Use that $a=a\cdot1$ and $1\cdot r=r$.

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Sorry, could you elaborate on this solution? I don't understand how it applies to the question. –  Mark Mar 18 '13 at 19:43
    
Under this ring (the reals), you are showing that the only other ideal besides {0} is the entire real line. Then all three properties are satisfied, which doesn't help me. Am I misinterpreting your solution? –  Mark Mar 18 '13 at 20:12
    
@Mark: No, I am assuming that the only ideal besides $\{0\}$ is the whole $R$ (which is not the reals, it is just a ring) and I am showing that any nonzero element has an inverse. Therefore II is true. –  P.. Mar 18 '13 at 20:56
    
You have proved that any non-zero element has a right inverse. In order to prove that it is invertible (which is correct) you need to go further... –  user26857 Mar 18 '13 at 22:31
    
How did you find that 1 was in your ring <a> ? –  Mark Mar 20 '13 at 0:02

Let $U$ be a right ideal in $R$. We have $U$ is a maximal ideal if and only if $R/U$ is a simple ring (i.e. If it has exactly two right ideals). Then is evident that $\{0\}$ is a maximal ideal in $R$.

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This has a serious problem (at least, with wording). $R/U$ is not always going to be a ring. $U$ is maximal right ideal iff $R/U$ is a simple module. –  rschwieb Mar 18 '13 at 19:41

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