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I was doing some abstract algebra exercises and I noticed that we if we assume there exists some homomorphism $\phi : \mathbb{Q}[\sqrt{n}] \rightarrow \mathbb{Q}[\sqrt{m}]$. Then if we want an isomorphism then $\phi$ must be surjective which means that $\phi(1) = 1$. Which means that $\phi (a) = a $ for some $a\in$ $\mathbb{Q}$. Then for some $(a + b\sqrt{n})$ we get that $$\phi(a + b\sqrt{n}) = a +b\phi(\sqrt{n})$$ Now we also have that, $$\phi (n) = n = \phi ((\sqrt{n})^2) = (\phi(\sqrt{n}))^2$$ Then if $\phi(\sqrt{n}) = c + d\sqrt{m}$ we get that $$n = (\phi(\sqrt{n}))^2 = (c + d\sqrt{m})^2$$ $$ n = c^2 + 2\sqrt{m}cd + md^2$$ we get that $cd = 0$ so we have that if $c = 0 $ $$ n = md^2 \Rightarrow d = \sqrt{\dfrac{n}{m}}$$ which is a contradiction as $d\in \mathbb{Q}$ similarly we get that if $ d = 0$ $$ n = c^2 \Rightarrow c = \sqrt{n} $$ which again contradicts that $c\in \mathbb{Q}$ which means that we cannot construct an isomorphism. Is there something that I am assuming that I shouldn't be or is this correct?

Viktor Pavlovic
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    Well, neither $2$ nor $8$ is a square, and $\Bbb Q(\sqrt2,)=\Bbb Q(\sqrt8,)$. I think you want to assume that $m$ and $n$ are distinct square-free integers. – Lubin Dec 08 '18 at 05:00
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    On the fifth line from the bottom you should have $d^2=n/m$. And this is the true condition. Neither $2$ nor $8$ is a square, but $\Bbb{Q}(\sqrt2)=\Bbb{Q}(\sqrt8)$. If $m,n$ are both square-free integers then the conclusion holds. – Jyrki Lahtonen Dec 08 '18 at 05:01
  • Well. They become isomorphic in many cases for $m,n$ if you view them as $\mathbb{Q}$- Vector spaces. – C.S. Dec 08 '18 at 05:01

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