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Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$?

Is it correct that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are isomorphic as fields?

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marked as duplicate by Jonas Meyer, Bill Dubuque, Qiaochu Yuan Mar 9 '11 at 21:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

no, since $3\neq 2$ – yoyo Mar 9 '11 at 21:34
Is there a square root of $2$ in $\mathbb{Q}(\sqrt{3})$? – Yuval Filmus Mar 9 '11 at 21:36
@yoyo: that is not enough. For example, it is true that $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{8})$ even though $2 \neq 8$. – Qiaochu Yuan Mar 9 '11 at 21:47
Surely a duplicate. – Bill Dubuque Mar 9 '11 at 21:48
Don't call me Shirley – The Chaz 2.0 Mar 9 '11 at 22:28

No. If $f$ were an isomorphism between $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ then $f$ is the identity on $\mathbb{Q}$ (it maps 1 to 1, so $\mathbb{Z}$ to itself, etc.). But what is $a = f(\sqrt(2))$? $\sqrt{2}$ satisfies the equation $x^2 - 2 = 0$ in the domain, so applying $f$ to this equation applies that $a$ also satisfies that same equation (same, as the coefficients are fixed under $f$). Now try to show that this cannot be, i.e. no $y$ of the form $c + d\sqrt{3}$ can satisfy this equation.

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