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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.

Mark
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    Your candidate $R$ is not a ring. – Michael Joyce Mar 18 '13 at 18:58
  • $R$ is not a ring since for example $I+I=2I\notin R$. –  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
  • It might be a seemingly impossible question, but that is not an appropriate title for this website because it gives no indication of what the question will ask inside. Could you update it to something else? Maybe "What is true for a ring with exactly two right ideals?" or something similar... – Tyler Mar 18 '13 at 19:02
  • Hint: Try the ring of quaternions to see that I. does not need to hold. – Jyrki Lahtonen Mar 18 '13 at 19:02
  • As an example for a ring that satisfies II, but not I, consider the quaternions. For a finite ring that satisfies the property, take any finite field (e.g., $\mathbb Z/2Z$. As for proving II, I guess it should not be too hard to adapt the standard proof that a commutative ring with exactly two ideals is a field. – Johannes Kloos Mar 18 '13 at 19:07
  • How can I see that the Quaternions only have two ideals? – Mark Mar 18 '13 at 19:22
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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 Answers2

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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 $1\cdot s=s$ and $a=a\cdot1$.

Sally G
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P..
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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... –  Mar 18 '13 at 22:31
  • How did you find that 1 was in your ring ? – Mark Mar 20 '13 at 00:02
  • For example, the ideal {n*5} = {..., -10, -5, 0, 5, 10, ...} in the integers does not contain 1. – Mark Mar 20 '13 at 00:11
  • @Mark: I edited to clear things up. If you need more help let me know. And you can't give $\mathbb Z$ as an example since the hypothesis of the question say that the ring $R$ has only two right ideals and $\mathbb Z$ has more than two! – P.. Mar 20 '13 at 06:53
  • Thanks @YACP! I gave another hint. – P.. Mar 20 '13 at 06:54
  • Why do we need to show that a = r? My intuition tells me that this is false in general, or else every element is its own inverse. Maybe you wanted to show that s = a? In which case we can do a = a1 = ars = 1s = s? – Mark Mar 20 '13 at 23:58
  • @Mark: You are right! That was a typo, now corrected. – P.. Mar 21 '13 at 06:05
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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 $R$.

manuel
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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