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**TheEmptySet** I need to prove that $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

So here is my thoughts on this:

Since $\displaystyle \mathbb{Q}$ is the prime subfield is must be fixed by any isomorphism from $\displaystyle \tau : \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{3})$

Since $\displaystyle \{1, \sqrt{2} \}$ is a basis for $\displaystyle \mathbb{Q}(\sqrt{2})$ and $\displaystyle \{1, \sqrt{3} \}$ is a basis for $\displaystyle \mathbb{Q}(\sqrt{3})$

The isomorphism must be of the form

$\displaystyle \tau(a+b\sqrt{2})=a+b{\sqrt{3}}$ or $\displaystyle \tau(a+b\sqrt{2})=a-b{\sqrt{3}}$

Both of these only work for the addative group structure, but don't preserve multiplication. So they are not isomorphisms.

Therefore $\displaystyle \mathbb{Q}(\sqrt{2}) \not \cong \mathbb{Q}(\sqrt{3})$

I feel unsatisfied with the above proof. Another thought that I had was since \mathbb{Q}(\sqrt{2}) has only two automorphism showing that neither of those could be made into an isomorphism to $\displaystyle \mathbb{Q}(\sqrt{3}) $ but that came up empty.

Any thoughts or comments would be awsome.

Thanks

TES