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How to prove that $\Bbb Q(\sqrt 2)$ and $\Bbb Q(\sqrt 3)$ are not isomorphic. I thought that they are but I got this problem in Dummit Foote Section 14.1. Question no 4. As they extension over $\Bbb Q$ by the polynomials $x^2-2$ and $x^2-3$ resp.

Ri-Li
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    Hint: if $\phi: \mathbb Q(\sqrt 2)\to \mathbb Q(\sqrt 3)$ were an isomorphism, show that $\left(\phi(\sqrt 2)\right)^2=2$. Derive a contradiction. Note: I am assuming that you mean "isomorphic as extensions of $\mathbb Q$ but you should specify that. As vector spaces over $\mathbb Q$, say, they are isomorphic. – lulu Dec 06 '18 at 10:59

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Hint:

Take $\;w=a+b\sqrt2\in\Bbb Q(\sqrt2)\;$ s.t. $\;\phi w=\sqrt3\in\Bbb Q(\sqrt3)\;$ , with $\;\phi\;$ an isomorphism. This means that

$$3=\phi w^2=\phi(a^2+2b^2+2ab\sqrt2)=a^2+2b^2+2ab\phi\sqrt2\implies\phi\sqrt2\in\Bbb Q$$

and now get a contradiction...

DonAntonio
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