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.
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 $\displaystyle F:Q[\sqrt{y} ]\rightarrow \frac{Q[x]}{x^2-y}$ that is one-to-one, onto, and has an inverse. That is the more explicit objective right?
What is x defined as though?
define the map $\displaystyle f: \mathbb{Q}[x] \longrightarrow \mathbb{Q}[y] $ by $\displaystyle f(p(x))=p(\sqrt{y}).$ clearly $\displaystyle f$ is an onto ring homomorphism. so we only need to prove that $\displaystyle \ker f = <x^2 - y>,$ which is quite straightforward:
if $\displaystyle p(x) \in <x^2 - y>,$ then obviously $\displaystyle p(\sqrt{y})=0,$ i.e. $\displaystyle p(x) \in \ker f.$ suppose now that $\displaystyle p(x) \in \ker f.$ we have: $\displaystyle p(x)=(x^2 -y)q(x) + \alpha x + \beta,$ for some $\displaystyle \alpha, \beta \in \mathbb{Q}.$ since $\displaystyle p(\sqrt{y})=0,$
we must have: $\displaystyle \alpha \sqrt{y} + \beta = 0,$ which gives us: $\displaystyle \alpha = \beta = 0,$ because $\displaystyle y$ is not a square of an integer. thus: $\displaystyle p(x)=(x^2 - y)q(x) \in <x^2 -y>. \ \ \Box$
This is an interesting problem that $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \mathbb{Q}(\sqrt{3})$ are isomorphic as vector spaces but not as fields. The way you show they are not isomorphic is by assuming that $\displaystyle \phi: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$ is an isomorphism. First $\displaystyle \phi ( 1 ) = \phi (1)$. Therefore, $\displaystyle \phi (n ) = \phi (1 + ... + 1) = \phi (1 ) + ... + \phi (1 ) = n$ for $\displaystyle n\in \mathbb{Z}^+$. It is easy to show $\displaystyle \phi ( n ) = n$ for $\displaystyle n\in \mathbb{Z}$ by considering the negatives. Next $\displaystyle \phi(a/b) = \phi(a)/\phi(b) = a/b$. Thus, $\displaystyle \phi|_{\mathbb{Q}}$ is an identity map. Now $\displaystyle \sqrt{2}^2 = 2 \implies \phi(\sqrt{2})^2 = 2 \implies \phi(\sqrt{2}) = \pm \sqrt{2}$. In both cases we never get an onto mapping because nothing gets mapped to $\displaystyle \sqrt{3}$. A contradiction.