Let y be an integer, which is not a square of another integer.

How do I prove that Q[sqrt(y)] = {a + bsqrt(y) : a,b in Q} is isomorphic to Q[x]/(x^2 - y) where Q is the set of rational numbers?

Thanks for your help.

Printable View

- Nov 3rd 2008, 03:34 PMagentZEROisomorphism
Let y be an integer, which is not a square of another integer.

How do I prove that Q[sqrt(y)] = {a + bsqrt(y) : a,b in Q} is isomorphic to Q[x]/(x^2 - y) where Q is the set of rational numbers?

Thanks for your help. - Nov 3rd 2008, 04:00 PMJameson
I think someone is replying to this now who should be able to help, so since I'm kind of learning this topic as well, I'll ask a question.

This proof requires that we show that there exists a map that is one-to-one, onto, and has an inverse. That is the more explicit objective right?

What is x defined as though? - Nov 3rd 2008, 04:08 PMNonCommAlg
- Nov 3rd 2008, 04:19 PMJameson
Oh man was I wrong. Sigh. So much to learn.

- Nov 4th 2008, 09:37 AMagentZERO
How do I use this to prove that Q[sqrt(2)] is not isomorphic to Q[sqrt(3)]? What property of a ring fails?

Thanks for the response from before. - Nov 4th 2008, 12:32 PMThePerfectHacker
This is an interesting problem that and are isomorphic as vector spaces but not as fields. The way you show they are not isomorphic is by assuming that is an isomorphism. First . Therefore, for . It is easy to show for by considering the negatives. Next . Thus, is an identity map. Now . In both cases we never get an onto mapping because nothing gets mapped to . A contradiction.