## fr0598 #66

Forum for the GRE subject test in mathematics.
rhnsrbh
Posts: 5
Joined: Sat Mar 27, 2010 5:03 am

### fr0598 #66

Hey i have problem with this question
" Let R be a ring with a 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
III R is infinite.

By the way, what s a right ideal???

enork
Posts: 33
Joined: Fri Sep 18, 2009 3:16 am

### Re: fr0598 #66

They helpfully define what a right ideal is in the question.

To answer the question you really need to know a thing or two about rings. In particular, every ring R automatically has two right ideals: {0} and the entire ring R, but may potentially have many more. You can check that those two sets satisfy the definition.

Any element u of a ring can generate a right ideal uR, which is the set of elements of the form ur, for all r in R (these are called principal ideals). For u != 0, the right ideal uR is clearly not {0}, so if {0} and R are the only two right ideals then we must have uR = R. Since R contains 1, there's some element v in R such that uv = 1, so u has an inverse, and R is a division ring. You can also check that the converse is true: a division ring has only two right ideals. Since the other two choices don't depend on R being a division ring, they aren't required to be true.

Mathemagician
Posts: 10
Joined: Thu Aug 05, 2010 2:29 pm

### Re: fr0598 #66

In particular, any quaternion is non-commutative with only trivial right ideals and any Galois field is finite with only trivial right ideals.