## Q. 66 GR0568

Forum for the GRE subject test in mathematics.
amateur
Posts: 42
Joined: Wed Sep 10, 2008 9:41 am

### Q. 66 GR0568

66. Let R be a ring with multiplicative identity. If U is an additive subgroup of R such that ur "belongs to" 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 inverses)
III. R is infinite

I tried it the following way (Comments are welcome).

Since the question talks of right ideals, we cannot conclusively say that R is commutative.

Since R has exactly two right ideals, they must be {0} (additive identity) and R itself. Using #5 in

http://en.wikipedia.org/wiki/Ideal_(ring_theory)#Examples

we can conclude that R is a division ring or a field.

Since the commutativity of R cannot be established, it is a division ring.

Finally the division ring of quaternions is proof enough that R need not be infinite.

Hence only II is acceptable (I wonder how I would be able to solve a similar question in an exam with 2.5 min/question on the average)!

moo5003
Posts: 36
Joined: Mon Oct 06, 2008 7:33 pm
Well you know III is wrong since the ring (0,1) satisfies all properties.

So the only real question is if R must be communative or a division ring for this property to hold. I already knew of a few rings that had only two ideals that were not commutative so I just put down II, however you could see this in a very straightforward way.

Suppose R is not a division ring, then there exists some u in U such that the right ideal generated by u does not equal the entire set R (Otherwise there exists some r in R such that u*r = identity). Thus R is more then 2 right ideals.

I think a great way to prepare is just to familiarize yourself with common group/rings. Just so you can come up with counter examples quicker.

Nameless
Posts: 128
Joined: Sun Aug 31, 2008 4:42 pm
66. Let R be a ring with multiplicative identity. If U is an additive subgroup of R such that ur "belongs to" 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 inverses)
III. R is infinite

I is incorrect since, For example, the division ring : Hamilton's Quaternion need not to be commutative

II is incorrect since Z_2 : is finite field even thought Z_2 have only two right ideals

Charles.Rambo
Posts: 8
Joined: Wed May 22, 2013 7:33 pm

### Re: Q. 66 GR0568

Hey guys, I've solved all the math GRE form 68 problems. You can see my solutions at

http://rambotutoring.com/GRE-math-subject-test-68-solutions.pdf

-Charles

redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am

### Re: Q. 66 GR0568

Yet another set of solutions can be found here: http://www.gigauploader.com/file/0770203183169331

Here's the post which describes the solutions:
viewtopic.php?f=1&t=1378

berkbelt
Posts: 15
Joined: Sun Jun 02, 2013 12:19 pm

### Re: Q. 66 GR0568

Charles.Rambo wrote:Hey guys, I've solved all the math GRE form 68 problems. You can see my solutions at

http://rambotutoring.com/GRE-math-subject-test-68-solutions.pdf

-Charles

I found a mistake in your calculations. Problem 39 should have a limit of cos(70), not 2 cos(70).

Also, I have a much simpler way of doing 8 that does not require the use of trigonometric identities. Just use the basic formula of a triangle in your picture: 1/2 * base * height = 1/2 * 1 * sin(theta).

Also, in 45, in the middle of your solution, you meant 1, not 6 (in regards to the number of points in the last sector).

Also, in 51, it is much easier than you have indicated. Just telescope the sum after performing the integral and you get a simple 1 + x + x^2 +.... series which is easily evaluated.

Otherwise, excellent job!

Charles.Rambo
Posts: 8
Joined: Wed May 22, 2013 7:33 pm

### Re: Q. 66 GR0568

Thanks! I think I fixed most of those... I'll check soon.