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Possible Duplicate:
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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    no, since $3\neq 2$2011-03-09
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    Is there a square root of $2$ in $\mathbb{Q}(\sqrt{3})$?2011-03-09
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    @yoyo: that is not enough. For example, it is true that $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{8})$ even though $2 \neq 8$.2011-03-09
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    Surely a duplicate.2011-03-09
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    Don't call me Shirley2011-03-09

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