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How can I show that $\mathbb{Q}(\sqrt{3})$ is isomorphic to $\mathbb{Q(\sqrt{-3})}$. (or possibly disprove it)?

What I know:

I don't know how to begin if it is the case that they are not isomorphic.

In the case that they are isomorphic:

I can show that they are isomorphic to one another by producing a bijective homomorphism.

I consider $\phi:\mathbb{Q}(\sqrt{3})\to\mathbb{Q}(\sqrt{-3})$ however I am having trouble thinking of a mapping for the elements that works.

$a+\sqrt{3}b$ maps to ??

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It seems that your problem is in checking that your map is a homomorphism.

I claim that no map $\varphi: \mathbb Q(\sqrt{-3}) \to \mathbb Q(\sqrt 3)$ can be a field homomorphism.

For assume such a map existed, then it would have to fix $\mathbb Q$ and we have that $-3 = \varphi(-3) = \varphi(\sqrt{-3} \sqrt{-3}) = \varphi(\sqrt{-3}) \varphi(\sqrt{-3})$ implying that $\varphi(\sqrt{-3}) \in \mathbb Q(\sqrt 3)$ squares to $-3$ which is a contradiction as $\mathbb Q(\sqrt 3) \subset \mathbb R$.

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  • $\begingroup$ I would like to clarify one thing if possible. My book exercise asks to show that $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{-3})$ are ring-isomorphic. I assume that if $\varphi$ is a field homomorphism, then $\varphi$ is also a ring homomorphism since all fields are rings. $\endgroup$ – Nathan C. Mar 26 '15 at 22:15
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    $\begingroup$ Indeed. A field homomorphism is just a ring homomorphism between fields. $\endgroup$ – kahen Mar 26 '15 at 22:16
  • $\begingroup$ OK, thanks for your help. $\endgroup$ – Nathan C. Mar 26 '15 at 22:17
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Note that $3 = \phi(3) = \phi((0+\sqrt{3})(0 + \sqrt{3}))$, but:

$-3 = (0 + \sqrt{-3})(0 + \sqrt{-3}) = \phi((0 + \sqrt{3}))\phi((0 + \sqrt{3}))$, so the $\phi$ you exhibited is not a ring-homomorphism.

To put this situation a different way: $x^2 + 3$ splits in one field, but not the other, so how can they be isomorphic?

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The two are not isomorphic, suppose there is an ring isomorphism $\phi:\mathbb{Q}(\sqrt{3})\to\mathbb{Q}(\sqrt{-3})$,then $\phi(1)=1, \phi(0)=0$, which implies that $\phi(x)=x$ for all $x\in\mathbb Q$. then since $\sqrt{3}^2-3=0$, we must have $(\phi(\sqrt{3}))^2-3=0$, but there is no solution of $x^2-3=0$ in $\mathbb{Q}(\sqrt{-3})$ (every element in $\mathbb{Q}(\sqrt{-3})$ is of form $a+b\sqrt{-3}$ with $a,b\in\mathbb Q$, if $b\neq0$, the corresponding minimal polynomial is $(x-a-b\sqrt{-3})(x-a+b\sqrt{-3})$, which has no real root).

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  • $\begingroup$ The final claim requires proof! $\ \ $ $\endgroup$ – Bill Dubuque Mar 26 '15 at 22:17
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    $\begingroup$ It would be easier to reverse the roles of the two fields and prove that $x^2+3=0$ has no solution in $\mathbb Q(\sqrt 3)$. $\endgroup$ – Andreas Blass Mar 26 '15 at 22:31
  • $\begingroup$ @AndreasBlass I see no difference, just as I've added. $\endgroup$ – Censi LI Mar 26 '15 at 22:55
  • $\begingroup$ The easier proof that I had in mind is that $\mathbb Q(\sqrt3)$ consists of real numbers, none of which have square equal to $-3$. $\endgroup$ – Andreas Blass Mar 26 '15 at 22:56
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    $\begingroup$ @AndreasBlass I agree that your proof is easy. But I don't it's harder to prove that the only real numbers contained in $\mathbb{Q}(\sqrt{-3})$ is the rational numbers. $\endgroup$ – Censi LI Mar 26 '15 at 23:07

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