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

I have no idea about field theory much, I am studying it now, I have a doubt as a vector space over $\mathbb{Q}$, $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are isomorphic(I hope so) and as a field are they isomorphic?

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marked as duplicate by Arturo Magidin, Nate Eldredge, Jyrki Lahtonen, Chris Eagle, sdcvvc Aug 27 '12 at 16:04

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.

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This is a repeat of this question. –  Arturo Magidin Apr 19 '12 at 18:50

2 Answers 2

up vote 4 down vote accepted

Note that any isomorphism between two extensions of $\mathbb Q$ must preserve $\mathbb Q$.

This means that an isomorphism of fields between $\mathbb Q[\sqrt 2]$ and $\mathbb Q[\sqrt 3]$ is determined by where $\sqrt 2$ is mapped to.

Suppose that $f$ was such isomorphism then $f(\sqrt 2)^2=f(\sqrt 2^2)=f(2)=2$. However there is no element in $\mathbb Q[\sqrt 3]$ whose square is $2$, therefore there is no isomorphism as fields.

As vector spaces, however, $\mathbb Q[\sqrt 2]$ is spanned by $\{1,\sqrt 2\}$ and similarly $\mathbb Q[\sqrt 3]$ is spanned by $\{1,\sqrt 3\}$ so it is fairly easy to find an isomorphism between those vector spaces.

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Yes, $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt3)$and are isomorphic as vector spaces, since they are both 2-dimensional over $\mathbb{Q}$. They are not isomorphic as fields, however. $\mathbb{Q}(\sqrt2)$ does not contain an element whose square is equal to 3. This fact is not immediately obvious, though. See this previous post.

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