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$\mathbf{Q}(\sqrt{2} )$ and $\mathbf{Q}(\sqrt{3}) $ are isomorphic as fields ? I think they are ring isom with ring hom which sends $\sqrt{2}$ to $\sqrt{3}$.

Then, number of quadratic extensions of $\mathbf{Q}$ is just one up to field isomorphisms ?? (but there are infinitely many number of quadratic extensions as sets)

Is my understanding right?

Thank you for your correction.

Pont
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    One of them contains a square root of $2$, the other doesn't. – lulu Sep 15 '21 at 16:06
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    The map sending sqrt(2) to sqrt(3) is an isomorphism of Q-vector spaces, not of fields. This map does not preserve multiplication. – D_S Sep 15 '21 at 16:10
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    When you define a ring homomorphism by defining where it sends generators, you need to make sure that any relation satisfied by generators must also correspond to a "homomorphic" relation satisfied by their images. If you map $\sqrt{2}$ to $\sqrt{3}$, then you have a problem in extending this to a ring homomorphism, since $\sqrt{2}$ satisfies $x^2 - 2 = 0$, while $\sqrt{3}$ does not. – Malkoun Sep 15 '21 at 16:11

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