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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?

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1  
Well, most rings have lots of ideals, so having exactly two is a rather special property! –  Mariano Suárez-Alvarez Jul 4 '11 at 17:39
    
Why (I) is not necessarily true? –  Jack Jul 4 '11 at 19:26
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Jack: because there exist non-commutative division rings, like the real quaternions. –  Mariano Suárez-Alvarez Jul 4 '11 at 19:36
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@Jack en.wikipedia.org/wiki/Quaternion –  Bill Dubuque Jul 4 '11 at 19:36
    
I think the ring that has such property is related to "simple ring". –  Jack Jul 4 '11 at 19:39

5 Answers 5

up vote 5 down vote accepted

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 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 $I=R$.

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Is your $aR$ called "principal ideal generated by $a$"? –  Jack Jul 4 '11 at 19:11
    
yep ... it is also the smallest ideal containing $a$. –  Peter Patzt Jul 4 '11 at 19:14
    
notice that we need $R$ to contain the multiplicative identity $1$. –  Peter Patzt Jul 4 '11 at 19:26
    
@Why is II ruled out? Isn't II the answer? –  Jack Jul 5 '11 at 2:24
    
edited, just a typo ;) –  Peter Patzt Jul 5 '11 at 8:33

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...

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Why "any right ideal is equal either to the zero ideal or the whole ring"? –  Jack Jul 4 '11 at 17:44
    
@Jack: because the zero ideal and the whole ring are already two ideals. I f you have an ideal, then,. it must be equal to one of those two: there are only two! –  Mariano Suárez-Alvarez Jul 4 '11 at 18:04
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@Jack Pigeonhole: if you place the three ideals $0,J,1$ into two ideal boxes then $\cdots$ –  Bill Dubuque Jul 4 '11 at 18:06
    
@Mariano: Fair enough. Since any ring has these two trivial right ideals, and when it has exactly these two, then, any right ideal is equal either to the zero ideal or the whole ring. –  Jack Jul 4 '11 at 18:47
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because if $1$ is in the ideal it is the whole ring. ($1R = R$) but if every non-zero element has an inverse, a non-zero ideal must contain $1$ too, as the inverse of an element of the ideal is also multiplicated times the element. –  Peter Patzt Jul 4 '11 at 19:17

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:

  1. 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

  2. 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.

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The quaternions, $\mathbb H$, is your desired example. They are a non-commutative division ring and a $4$-dimensional real algebra. –  kahen May 23 '13 at 5:18

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.

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It is a perfectly acceptible definition of group that it is a set with an associative product, a right identity, and right inverses exist. An early exercise in many abstract algebra courses is to show this implies left identity and left inverses. –  Barbara Osofsky Jan 26 '13 at 18:03

The solutions here are pretty good. But I just wanted to mention that I've solved all the math GRE form 68 problems. You can view my solutions at http://rambotutoring.com/GRE-math-subject-test-68-solutions.pdf

This is problem 66 in the booklet, by the way.

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