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Are the fields $\mathbb{Q}(i)$ and $\mathbb{Q}(2i)$ isomorphic? I'm confused since they seem to be equal as sets but $\mathbb{Q}(i)\cong \mathbb{Q}[X]/(X^2+1)$ but $\mathbb{Q}(2i)\cong \mathbb{Q}[X]/(X^2+4)$.

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They are isomorphic; the isomorphism just doesn't send the first $X$ to the second $X$. (It would be a good idea to use a different letter in the second presentation for this reason.) –  Qiaochu Yuan Sep 4 '12 at 20:01
    
It's easier to see if you don't use the same symbol, $X$. You want to show that $\mathbb Q(X)/(X^2+1) \cong \mathbb Q(Y)/(Y^2+4)$ –  Thomas Andrews Sep 4 '12 at 20:44
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2 Answers 2

They are the same field.

Clearly $\mathbb{Q}(2i) \subset \mathbb{Q}(i)$

Note that $i \in \mathbb{Q}(2i)$ since $i = \frac{1}{2}(2i)$. Since $i \in \mathbb{Q}(2i)$, one has that $\mathbb{Q}(i) \subset \mathbb{Q}(2i)$.

Hence $\mathbb{Q}(2i) = \mathbb{Q}(i)$.


If they are equal as sets then they are certainly isomorphic. It is some sense better than being isomorphic since they are actually the same thing.

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The identity map is certainly an isomorphism :)

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