fr0598 #66

Forum for the GRE subject test in mathematics.
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fr0598 #66

Post by rhnsrbh » Mon Mar 29, 2010 6:16 am

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???

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Re: fr0598 #66

Post by enork » Thu Apr 01, 2010 12:59 pm

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.

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Re: fr0598 #66

Post by Mathemagician » Fri Oct 22, 2010 12:16 pm

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

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