1. ## isomorphism

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?

2. Originally Posted by agentZERO
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?

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 $F:Q[\sqrt{y} ]\rightarrow \frac{Q[x]}{x^2-y}$ that is one-to-one, onto, and has an inverse. That is the more explicit objective right?

What is x defined as though?

3. Originally Posted by agentZERO
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?

define the map $f: \mathbb{Q}[x] \longrightarrow \mathbb{Q}[y]$ by $f(p(x))=p(\sqrt{y}).$ clearly $f$ is an onto ring homomorphism. so we only need to prove that $\ker f = ,$ which is quite straightforward:

if $p(x) \in ,$ then obviously $p(\sqrt{y})=0,$ i.e. $p(x) \in \ker f.$ suppose now that $p(x) \in \ker f.$ we have: $p(x)=(x^2 -y)q(x) + \alpha x + \beta,$ for some $\alpha, \beta \in \mathbb{Q}.$ since $p(\sqrt{y})=0,$

we must have: $\alpha \sqrt{y} + \beta = 0,$ which gives us: $\alpha = \beta = 0,$ because $y$ is not a square of an integer. thus: $p(x)=(x^2 - y)q(x) \in . \ \ \Box$

4. Oh man was I wrong. Sigh. So much to learn.

5. 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.

6. Originally Posted by agentZERO
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.
This is an interesting problem that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are isomorphic as vector spaces but not as fields. The way you show they are not isomorphic is by assuming that $\phi: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$ is an isomorphism. First $\phi ( 1 ) = \phi (1)$. Therefore, $\phi (n ) = \phi (1 + ... + 1) = \phi (1 ) + ... + \phi (1 ) = n$ for $n\in \mathbb{Z}^+$. It is easy to show $\phi ( n ) = n$ for $n\in \mathbb{Z}$ by considering the negatives. Next $\phi(a/b) = \phi(a)/\phi(b) = a/b$. Thus, $\phi|_{\mathbb{Q}}$ is an identity map. Now $\sqrt{2}^2 = 2 \implies \phi(\sqrt{2})^2 = 2 \implies \phi(\sqrt{2}) = \pm \sqrt{2}$. In both cases we never get an onto mapping because nothing gets mapped to $\sqrt{3}$. A contradiction.